3.79 \(\int x^3 (A+B x) (b x+c x^2)^{3/2} \, dx\)

Optimal. Leaf size=238 \[ -\frac{9 b^5 (b+2 c x) \sqrt{b x+c x^2} (11 b B-16 A c)}{16384 c^6}+\frac{3 b^3 (b+2 c x) \left (b x+c x^2\right )^{3/2} (11 b B-16 A c)}{2048 c^5}-\frac{3 b^2 \left (b x+c x^2\right )^{5/2} (11 b B-16 A c)}{640 c^4}+\frac{9 b^7 (11 b B-16 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{16384 c^{13/2}}+\frac{3 b x \left (b x+c x^2\right )^{5/2} (11 b B-16 A c)}{448 c^3}-\frac{x^2 \left (b x+c x^2\right )^{5/2} (11 b B-16 A c)}{112 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{5/2}}{8 c} \]

[Out]

(-9*b^5*(11*b*B - 16*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(16384*c^6) + (3*b^3*(11*b*B - 16*A*c)*(b + 2*c*x)*(b
*x + c*x^2)^(3/2))/(2048*c^5) - (3*b^2*(11*b*B - 16*A*c)*(b*x + c*x^2)^(5/2))/(640*c^4) + (3*b*(11*b*B - 16*A*
c)*x*(b*x + c*x^2)^(5/2))/(448*c^3) - ((11*b*B - 16*A*c)*x^2*(b*x + c*x^2)^(5/2))/(112*c^2) + (B*x^3*(b*x + c*
x^2)^(5/2))/(8*c) + (9*b^7*(11*b*B - 16*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(16384*c^(13/2))

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Rubi [A]  time = 0.247669, antiderivative size = 238, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {794, 670, 640, 612, 620, 206} \[ -\frac{9 b^5 (b+2 c x) \sqrt{b x+c x^2} (11 b B-16 A c)}{16384 c^6}+\frac{3 b^3 (b+2 c x) \left (b x+c x^2\right )^{3/2} (11 b B-16 A c)}{2048 c^5}-\frac{3 b^2 \left (b x+c x^2\right )^{5/2} (11 b B-16 A c)}{640 c^4}+\frac{9 b^7 (11 b B-16 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{16384 c^{13/2}}+\frac{3 b x \left (b x+c x^2\right )^{5/2} (11 b B-16 A c)}{448 c^3}-\frac{x^2 \left (b x+c x^2\right )^{5/2} (11 b B-16 A c)}{112 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{5/2}}{8 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*b^5*(11*b*B - 16*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(16384*c^6) + (3*b^3*(11*b*B - 16*A*c)*(b + 2*c*x)*(b
*x + c*x^2)^(3/2))/(2048*c^5) - (3*b^2*(11*b*B - 16*A*c)*(b*x + c*x^2)^(5/2))/(640*c^4) + (3*b*(11*b*B - 16*A*
c)*x*(b*x + c*x^2)^(5/2))/(448*c^3) - ((11*b*B - 16*A*c)*x^2*(b*x + c*x^2)^(5/2))/(112*c^2) + (B*x^3*(b*x + c*
x^2)^(5/2))/(8*c) + (9*b^7*(11*b*B - 16*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(16384*c^(13/2))

Rule 794

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

Rule 670

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[((m + p)*(2*c*d - b*e))/(c*(m + 2*p + 1)), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

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 x^3 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx &=\frac{B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}+\frac{\left (3 (-b B+A c)+\frac{5}{2} (-b B+2 A c)\right ) \int x^3 \left (b x+c x^2\right )^{3/2} \, dx}{8 c}\\ &=-\frac{(11 b B-16 A c) x^2 \left (b x+c x^2\right )^{5/2}}{112 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}+\frac{(9 b (11 b B-16 A c)) \int x^2 \left (b x+c x^2\right )^{3/2} \, dx}{224 c^2}\\ &=\frac{3 b (11 b B-16 A c) x \left (b x+c x^2\right )^{5/2}}{448 c^3}-\frac{(11 b B-16 A c) x^2 \left (b x+c x^2\right )^{5/2}}{112 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}-\frac{\left (3 b^2 (11 b B-16 A c)\right ) \int x \left (b x+c x^2\right )^{3/2} \, dx}{128 c^3}\\ &=-\frac{3 b^2 (11 b B-16 A c) \left (b x+c x^2\right )^{5/2}}{640 c^4}+\frac{3 b (11 b B-16 A c) x \left (b x+c x^2\right )^{5/2}}{448 c^3}-\frac{(11 b B-16 A c) x^2 \left (b x+c x^2\right )^{5/2}}{112 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}+\frac{\left (3 b^3 (11 b B-16 A c)\right ) \int \left (b x+c x^2\right )^{3/2} \, dx}{256 c^4}\\ &=\frac{3 b^3 (11 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{2048 c^5}-\frac{3 b^2 (11 b B-16 A c) \left (b x+c x^2\right )^{5/2}}{640 c^4}+\frac{3 b (11 b B-16 A c) x \left (b x+c x^2\right )^{5/2}}{448 c^3}-\frac{(11 b B-16 A c) x^2 \left (b x+c x^2\right )^{5/2}}{112 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}-\frac{\left (9 b^5 (11 b B-16 A c)\right ) \int \sqrt{b x+c x^2} \, dx}{4096 c^5}\\ &=-\frac{9 b^5 (11 b B-16 A c) (b+2 c x) \sqrt{b x+c x^2}}{16384 c^6}+\frac{3 b^3 (11 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{2048 c^5}-\frac{3 b^2 (11 b B-16 A c) \left (b x+c x^2\right )^{5/2}}{640 c^4}+\frac{3 b (11 b B-16 A c) x \left (b x+c x^2\right )^{5/2}}{448 c^3}-\frac{(11 b B-16 A c) x^2 \left (b x+c x^2\right )^{5/2}}{112 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}+\frac{\left (9 b^7 (11 b B-16 A c)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{32768 c^6}\\ &=-\frac{9 b^5 (11 b B-16 A c) (b+2 c x) \sqrt{b x+c x^2}}{16384 c^6}+\frac{3 b^3 (11 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{2048 c^5}-\frac{3 b^2 (11 b B-16 A c) \left (b x+c x^2\right )^{5/2}}{640 c^4}+\frac{3 b (11 b B-16 A c) x \left (b x+c x^2\right )^{5/2}}{448 c^3}-\frac{(11 b B-16 A c) x^2 \left (b x+c x^2\right )^{5/2}}{112 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}+\frac{\left (9 b^7 (11 b B-16 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{16384 c^6}\\ &=-\frac{9 b^5 (11 b B-16 A c) (b+2 c x) \sqrt{b x+c x^2}}{16384 c^6}+\frac{3 b^3 (11 b B-16 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{2048 c^5}-\frac{3 b^2 (11 b B-16 A c) \left (b x+c x^2\right )^{5/2}}{640 c^4}+\frac{3 b (11 b B-16 A c) x \left (b x+c x^2\right )^{5/2}}{448 c^3}-\frac{(11 b B-16 A c) x^2 \left (b x+c x^2\right )^{5/2}}{112 c^2}+\frac{B x^3 \left (b x+c x^2\right )^{5/2}}{8 c}+\frac{9 b^7 (11 b B-16 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{16384 c^{13/2}}\\ \end{align*}

Mathematica [A]  time = 0.46364, size = 179, normalized size = 0.75 \[ \frac{x^5 \sqrt{x (b+c x)} \left (\frac{11 (11 b B-16 A c) \left (315 b^{13/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )-\sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \left (168 b^4 c^2 x^2-144 b^3 c^3 x^3+128 b^2 c^4 x^4-210 b^5 c x+315 b^6+6400 b c^5 x^5+5120 c^6 x^6\right )\right )}{71680 c^{11/2} x^{11/2} \sqrt{\frac{c x}{b}+1}}+11 B (b+c x)^2\right )}{88 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^5*Sqrt[x*(b + c*x)]*(11*B*(b + c*x)^2 + (11*(11*b*B - 16*A*c)*(-(Sqrt[c]*Sqrt[x]*Sqrt[1 + (c*x)/b]*(315*b^6
- 210*b^5*c*x + 168*b^4*c^2*x^2 - 144*b^3*c^3*x^3 + 128*b^2*c^4*x^4 + 6400*b*c^5*x^5 + 5120*c^6*x^6)) + 315*b
^(13/2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]]))/(71680*c^(11/2)*x^(11/2)*Sqrt[1 + (c*x)/b])))/(88*c)

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Maple [A]  time = 0.01, size = 373, normalized size = 1.6 \begin{align*}{\frac{B{x}^{3}}{8\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{11\,Bb{x}^{2}}{112\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{33\,{b}^{2}Bx}{448\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{33\,{b}^{3}B}{640\,{c}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{33\,{b}^{4}Bx}{1024\,{c}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{33\,B{b}^{5}}{2048\,{c}^{5}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{99\,{b}^{6}Bx}{8192\,{c}^{5}}\sqrt{c{x}^{2}+bx}}-{\frac{99\,B{b}^{7}}{16384\,{c}^{6}}\sqrt{c{x}^{2}+bx}}+{\frac{99\,B{b}^{8}}{32768}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{13}{2}}}}+{\frac{A{x}^{2}}{7\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{3\,Abx}{28\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{3\,A{b}^{2}}{40\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{3\,A{b}^{3}x}{64\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{3\,A{b}^{4}}{128\,{c}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{9\,A{b}^{5}x}{512\,{c}^{4}}\sqrt{c{x}^{2}+bx}}+{\frac{9\,A{b}^{6}}{1024\,{c}^{5}}\sqrt{c{x}^{2}+bx}}-{\frac{9\,A{b}^{7}}{2048}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*B*x^3*(c*x^2+b*x)^(5/2)/c-11/112*B*b/c^2*x^2*(c*x^2+b*x)^(5/2)+33/448*B*b^2/c^3*x*(c*x^2+b*x)^(5/2)-33/640
*B*b^3/c^4*(c*x^2+b*x)^(5/2)+33/1024*B*b^4/c^4*(c*x^2+b*x)^(3/2)*x+33/2048*B*b^5/c^5*(c*x^2+b*x)^(3/2)-99/8192
*B*b^6/c^5*(c*x^2+b*x)^(1/2)*x-99/16384*B*b^7/c^6*(c*x^2+b*x)^(1/2)+99/32768*B*b^8/c^(13/2)*ln((1/2*b+c*x)/c^(
1/2)+(c*x^2+b*x)^(1/2))+1/7*A*x^2*(c*x^2+b*x)^(5/2)/c-3/28*A*b/c^2*x*(c*x^2+b*x)^(5/2)+3/40*A*b^2/c^3*(c*x^2+b
*x)^(5/2)-3/64*A*b^3/c^3*(c*x^2+b*x)^(3/2)*x-3/128*A*b^4/c^4*(c*x^2+b*x)^(3/2)+9/512*A*b^5/c^4*(c*x^2+b*x)^(1/
2)*x+9/1024*A*b^6/c^5*(c*x^2+b*x)^(1/2)-9/2048*A*b^7/c^(11/2)*ln((1/2*b+c*x)/c^(1/2)+(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(x^3*(B*x+A)*(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.9895, size = 1077, normalized size = 4.53 \begin{align*} \left [-\frac{315 \,{\left (11 \, B b^{8} - 16 \, A b^{7} c\right )} \sqrt{c} \log \left (2 \, c x + b - 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (71680 \, B c^{8} x^{7} - 3465 \, B b^{7} c + 5040 \, A b^{6} c^{2} + 5120 \,{\left (17 \, B b c^{7} + 16 \, A c^{8}\right )} x^{6} + 1280 \,{\left (B b^{2} c^{6} + 80 \, A b c^{7}\right )} x^{5} - 128 \,{\left (11 \, B b^{3} c^{5} - 16 \, A b^{2} c^{6}\right )} x^{4} + 144 \,{\left (11 \, B b^{4} c^{4} - 16 \, A b^{3} c^{5}\right )} x^{3} - 168 \,{\left (11 \, B b^{5} c^{3} - 16 \, A b^{4} c^{4}\right )} x^{2} + 210 \,{\left (11 \, B b^{6} c^{2} - 16 \, A b^{5} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{1146880 \, c^{7}}, -\frac{315 \,{\left (11 \, B b^{8} - 16 \, A b^{7} c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (71680 \, B c^{8} x^{7} - 3465 \, B b^{7} c + 5040 \, A b^{6} c^{2} + 5120 \,{\left (17 \, B b c^{7} + 16 \, A c^{8}\right )} x^{6} + 1280 \,{\left (B b^{2} c^{6} + 80 \, A b c^{7}\right )} x^{5} - 128 \,{\left (11 \, B b^{3} c^{5} - 16 \, A b^{2} c^{6}\right )} x^{4} + 144 \,{\left (11 \, B b^{4} c^{4} - 16 \, A b^{3} c^{5}\right )} x^{3} - 168 \,{\left (11 \, B b^{5} c^{3} - 16 \, A b^{4} c^{4}\right )} x^{2} + 210 \,{\left (11 \, B b^{6} c^{2} - 16 \, A b^{5} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{573440 \, c^{7}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/1146880*(315*(11*B*b^8 - 16*A*b^7*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(71680*B*c^8
*x^7 - 3465*B*b^7*c + 5040*A*b^6*c^2 + 5120*(17*B*b*c^7 + 16*A*c^8)*x^6 + 1280*(B*b^2*c^6 + 80*A*b*c^7)*x^5 -
128*(11*B*b^3*c^5 - 16*A*b^2*c^6)*x^4 + 144*(11*B*b^4*c^4 - 16*A*b^3*c^5)*x^3 - 168*(11*B*b^5*c^3 - 16*A*b^4*c
^4)*x^2 + 210*(11*B*b^6*c^2 - 16*A*b^5*c^3)*x)*sqrt(c*x^2 + b*x))/c^7, -1/573440*(315*(11*B*b^8 - 16*A*b^7*c)*
sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (71680*B*c^8*x^7 - 3465*B*b^7*c + 5040*A*b^6*c^2 + 5120*(1
7*B*b*c^7 + 16*A*c^8)*x^6 + 1280*(B*b^2*c^6 + 80*A*b*c^7)*x^5 - 128*(11*B*b^3*c^5 - 16*A*b^2*c^6)*x^4 + 144*(1
1*B*b^4*c^4 - 16*A*b^3*c^5)*x^3 - 168*(11*B*b^5*c^3 - 16*A*b^4*c^4)*x^2 + 210*(11*B*b^6*c^2 - 16*A*b^5*c^3)*x)
*sqrt(c*x^2 + b*x))/c^7]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18136, size = 336, normalized size = 1.41 \begin{align*} \frac{1}{573440} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (10 \,{\left (4 \,{\left (14 \, B c x + \frac{17 \, B b c^{7} + 16 \, A c^{8}}{c^{7}}\right )} x + \frac{B b^{2} c^{6} + 80 \, A b c^{7}}{c^{7}}\right )} x - \frac{11 \, B b^{3} c^{5} - 16 \, A b^{2} c^{6}}{c^{7}}\right )} x + \frac{9 \,{\left (11 \, B b^{4} c^{4} - 16 \, A b^{3} c^{5}\right )}}{c^{7}}\right )} x - \frac{21 \,{\left (11 \, B b^{5} c^{3} - 16 \, A b^{4} c^{4}\right )}}{c^{7}}\right )} x + \frac{105 \,{\left (11 \, B b^{6} c^{2} - 16 \, A b^{5} c^{3}\right )}}{c^{7}}\right )} x - \frac{315 \,{\left (11 \, B b^{7} c - 16 \, A b^{6} c^{2}\right )}}{c^{7}}\right )} - \frac{9 \,{\left (11 \, B b^{8} - 16 \, A b^{7} c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{32768 \, c^{\frac{13}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/573440*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*(10*(4*(14*B*c*x + (17*B*b*c^7 + 16*A*c^8)/c^7)*x + (B*b^2*c^6 + 80*A*b
*c^7)/c^7)*x - (11*B*b^3*c^5 - 16*A*b^2*c^6)/c^7)*x + 9*(11*B*b^4*c^4 - 16*A*b^3*c^5)/c^7)*x - 21*(11*B*b^5*c^
3 - 16*A*b^4*c^4)/c^7)*x + 105*(11*B*b^6*c^2 - 16*A*b^5*c^3)/c^7)*x - 315*(11*B*b^7*c - 16*A*b^6*c^2)/c^7) - 9
/32768*(11*B*b^8 - 16*A*b^7*c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(13/2)