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

Optimal. Leaf size=304 \[ -2 a^{3/2} c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {1}{96} \sqrt {a+b x} (c+d x)^{3/2} \left (\frac {3 a^2 d}{b}+50 a c-\frac {5 b c^2}{d}\right )-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (3 a^3 d^3-17 a^2 b c d^2-55 a b^2 c^2 d+5 b^3 c^3\right )}{64 b^2 d}-\frac {\left (-3 a^4 d^4+20 a^3 b c d^3-90 a^2 b^2 c^2 d^2-60 a b^3 c^3 d+5 b^4 c^4\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{5/2} d^{3/2}}+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}+\frac {\sqrt {a+b x} (c+d x)^{5/2} (3 a d+5 b c)}{24 d} \]

[Out]

1/4*(b*x+a)^(3/2)*(d*x+c)^(5/2)-2*a^(3/2)*c^(5/2)*arctanh(c^(1/2)*(b*x+a)^(1/2)/a^(1/2)/(d*x+c)^(1/2))-1/64*(-
3*a^4*d^4+20*a^3*b*c*d^3-90*a^2*b^2*c^2*d^2-60*a*b^3*c^3*d+5*b^4*c^4)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d
*x+c)^(1/2))/b^(5/2)/d^(3/2)+1/96*(50*a*c-5*b*c^2/d+3*a^2*d/b)*(d*x+c)^(3/2)*(b*x+a)^(1/2)+1/24*(3*a*d+5*b*c)*
(d*x+c)^(5/2)*(b*x+a)^(1/2)/d-1/64*(3*a^3*d^3-17*a^2*b*c*d^2-55*a*b^2*c^2*d+5*b^3*c^3)*(b*x+a)^(1/2)*(d*x+c)^(
1/2)/b^2/d

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Rubi [A]  time = 0.29, antiderivative size = 304, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {101, 154, 157, 63, 217, 206, 93, 208} \[ -\frac {\sqrt {a+b x} \sqrt {c+d x} \left (-17 a^2 b c d^2+3 a^3 d^3-55 a b^2 c^2 d+5 b^3 c^3\right )}{64 b^2 d}-\frac {\left (-90 a^2 b^2 c^2 d^2+20 a^3 b c d^3-3 a^4 d^4-60 a b^3 c^3 d+5 b^4 c^4\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{5/2} d^{3/2}}+\frac {1}{96} \sqrt {a+b x} (c+d x)^{3/2} \left (\frac {3 a^2 d}{b}+50 a c-\frac {5 b c^2}{d}\right )-2 a^{3/2} c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}+\frac {\sqrt {a+b x} (c+d x)^{5/2} (3 a d+5 b c)}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((5*b^3*c^3 - 55*a*b^2*c^2*d - 17*a^2*b*c*d^2 + 3*a^3*d^3)*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b^2*d) + ((50*a*c
- (5*b*c^2)/d + (3*a^2*d)/b)*Sqrt[a + b*x]*(c + d*x)^(3/2))/96 + ((5*b*c + 3*a*d)*Sqrt[a + b*x]*(c + d*x)^(5/
2))/(24*d) + ((a + b*x)^(3/2)*(c + d*x)^(5/2))/4 - 2*a^(3/2)*c^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*
Sqrt[c + d*x])] - ((5*b^4*c^4 - 60*a*b^3*c^3*d - 90*a^2*b^2*c^2*d^2 + 20*a^3*b*c*d^3 - 3*a^4*d^4)*ArcTanh[(Sqr
t[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(5/2)*d^(3/2))

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 101

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a +
b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1))/(f*(m + n + p + 1)), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[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[(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)^{3/2} (c+d x)^{5/2}}{x} \, dx &=\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}-\frac {1}{4} \int \frac {\sqrt {a+b x} (c+d x)^{3/2} \left (-4 a c+\frac {1}{2} (-5 b c-3 a d) x\right )}{x} \, dx\\ &=\frac {(5 b c+3 a d) \sqrt {a+b x} (c+d x)^{5/2}}{24 d}+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}-\frac {\int \frac {(c+d x)^{3/2} \left (-12 a^2 c d+\frac {1}{4} \left (5 b^2 c^2-50 a b c d-3 a^2 d^2\right ) x\right )}{x \sqrt {a+b x}} \, dx}{12 d}\\ &=\frac {1}{96} \left (50 a c-\frac {5 b c^2}{d}+\frac {3 a^2 d}{b}\right ) \sqrt {a+b x} (c+d x)^{3/2}+\frac {(5 b c+3 a d) \sqrt {a+b x} (c+d x)^{5/2}}{24 d}+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}-\frac {\int \frac {\sqrt {c+d x} \left (-24 a^2 b c^2 d+\frac {3}{8} \left (5 b^3 c^3-55 a b^2 c^2 d-17 a^2 b c d^2+3 a^3 d^3\right ) x\right )}{x \sqrt {a+b x}} \, dx}{24 b d}\\ &=-\frac {\left (5 b^3 c^3-55 a b^2 c^2 d-17 a^2 b c d^2+3 a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^2 d}+\frac {1}{96} \left (50 a c-\frac {5 b c^2}{d}+\frac {3 a^2 d}{b}\right ) \sqrt {a+b x} (c+d x)^{3/2}+\frac {(5 b c+3 a d) \sqrt {a+b x} (c+d x)^{5/2}}{24 d}+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}-\frac {\int \frac {-24 a^2 b^2 c^3 d+\frac {3}{16} \left (5 b^4 c^4-60 a b^3 c^3 d-90 a^2 b^2 c^2 d^2+20 a^3 b c d^3-3 a^4 d^4\right ) x}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{24 b^2 d}\\ &=-\frac {\left (5 b^3 c^3-55 a b^2 c^2 d-17 a^2 b c d^2+3 a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^2 d}+\frac {1}{96} \left (50 a c-\frac {5 b c^2}{d}+\frac {3 a^2 d}{b}\right ) \sqrt {a+b x} (c+d x)^{3/2}+\frac {(5 b c+3 a d) \sqrt {a+b x} (c+d x)^{5/2}}{24 d}+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}+\left (a^2 c^3\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx-\frac {\left (5 b^4 c^4-60 a b^3 c^3 d-90 a^2 b^2 c^2 d^2+20 a^3 b c d^3-3 a^4 d^4\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{128 b^2 d}\\ &=-\frac {\left (5 b^3 c^3-55 a b^2 c^2 d-17 a^2 b c d^2+3 a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^2 d}+\frac {1}{96} \left (50 a c-\frac {5 b c^2}{d}+\frac {3 a^2 d}{b}\right ) \sqrt {a+b x} (c+d x)^{3/2}+\frac {(5 b c+3 a d) \sqrt {a+b x} (c+d x)^{5/2}}{24 d}+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}+\left (2 a^2 c^3\right ) \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^4 c^4-60 a b^3 c^3 d-90 a^2 b^2 c^2 d^2+20 a^3 b c d^3-3 a^4 d^4\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 )}{64 b^3 d}\\ &=-\frac {\left (5 b^3 c^3-55 a b^2 c^2 d-17 a^2 b c d^2+3 a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^2 d}+\frac {1}{96} \left (50 a c-\frac {5 b c^2}{d}+\frac {3 a^2 d}{b}\right ) \sqrt {a+b x} (c+d x)^{3/2}+\frac {(5 b c+3 a d) \sqrt {a+b x} (c+d x)^{5/2}}{24 d}+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}-2 a^{3/2} c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )-\frac {\left (5 b^4 c^4-60 a b^3 c^3 d-90 a^2 b^2 c^2 d^2+20 a^3 b c d^3-3 a^4 d^4\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 )}{64 b^3 d}\\ &=-\frac {\left (5 b^3 c^3-55 a b^2 c^2 d-17 a^2 b c d^2+3 a^3 d^3\right ) \sqrt {a+b x} \sqrt {c+d x}}{64 b^2 d}+\frac {1}{96} \left (50 a c-\frac {5 b c^2}{d}+\frac {3 a^2 d}{b}\right ) \sqrt {a+b x} (c+d x)^{3/2}+\frac {(5 b c+3 a d) \sqrt {a+b x} (c+d x)^{5/2}}{24 d}+\frac {1}{4} (a+b x)^{3/2} (c+d x)^{5/2}-2 a^{3/2} c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )-\frac {\left (5 b^4 c^4-60 a b^3 c^3 d-90 a^2 b^2 c^2 d^2+20 a^3 b c d^3-3 a^4 d^4\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{5/2} d^{3/2}}\\ \end {align*}

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Mathematica [B]  time = 2.93, size = 910, normalized size = 2.99 \[ \frac {\sqrt {c+d x} \left (-15 b^4 (c+d x)^2 \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right ) c^4+15 b \sqrt {d} (b c-a d)^{5/2} \sqrt {a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/2} c^3+180 a b^3 d (c+d x)^2 \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right ) c^3-384 a^{3/2} b d^{3/2} (b c-a d)^{3/2} \sqrt {c+d x} \left (\frac {b (c+d x)}{b c-a d}\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right ) c^{5/2}+337 a d^{3/2} (b c-a d)^{5/2} \sqrt {a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/2} c^2+118 b d^{3/2} (b c-a d)^{5/2} x \sqrt {a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/2} c^2+270 a^2 b^2 d^2 (c+d x)^2 \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right ) c^2+\frac {57 a^2 d^{5/2} (b c-a d)^{5/2} \sqrt {a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/2} c}{b}+136 b d^{5/2} (b c-a d)^{5/2} x^2 \sqrt {a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/2} c+244 a d^{5/2} (b c-a d)^{5/2} x \sqrt {a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/2} c-60 a^3 b d^3 (c+d x)^2 \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right ) c+48 b d^{7/2} (b c-a d)^{5/2} x^3 \sqrt {a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/2}+72 a d^{7/2} (b c-a d)^{5/2} x^2 \sqrt {a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/2}+\frac {6 a^2 d^{7/2} (b c-a d)^{5/2} x \sqrt {a+b x} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/2}}{b}+9 a^4 d^4 (c+d x)^2 \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )-9 a^3 d^{7/2} \sqrt {b c-a d} \sqrt {a+b x} (c+d x)^2 \sqrt {\frac {b (c+d x)}{b c-a d}}\right )}{192 d^{3/2} (b c-a d)^{5/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*(-9*a^3*d^(7/2)*Sqrt[b*c - a*d]*Sqrt[a + b*x]*(c + d*x)^2*Sqrt[(b*(c + d*x))/(b*c - a*d)] + 15*
b*c^3*Sqrt[d]*(b*c - a*d)^(5/2)*Sqrt[a + b*x]*((b*(c + d*x))/(b*c - a*d))^(5/2) + 337*a*c^2*d^(3/2)*(b*c - a*d
)^(5/2)*Sqrt[a + b*x]*((b*(c + d*x))/(b*c - a*d))^(5/2) + (57*a^2*c*d^(5/2)*(b*c - a*d)^(5/2)*Sqrt[a + b*x]*((
b*(c + d*x))/(b*c - a*d))^(5/2))/b + 118*b*c^2*d^(3/2)*(b*c - a*d)^(5/2)*x*Sqrt[a + b*x]*((b*(c + d*x))/(b*c -
a*d))^(5/2) + 244*a*c*d^(5/2)*(b*c - a*d)^(5/2)*x*Sqrt[a + b*x]*((b*(c + d*x))/(b*c - a*d))^(5/2) + (6*a^2*d^
(7/2)*(b*c - a*d)^(5/2)*x*Sqrt[a + b*x]*((b*(c + d*x))/(b*c - a*d))^(5/2))/b + 136*b*c*d^(5/2)*(b*c - a*d)^(5/
2)*x^2*Sqrt[a + b*x]*((b*(c + d*x))/(b*c - a*d))^(5/2) + 72*a*d^(7/2)*(b*c - a*d)^(5/2)*x^2*Sqrt[a + b*x]*((b*
(c + d*x))/(b*c - a*d))^(5/2) + 48*b*d^(7/2)*(b*c - a*d)^(5/2)*x^3*Sqrt[a + b*x]*((b*(c + d*x))/(b*c - a*d))^(
5/2) - 15*b^4*c^4*(c + d*x)^2*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]] + 180*a*b^3*c^3*d*(c + d*x)^2*A
rcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]] + 270*a^2*b^2*c^2*d^2*(c + d*x)^2*ArcSinh[(Sqrt[d]*Sqrt[a + b*
x])/Sqrt[b*c - a*d]] - 60*a^3*b*c*d^3*(c + d*x)^2*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]] + 9*a^4*d^4
*(c + d*x)^2*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]] - 384*a^(3/2)*b*c^(5/2)*d^(3/2)*(b*c - a*d)^(3/2
)*Sqrt[c + d*x]*((b*(c + d*x))/(b*c - a*d))^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(
192*d^(3/2)*(b*c - a*d)^(5/2)*((b*(c + d*x))/(b*c - a*d))^(5/2))

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fricas [A]  time = 33.40, size = 1481, normalized size = 4.87 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/768*(384*sqrt(a*c)*a*b^3*c^2*d^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) - 3*(5*b^4*c^4 - 60*a*b^3*c^3*d -
90*a^2*b^2*c^2*d^2 + 20*a^3*b*c*d^3 - 3*a^4*d^4)*sqrt(b*d)*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*(48*b^4*d^4*x^3
+ 15*b^4*c^3*d + 337*a*b^3*c^2*d^2 + 57*a^2*b^2*c*d^3 - 9*a^3*b*d^4 + 8*(17*b^4*c*d^3 + 9*a*b^3*d^4)*x^2 + 2*
(59*b^4*c^2*d^2 + 122*a*b^3*c*d^3 + 3*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d^2), 1/384*(192*sqrt(
a*c)*a*b^3*c^2*d^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) + 3*(5*b^4*c^4 - 60*a*b^3*c^3*d - 90*a^2*b^2*c^2*d
^2 + 20*a^3*b*c*d^3 - 3*a^4*d^4)*sqrt(-b*d)*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*(48*b^4*d^4*x^3 + 15*b^4*c^3*d + 337*a*b^3*c^2*d^2
+ 57*a^2*b^2*c*d^3 - 9*a^3*b*d^4 + 8*(17*b^4*c*d^3 + 9*a*b^3*d^4)*x^2 + 2*(59*b^4*c^2*d^2 + 122*a*b^3*c*d^3 +
3*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d^2), 1/768*(768*sqrt(-a*c)*a*b^3*c^2*d^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)) - 3
*(5*b^4*c^4 - 60*a*b^3*c^3*d - 90*a^2*b^2*c^2*d^2 + 20*a^3*b*c*d^3 - 3*a^4*d^4)*sqrt(b*d)*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*(48*b^4*d^4*x^3 + 15*b^4*c^3*d + 337*a*b^3*c^2*d^2 + 57*a^2*b^2*c*d^3 - 9*a^3*b*d^4 + 8*(17*b^4
*c*d^3 + 9*a*b^3*d^4)*x^2 + 2*(59*b^4*c^2*d^2 + 122*a*b^3*c*d^3 + 3*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c
))/(b^3*d^2), 1/384*(384*sqrt(-a*c)*a*b^3*c^2*d^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)) + 3*(5*b^4*c^4 - 60*a*b^3*c^3*d - 90*a^2*b^2*c^
2*d^2 + 20*a^3*b*c*d^3 - 3*a^4*d^4)*sqrt(-b*d)*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*(48*b^4*d^4*x^3 + 15*b^4*c^3*d + 337*a*b^3*c^2*d
^2 + 57*a^2*b^2*c*d^3 - 9*a^3*b*d^4 + 8*(17*b^4*c*d^3 + 9*a*b^3*d^4)*x^2 + 2*(59*b^4*c^2*d^2 + 122*a*b^3*c*d^3
+ 3*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 1.24

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maple [B]  time = 0.02, size = 828, normalized size = 2.72 \[ \frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (9 \sqrt {a c}\, a^{4} d^{4} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-60 \sqrt {a c}\, a^{3} b c \,d^{3} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-384 \sqrt {b d}\, a^{2} b^{2} c^{3} d \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}}{x}\right )+270 \sqrt {a c}\, a^{2} b^{2} c^{2} d^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+180 \sqrt {a c}\, a \,b^{3} c^{3} d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-15 \sqrt {a c}\, b^{4} c^{4} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+96 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{3} d^{3} x^{3}+144 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, a \,b^{2} d^{3} x^{2}+272 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{3} c \,d^{2} x^{2}+12 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{2} b \,d^{3} x +488 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a \,b^{2} c \,d^{2} x +236 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b^{3} c^{2} d x -18 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{3} d^{3}+114 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a^{2} b c \,d^{2}+674 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, a \,b^{2} c^{2} d +30 \sqrt {b d}\, \sqrt {a c}\, \sqrt {b d \,x^{2}+a d x +b c x +a c}\, b^{3} c^{3}\right )}{384 \sqrt {b d \,x^{2}+a d x +b c x +a c}\, \sqrt {b d}\, \sqrt {a c}\, b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(96*x^3*b^3*d^3*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)+144*
x^2*a*b^2*d^3*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)+272*x^2*b^3*c*d^2*(b*d*x^2+a*d*x+b*c*x+a
*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)+9*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)*a^4*d^4-60*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)*a^3*b*c*d^3+270*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)*a^2*b^2*c^2*d^2+180*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)*a*b^3*c^3*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)*b^4*c^4-384*(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)*a^2*b^2*c^3*d+12*(b*d)^(1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a^2*b*d^3+488*(b*d)^(
1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a*b^2*c*d^2+236*(b*d)^(1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*
x+a*c)^(1/2)*x*b^3*c^2*d-18*(b*d)^(1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^3*d^3+114*(b*d)^(1/2)*(a
*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^2*b*c*d^2+674*(b*d)^(1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1
/2)*a*b^2*c^2*d+30*(b*d)^(1/2)*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*b^3*c^3)/b^2/d/(b*d*x^2+a*d*x+b*c*x
+a*c)^(1/2)/(b*d)^(1/2)/(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)^(3/2)*(d*x+c)^(5/2)/x,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 \[ \int \frac {{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{5/2}}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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