3.7.71 \(\int \frac {(a+b x)^{5/2} (c+d x)^{5/2}}{x^5} \, dx\) [671]

Optimal. Leaf size=380 \[ \frac {5 d \left (b^3 c^3+45 a b^2 c^2 d+19 a^2 b c d^2-a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 a c^2}-\frac {5 (3 b c+a d) \left (b^2 c^2+24 a b c d-a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{192 a c^2 x}-\frac {5 \left (3 b^2 c^2+14 a b c d-a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{96 c^2 x^2}-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{24 c x^3}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{4 x^4}+\frac {5 \left (b^4 c^4-20 a b^3 c^3 d-90 a^2 b^2 c^2 d^2-20 a^3 b c d^3+a^4 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{64 a^{3/2} c^{3/2}}+5 b^{3/2} d^{3/2} (b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right ) \]

[Out]

-5/24*(a*d+b*c)*(b*x+a)^(3/2)*(d*x+c)^(5/2)/c/x^3-1/4*(b*x+a)^(5/2)*(d*x+c)^(5/2)/x^4+5/64*(a^4*d^4-20*a^3*b*c
*d^3-90*a^2*b^2*c^2*d^2-20*a*b^3*c^3*d+b^4*c^4)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))/a^(3/2)/c
^(3/2)+5*b^(3/2)*d^(3/2)*(a*d+b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))-5/192*(a*d+3*b*c)*(-a^
2*d^2+24*a*b*c*d+b^2*c^2)*(d*x+c)^(3/2)*(b*x+a)^(1/2)/a/c^2/x-5/96*(-a^2*d^2+14*a*b*c*d+3*b^2*c^2)*(d*x+c)^(5/
2)*(b*x+a)^(1/2)/c^2/x^2+5/64*d*(-a^3*d^3+19*a^2*b*c*d^2+45*a*b^2*c^2*d+b^3*c^3)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a
/c^2

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Rubi [A]
time = 0.30, antiderivative size = 380, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {99, 154, 159, 163, 65, 223, 212, 95, 214} \begin {gather*} -\frac {5 \sqrt {a+b x} (c+d x)^{5/2} \left (-a^2 d^2+14 a b c d+3 b^2 c^2\right )}{96 c^2 x^2}-\frac {5 \sqrt {a+b x} (c+d x)^{3/2} (a d+3 b c) \left (-a^2 d^2+24 a b c d+b^2 c^2\right )}{192 a c^2 x}+\frac {5 d \sqrt {a+b x} \sqrt {c+d x} \left (-a^3 d^3+19 a^2 b c d^2+45 a b^2 c^2 d+b^3 c^3\right )}{64 a c^2}+\frac {5 \left (a^4 d^4-20 a^3 b c d^3-90 a^2 b^2 c^2 d^2-20 a b^3 c^3 d+b^4 c^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{64 a^{3/2} c^{3/2}}+5 b^{3/2} d^{3/2} (a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{4 x^4}-\frac {5 (a+b x)^{3/2} (c+d x)^{5/2} (a d+b c)}{24 c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(c + d*x)^(5/2))/x^5,x]

[Out]

(5*d*(b^3*c^3 + 45*a*b^2*c^2*d + 19*a^2*b*c*d^2 - a^3*d^3)*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*a*c^2) - (5*(3*b*c
+ a*d)*(b^2*c^2 + 24*a*b*c*d - a^2*d^2)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(192*a*c^2*x) - (5*(3*b^2*c^2 + 14*a*b
*c*d - a^2*d^2)*Sqrt[a + b*x]*(c + d*x)^(5/2))/(96*c^2*x^2) - (5*(b*c + a*d)*(a + b*x)^(3/2)*(c + d*x)^(5/2))/
(24*c*x^3) - ((a + b*x)^(5/2)*(c + d*x)^(5/2))/(4*x^4) + (5*(b^4*c^4 - 20*a*b^3*c^3*d - 90*a^2*b^2*c^2*d^2 - 2
0*a^3*b*c*d^3 + a^4*d^4)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(64*a^(3/2)*c^(3/2)) + 5*b^
(3/2)*d^(3/2)*(b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]

Rule 159

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 163

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {(a+b x)^{5/2} (c+d x)^{5/2}}{x^5} \, dx &=-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{4 x^4}+\frac {1}{4} \int \frac {(a+b x)^{3/2} (c+d x)^{3/2} \left (\frac {5}{2} (b c+a d)+5 b d x\right )}{x^4} \, dx\\ &=-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{24 c x^3}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{4 x^4}+\frac {\int \frac {\sqrt {a+b x} (c+d x)^{3/2} \left (\frac {5}{4} \left (3 b^2 c^2+14 a b c d-a^2 d^2\right )+\frac {5}{2} b d (7 b c+a d) x\right )}{x^3} \, dx}{12 c}\\ &=-\frac {5 \left (3 b^2 c^2+14 a b c d-a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{96 c^2 x^2}-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{24 c x^3}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{4 x^4}+\frac {\int \frac {(c+d x)^{3/2} \left (\frac {5}{8} (3 b c+a d) \left (b^2 c^2+24 a b c d-a^2 d^2\right )+\frac {5}{4} b d \left (31 b^2 c^2+18 a b c d-a^2 d^2\right ) x\right )}{x^2 \sqrt {a+b x}} \, dx}{24 c^2}\\ &=-\frac {5 (3 b c+a d) \left (b^2 c^2+24 a b c d-a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{192 a c^2 x}-\frac {5 \left (3 b^2 c^2+14 a b c d-a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{96 c^2 x^2}-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{24 c x^3}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{4 x^4}+\frac {\int \frac {\sqrt {c+d x} \left (-\frac {15}{16} \left (b^4 c^4-20 a b^3 c^3 d-90 a^2 b^2 c^2 d^2-20 a^3 b c d^3+a^4 d^4\right )+\frac {15}{8} b d \left (b^3 c^3+45 a b^2 c^2 d+19 a^2 b c d^2-a^3 d^3\right ) x\right )}{x \sqrt {a+b x}} \, dx}{24 a c^2}\\ &=\frac {5 d \left (b^3 c^3+45 a b^2 c^2 d+19 a^2 b c d^2-a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 a c^2}-\frac {5 (3 b c+a d) \left (b^2 c^2+24 a b c d-a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{192 a c^2 x}-\frac {5 \left (3 b^2 c^2+14 a b c d-a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{96 c^2 x^2}-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{24 c x^3}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{4 x^4}+\frac {\int \frac {-\frac {15}{16} b c \left (b^4 c^4-20 a b^3 c^3 d-90 a^2 b^2 c^2 d^2-20 a^3 b c d^3+a^4 d^4\right )+60 a b^3 c^2 d^2 (b c+a d) x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{24 a b c^2}\\ &=\frac {5 d \left (b^3 c^3+45 a b^2 c^2 d+19 a^2 b c d^2-a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 a c^2}-\frac {5 (3 b c+a d) \left (b^2 c^2+24 a b c d-a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{192 a c^2 x}-\frac {5 \left (3 b^2 c^2+14 a b c d-a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{96 c^2 x^2}-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{24 c x^3}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{4 x^4}+\frac {1}{2} \left (5 b^2 d^2 (b c+a d)\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx-\frac {\left (5 \left (b^4 c^4-20 a b^3 c^3 d-90 a^2 b^2 c^2 d^2-20 a^3 b c d^3+a^4 d^4\right )\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{128 a c}\\ &=\frac {5 d \left (b^3 c^3+45 a b^2 c^2 d+19 a^2 b c d^2-a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 a c^2}-\frac {5 (3 b c+a d) \left (b^2 c^2+24 a b c d-a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{192 a c^2 x}-\frac {5 \left (3 b^2 c^2+14 a b c d-a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{96 c^2 x^2}-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{24 c x^3}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{4 x^4}+\left (5 b d^2 (b c+a d)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )-\frac {\left (5 \left (b^4 c^4-20 a b^3 c^3 d-90 a^2 b^2 c^2 d^2-20 a^3 b c d^3+a^4 d^4\right )\right ) \text {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{64 a c}\\ &=\frac {5 d \left (b^3 c^3+45 a b^2 c^2 d+19 a^2 b c d^2-a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 a c^2}-\frac {5 (3 b c+a d) \left (b^2 c^2+24 a b c d-a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{192 a c^2 x}-\frac {5 \left (3 b^2 c^2+14 a b c d-a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{96 c^2 x^2}-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{24 c x^3}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{4 x^4}+\frac {5 \left (b^4 c^4-20 a b^3 c^3 d-90 a^2 b^2 c^2 d^2-20 a^3 b c d^3+a^4 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{64 a^{3/2} c^{3/2}}+\left (5 b d^2 (b c+a d)\right ) \text {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )\\ &=\frac {5 d \left (b^3 c^3+45 a b^2 c^2 d+19 a^2 b c d^2-a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 a c^2}-\frac {5 (3 b c+a d) \left (b^2 c^2+24 a b c d-a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{192 a c^2 x}-\frac {5 \left (3 b^2 c^2+14 a b c d-a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{5/2}}{96 c^2 x^2}-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{24 c x^3}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{4 x^4}+\frac {5 \left (b^4 c^4-20 a b^3 c^3 d-90 a^2 b^2 c^2 d^2-20 a^3 b c d^3+a^4 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{64 a^{3/2} c^{3/2}}+5 b^{3/2} d^{3/2} (b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )\\ \end {align*}

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Mathematica [A]
time = 1.00, size = 274, normalized size = 0.72 \begin {gather*} -\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 b^3 c^3 x^3+a b^2 c x^2 \left (118 c^2+601 c d x-192 d^2 x^2\right )+a^2 b c x \left (136 c^2+452 c d x+601 d^2 x^2\right )+a^3 \left (48 c^3+136 c^2 d x+118 c d^2 x^2+15 d^3 x^3\right )\right )}{192 a c x^4}+\frac {5 \left (b^4 c^4-20 a b^3 c^3 d-90 a^2 b^2 c^2 d^2-20 a^3 b c d^3+a^4 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{64 a^{3/2} c^{3/2}}+5 b^{3/2} d^{3/2} (b c+a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*(c + d*x)^(5/2))/x^5,x]

[Out]

-1/192*(Sqrt[a + b*x]*Sqrt[c + d*x]*(15*b^3*c^3*x^3 + a*b^2*c*x^2*(118*c^2 + 601*c*d*x - 192*d^2*x^2) + a^2*b*
c*x*(136*c^2 + 452*c*d*x + 601*d^2*x^2) + a^3*(48*c^3 + 136*c^2*d*x + 118*c*d^2*x^2 + 15*d^3*x^3)))/(a*c*x^4)
+ (5*(b^4*c^4 - 20*a*b^3*c^3*d - 90*a^2*b^2*c^2*d^2 - 20*a^3*b*c*d^3 + a^4*d^4)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x]
)/(Sqrt[c]*Sqrt[a + b*x])])/(64*a^(3/2)*c^(3/2)) + 5*b^(3/2)*d^(3/2)*(b*c + a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x
])/(Sqrt[d]*Sqrt[a + b*x])]

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(828\) vs. \(2(324)=648\).
time = 0.07, size = 829, normalized size = 2.18

method	result	size
default	\(\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{4} d^{4} x^{4} \sqrt {b d}-300 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{3} b c \,d^{3} x^{4} \sqrt {b d}-1350 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a^{2} b^{2} c^{2} d^{2} x^{4} \sqrt {b d}-300 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) a \,b^{3} c^{3} d \,x^{4} \sqrt {b d}+15 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}+2 a c}{x}\right ) b^{4} c^{4} x^{4} \sqrt {b d}+960 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b^{2} c \,d^{3} x^{4} \sqrt {a c}+960 \ln \left (\frac {2 b d x +2 \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{3} c^{2} d^{2} x^{4} \sqrt {a c}+384 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a \,b^{2} c \,d^{2} x^{4}-30 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{3} d^{3} x^{3}-1202 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{2} b c \,d^{2} x^{3}-1202 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a \,b^{2} c^{2} d \,x^{3}-30 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, b^{3} c^{3} x^{3}-236 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{3} c \,d^{2} x^{2}-904 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{2} b \,c^{2} d \,x^{2}-236 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a \,b^{2} c^{3} x^{2}-272 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{3} c^{2} d x -272 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{2} b \,c^{3} x -96 \sqrt {b d}\, \sqrt {a c}\, \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, a^{3} c^{3}\right )}{384 a c \sqrt {\left (d x +c \right ) \left (b x +a \right )}\, x^{4} \sqrt {b d}\, \sqrt {a c}}\)	\(829\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(d*x+c)^(5/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

1/384*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a/c*(15*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+2*a*c)/x)*a^4*
d^4*x^4*(b*d)^(1/2)-300*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+2*a*c)/x)*a^3*b*c*d^3*x^4*(b*d)^
(1/2)-1350*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+2*a*c)/x)*a^2*b^2*c^2*d^2*x^4*(b*d)^(1/2)-300
*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+2*a*c)/x)*a*b^3*c^3*d*x^4*(b*d)^(1/2)+15*ln((a*d*x+b*c*
x+2*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)+2*a*c)/x)*b^4*c^4*x^4*(b*d)^(1/2)+960*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+
a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b^2*c*d^3*x^4*(a*c)^(1/2)+960*ln(1/2*(2*b*d*x+2*((d*x+c)*(b*x+
a))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b^3*c^2*d^2*x^4*(a*c)^(1/2)+384*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)
*(b*x+a))^(1/2)*a*b^2*c*d^2*x^4-30*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a^3*d^3*x^3-1202*(b*d)^(1/2
)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a^2*b*c*d^2*x^3-1202*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a*b
^2*c^2*d*x^3-30*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*b^3*c^3*x^3-236*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+
c)*(b*x+a))^(1/2)*a^3*c*d^2*x^2-904*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a^2*b*c^2*d*x^2-236*(b*d)^
(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a*b^2*c^3*x^2-272*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a^
3*c^2*d*x-272*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*(b*x+a))^(1/2)*a^2*b*c^3*x-96*(b*d)^(1/2)*(a*c)^(1/2)*((d*x+c)*
(b*x+a))^(1/2)*a^3*c^3)/((d*x+c)*(b*x+a))^(1/2)/x^4/(b*d)^(1/2)/(a*c)^(1/2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(5/2)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more detail

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Fricas [A]
time = 7.87, size = 1633, normalized size = 4.30 \begin {gather*} \left [\frac {960 \, {\left (a^{2} b^{2} c^{3} d + a^{3} b c^{2} d^{2}\right )} \sqrt {b d} x^{4} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 15 \, {\left (b^{4} c^{4} - 20 \, a b^{3} c^{3} d - 90 \, a^{2} b^{2} c^{2} d^{2} - 20 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {a c} x^{4} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} + 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \, {\left (192 \, a^{2} b^{2} c^{2} d^{2} x^{4} - 48 \, a^{4} c^{4} - {\left (15 \, a b^{3} c^{4} + 601 \, a^{2} b^{2} c^{3} d + 601 \, a^{3} b c^{2} d^{2} + 15 \, a^{4} c d^{3}\right )} x^{3} - 2 \, {\left (59 \, a^{2} b^{2} c^{4} + 226 \, a^{3} b c^{3} d + 59 \, a^{4} c^{2} d^{2}\right )} x^{2} - 136 \, {\left (a^{3} b c^{4} + a^{4} c^{3} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{768 \, a^{2} c^{2} x^{4}}, -\frac {1920 \, {\left (a^{2} b^{2} c^{3} d + a^{3} b c^{2} d^{2}\right )} \sqrt {-b d} x^{4} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 15 \, {\left (b^{4} c^{4} - 20 \, a b^{3} c^{3} d - 90 \, a^{2} b^{2} c^{2} d^{2} - 20 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {a c} x^{4} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} + 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (192 \, a^{2} b^{2} c^{2} d^{2} x^{4} - 48 \, a^{4} c^{4} - {\left (15 \, a b^{3} c^{4} + 601 \, a^{2} b^{2} c^{3} d + 601 \, a^{3} b c^{2} d^{2} + 15 \, a^{4} c d^{3}\right )} x^{3} - 2 \, {\left (59 \, a^{2} b^{2} c^{4} + 226 \, a^{3} b c^{3} d + 59 \, a^{4} c^{2} d^{2}\right )} x^{2} - 136 \, {\left (a^{3} b c^{4} + a^{4} c^{3} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{768 \, a^{2} c^{2} x^{4}}, -\frac {15 \, {\left (b^{4} c^{4} - 20 \, a b^{3} c^{3} d - 90 \, a^{2} b^{2} c^{2} d^{2} - 20 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {-a c} x^{4} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - 480 \, {\left (a^{2} b^{2} c^{3} d + a^{3} b c^{2} d^{2}\right )} \sqrt {b d} x^{4} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 2 \, {\left (192 \, a^{2} b^{2} c^{2} d^{2} x^{4} - 48 \, a^{4} c^{4} - {\left (15 \, a b^{3} c^{4} + 601 \, a^{2} b^{2} c^{3} d + 601 \, a^{3} b c^{2} d^{2} + 15 \, a^{4} c d^{3}\right )} x^{3} - 2 \, {\left (59 \, a^{2} b^{2} c^{4} + 226 \, a^{3} b c^{3} d + 59 \, a^{4} c^{2} d^{2}\right )} x^{2} - 136 \, {\left (a^{3} b c^{4} + a^{4} c^{3} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{384 \, a^{2} c^{2} x^{4}}, -\frac {15 \, {\left (b^{4} c^{4} - 20 \, a b^{3} c^{3} d - 90 \, a^{2} b^{2} c^{2} d^{2} - 20 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {-a c} x^{4} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) + 960 \, {\left (a^{2} b^{2} c^{3} d + a^{3} b c^{2} d^{2}\right )} \sqrt {-b d} x^{4} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (192 \, a^{2} b^{2} c^{2} d^{2} x^{4} - 48 \, a^{4} c^{4} - {\left (15 \, a b^{3} c^{4} + 601 \, a^{2} b^{2} c^{3} d + 601 \, a^{3} b c^{2} d^{2} + 15 \, a^{4} c d^{3}\right )} x^{3} - 2 \, {\left (59 \, a^{2} b^{2} c^{4} + 226 \, a^{3} b c^{3} d + 59 \, a^{4} c^{2} d^{2}\right )} x^{2} - 136 \, {\left (a^{3} b c^{4} + a^{4} c^{3} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{384 \, a^{2} c^{2} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(5/2)/x^5,x, algorithm="fricas")

[Out]

[1/768*(960*(a^2*b^2*c^3*d + a^3*b*c^2*d^2)*sqrt(b*d)*x^4*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 +
4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 15*(b^4*c^4 - 20*a*
b^3*c^3*d - 90*a^2*b^2*c^2*d^2 - 20*a^3*b*c*d^3 + a^4*d^4)*sqrt(a*c)*x^4*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d
+ a^2*d^2)*x^2 + 4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x
^2) + 4*(192*a^2*b^2*c^2*d^2*x^4 - 48*a^4*c^4 - (15*a*b^3*c^4 + 601*a^2*b^2*c^3*d + 601*a^3*b*c^2*d^2 + 15*a^4
*c*d^3)*x^3 - 2*(59*a^2*b^2*c^4 + 226*a^3*b*c^3*d + 59*a^4*c^2*d^2)*x^2 - 136*(a^3*b*c^4 + a^4*c^3*d)*x)*sqrt(
b*x + a)*sqrt(d*x + c))/(a^2*c^2*x^4), -1/768*(1920*(a^2*b^2*c^3*d + a^3*b*c^2*d^2)*sqrt(-b*d)*x^4*arctan(1/2*
(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x))
- 15*(b^4*c^4 - 20*a*b^3*c^3*d - 90*a^2*b^2*c^2*d^2 - 20*a^3*b*c*d^3 + a^4*d^4)*sqrt(a*c)*x^4*log((8*a^2*c^2 +
(b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 + 4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*
b*c^2 + a^2*c*d)*x)/x^2) - 4*(192*a^2*b^2*c^2*d^2*x^4 - 48*a^4*c^4 - (15*a*b^3*c^4 + 601*a^2*b^2*c^3*d + 601*a
^3*b*c^2*d^2 + 15*a^4*c*d^3)*x^3 - 2*(59*a^2*b^2*c^4 + 226*a^3*b*c^3*d + 59*a^4*c^2*d^2)*x^2 - 136*(a^3*b*c^4
+ a^4*c^3*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^2*x^4), -1/384*(15*(b^4*c^4 - 20*a*b^3*c^3*d - 90*a^2*b^2*
c^2*d^2 - 20*a^3*b*c*d^3 + a^4*d^4)*sqrt(-a*c)*x^4*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)
*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) - 480*(a^2*b^2*c^3*d + a^3*b*c^2*d^2)*sqrt(b*d
)*x^4*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt
(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 2*(192*a^2*b^2*c^2*d^2*x^4 - 48*a^4*c^4 - (15*a*b^3*c^4 + 601*a^2*b^2*c
^3*d + 601*a^3*b*c^2*d^2 + 15*a^4*c*d^3)*x^3 - 2*(59*a^2*b^2*c^4 + 226*a^3*b*c^3*d + 59*a^4*c^2*d^2)*x^2 - 136
*(a^3*b*c^4 + a^4*c^3*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^2*x^4), -1/384*(15*(b^4*c^4 - 20*a*b^3*c^3*d -
90*a^2*b^2*c^2*d^2 - 20*a^3*b*c*d^3 + a^4*d^4)*sqrt(-a*c)*x^4*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*s
qrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) + 960*(a^2*b^2*c^3*d + a^3*b*c^2*d
^2)*sqrt(-b*d)*x^4*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*
c*d + (b^2*c*d + a*b*d^2)*x)) - 2*(192*a^2*b^2*c^2*d^2*x^4 - 48*a^4*c^4 - (15*a*b^3*c^4 + 601*a^2*b^2*c^3*d +
601*a^3*b*c^2*d^2 + 15*a^4*c*d^3)*x^3 - 2*(59*a^2*b^2*c^4 + 226*a^3*b*c^3*d + 59*a^4*c^2*d^2)*x^2 - 136*(a^3*b
*c^4 + a^4*c^3*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*c^2*x^4)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {5}{2}}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)*(d*x+c)**(5/2)/x**5,x)

[Out]

Integral((a + b*x)**(5/2)*(c + d*x)**(5/2)/x**5, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 3937 vs. \(2 (324) = 648\).
time = 4.89, size = 3937, normalized size = 10.36 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(5/2)/x^5,x, algorithm="giac")

[Out]

1/192*(192*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*b*d^2*abs(b) - 480*(sqrt(b*d)*b^2*c*d*abs(b) + sq
rt(b*d)*a*b*d^2*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2) + 15*(sqrt(b*d)
*b^5*c^4*abs(b) - 20*sqrt(b*d)*a*b^4*c^3*d*abs(b) - 90*sqrt(b*d)*a^2*b^3*c^2*d^2*abs(b) - 20*sqrt(b*d)*a^3*b^2
*c*d^3*abs(b) + sqrt(b*d)*a^4*b*d^4*abs(b))*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c
+ (b*x + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*a*b*c) - 2*(15*sqrt(b*d)*b^19*c^11*abs(b) +
481*sqrt(b*d)*a*b^18*c^10*d*abs(b) - 3787*sqrt(b*d)*a^2*b^17*c^9*d^2*abs(b) + 11195*sqrt(b*d)*a^3*b^16*c^8*d^3
*abs(b) - 15898*sqrt(b*d)*a^4*b^15*c^7*d^4*abs(b) + 7994*sqrt(b*d)*a^5*b^14*c^6*d^5*abs(b) + 7994*sqrt(b*d)*a^
6*b^13*c^5*d^6*abs(b) - 15898*sqrt(b*d)*a^7*b^12*c^4*d^7*abs(b) + 11195*sqrt(b*d)*a^8*b^11*c^3*d^8*abs(b) - 37
87*sqrt(b*d)*a^9*b^10*c^2*d^9*abs(b) + 481*sqrt(b*d)*a^10*b^9*c*d^10*abs(b) + 15*sqrt(b*d)*a^11*b^8*d^11*abs(b
) - 105*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^17*c^10*abs(b) - 3446*sq
rt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^16*c^9*d*abs(b) + 15371*sqrt(b*d
)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^15*c^8*d^2*abs(b) - 18056*sqrt(b*d)*
(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^14*c^7*d^3*abs(b) - 8354*sqrt(b*d)*(sq
rt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^13*c^6*d^4*abs(b) + 29180*sqrt(b*d)*(sqrt
(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^5*b^12*c^5*d^5*abs(b) - 8354*sqrt(b*d)*(sqrt(b*
d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^6*b^11*c^4*d^6*abs(b) - 18056*sqrt(b*d)*(sqrt(b*d)
*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^7*b^10*c^3*d^7*abs(b) + 15371*sqrt(b*d)*(sqrt(b*d)*s
qrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^8*b^9*c^2*d^8*abs(b) - 3446*sqrt(b*d)*(sqrt(b*d)*sqrt(
b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^9*b^8*c*d^9*abs(b) - 105*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a
) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^10*b^7*d^10*abs(b) + 315*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqr
t(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^15*c^9*abs(b) + 10867*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^4*a*b^14*c^8*d*abs(b) - 19124*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x +
a)*b*d - a*b*d))^4*a^2*b^13*c^7*d^2*abs(b) - 10980*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a
)*b*d - a*b*d))^4*a^3*b^12*c^6*d^3*abs(b) + 18922*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*
b*d - a*b*d))^4*a^4*b^11*c^5*d^4*abs(b) + 18922*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*
d - a*b*d))^4*a^5*b^10*c^4*d^5*abs(b) - 10980*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d
- a*b*d))^4*a^6*b^9*c^3*d^6*abs(b) - 19124*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a
*b*d))^4*a^7*b^8*c^2*d^7*abs(b) + 10867*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*
d))^4*a^8*b^7*c*d^8*abs(b) + 315*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a
^9*b^6*d^9*abs(b) - 525*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^13*c^8*a
bs(b) - 19480*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^12*c^7*d*abs(b)
- 4460*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^2*b^11*c^6*d^2*abs(b) + 1
9288*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^3*b^10*c^5*d^3*abs(b) + 103
54*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^4*b^9*c^4*d^4*abs(b) + 19288*
sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^5*b^8*c^3*d^5*abs(b) - 4460*sqrt
(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^6*b^7*c^2*d^6*abs(b) - 19480*sqrt(b*
d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^7*b^6*c*d^7*abs(b) - 525*sqrt(b*d)*(sqr
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^8*b^5*d^8*abs(b) + 525*sqrt(b*d)*(sqrt(b*d)*sq
rt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*b^11*c^7*abs(b) + 21335*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a
) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a*b^10*c^6*d*abs(b) + 33325*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sq
rt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a^2*b^9*c^5*d^2*abs(b) + 23663*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(
b^2*c + (b*x + a)*b*d - a*b*d))^8*a^3*b^8*c^4*d^3*abs(b) + 23663*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2
*c + (b*x + a)*b*d - a*b*d))^8*a^4*b^7*c^3*d^4*abs(b) + 33325*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c
+ (b*x + a)*b*d - a*b*d))^8*a^5*b^6*c^2*d^5*abs(b) + 21335*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (
b*x + a)*b*d - a*b*d))^8*a^6*b^5*c*d^6*abs(b) + 525*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a
)*b*d - a*b*d))^8*a^7*b^4*d^7*abs(b) - 315*sqrt...

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{5/2}}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^(5/2)*(c + d*x)^(5/2))/x^5,x)

[Out]

int(((a + b*x)^(5/2)*(c + d*x)^(5/2))/x^5, x)

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