3.1207 \(\int \frac{(A+B x) (d+e x)^3}{(b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=220 \[ \frac{2 \left (x \left (2 b^2 c^2 d e (4 A e+3 B d)-8 b c^3 d^2 (3 A e+B d)+16 A c^4 d^3+2 b^3 B c d e^2-3 b^4 B e^3\right )+b c d^2 \left (-4 b c (2 A e+B d)+8 A c^2 d+b^2 B e\right )\right )}{3 b^4 c^2 \sqrt{b x+c x^2}}-\frac{2 (d+e x)^2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}+\frac{2 B e^3 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{5/2}} \]

[Out]

(-2*(d + e*x)^2*(A*b*c*d + (2*A*c^2*d + b^2*B*e - b*c*(B*d + A*e))*x))/(3*b^2*c*(b*x + c*x^2)^(3/2)) + (2*(b*c
*d^2*(8*A*c^2*d + b^2*B*e - 4*b*c*(B*d + 2*A*e)) + (16*A*c^4*d^3 + 2*b^3*B*c*d*e^2 - 3*b^4*B*e^3 - 8*b*c^3*d^2
*(B*d + 3*A*e) + 2*b^2*c^2*d*e*(3*B*d + 4*A*e))*x))/(3*b^4*c^2*Sqrt[b*x + c*x^2]) + (2*B*e^3*ArcTanh[(Sqrt[c]*
x)/Sqrt[b*x + c*x^2]])/c^(5/2)

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Rubi [A]  time = 0.202831, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {818, 777, 620, 206} \[ \frac{2 \left (x \left (2 b^2 c^2 d e (4 A e+3 B d)-8 b c^3 d^2 (3 A e+B d)+16 A c^4 d^3+2 b^3 B c d e^2-3 b^4 B e^3\right )+b c d^2 \left (-4 b c (2 A e+B d)+8 A c^2 d+b^2 B e\right )\right )}{3 b^4 c^2 \sqrt{b x+c x^2}}-\frac{2 (d+e x)^2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}+\frac{2 B e^3 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)^2*(A*b*c*d + (2*A*c^2*d + b^2*B*e - b*c*(B*d + A*e))*x))/(3*b^2*c*(b*x + c*x^2)^(3/2)) + (2*(b*c
*d^2*(8*A*c^2*d + b^2*B*e - 4*b*c*(B*d + 2*A*e)) + (16*A*c^4*d^3 + 2*b^3*B*c*d*e^2 - 3*b^4*B*e^3 - 8*b*c^3*d^2
*(B*d + 3*A*e) + 2*b^2*c^2*d*e*(3*B*d + 4*A*e))*x))/(3*b^4*c^2*Sqrt[b*x + c*x^2]) + (2*B*e^3*ArcTanh[(Sqrt[c]*
x)/Sqrt[b*x + c*x^2]])/c^(5/2)

Rule 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 777

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((2
*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (b^2*e*g - b*c*(e*f + d*g) + 2*c*(c*d*f - a*e*g))*x)*(a + b*x + c*x^2)^
(p + 1))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p
+ 3))/(c*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && N
eQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, 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])

Rubi steps

\begin{align*} \int \frac{(A+B x) (d+e x)^3}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac{2 (d+e x)^2 \left (A b c d+\left (2 A c^2 d+b^2 B e-b c (B d+A e)\right ) x\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}+\frac{2 \int \frac{(d+e x) \left (-\frac{1}{2} d \left (8 A c^2 d+b^2 B e-4 b c (B d+2 A e)\right )+\frac{3}{2} b^2 B e^2 x\right )}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 b^2 c}\\ &=-\frac{2 (d+e x)^2 \left (A b c d+\left (2 A c^2 d+b^2 B e-b c (B d+A e)\right ) x\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}+\frac{2 \left (b c d^2 \left (8 A c^2 d+b^2 B e-4 b c (B d+2 A e)\right )+\left (16 A c^4 d^3+2 b^3 B c d e^2-3 b^4 B e^3-8 b c^3 d^2 (B d+3 A e)+2 b^2 c^2 d e (3 B d+4 A e)\right ) x\right )}{3 b^4 c^2 \sqrt{b x+c x^2}}+\frac{\left (B e^3\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{c^2}\\ &=-\frac{2 (d+e x)^2 \left (A b c d+\left (2 A c^2 d+b^2 B e-b c (B d+A e)\right ) x\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}+\frac{2 \left (b c d^2 \left (8 A c^2 d+b^2 B e-4 b c (B d+2 A e)\right )+\left (16 A c^4 d^3+2 b^3 B c d e^2-3 b^4 B e^3-8 b c^3 d^2 (B d+3 A e)+2 b^2 c^2 d e (3 B d+4 A e)\right ) x\right )}{3 b^4 c^2 \sqrt{b x+c x^2}}+\frac{\left (2 B e^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{c^2}\\ &=-\frac{2 (d+e x)^2 \left (A b c d+\left (2 A c^2 d+b^2 B e-b c (B d+A e)\right ) x\right )}{3 b^2 c \left (b x+c x^2\right )^{3/2}}+\frac{2 \left (b c d^2 \left (8 A c^2 d+b^2 B e-4 b c (B d+2 A e)\right )+\left (16 A c^4 d^3+2 b^3 B c d e^2-3 b^4 B e^3-8 b c^3 d^2 (B d+3 A e)+2 b^2 c^2 d e (3 B d+4 A e)\right ) x\right )}{3 b^4 c^2 \sqrt{b x+c x^2}}+\frac{2 B e^3 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{c^{5/2}}\\ \end{align*}

Mathematica [C]  time = 2.64175, size = 422, normalized size = 1.92 \[ \frac{x \left (\frac{168 A b^2 \left (6 b^2 c d x \left (d^2-6 d e x+e^2 x^2\right )+b^3 \left (-9 d^2 e x-d^3+9 d e^2 x^2+e^3 x^3\right )+24 b c^2 d^2 x^2 (d-e x)+16 c^3 d^3 x^3\right )}{x}-\frac{B (b+c x) \sqrt{\frac{c x}{b}+1} \left (7 \left (b^2 \sqrt{-\frac{c x (b+c x)}{b^2}} \left (3 b^2 \left (180 d^2 e x+95 d^3+75 d e^2 x^2+14 e^3 x^3\right )-2 b c x \left (180 d^2 e x+95 d^3+75 d e^2 x^2+14 e^3 x^3\right )+8 c^2 x^2 \left (9 d^2 e x+28 d^3+6 d e^2 x^2+e^3 x^3\right )\right )-3 b^4 \left (180 d^2 e x+95 d^3+75 d e^2 x^2+14 e^3 x^3\right ) \sin ^{-1}\left (\sqrt{-\frac{c x}{b}}\right )-8 c^4 x^4 \sqrt{-\frac{c x}{b}} (d+e x)^2 (3 d+e x) \, _2F_1\left (\frac{3}{2},\frac{9}{2};\frac{11}{2};-\frac{c x}{b}\right )\right )-96 b^4 \left (-\frac{c x}{b}\right )^{7/2} (d+e x)^3 \text{HypergeometricPFQ}\left (\left \{\frac{1}{2},2,2,\frac{7}{2}\right \},\left \{1,1,\frac{9}{2}\right \},-\frac{c x}{b}\right )\right )}{\left (-\frac{c x}{b}\right )^{5/2}}\right )}{252 b^6 (x (b+c x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*((168*A*b^2*(16*c^3*d^3*x^3 + 24*b*c^2*d^2*x^2*(d - e*x) + 6*b^2*c*d*x*(d^2 - 6*d*e*x + e^2*x^2) + b^3*(-d^
3 - 9*d^2*e*x + 9*d*e^2*x^2 + e^3*x^3)))/x - (B*(b + c*x)*Sqrt[1 + (c*x)/b]*(7*(b^2*Sqrt[-((c*x*(b + c*x))/b^2
)]*(8*c^2*x^2*(28*d^3 + 9*d^2*e*x + 6*d*e^2*x^2 + e^3*x^3) + 3*b^2*(95*d^3 + 180*d^2*e*x + 75*d*e^2*x^2 + 14*e
^3*x^3) - 2*b*c*x*(95*d^3 + 180*d^2*e*x + 75*d*e^2*x^2 + 14*e^3*x^3)) - 3*b^4*(95*d^3 + 180*d^2*e*x + 75*d*e^2
*x^2 + 14*e^3*x^3)*ArcSin[Sqrt[-((c*x)/b)]] - 8*c^4*x^4*Sqrt[-((c*x)/b)]*(d + e*x)^2*(3*d + e*x)*Hypergeometri
c2F1[3/2, 9/2, 11/2, -((c*x)/b)]) - 96*b^4*(-((c*x)/b))^(7/2)*(d + e*x)^3*HypergeometricPFQ[{1/2, 2, 2, 7/2},
{1, 1, 9/2}, -((c*x)/b)]))/(-((c*x)/b))^(5/2)))/(252*b^6*(x*(b + c*x))^(3/2))

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Maple [B]  time = 0.009, size = 680, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^3/(c*x^2+b*x)^(5/2),x)

[Out]

1/c^2/(c*x^2+b*x)^(1/2)*B*d*e^2+2/3/b/(c*x^2+b*x)^(3/2)*x*B*d^3-b/c^2/(c*x^2+b*x)^(3/2)*x*B*d*e^2+2/b/c/(c*x^2
+b*x)^(1/2)*x*B*d*e^2-16/b^3*c/(c*x^2+b*x)^(1/2)*x*A*d^2*e+2/b/c/(c*x^2+b*x)^(1/2)*A*d*e^2+2/b/c/(c*x^2+b*x)^(
1/2)*B*d^2*e-1/3*b/c^2/(c*x^2+b*x)^(3/2)*x*A*e^3+2/3/b/c/(c*x^2+b*x)^(1/2)*x*A*e^3-3*x^2/c/(c*x^2+b*x)^(3/2)*B
*d*e^2+1/6*B*e^3*b^2/c^3/(c*x^2+b*x)^(3/2)*x+1/2*B*e^3*b/c^2*x^2/(c*x^2+b*x)^(3/2)-2/c/(c*x^2+b*x)^(3/2)*x*B*d
^2*e+4/b^2/(c*x^2+b*x)^(1/2)*x*A*d*e^2+4/b^2/(c*x^2+b*x)^(1/2)*x*B*d^2*e-2/c/(c*x^2+b*x)^(3/2)*x*A*d*e^2+2/b/(
c*x^2+b*x)^(3/2)*x*A*d^2*e-16/3/b^3*c/(c*x^2+b*x)^(1/2)*x*B*d^3-4/3*A*d^3/b^2/(c*x^2+b*x)^(3/2)*x*c+32/3*A*d^3
*c^2/b^4/(c*x^2+b*x)^(1/2)*x+1/3/c^2/(c*x^2+b*x)^(1/2)*A*e^3+B*e^3/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^
(1/2))-2/3*A*d^3/b/(c*x^2+b*x)^(3/2)-8/3/b^2/(c*x^2+b*x)^(1/2)*B*d^3-1/6*B*e^3*b/c^3/(c*x^2+b*x)^(1/2)-1/3*B*e
^3*x^3/c/(c*x^2+b*x)^(3/2)-8/b^2/(c*x^2+b*x)^(1/2)*A*d^2*e+16/3*A*d^3*c/b^3/(c*x^2+b*x)^(1/2)-7/3*B*e^3/c^2/(c
*x^2+b*x)^(1/2)*x-x^2/c/(c*x^2+b*x)^(3/2)*A*e^3

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.87766, size = 1361, normalized size = 6.19 \begin{align*} \left [\frac{3 \,{\left (B b^{4} c^{2} e^{3} x^{4} + 2 \, B b^{5} c e^{3} x^{3} + B b^{6} e^{3} x^{2}\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (A b^{3} c^{3} d^{3} +{\left (8 \,{\left (B b c^{5} - 2 \, A c^{6}\right )} d^{3} - 6 \,{\left (B b^{2} c^{4} - 4 \, A b c^{5}\right )} d^{2} e - 3 \,{\left (B b^{3} c^{3} + 2 \, A b^{2} c^{4}\right )} d e^{2} +{\left (4 \, B b^{4} c^{2} - A b^{3} c^{3}\right )} e^{3}\right )} x^{3} - 3 \,{\left (3 \, A b^{3} c^{3} d e^{2} - B b^{5} c e^{3} - 4 \,{\left (B b^{2} c^{4} - 2 \, A b c^{5}\right )} d^{3} + 3 \,{\left (B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} d^{2} e\right )} x^{2} + 3 \,{\left (3 \, A b^{3} c^{3} d^{2} e +{\left (B b^{3} c^{3} - 2 \, A b^{2} c^{4}\right )} d^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{3 \,{\left (b^{4} c^{5} x^{4} + 2 \, b^{5} c^{4} x^{3} + b^{6} c^{3} x^{2}\right )}}, -\frac{2 \,{\left (3 \,{\left (B b^{4} c^{2} e^{3} x^{4} + 2 \, B b^{5} c e^{3} x^{3} + B b^{6} e^{3} x^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (A b^{3} c^{3} d^{3} +{\left (8 \,{\left (B b c^{5} - 2 \, A c^{6}\right )} d^{3} - 6 \,{\left (B b^{2} c^{4} - 4 \, A b c^{5}\right )} d^{2} e - 3 \,{\left (B b^{3} c^{3} + 2 \, A b^{2} c^{4}\right )} d e^{2} +{\left (4 \, B b^{4} c^{2} - A b^{3} c^{3}\right )} e^{3}\right )} x^{3} - 3 \,{\left (3 \, A b^{3} c^{3} d e^{2} - B b^{5} c e^{3} - 4 \,{\left (B b^{2} c^{4} - 2 \, A b c^{5}\right )} d^{3} + 3 \,{\left (B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} d^{2} e\right )} x^{2} + 3 \,{\left (3 \, A b^{3} c^{3} d^{2} e +{\left (B b^{3} c^{3} - 2 \, A b^{2} c^{4}\right )} d^{3}\right )} x\right )} \sqrt{c x^{2} + b x}\right )}}{3 \,{\left (b^{4} c^{5} x^{4} + 2 \, b^{5} c^{4} x^{3} + b^{6} c^{3} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(3*(B*b^4*c^2*e^3*x^4 + 2*B*b^5*c*e^3*x^3 + B*b^6*e^3*x^2)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sq
rt(c)) - 2*(A*b^3*c^3*d^3 + (8*(B*b*c^5 - 2*A*c^6)*d^3 - 6*(B*b^2*c^4 - 4*A*b*c^5)*d^2*e - 3*(B*b^3*c^3 + 2*A*
b^2*c^4)*d*e^2 + (4*B*b^4*c^2 - A*b^3*c^3)*e^3)*x^3 - 3*(3*A*b^3*c^3*d*e^2 - B*b^5*c*e^3 - 4*(B*b^2*c^4 - 2*A*
b*c^5)*d^3 + 3*(B*b^3*c^3 - 4*A*b^2*c^4)*d^2*e)*x^2 + 3*(3*A*b^3*c^3*d^2*e + (B*b^3*c^3 - 2*A*b^2*c^4)*d^3)*x)
*sqrt(c*x^2 + b*x))/(b^4*c^5*x^4 + 2*b^5*c^4*x^3 + b^6*c^3*x^2), -2/3*(3*(B*b^4*c^2*e^3*x^4 + 2*B*b^5*c*e^3*x^
3 + B*b^6*e^3*x^2)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (A*b^3*c^3*d^3 + (8*(B*b*c^5 - 2*A*c^6)
*d^3 - 6*(B*b^2*c^4 - 4*A*b*c^5)*d^2*e - 3*(B*b^3*c^3 + 2*A*b^2*c^4)*d*e^2 + (4*B*b^4*c^2 - A*b^3*c^3)*e^3)*x^
3 - 3*(3*A*b^3*c^3*d*e^2 - B*b^5*c*e^3 - 4*(B*b^2*c^4 - 2*A*b*c^5)*d^3 + 3*(B*b^3*c^3 - 4*A*b^2*c^4)*d^2*e)*x^
2 + 3*(3*A*b^3*c^3*d^2*e + (B*b^3*c^3 - 2*A*b^2*c^4)*d^3)*x)*sqrt(c*x^2 + b*x))/(b^4*c^5*x^4 + 2*b^5*c^4*x^3 +
b^6*c^3*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (d + e x\right )^{3}}{\left (x \left (b + c x\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**3/(c*x**2+b*x)**(5/2),x)

[Out]

Integral((A + B*x)*(d + e*x)**3/(x*(b + c*x))**(5/2), x)

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Giac [A]  time = 1.65869, size = 390, normalized size = 1.77 \begin{align*} -\frac{B e^{3} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{c^{\frac{5}{2}}} - \frac{2 \,{\left (\frac{A d^{3}}{b} +{\left (x{\left (\frac{{\left (8 \, B b c^{4} d^{3} - 16 \, A c^{5} d^{3} - 6 \, B b^{2} c^{3} d^{2} e + 24 \, A b c^{4} d^{2} e - 3 \, B b^{3} c^{2} d e^{2} - 6 \, A b^{2} c^{3} d e^{2} + 4 \, B b^{4} c e^{3} - A b^{3} c^{2} e^{3}\right )} x}{b^{4} c^{2}} + \frac{3 \,{\left (4 \, B b^{2} c^{3} d^{3} - 8 \, A b c^{4} d^{3} - 3 \, B b^{3} c^{2} d^{2} e + 12 \, A b^{2} c^{3} d^{2} e - 3 \, A b^{3} c^{2} d e^{2} + B b^{5} e^{3}\right )}}{b^{4} c^{2}}\right )} + \frac{3 \,{\left (B b^{3} c^{2} d^{3} - 2 \, A b^{2} c^{3} d^{3} + 3 \, A b^{3} c^{2} d^{2} e\right )}}{b^{4} c^{2}}\right )} x\right )}}{3 \,{\left (c x^{2} + b x\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-B*e^3*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2) - 2/3*(A*d^3/b + (x*((8*B*b*c^4*d^3 -
16*A*c^5*d^3 - 6*B*b^2*c^3*d^2*e + 24*A*b*c^4*d^2*e - 3*B*b^3*c^2*d*e^2 - 6*A*b^2*c^3*d*e^2 + 4*B*b^4*c*e^3 -
A*b^3*c^2*e^3)*x/(b^4*c^2) + 3*(4*B*b^2*c^3*d^3 - 8*A*b*c^4*d^3 - 3*B*b^3*c^2*d^2*e + 12*A*b^2*c^3*d^2*e - 3*A
*b^3*c^2*d*e^2 + B*b^5*e^3)/(b^4*c^2)) + 3*(B*b^3*c^2*d^3 - 2*A*b^2*c^3*d^3 + 3*A*b^3*c^2*d^2*e)/(b^4*c^2))*x)
/(c*x^2 + b*x)^(3/2)