3.747 \(\int \frac{1}{x^3 \sqrt{a+b x} (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=194 \[ -\frac{3 \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{5/2} c^{7/2}}+\frac{d \sqrt{a+b x} (3 b c-5 a d) (3 a d+b c)}{4 a^2 c^3 \sqrt{c+d x} (b c-a d)}+\frac{\sqrt{a+b x} (5 a d+3 b c)}{4 a^2 c^2 x \sqrt{c+d x}}-\frac{\sqrt{a+b x}}{2 a c x^2 \sqrt{c+d x}} \]

[Out]

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

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Rubi [A]  time = 0.157093, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {103, 151, 152, 12, 93, 208} \[ -\frac{3 \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{5/2} c^{7/2}}+\frac{d \sqrt{a+b x} (3 b c-5 a d) (3 a d+b c)}{4 a^2 c^3 \sqrt{c+d x} (b c-a d)}+\frac{\sqrt{a+b x} (5 a d+3 b c)}{4 a^2 c^2 x \sqrt{c+d x}}-\frac{\sqrt{a+b x}}{2 a c x^2 \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[a + b*x]*(c + d*x)^(3/2)),x]

[Out]

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

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(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*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

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

Mathematica [A]  time = 0.171115, size = 186, normalized size = 0.96 \[ \frac{\frac{\sqrt{a} \sqrt{c} \sqrt{a+b x} \left (a^2 d \left (2 c^2-5 c d x-15 d^2 x^2\right )+2 a b c \left (-c^2+c d x+2 d^2 x^2\right )+3 b^2 c^2 x (c+d x)\right )}{x^2 \sqrt{c+d x}}-3 \left (3 a^2 b c d^2-5 a^3 d^3+a b^2 c^2 d+b^3 c^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{5/2} c^{7/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[a + b*x]*(c + d*x)^(3/2)),x]

[Out]

((Sqrt[a]*Sqrt[c]*Sqrt[a + b*x]*(3*b^2*c^2*x*(c + d*x) + a^2*d*(2*c^2 - 5*c*d*x - 15*d^2*x^2) + 2*a*b*c*(-c^2
+ c*d*x + 2*d^2*x^2)))/(x^2*Sqrt[c + d*x]) - 3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*ArcTanh[(Sq
rt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(5/2)*c^(7/2)*(b*c - a*d))

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Maple [B]  time = 0.03, size = 683, normalized size = 3.5 \begin{align*} -{\frac{1}{8\,{c}^{3}{a}^{2}{x}^{2} \left ( ad-bc \right ) }\sqrt{bx+a} \left ( 15\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}{a}^{3}{d}^{4}-9\,\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}bc{d}^{3}-3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}a{b}^{2}{c}^{2}{d}^{2}-3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}{b}^{3}{c}^{3}d+15\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}{a}^{3}c{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}^{2}{a}^{2}b{c}^{2}{d}^{2}-3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}a{b}^{2}{c}^{3}d-3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}{b}^{3}{c}^{4}-30\,{x}^{2}{a}^{2}{d}^{3}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}+8\,{x}^{2}abc{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}+6\,{x}^{2}{b}^{2}{c}^{2}d\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}-10\,x{a}^{2}c{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}+4\,xab{c}^{2}d\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}+6\,x{b}^{2}{c}^{3}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}+4\,{a}^{2}{c}^{2}d\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}-4\,ab{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 ) }}}{\frac{1}{\sqrt{dx+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(d*x+c)^(3/2)/(b*x+a)^(1/2),x)

[Out]

-1/8*(b*x+a)^(1/2)/a^2/c^3*(15*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^3*d^4-9*l
n((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*b*c*d^3-3*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^2*c^2*d^2-3*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^3*c^3*d+15*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^3*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^2*a^2*b*c^2*d^2-3*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^2*c^3*d-3*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^3*c^4-30*x^2*a^2*d^3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+8*x^2*a*b*c*d^2*((b*x+a)*(d*x+c))^(1/2)*(
a*c)^(1/2)+6*x^2*b^2*c^2*d*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-10*x*a^2*c*d^2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1
/2)+4*x*a*b*c^2*d*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+6*x*b^2*c^3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+4*a^2*c^
2*d*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-4*a*b*c^3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2))/x^2/(a*c)^(1/2)/(a*d-b*
c)/((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b x + a}{\left (d x + c\right )}^{\frac{3}{2}} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*x + a)*(d*x + c)^(3/2)*x^3), x)

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Fricas [A]  time = 9.13918, size = 1382, normalized size = 7.12 \begin{align*} \left [\frac{3 \,{\left ({\left (b^{3} c^{3} d + a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - 5 \, a^{3} d^{4}\right )} x^{3} +{\left (b^{3} c^{4} + a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - 5 \, a^{3} c d^{3}\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 (2 \, a^{2} b c^{4} - 2 \, a^{3} c^{3} d -{\left (3 \, a b^{2} c^{3} d + 4 \, a^{2} b c^{2} d^{2} - 15 \, a^{3} c d^{3}\right )} x^{2} -{\left (3 \, a b^{2} c^{4} + 2 \, a^{2} b c^{3} d - 5 \, a^{3} c^{2} d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{16 \,{\left ({\left (a^{3} b c^{5} d - a^{4} c^{4} d^{2}\right )} x^{3} +{\left (a^{3} b c^{6} - a^{4} c^{5} d\right )} x^{2}\right )}}, \frac{3 \,{\left ({\left (b^{3} c^{3} d + a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - 5 \, a^{3} d^{4}\right )} x^{3} +{\left (b^{3} c^{4} + a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - 5 \, a^{3} c d^{3}\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 (2 \, a^{2} b c^{4} - 2 \, a^{3} c^{3} d -{\left (3 \, a b^{2} c^{3} d + 4 \, a^{2} b c^{2} d^{2} - 15 \, a^{3} c d^{3}\right )} x^{2} -{\left (3 \, a b^{2} c^{4} + 2 \, a^{2} b c^{3} d - 5 \, a^{3} c^{2} d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{8 \,{\left ({\left (a^{3} b c^{5} d - a^{4} c^{4} d^{2}\right )} x^{3} +{\left (a^{3} b c^{6} - a^{4} c^{5} d\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(3*((b^3*c^3*d + a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - 5*a^3*d^4)*x^3 + (b^3*c^4 + a*b^2*c^3*d + 3*a^2*b*c^2*d
^2 - 5*a^3*c*d^3)*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*(2*a^2*b*c^4 - 2*a^3*c^3*d - (3*
a*b^2*c^3*d + 4*a^2*b*c^2*d^2 - 15*a^3*c*d^3)*x^2 - (3*a*b^2*c^4 + 2*a^2*b*c^3*d - 5*a^3*c^2*d^2)*x)*sqrt(b*x
+ a)*sqrt(d*x + c))/((a^3*b*c^5*d - a^4*c^4*d^2)*x^3 + (a^3*b*c^6 - a^4*c^5*d)*x^2), 1/8*(3*((b^3*c^3*d + a*b^
2*c^2*d^2 + 3*a^2*b*c*d^3 - 5*a^3*d^4)*x^3 + (b^3*c^4 + a*b^2*c^3*d + 3*a^2*b*c^2*d^2 - 5*a^3*c*d^3)*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*(2*a^2*b*c^4 - 2*a^3*c^3*d - (3*a*b^2*c^3*d + 4*a^2*b*c^2*d^2 - 15*a^3*c*d^3)*x^2 - (3
*a*b^2*c^4 + 2*a^2*b*c^3*d - 5*a^3*c^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/((a^3*b*c^5*d - a^4*c^4*d^2)*x^3 +
(a^3*b*c^6 - a^4*c^5*d)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \sqrt{a + b x} \left (c + d x\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(d*x+c)**(3/2)/(b*x+a)**(1/2),x)

[Out]

Integral(1/(x**3*sqrt(a + b*x)*(c + d*x)**(3/2)), x)

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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(1/x^3/(d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError