3.1199 \(\int \frac{(A+B x) (d+e x)^3}{(b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=238 \[ \frac{e \sqrt{b x+c x^2} \left (2 c e x \left (-4 b c (A e+B d)+8 A c^2 d+5 b^2 B e\right )+12 b^2 c e (A e+3 B d)-8 b c^2 d (3 A e+2 B d)+32 A c^3 d^2-15 b^3 B e^2\right )}{4 b^2 c^3}+\frac{3 e \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) \left (4 A c e (2 c d-b e)+B \left (5 b^2 e^2-12 b c d e+8 c^2 d^2\right )\right )}{4 c^{7/2}}-\frac{2 (d+e x)^2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \sqrt{b x+c x^2}} \]

[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))/(b^2*c*Sqrt[b*x + c*x^2]) + (e*(32*A*c^
3*d^2 - 15*b^3*B*e^2 + 12*b^2*c*e*(3*B*d + A*e) - 8*b*c^2*d*(2*B*d + 3*A*e) + 2*c*e*(8*A*c^2*d + 5*b^2*B*e - 4
*b*c*(B*d + A*e))*x)*Sqrt[b*x + c*x^2])/(4*b^2*c^3) + (3*e*(4*A*c*e*(2*c*d - b*e) + B*(8*c^2*d^2 - 12*b*c*d*e
+ 5*b^2*e^2))*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(7/2))

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Rubi [A]  time = 0.21506, antiderivative size = 238, 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, 779, 620, 206} \[ \frac{e \sqrt{b x+c x^2} \left (2 c e x \left (-4 b c (A e+B d)+8 A c^2 d+5 b^2 B e\right )+12 b^2 c e (A e+3 B d)-8 b c^2 d (3 A e+2 B d)+32 A c^3 d^2-15 b^3 B e^2\right )}{4 b^2 c^3}+\frac{3 e \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right ) \left (4 A c e (2 c d-b e)+B \left (5 b^2 e^2-12 b c d e+8 c^2 d^2\right )\right )}{4 c^{7/2}}-\frac{2 (d+e x)^2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \sqrt{b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^3)/(b*x + c*x^2)^(3/2),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))/(b^2*c*Sqrt[b*x + c*x^2]) + (e*(32*A*c^
3*d^2 - 15*b^3*B*e^2 + 12*b^2*c*e*(3*B*d + A*e) - 8*b*c^2*d*(2*B*d + 3*A*e) + 2*c*e*(8*A*c^2*d + 5*b^2*B*e - 4
*b*c*(B*d + A*e))*x)*Sqrt[b*x + c*x^2])/(4*b^2*c^3) + (3*e*(4*A*c*e*(2*c*d - b*e) + B*(8*c^2*d^2 - 12*b*c*d*e
+ 5*b^2*e^2))*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*c^(7/2))

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 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), 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))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[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 )^{3/2}} \, 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 )}{b^2 c \sqrt{b x+c x^2}}+\frac{2 \int \frac{(d+e x) \left (\frac{1}{2} b (b B+4 A c) d e+\frac{1}{2} e \left (8 A c^2 d+5 b^2 B e-4 b c (B d+A e)\right ) x\right )}{\sqrt{b x+c x^2}} \, dx}{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 )}{b^2 c \sqrt{b x+c x^2}}+\frac{e \left (32 A c^3 d^2-15 b^3 B e^2+12 b^2 c e (3 B d+A e)-8 b c^2 d (2 B d+3 A e)+2 c e \left (8 A c^2 d+5 b^2 B e-4 b c (B d+A e)\right ) x\right ) \sqrt{b x+c x^2}}{4 b^2 c^3}+\frac{\left (3 e \left (4 A c e (2 c d-b e)+B \left (8 c^2 d^2-12 b c d e+5 b^2 e^2\right )\right )\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{8 c^3}\\ &=-\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 )}{b^2 c \sqrt{b x+c x^2}}+\frac{e \left (32 A c^3 d^2-15 b^3 B e^2+12 b^2 c e (3 B d+A e)-8 b c^2 d (2 B d+3 A e)+2 c e \left (8 A c^2 d+5 b^2 B e-4 b c (B d+A e)\right ) x\right ) \sqrt{b x+c x^2}}{4 b^2 c^3}+\frac{\left (3 e \left (4 A c e (2 c d-b e)+B \left (8 c^2 d^2-12 b c d e+5 b^2 e^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{4 c^3}\\ &=-\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 )}{b^2 c \sqrt{b x+c x^2}}+\frac{e \left (32 A c^3 d^2-15 b^3 B e^2+12 b^2 c e (3 B d+A e)-8 b c^2 d (2 B d+3 A e)+2 c e \left (8 A c^2 d+5 b^2 B e-4 b c (B d+A e)\right ) x\right ) \sqrt{b x+c x^2}}{4 b^2 c^3}+\frac{3 e \left (4 A c e (2 c d-b e)+B \left (8 c^2 d^2-12 b c d e+5 b^2 e^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.245299, size = 229, normalized size = 0.96 \[ \frac{\sqrt{c} \left (4 A c \left (b^2 c e^2 x (e x-6 d)+3 b^3 e^3 x-2 b c^2 d^2 (d-3 e x)-4 c^3 d^3 x\right )+b B x \left (b^2 c e^2 (36 d-5 e x)-15 b^3 e^3+2 b c^2 e \left (-12 d^2+6 d e x+e^2 x^2\right )+8 c^3 d^3\right )\right )+3 b^{5/2} e \sqrt{x} \sqrt{\frac{c x}{b}+1} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right ) \left (4 A c e (2 c d-b e)+B \left (5 b^2 e^2-12 b c d e+8 c^2 d^2\right )\right )}{4 b^2 c^{7/2} \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*(4*A*c*(-4*c^3*d^3*x + 3*b^3*e^3*x - 2*b*c^2*d^2*(d - 3*e*x) + b^2*c*e^2*x*(-6*d + e*x)) + b*B*x*(8*c
^3*d^3 - 15*b^3*e^3 + b^2*c*e^2*(36*d - 5*e*x) + 2*b*c^2*e*(-12*d^2 + 6*d*e*x + e^2*x^2))) + 3*b^(5/2)*e*(4*A*
c*e*(2*c*d - b*e) + B*(8*c^2*d^2 - 12*b*c*d*e + 5*b^2*e^2))*Sqrt[x]*Sqrt[1 + (c*x)/b]*ArcSinh[(Sqrt[c]*Sqrt[x]
)/Sqrt[b]])/(4*b^2*c^(7/2)*Sqrt[x*(b + c*x)])

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Maple [B]  time = 0.011, size = 450, normalized size = 1.9 \begin{align*}{\frac{B{e}^{3}{x}^{3}}{2\,c}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{5\,B{e}^{3}b{x}^{2}}{4\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}-{\frac{15\,B{e}^{3}{b}^{2}x}{4\,{c}^{3}}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+{\frac{15\,B{e}^{3}{b}^{2}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}}+{\frac{{x}^{2}A{e}^{3}}{c}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+3\,{\frac{{x}^{2}Bd{e}^{2}}{c\sqrt{c{x}^{2}+bx}}}+3\,{\frac{Abx{e}^{3}}{{c}^{2}\sqrt{c{x}^{2}+bx}}}+9\,{\frac{Bbdx{e}^{2}}{{c}^{2}\sqrt{c{x}^{2}+bx}}}-{\frac{3\,Ab{e}^{3}}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}-{\frac{9\,bBd{e}^{2}}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}-6\,{\frac{xAd{e}^{2}}{c\sqrt{c{x}^{2}+bx}}}-6\,{\frac{Bx{d}^{2}e}{c\sqrt{c{x}^{2}+bx}}}+3\,{\frac{Ad{e}^{2}}{{c}^{3/2}}\ln \left ({\frac{b/2+cx}{\sqrt{c}}}+\sqrt{c{x}^{2}+bx} \right ) }+3\,{\frac{B{d}^{2}e}{{c}^{3/2}}\ln \left ({\frac{b/2+cx}{\sqrt{c}}}+\sqrt{c{x}^{2}+bx} \right ) }+6\,{\frac{xA{d}^{2}e}{b\sqrt{c{x}^{2}+bx}}}+2\,{\frac{Bx{d}^{3}}{b\sqrt{c{x}^{2}+bx}}}-2\,{\frac{A{d}^{3} \left ( 2\,cx+b \right ) }{{b}^{2}\sqrt{c{x}^{2}+bx}}} \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)^(3/2),x)

[Out]

1/2*B*e^3*x^3/c/(c*x^2+b*x)^(1/2)-5/4*B*e^3*b/c^2*x^2/(c*x^2+b*x)^(1/2)-15/4*B*e^3*b^2/c^3/(c*x^2+b*x)^(1/2)*x
+15/8*B*e^3*b^2/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+x^2/c/(c*x^2+b*x)^(1/2)*A*e^3+3*x^2/c/(c*x^2
+b*x)^(1/2)*B*d*e^2+3*b/c^2/(c*x^2+b*x)^(1/2)*x*A*e^3+9*b/c^2/(c*x^2+b*x)^(1/2)*x*B*d*e^2-3/2*b/c^(5/2)*ln((1/
2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))*A*e^3-9/2*b/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))*B*d*e^2-6/c/
(c*x^2+b*x)^(1/2)*x*A*d*e^2-6/c/(c*x^2+b*x)^(1/2)*x*B*d^2*e+3/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2)
)*A*d*e^2+3/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))*B*d^2*e+6/b/(c*x^2+b*x)^(1/2)*x*A*d^2*e+2/b/(c*x
^2+b*x)^(1/2)*x*B*d^3-2*A*d^3*(2*c*x+b)/b^2/(c*x^2+b*x)^(1/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62401, size = 1434, normalized size = 6.03 \begin{align*} \left [\frac{3 \,{\left ({\left (8 \, B b^{2} c^{3} d^{2} e - 4 \,{\left (3 \, B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d e^{2} +{\left (5 \, B b^{4} c - 4 \, A b^{3} c^{2}\right )} e^{3}\right )} x^{2} +{\left (8 \, B b^{3} c^{2} d^{2} e - 4 \,{\left (3 \, B b^{4} c - 2 \, A b^{3} c^{2}\right )} d e^{2} +{\left (5 \, B b^{5} - 4 \, A b^{4} c\right )} e^{3}\right )} x\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (2 \, B b^{2} c^{3} e^{3} x^{3} - 8 \, A b c^{4} d^{3} +{\left (12 \, B b^{2} c^{3} d e^{2} -{\left (5 \, B b^{3} c^{2} - 4 \, A b^{2} c^{3}\right )} e^{3}\right )} x^{2} +{\left (8 \,{\left (B b c^{4} - 2 \, A c^{5}\right )} d^{3} - 24 \,{\left (B b^{2} c^{3} - A b c^{4}\right )} d^{2} e + 12 \,{\left (3 \, B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d e^{2} - 3 \,{\left (5 \, B b^{4} c - 4 \, A b^{3} c^{2}\right )} e^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{8 \,{\left (b^{2} c^{5} x^{2} + b^{3} c^{4} x\right )}}, -\frac{3 \,{\left ({\left (8 \, B b^{2} c^{3} d^{2} e - 4 \,{\left (3 \, B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d e^{2} +{\left (5 \, B b^{4} c - 4 \, A b^{3} c^{2}\right )} e^{3}\right )} x^{2} +{\left (8 \, B b^{3} c^{2} d^{2} e - 4 \,{\left (3 \, B b^{4} c - 2 \, A b^{3} c^{2}\right )} d e^{2} +{\left (5 \, B b^{5} - 4 \, A b^{4} c\right )} e^{3}\right )} x\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (2 \, B b^{2} c^{3} e^{3} x^{3} - 8 \, A b c^{4} d^{3} +{\left (12 \, B b^{2} c^{3} d e^{2} -{\left (5 \, B b^{3} c^{2} - 4 \, A b^{2} c^{3}\right )} e^{3}\right )} x^{2} +{\left (8 \,{\left (B b c^{4} - 2 \, A c^{5}\right )} d^{3} - 24 \,{\left (B b^{2} c^{3} - A b c^{4}\right )} d^{2} e + 12 \,{\left (3 \, B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d e^{2} - 3 \,{\left (5 \, B b^{4} c - 4 \, A b^{3} c^{2}\right )} e^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{4 \,{\left (b^{2} c^{5} x^{2} + b^{3} c^{4} x\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)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(3*((8*B*b^2*c^3*d^2*e - 4*(3*B*b^3*c^2 - 2*A*b^2*c^3)*d*e^2 + (5*B*b^4*c - 4*A*b^3*c^2)*e^3)*x^2 + (8*B*
b^3*c^2*d^2*e - 4*(3*B*b^4*c - 2*A*b^3*c^2)*d*e^2 + (5*B*b^5 - 4*A*b^4*c)*e^3)*x)*sqrt(c)*log(2*c*x + b + 2*sq
rt(c*x^2 + b*x)*sqrt(c)) + 2*(2*B*b^2*c^3*e^3*x^3 - 8*A*b*c^4*d^3 + (12*B*b^2*c^3*d*e^2 - (5*B*b^3*c^2 - 4*A*b
^2*c^3)*e^3)*x^2 + (8*(B*b*c^4 - 2*A*c^5)*d^3 - 24*(B*b^2*c^3 - A*b*c^4)*d^2*e + 12*(3*B*b^3*c^2 - 2*A*b^2*c^3
)*d*e^2 - 3*(5*B*b^4*c - 4*A*b^3*c^2)*e^3)*x)*sqrt(c*x^2 + b*x))/(b^2*c^5*x^2 + b^3*c^4*x), -1/4*(3*((8*B*b^2*
c^3*d^2*e - 4*(3*B*b^3*c^2 - 2*A*b^2*c^3)*d*e^2 + (5*B*b^4*c - 4*A*b^3*c^2)*e^3)*x^2 + (8*B*b^3*c^2*d^2*e - 4*
(3*B*b^4*c - 2*A*b^3*c^2)*d*e^2 + (5*B*b^5 - 4*A*b^4*c)*e^3)*x)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*
x)) - (2*B*b^2*c^3*e^3*x^3 - 8*A*b*c^4*d^3 + (12*B*b^2*c^3*d*e^2 - (5*B*b^3*c^2 - 4*A*b^2*c^3)*e^3)*x^2 + (8*(
B*b*c^4 - 2*A*c^5)*d^3 - 24*(B*b^2*c^3 - A*b*c^4)*d^2*e + 12*(3*B*b^3*c^2 - 2*A*b^2*c^3)*d*e^2 - 3*(5*B*b^4*c
- 4*A*b^3*c^2)*e^3)*x)*sqrt(c*x^2 + b*x))/(b^2*c^5*x^2 + b^3*c^4*x)]

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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{3}{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)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.34671, size = 339, normalized size = 1.42 \begin{align*} -\frac{\frac{8 \, A d^{3}}{b} -{\left ({\left (\frac{2 \, B x e^{3}}{c} + \frac{12 \, B b^{2} c^{2} d e^{2} - 5 \, B b^{3} c e^{3} + 4 \, A b^{2} c^{2} e^{3}}{b^{2} c^{3}}\right )} x + \frac{8 \, B b c^{3} d^{3} - 16 \, A c^{4} d^{3} - 24 \, B b^{2} c^{2} d^{2} e + 24 \, A b c^{3} d^{2} e + 36 \, B b^{3} c d e^{2} - 24 \, A b^{2} c^{2} d e^{2} - 15 \, B b^{4} e^{3} + 12 \, A b^{3} c e^{3}}{b^{2} c^{3}}\right )} x}{4 \, \sqrt{c x^{2} + b x}} - \frac{3 \,{\left (8 \, B c^{2} d^{2} e - 12 \, B b c d e^{2} + 8 \, A c^{2} d e^{2} + 5 \, B b^{2} e^{3} - 4 \, A b c e^{3}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{8 \, c^{\frac{7}{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)^(3/2),x, algorithm="giac")

[Out]

-1/4*(8*A*d^3/b - ((2*B*x*e^3/c + (12*B*b^2*c^2*d*e^2 - 5*B*b^3*c*e^3 + 4*A*b^2*c^2*e^3)/(b^2*c^3))*x + (8*B*b
*c^3*d^3 - 16*A*c^4*d^3 - 24*B*b^2*c^2*d^2*e + 24*A*b*c^3*d^2*e + 36*B*b^3*c*d*e^2 - 24*A*b^2*c^2*d*e^2 - 15*B
*b^4*e^3 + 12*A*b^3*c*e^3)/(b^2*c^3))*x)/sqrt(c*x^2 + b*x) - 3/8*(8*B*c^2*d^2*e - 12*B*b*c*d*e^2 + 8*A*c^2*d*e
^2 + 5*B*b^2*e^3 - 4*A*b*c*e^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2)