3.669 \(\int \frac {(a+b x)^{5/2} (c+d x)^{5/2}}{x^3} \, dx\)

Optimal. Leaf size=319 \[ \frac {5 \sqrt {a+b x} (c+d x)^{3/2} \left (3 a^2 d^2+8 a b c d+b^2 c^2\right )}{12 c}+\frac {5}{8} \sqrt {a+b x} \sqrt {c+d x} \left (5 a^2 d^2+10 a b c d+b^2 c^2\right )+\frac {5 (a d+b c) \left (a^2 d^2+14 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} \sqrt {d}}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{2 x^2}-\frac {5 (a+b x)^{3/2} (c+d x)^{5/2} (a d+b c)}{4 c x}+\frac {5 b \sqrt {a+b x} (c+d x)^{5/2} (3 a d+5 b c)}{12 c}-\frac {5}{4} \sqrt {a} \sqrt {c} (a d+3 b c) (3 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right ) \]

[Out]

-5/4*(a*d+b*c)*(b*x+a)^(3/2)*(d*x+c)^(5/2)/c/x-1/2*(b*x+a)^(5/2)*(d*x+c)^(5/2)/x^2-5/4*(a*d+3*b*c)*(3*a*d+b*c)
*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))*a^(1/2)*c^(1/2)+5/8*(a*d+b*c)*(a^2*d^2+14*a*b*c*d+b^2*c^
2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(1/2)/d^(1/2)+5/12*(3*a^2*d^2+8*a*b*c*d+b^2*c^2)*(d*
x+c)^(3/2)*(b*x+a)^(1/2)/c+5/12*b*(3*a*d+5*b*c)*(d*x+c)^(5/2)*(b*x+a)^(1/2)/c+5/8*(5*a^2*d^2+10*a*b*c*d+b^2*c^
2)*(b*x+a)^(1/2)*(d*x+c)^(1/2)

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Rubi [A]  time = 0.39, antiderivative size = 319, 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 = {97, 149, 154, 157, 63, 217, 206, 93, 208} \[ \frac {5 \sqrt {a+b x} (c+d x)^{3/2} \left (3 a^2 d^2+8 a b c d+b^2 c^2\right )}{12 c}+\frac {5}{8} \sqrt {a+b x} \sqrt {c+d x} \left (5 a^2 d^2+10 a b c d+b^2 c^2\right )+\frac {5 (a d+b c) \left (a^2 d^2+14 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} \sqrt {d}}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{2 x^2}-\frac {5 (a+b x)^{3/2} (c+d x)^{5/2} (a d+b c)}{4 c x}+\frac {5 b \sqrt {a+b x} (c+d x)^{5/2} (3 a d+5 b c)}{12 c}-\frac {5}{4} \sqrt {a} \sqrt {c} (a d+3 b c) (3 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(c + d*x)^(5/2))/x^3,x]

[Out]

(5*(b^2*c^2 + 10*a*b*c*d + 5*a^2*d^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/8 + (5*(b^2*c^2 + 8*a*b*c*d + 3*a^2*d^2)*Sq
rt[a + b*x]*(c + d*x)^(3/2))/(12*c) + (5*b*(5*b*c + 3*a*d)*Sqrt[a + b*x]*(c + d*x)^(5/2))/(12*c) - (5*(b*c + a
*d)*(a + b*x)^(3/2)*(c + d*x)^(5/2))/(4*c*x) - ((a + b*x)^(5/2)*(c + d*x)^(5/2))/(2*x^2) - (5*Sqrt[a]*Sqrt[c]*
(3*b*c + a*d)*(b*c + 3*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/4 + (5*(b*c + a*d)*(b^2*
c^2 + 14*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*Sqrt[b]*Sqrt[d])

Rule 63

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 93

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 97

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 149

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] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

Rule 154

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 157

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 206

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

Rule 208

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

Rule 217

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^3} \, dx &=-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{2 x^2}+\frac {1}{2} \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^2} \, dx\\ &=-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{2 x^2}+\frac {\int \frac {\sqrt {a+b x} (c+d x)^{3/2} \left (\frac {5}{4} (3 b c+a d) (b c+3 a d)+\frac {5}{2} b d (5 b c+3 a d) x\right )}{x} \, dx}{2 c}\\ &=\frac {5 b (5 b c+3 a d) \sqrt {a+b x} (c+d x)^{5/2}}{12 c}-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{2 x^2}+\frac {\int \frac {(c+d x)^{3/2} \left (\frac {15}{4} a d (3 b c+a d) (b c+3 a d)+5 b d \left (b^2 c^2+8 a b c d+3 a^2 d^2\right ) x\right )}{x \sqrt {a+b x}} \, dx}{6 c d}\\ &=\frac {5 \left (b^2 c^2+8 a b c d+3 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 c}+\frac {5 b (5 b c+3 a d) \sqrt {a+b x} (c+d x)^{5/2}}{12 c}-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{2 x^2}+\frac {\int \frac {\sqrt {c+d x} \left (\frac {15}{2} a b c d (3 b c+a d) (b c+3 a d)+\frac {15}{2} b^2 c d \left (b^2 c^2+10 a b c d+5 a^2 d^2\right ) x\right )}{x \sqrt {a+b x}} \, dx}{12 b c d}\\ &=\frac {5}{8} \left (b^2 c^2+10 a b c d+5 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}+\frac {5 \left (b^2 c^2+8 a b c d+3 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 c}+\frac {5 b (5 b c+3 a d) \sqrt {a+b x} (c+d x)^{5/2}}{12 c}-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{2 x^2}+\frac {\int \frac {\frac {15}{2} a b^2 c^2 d (3 b c+a d) (b c+3 a d)+\frac {15}{4} b^2 c d (b c+a d) \left (b^2 c^2+14 a b c d+a^2 d^2\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{12 b^2 c d}\\ &=\frac {5}{8} \left (b^2 c^2+10 a b c d+5 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}+\frac {5 \left (b^2 c^2+8 a b c d+3 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 c}+\frac {5 b (5 b c+3 a d) \sqrt {a+b x} (c+d x)^{5/2}}{12 c}-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{2 x^2}+\frac {1}{8} (5 a c (3 b c+a d) (b c+3 a d)) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx+\frac {1}{16} \left (5 (b c+a d) \left (b^2 c^2+14 a b c d+a^2 d^2\right )\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx\\ &=\frac {5}{8} \left (b^2 c^2+10 a b c d+5 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}+\frac {5 \left (b^2 c^2+8 a b c d+3 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 c}+\frac {5 b (5 b c+3 a d) \sqrt {a+b x} (c+d x)^{5/2}}{12 c}-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{2 x^2}+\frac {1}{4} (5 a c (3 b c+a d) (b c+3 a d)) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )+\frac {\left (5 (b c+a d) \left (b^2 c^2+14 a b c d+a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{8 b}\\ &=\frac {5}{8} \left (b^2 c^2+10 a b c d+5 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}+\frac {5 \left (b^2 c^2+8 a b c d+3 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 c}+\frac {5 b (5 b c+3 a d) \sqrt {a+b x} (c+d x)^{5/2}}{12 c}-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{2 x^2}-\frac {5}{4} \sqrt {a} \sqrt {c} (3 b c+a d) (b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {\left (5 (b c+a d) \left (b^2 c^2+14 a b c d+a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{8 b}\\ &=\frac {5}{8} \left (b^2 c^2+10 a b c d+5 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}+\frac {5 \left (b^2 c^2+8 a b c d+3 a^2 d^2\right ) \sqrt {a+b x} (c+d x)^{3/2}}{12 c}+\frac {5 b (5 b c+3 a d) \sqrt {a+b x} (c+d x)^{5/2}}{12 c}-\frac {5 (b c+a d) (a+b x)^{3/2} (c+d x)^{5/2}}{4 c x}-\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{2 x^2}-\frac {5}{4} \sqrt {a} \sqrt {c} (3 b c+a d) (b c+3 a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {5 (b c+a d) \left (b^2 c^2+14 a b c d+a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 \sqrt {b} \sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 2.79, size = 282, normalized size = 0.88 \[ \frac {\sqrt {a+b x} \sqrt {c+d x} \left (-3 a^2 \left (4 c^2+18 c d x-11 d^2 x^2\right )+2 a b x \left (-27 c^2+61 c d x+13 d^2 x^2\right )+b^2 x^2 \left (33 c^2+26 c d x+8 d^2 x^2\right )\right )}{24 x^2}-\frac {5}{4} \sqrt {a} \sqrt {c} \left (3 a^2 d^2+10 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {5 \sqrt {c+d x} \left (a^3 d^3+15 a^2 b c d^2+15 a b^2 c^2 d+b^3 c^3\right ) \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{8 \sqrt {d} \sqrt {b c-a d} \sqrt {\frac {b (c+d x)}{b c-a d}}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*(c + d*x)^(5/2))/x^3,x]

[Out]

(Sqrt[a + b*x]*Sqrt[c + d*x]*(-3*a^2*(4*c^2 + 18*c*d*x - 11*d^2*x^2) + b^2*x^2*(33*c^2 + 26*c*d*x + 8*d^2*x^2)
+ 2*a*b*x*(-27*c^2 + 61*c*d*x + 13*d^2*x^2)))/(24*x^2) + (5*(b^3*c^3 + 15*a*b^2*c^2*d + 15*a^2*b*c*d^2 + a^3*
d^3)*Sqrt[c + d*x]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(8*Sqrt[d]*Sqrt[b*c - a*d]*Sqrt[(b*(c + d
*x))/(b*c - a*d)]) - (5*Sqrt[a]*Sqrt[c]*(3*b^2*c^2 + 10*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(
Sqrt[a]*Sqrt[c + d*x])])/4

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fricas [A]  time = 17.47, size = 1469, normalized size = 4.61 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(5/2)/x^3,x, algorithm="fricas")

[Out]

[1/96*(15*(b^3*c^3 + 15*a*b^2*c^2*d + 15*a^2*b*c*d^2 + a^3*d^3)*sqrt(b*d)*x^2*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) +
30*(3*b^3*c^2*d + 10*a*b^2*c*d^2 + 3*a^2*b*d^3)*sqrt(a*c)*x^2*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*(8
*b^3*d^3*x^4 - 12*a^2*b*c^2*d + 26*(b^3*c*d^2 + a*b^2*d^3)*x^3 + (33*b^3*c^2*d + 122*a*b^2*c*d^2 + 33*a^2*b*d^
3)*x^2 - 54*(a*b^2*c^2*d + a^2*b*c*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d*x^2), -1/48*(15*(b^3*c^3 + 15*a*b
^2*c^2*d + 15*a^2*b*c*d^2 + a^3*d^3)*sqrt(-b*d)*x^2*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*(3*b^3*c^2*d + 10*a*b^2*c*d^2 + 3*a^2*b*d^
3)*sqrt(a*c)*x^2*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)*sq
rt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 2*(8*b^3*d^3*x^4 - 12*a^2*b*c^2*d + 26*(b^3*c*d^2
+ a*b^2*d^3)*x^3 + (33*b^3*c^2*d + 122*a*b^2*c*d^2 + 33*a^2*b*d^3)*x^2 - 54*(a*b^2*c^2*d + a^2*b*c*d^2)*x)*sqr
t(b*x + a)*sqrt(d*x + c))/(b*d*x^2), 1/96*(60*(3*b^3*c^2*d + 10*a*b^2*c*d^2 + 3*a^2*b*d^3)*sqrt(-a*c)*x^2*arct
an(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)) + 15*(b^3*c^3 + 15*a*b^2*c^2*d + 15*a^2*b*c*d^2 + a^3*d^3)*sqrt(b*d)*x^2*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) + 4*(8*b^3*d^3*x^4 - 12*a^2*b*c^2*d + 26*(b^3*c*d^2 + a*b^2*d^3)*x^3 + (33*b^3*c^2*d + 122*a*b^2*c*d^2 + 33
*a^2*b*d^3)*x^2 - 54*(a*b^2*c^2*d + a^2*b*c*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d*x^2), 1/48*(30*(3*b^3*c^
2*d + 10*a*b^2*c*d^2 + 3*a^2*b*d^3)*sqrt(-a*c)*x^2*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)) - 15*(b^3*c^3 + 15*a*b^2*c^2*d + 15*a^2*b*c*d^
2 + a^3*d^3)*sqrt(-b*d)*x^2*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*(8*b^3*d^3*x^4 - 12*a^2*b*c^2*d + 26*(b^3*c*d^2 + a*b^2*d^3)*x^3 +
(33*b^3*c^2*d + 122*a*b^2*c*d^2 + 33*a^2*b*d^3)*x^2 - 54*(a*b^2*c^2*d + a^2*b*c*d^2)*x)*sqrt(b*x + a)*sqrt(d*x
+ c))/(b*d*x^2)]

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giac [B]  time = 20.40, size = 1336, normalized size = 4.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(5/2)/x^3,x, algorithm="giac")

[Out]

1/48*(2*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(4*(b*x + a)*d^2*abs(b)/b + (13*b*c*d^5*abs(b) + 5*a*d^6*abs(b)
)/(b*d^4))*(b*x + a) + 3*(11*b^2*c^2*d^4*abs(b) + 32*a*b*c*d^5*abs(b) + 5*a^2*d^6*abs(b))/(b*d^4))*sqrt(b*x +
a) - 60*(3*sqrt(b*d)*a*b^3*c^3*abs(b) + 10*sqrt(b*d)*a^2*b^2*c^2*d*abs(b) + 3*sqrt(b*d)*a^3*b*c*d^2*abs(b))*ar
ctan(-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)*b) - 15*(sqrt(b*d)*b^3*c^3*abs(b) + 15*sqrt(b*d)*a*b^2*c^2*d*abs(b) + 15*sqrt(b*d)*a^2*b*c
*d^2*abs(b) + sqrt(b*d)*a^3*d^3*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)
/(b*d) - 24*(9*sqrt(b*d)*a*b^9*c^6*abs(b) - 27*sqrt(b*d)*a^2*b^8*c^5*d*abs(b) + 18*sqrt(b*d)*a^3*b^7*c^4*d^2*a
bs(b) + 18*sqrt(b*d)*a^4*b^6*c^3*d^3*abs(b) - 27*sqrt(b*d)*a^5*b^5*c^2*d^4*abs(b) + 9*sqrt(b*d)*a^6*b^4*c*d^5*
abs(b) - 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^7*c^5*abs(b) + 4*s
qrt(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^6*c^4*d*abs(b) + 46*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^5*c^3*d^2*abs(b) + 4*sqrt(b*d)*(sqrt(
b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^4*c^2*d^3*abs(b) - 27*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^5*b^3*c*d^4*abs(b) + 27*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^5*c^4*abs(b) + 45*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^4*c^3*d*abs(b) + 45*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^3*c^2*d^2*abs(b) + 27*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^2*c*d^3*abs(b) - 9*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^3*c^3*abs(b) - 22*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^2*c^2*d*abs(b) - 9*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*c*d^
2*abs(b))/(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b
*d))^2*b^2*c - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*d)*sqrt(b*x
+ a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)^2)/b

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maple [B]  time = 0.02, size = 850, normalized size = 2.66 \[ \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (-90 \sqrt {b d}\, a^{3} c \,d^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+15 \sqrt {a c}\, a^{3} d^{3} x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-300 \sqrt {b d}\, a^{2} b \,c^{2} d \,x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+225 \sqrt {a c}\, a^{2} b c \,d^{2} x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-90 \sqrt {b d}\, a \,b^{2} c^{3} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+225 \sqrt {a c}\, a \,b^{2} c^{2} d \,x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+15 \sqrt {a c}\, b^{3} c^{3} x^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+16 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{2} d^{2} x^{4}+52 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a b \,d^{2} x^{3}+52 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{2} c d \,x^{3}+66 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} d^{2} x^{2}+244 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a b c d \,x^{2}+66 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{2} c^{2} x^{2}-108 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} c d x -108 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a b \,c^{2} x -24 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a^{2} c^{2}\right )}{48 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(d*x+c)^(5/2)/x^3,x)

[Out]

1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(16*x^4*b^2*d^2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)+15*ln
(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*(a*c)^(1/2)*x^2*a^3*d^3+225*
ln(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*(a*c)^(1/2)*x^2*a^2*b*c*d^
2+225*ln(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*(a*c)^(1/2)*x^2*a*b^
2*c^2*d+15*ln(1/2*(2*b*d*x+a*d+b*c+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*(a*c)^(1/2)*x^2
*b^3*c^3-90*(b*d)^(1/2)*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2))/x)*x^2*a^3*c*d^2-
300*(b*d)^(1/2)*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2))/x)*x^2*a^2*b*c^2*d-90*(b*
d)^(1/2)*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2))/x)*x^2*a*b^2*c^3+52*(b*d*x^2+a*d
*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a*b*d^2*x^3+52*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(
1/2)*b^2*c*d*x^3+66*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^2*d^2*x^2+244*(b*d*x^2+a*d*x+b*c
*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a*b*c*d*x^2+66*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*b
^2*c^2*x^2-108*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^2*c*d*x-108*(b*d*x^2+a*d*x+b*c*x+a*c)
^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a*b*c^2*x-24*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*a^2*c^2)/(
b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)/(b*d)^(1/2)/x^2/(a*c)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(d*x+c)^(5/2)/x^3,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 details)Is a*d-b*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{5/2}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^(5/2)*(c + d*x)^(5/2))/x^3,x)

[Out]

int(((a + b*x)^(5/2)*(c + d*x)^(5/2))/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {5}{2}}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)*(d*x+c)**(5/2)/x**3,x)

[Out]

Integral((a + b*x)**(5/2)*(c + d*x)**(5/2)/x**3, x)

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