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

Optimal. Leaf size=204 \[ -\frac{\left (35 a^2 d^2-30 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 \sqrt{a} c^{9/2}}-\frac{5 d \sqrt{a+b x} (11 b c-21 a d)}{12 c^4 \sqrt{c+d x}}-\frac{d \sqrt{a+b x} (23 b c-35 a d)}{12 c^3 (c+d x)^{3/2}}-\frac{\sqrt{a+b x} (5 b c-7 a d)}{4 c^2 x (c+d x)^{3/2}}-\frac{a \sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}} \]

[Out]

-(d*(23*b*c - 35*a*d)*Sqrt[a + b*x])/(12*c^3*(c + d*x)^(3/2)) - (a*Sqrt[a + b*x])/(2*c*x^2*(c + d*x)^(3/2)) -
((5*b*c - 7*a*d)*Sqrt[a + b*x])/(4*c^2*x*(c + d*x)^(3/2)) - (5*d*(11*b*c - 21*a*d)*Sqrt[a + b*x])/(12*c^4*Sqrt
[c + d*x]) - ((3*b^2*c^2 - 30*a*b*c*d + 35*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/
(4*Sqrt[a]*c^(9/2))

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Rubi [A]  time = 0.231078, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {98, 151, 152, 12, 93, 208} \[ -\frac{\left (35 a^2 d^2-30 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 \sqrt{a} c^{9/2}}-\frac{5 d \sqrt{a+b x} (11 b c-21 a d)}{12 c^4 \sqrt{c+d x}}-\frac{d \sqrt{a+b x} (23 b c-35 a d)}{12 c^3 (c+d x)^{3/2}}-\frac{\sqrt{a+b x} (5 b c-7 a d)}{4 c^2 x (c+d x)^{3/2}}-\frac{a \sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(23*b*c - 35*a*d)*Sqrt[a + b*x])/(12*c^3*(c + d*x)^(3/2)) - (a*Sqrt[a + b*x])/(2*c*x^2*(c + d*x)^(3/2)) -
((5*b*c - 7*a*d)*Sqrt[a + b*x])/(4*c^2*x*(c + d*x)^(3/2)) - (5*d*(11*b*c - 21*a*d)*Sqrt[a + b*x])/(12*c^4*Sqrt
[c + d*x]) - ((3*b^2*c^2 - 30*a*b*c*d + 35*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/
(4*Sqrt[a]*c^(9/2))

Rule 98

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

Rule 151

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 + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 152

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 + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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]

Rubi steps

\begin{align*} \int \frac{(a+b x)^{3/2}}{x^3 (c+d x)^{5/2}} \, dx &=-\frac{a \sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac{\int \frac{-\frac{1}{2} a (5 b c-7 a d)-b (2 b c-3 a d) x}{x^2 \sqrt{a+b x} (c+d x)^{5/2}} \, dx}{2 c}\\ &=-\frac{a \sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac{(5 b c-7 a d) \sqrt{a+b x}}{4 c^2 x (c+d x)^{3/2}}+\frac{\int \frac{\frac{1}{4} a \left (3 b^2 c^2-30 a b c d+35 a^2 d^2\right )-a b d (5 b c-7 a d) x}{x \sqrt{a+b x} (c+d x)^{5/2}} \, dx}{2 a c^2}\\ &=-\frac{d (23 b c-35 a d) \sqrt{a+b x}}{12 c^3 (c+d x)^{3/2}}-\frac{a \sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac{(5 b c-7 a d) \sqrt{a+b x}}{4 c^2 x (c+d x)^{3/2}}-\frac{\int \frac{-\frac{3}{8} a (b c-a d) \left (3 b^2 c^2-30 a b c d+35 a^2 d^2\right )+\frac{1}{4} a b d (23 b c-35 a d) (b c-a d) x}{x \sqrt{a+b x} (c+d x)^{3/2}} \, dx}{3 a c^3 (b c-a d)}\\ &=-\frac{d (23 b c-35 a d) \sqrt{a+b x}}{12 c^3 (c+d x)^{3/2}}-\frac{a \sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac{(5 b c-7 a d) \sqrt{a+b x}}{4 c^2 x (c+d x)^{3/2}}-\frac{5 d (11 b c-21 a d) \sqrt{a+b x}}{12 c^4 \sqrt{c+d x}}+\frac{2 \int \frac{3 a (b c-a d)^2 \left (3 b^2 c^2-30 a b c d+35 a^2 d^2\right )}{16 x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{3 a c^4 (b c-a d)^2}\\ &=-\frac{d (23 b c-35 a d) \sqrt{a+b x}}{12 c^3 (c+d x)^{3/2}}-\frac{a \sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac{(5 b c-7 a d) \sqrt{a+b x}}{4 c^2 x (c+d x)^{3/2}}-\frac{5 d (11 b c-21 a d) \sqrt{a+b x}}{12 c^4 \sqrt{c+d x}}+\frac{\left (3 b^2 c^2-30 a b c d+35 a^2 d^2\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 c^4}\\ &=-\frac{d (23 b c-35 a d) \sqrt{a+b x}}{12 c^3 (c+d x)^{3/2}}-\frac{a \sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac{(5 b c-7 a d) \sqrt{a+b x}}{4 c^2 x (c+d x)^{3/2}}-\frac{5 d (11 b c-21 a d) \sqrt{a+b x}}{12 c^4 \sqrt{c+d x}}+\frac{\left (3 b^2 c^2-30 a b c d+35 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{4 c^4}\\ &=-\frac{d (23 b c-35 a d) \sqrt{a+b x}}{12 c^3 (c+d x)^{3/2}}-\frac{a \sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac{(5 b c-7 a d) \sqrt{a+b x}}{4 c^2 x (c+d x)^{3/2}}-\frac{5 d (11 b c-21 a d) \sqrt{a+b x}}{12 c^4 \sqrt{c+d x}}-\frac{\left (3 b^2 c^2-30 a b c d+35 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 \sqrt{a} c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.215982, size = 171, normalized size = 0.84 \[ -\frac{-x^2 \left (35 a^2 d^2-30 a b c d+3 b^2 c^2\right ) \left (\sqrt{c} \sqrt{a+b x} (4 a c+3 a d x+b c x)-3 a^{3/2} (c+d x)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )\right )+3 c^{5/2} x (a+b x)^{5/2} (b c-7 a d)+6 a c^{7/2} (a+b x)^{5/2}}{12 a^2 c^{9/2} x^2 (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(6*a*c^(7/2)*(a + b*x)^(5/2) + 3*c^(5/2)*(b*c - 7*a*d)*x*(a + b*x)^(5/2) - (3*b^2*c^2 - 30*a*b*c*d + 35*a^2*d
^2)*x^2*(Sqrt[c]*Sqrt[a + b*x]*(4*a*c + b*c*x + 3*a*d*x) - 3*a^(3/2)*(c + d*x)^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[a +
b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(12*a^2*c^(9/2)*x^2*(c + d*x)^(3/2))

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Maple [B]  time = 0.03, size = 679, normalized size = 3.3 \begin{align*} -{\frac{1}{24\,{c}^{4}{x}^{2}}\sqrt{bx+a} \left ( 105\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{4}{a}^{2}{d}^{4}-90\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{4}abc{d}^{3}+9\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{4}{b}^{2}{c}^{2}{d}^{2}+210\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}{a}^{2}c{d}^{3}-180\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}ab{c}^{2}{d}^{2}+18\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}{b}^{2}{c}^{3}d+105\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}{a}^{2}{c}^{2}{d}^{2}-90\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}ab{c}^{3}d+9\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}{b}^{2}{c}^{4}-210\,{x}^{3}a{d}^{3}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}+110\,{x}^{3}bc{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}-280\,{x}^{2}ac{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}+156\,{x}^{2}b{c}^{2}d\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}-42\,xa{c}^{2}d\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}+30\,xb{c}^{3}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}+12\,a{c}^{3}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}} \left ( dx+c \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(b*x+a)^(1/2)/c^4*(105*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^4*a^2*d^4-90*ln
((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^4*a*b*c*d^3+9*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((
b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^4*b^2*c^2*d^2+210*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*
c)/x)*x^3*a^2*c*d^3-180*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^3*a*b*c^2*d^2+18*ln(
(a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^3*b^2*c^3*d+105*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(
(b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*a^2*c^2*d^2-90*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*
c)/x)*x^2*a*b*c^3*d+9*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*b^2*c^4-210*x^3*a*d^
3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+110*x^3*b*c*d^2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-280*x^2*a*c*d^2*((b*
x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+156*x^2*b*c^2*d*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-42*x*a*c^2*d*((b*x+a)*(d*x
+c))^(1/2)*(a*c)^(1/2)+30*x*b*c^3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+12*a*c^3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(
1/2))/x^2/(a*c)^(1/2)/((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 25.3984, size = 1386, normalized size = 6.79 \begin{align*} \left [\frac{3 \,{\left ({\left (3 \, b^{2} c^{2} d^{2} - 30 \, a b c d^{3} + 35 \, a^{2} d^{4}\right )} x^{4} + 2 \,{\left (3 \, b^{2} c^{3} d - 30 \, a b c^{2} d^{2} + 35 \, a^{2} c d^{3}\right )} x^{3} +{\left (3 \, b^{2} c^{4} - 30 \, a b c^{3} d + 35 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt{a c} \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 (6 \, a^{2} c^{4} + 5 \,{\left (11 \, a b c^{2} d^{2} - 21 \, a^{2} c d^{3}\right )} x^{3} + 2 \,{\left (39 \, a b c^{3} d - 70 \, a^{2} c^{2} d^{2}\right )} x^{2} + 3 \,{\left (5 \, a b c^{4} - 7 \, a^{2} c^{3} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{48 \,{\left (a c^{5} d^{2} x^{4} + 2 \, a c^{6} d x^{3} + a c^{7} x^{2}\right )}}, \frac{3 \,{\left ({\left (3 \, b^{2} c^{2} d^{2} - 30 \, a b c d^{3} + 35 \, a^{2} d^{4}\right )} x^{4} + 2 \,{\left (3 \, b^{2} c^{3} d - 30 \, a b c^{2} d^{2} + 35 \, a^{2} c d^{3}\right )} x^{3} +{\left (3 \, b^{2} c^{4} - 30 \, a b c^{3} d + 35 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt{-a c} \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 ) - 2 \,{\left (6 \, a^{2} c^{4} + 5 \,{\left (11 \, a b c^{2} d^{2} - 21 \, a^{2} c d^{3}\right )} x^{3} + 2 \,{\left (39 \, a b c^{3} d - 70 \, a^{2} c^{2} d^{2}\right )} x^{2} + 3 \,{\left (5 \, a b c^{4} - 7 \, a^{2} c^{3} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{24 \,{\left (a c^{5} d^{2} x^{4} + 2 \, a c^{6} d x^{3} + a c^{7} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(3*((3*b^2*c^2*d^2 - 30*a*b*c*d^3 + 35*a^2*d^4)*x^4 + 2*(3*b^2*c^3*d - 30*a*b*c^2*d^2 + 35*a^2*c*d^3)*x^
3 + (3*b^2*c^4 - 30*a*b*c^3*d + 35*a^2*c^2*d^2)*x^2)*sqrt(a*c)*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*(6
*a^2*c^4 + 5*(11*a*b*c^2*d^2 - 21*a^2*c*d^3)*x^3 + 2*(39*a*b*c^3*d - 70*a^2*c^2*d^2)*x^2 + 3*(5*a*b*c^4 - 7*a^
2*c^3*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^5*d^2*x^4 + 2*a*c^6*d*x^3 + a*c^7*x^2), 1/24*(3*((3*b^2*c^2*d^2
- 30*a*b*c*d^3 + 35*a^2*d^4)*x^4 + 2*(3*b^2*c^3*d - 30*a*b*c^2*d^2 + 35*a^2*c*d^3)*x^3 + (3*b^2*c^4 - 30*a*b*c
^3*d + 35*a^2*c^2*d^2)*x^2)*sqrt(-a*c)*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)) - 2*(6*a^2*c^4 + 5*(11*a*b*c^2*d^2 - 21*a^2*c*d^3)*x^3 + 2
*(39*a*b*c^3*d - 70*a^2*c^2*d^2)*x^2 + 3*(5*a*b*c^4 - 7*a^2*c^3*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^5*d^2*
x^4 + 2*a*c^6*d*x^3 + a*c^7*x^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError