3.94 \(\int x^2 (A+B x) (b x+c x^2)^{5/2} \, dx\)

Optimal. Leaf size=241 \[ -\frac{5 b^6 (b+2 c x) \sqrt{b x+c x^2} (11 b B-18 A c)}{32768 c^6}+\frac{5 b^4 (b+2 c x) \left (b x+c x^2\right )^{3/2} (11 b B-18 A c)}{12288 c^5}-\frac{b^2 (b+2 c x) \left (b x+c x^2\right )^{5/2} (11 b B-18 A c)}{768 c^4}+\frac{5 b^8 (11 b B-18 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{32768 c^{13/2}}+\frac{b \left (b x+c x^2\right )^{7/2} (11 b B-18 A c)}{224 c^3}-\frac{x \left (b x+c x^2\right )^{7/2} (11 b B-18 A c)}{144 c^2}+\frac{B x^2 \left (b x+c x^2\right )^{7/2}}{9 c} \]

[Out]

(-5*b^6*(11*b*B - 18*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(32768*c^6) + (5*b^4*(11*b*B - 18*A*c)*(b + 2*c*x)*(b
*x + c*x^2)^(3/2))/(12288*c^5) - (b^2*(11*b*B - 18*A*c)*(b + 2*c*x)*(b*x + c*x^2)^(5/2))/(768*c^4) + (b*(11*b*
B - 18*A*c)*(b*x + c*x^2)^(7/2))/(224*c^3) - ((11*b*B - 18*A*c)*x*(b*x + c*x^2)^(7/2))/(144*c^2) + (B*x^2*(b*x
+ c*x^2)^(7/2))/(9*c) + (5*b^8*(11*b*B - 18*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(32768*c^(13/2))

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Rubi [A]  time = 0.247218, antiderivative size = 241, 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} \[ -\frac{5 b^6 (b+2 c x) \sqrt{b x+c x^2} (11 b B-18 A c)}{32768 c^6}+\frac{5 b^4 (b+2 c x) \left (b x+c x^2\right )^{3/2} (11 b B-18 A c)}{12288 c^5}-\frac{b^2 (b+2 c x) \left (b x+c x^2\right )^{5/2} (11 b B-18 A c)}{768 c^4}+\frac{5 b^8 (11 b B-18 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{32768 c^{13/2}}+\frac{b \left (b x+c x^2\right )^{7/2} (11 b B-18 A c)}{224 c^3}-\frac{x \left (b x+c x^2\right )^{7/2} (11 b B-18 A c)}{144 c^2}+\frac{B x^2 \left (b x+c x^2\right )^{7/2}}{9 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b^6*(11*b*B - 18*A*c)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(32768*c^6) + (5*b^4*(11*b*B - 18*A*c)*(b + 2*c*x)*(b
*x + c*x^2)^(3/2))/(12288*c^5) - (b^2*(11*b*B - 18*A*c)*(b + 2*c*x)*(b*x + c*x^2)^(5/2))/(768*c^4) + (b*(11*b*
B - 18*A*c)*(b*x + c*x^2)^(7/2))/(224*c^3) - ((11*b*B - 18*A*c)*x*(b*x + c*x^2)^(7/2))/(144*c^2) + (B*x^2*(b*x
+ c*x^2)^(7/2))/(9*c) + (5*b^8*(11*b*B - 18*A*c)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(32768*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^2 (A+B x) \left (b x+c x^2\right )^{5/2} \, dx &=\frac{B x^2 \left (b x+c x^2\right )^{7/2}}{9 c}+\frac{\left (2 (-b B+A c)+\frac{7}{2} (-b B+2 A c)\right ) \int x^2 \left (b x+c x^2\right )^{5/2} \, dx}{9 c}\\ &=-\frac{(11 b B-18 A c) x \left (b x+c x^2\right )^{7/2}}{144 c^2}+\frac{B x^2 \left (b x+c x^2\right )^{7/2}}{9 c}+\frac{(b (11 b B-18 A c)) \int x \left (b x+c x^2\right )^{5/2} \, dx}{32 c^2}\\ &=\frac{b (11 b B-18 A c) \left (b x+c x^2\right )^{7/2}}{224 c^3}-\frac{(11 b B-18 A c) x \left (b x+c x^2\right )^{7/2}}{144 c^2}+\frac{B x^2 \left (b x+c x^2\right )^{7/2}}{9 c}-\frac{\left (b^2 (11 b B-18 A c)\right ) \int \left (b x+c x^2\right )^{5/2} \, dx}{64 c^3}\\ &=-\frac{b^2 (11 b B-18 A c) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{768 c^4}+\frac{b (11 b B-18 A c) \left (b x+c x^2\right )^{7/2}}{224 c^3}-\frac{(11 b B-18 A c) x \left (b x+c x^2\right )^{7/2}}{144 c^2}+\frac{B x^2 \left (b x+c x^2\right )^{7/2}}{9 c}+\frac{\left (5 b^4 (11 b B-18 A c)\right ) \int \left (b x+c x^2\right )^{3/2} \, dx}{1536 c^4}\\ &=\frac{5 b^4 (11 b B-18 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{12288 c^5}-\frac{b^2 (11 b B-18 A c) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{768 c^4}+\frac{b (11 b B-18 A c) \left (b x+c x^2\right )^{7/2}}{224 c^3}-\frac{(11 b B-18 A c) x \left (b x+c x^2\right )^{7/2}}{144 c^2}+\frac{B x^2 \left (b x+c x^2\right )^{7/2}}{9 c}-\frac{\left (5 b^6 (11 b B-18 A c)\right ) \int \sqrt{b x+c x^2} \, dx}{8192 c^5}\\ &=-\frac{5 b^6 (11 b B-18 A c) (b+2 c x) \sqrt{b x+c x^2}}{32768 c^6}+\frac{5 b^4 (11 b B-18 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{12288 c^5}-\frac{b^2 (11 b B-18 A c) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{768 c^4}+\frac{b (11 b B-18 A c) \left (b x+c x^2\right )^{7/2}}{224 c^3}-\frac{(11 b B-18 A c) x \left (b x+c x^2\right )^{7/2}}{144 c^2}+\frac{B x^2 \left (b x+c x^2\right )^{7/2}}{9 c}+\frac{\left (5 b^8 (11 b B-18 A c)\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{65536 c^6}\\ &=-\frac{5 b^6 (11 b B-18 A c) (b+2 c x) \sqrt{b x+c x^2}}{32768 c^6}+\frac{5 b^4 (11 b B-18 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{12288 c^5}-\frac{b^2 (11 b B-18 A c) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{768 c^4}+\frac{b (11 b B-18 A c) \left (b x+c x^2\right )^{7/2}}{224 c^3}-\frac{(11 b B-18 A c) x \left (b x+c x^2\right )^{7/2}}{144 c^2}+\frac{B x^2 \left (b x+c x^2\right )^{7/2}}{9 c}+\frac{\left (5 b^8 (11 b B-18 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{32768 c^6}\\ &=-\frac{5 b^6 (11 b B-18 A c) (b+2 c x) \sqrt{b x+c x^2}}{32768 c^6}+\frac{5 b^4 (11 b B-18 A c) (b+2 c x) \left (b x+c x^2\right )^{3/2}}{12288 c^5}-\frac{b^2 (11 b B-18 A c) (b+2 c x) \left (b x+c x^2\right )^{5/2}}{768 c^4}+\frac{b (11 b B-18 A c) \left (b x+c x^2\right )^{7/2}}{224 c^3}-\frac{(11 b B-18 A c) x \left (b x+c x^2\right )^{7/2}}{144 c^2}+\frac{B x^2 \left (b x+c x^2\right )^{7/2}}{9 c}+\frac{5 b^8 (11 b B-18 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{32768 c^{13/2}}\\ \end{align*}

Mathematica [A]  time = 0.410069, size = 197, normalized size = 0.82 \[ \frac{x^3 (x (b+c x))^{5/2} \left (\frac{11 (11 b B-18 A c) \left (315 b^{15/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )-\sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1} \left (168 b^5 c^2 x^2-144 b^4 c^3 x^3+128 b^3 c^4 x^4+20736 b^2 c^5 x^5-210 b^6 c x+315 b^7+33792 b c^6 x^6+14336 c^7 x^7\right )\right )}{229376 c^{11/2} x^{11/2} \sqrt{\frac{c x}{b}+1}}+11 B (b+c x)^3\right )}{99 c (b+c x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(x*(b + c*x))^(5/2)*(11*B*(b + c*x)^3 + (11*(11*b*B - 18*A*c)*(-(Sqrt[c]*Sqrt[x]*Sqrt[1 + (c*x)/b]*(315*b
^7 - 210*b^6*c*x + 168*b^5*c^2*x^2 - 144*b^4*c^3*x^3 + 128*b^3*c^4*x^4 + 20736*b^2*c^5*x^5 + 33792*b*c^6*x^6 +
14336*c^7*x^7)) + 315*b^(15/2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]]))/(229376*c^(11/2)*x^(11/2)*Sqrt[1 + (c*x)/
b])))/(99*c*(b + c*x)^2)

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Maple [A]  time = 0.008, size = 409, normalized size = 1.7 \begin{align*}{\frac{B{x}^{2}}{9\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{11\,bBx}{144\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}+{\frac{11\,{b}^{2}B}{224\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{11\,{b}^{3}Bx}{384\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{11\,{b}^{4}B}{768\,{c}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{55\,B{b}^{5}x}{6144\,{c}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{55\,B{b}^{6}}{12288\,{c}^{5}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{55\,B{b}^{7}x}{16384\,{c}^{5}}\sqrt{c{x}^{2}+bx}}-{\frac{55\,B{b}^{8}}{32768\,{c}^{6}}\sqrt{c{x}^{2}+bx}}+{\frac{55\,B{b}^{9}}{65536}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{13}{2}}}}+{\frac{Ax}{8\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{9\,Ab}{112\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}+{\frac{3\,A{b}^{2}x}{64\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{3\,A{b}^{3}}{128\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{15\,A{b}^{4}x}{1024\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{15\,A{b}^{5}}{2048\,{c}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{45\,A{b}^{6}x}{8192\,{c}^{4}}\sqrt{c{x}^{2}+bx}}+{\frac{45\,A{b}^{7}}{16384\,{c}^{5}}\sqrt{c{x}^{2}+bx}}-{\frac{45\,A{b}^{8}}{32768}\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^2*(B*x+A)*(c*x^2+b*x)^(5/2),x)

[Out]

1/9*B*x^2*(c*x^2+b*x)^(7/2)/c-11/144*B*b/c^2*x*(c*x^2+b*x)^(7/2)+11/224*B*b^2/c^3*(c*x^2+b*x)^(7/2)-11/384*B*b
^3/c^3*(c*x^2+b*x)^(5/2)*x-11/768*B*b^4/c^4*(c*x^2+b*x)^(5/2)+55/6144*B*b^5/c^4*(c*x^2+b*x)^(3/2)*x+55/12288*B
*b^6/c^5*(c*x^2+b*x)^(3/2)-55/16384*B*b^7/c^5*(c*x^2+b*x)^(1/2)*x-55/32768*B*b^8/c^6*(c*x^2+b*x)^(1/2)+55/6553
6*B*b^9/c^(13/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/8*A*x*(c*x^2+b*x)^(7/2)/c-9/112*A*b/c^2*(c*x^2+b*
x)^(7/2)+3/64*A*b^2/c^2*(c*x^2+b*x)^(5/2)*x+3/128*A*b^3/c^3*(c*x^2+b*x)^(5/2)-15/1024*A*b^4/c^3*(c*x^2+b*x)^(3
/2)*x-15/2048*A*b^5/c^4*(c*x^2+b*x)^(3/2)+45/8192*A*b^6/c^4*(c*x^2+b*x)^(1/2)*x+45/16384*A*b^7/c^5*(c*x^2+b*x)
^(1/2)-45/32768*A*b^8/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^2*(B*x+A)*(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.97569, size = 1208, normalized size = 5.01 \begin{align*} \left [-\frac{315 \,{\left (11 \, B b^{9} - 18 \, A b^{8} c\right )} \sqrt{c} \log \left (2 \, c x + b - 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (229376 \, B c^{9} x^{8} - 3465 \, B b^{8} c + 5670 \, A b^{7} c^{2} + 14336 \,{\left (37 \, B b c^{8} + 18 \, A c^{9}\right )} x^{7} + 3072 \,{\left (103 \, B b^{2} c^{7} + 198 \, A b c^{8}\right )} x^{6} + 256 \,{\left (5 \, B b^{3} c^{6} + 1458 \, A b^{2} c^{7}\right )} x^{5} - 128 \,{\left (11 \, B b^{4} c^{5} - 18 \, A b^{3} c^{6}\right )} x^{4} + 144 \,{\left (11 \, B b^{5} c^{4} - 18 \, A b^{4} c^{5}\right )} x^{3} - 168 \,{\left (11 \, B b^{6} c^{3} - 18 \, A b^{5} c^{4}\right )} x^{2} + 210 \,{\left (11 \, B b^{7} c^{2} - 18 \, A b^{6} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{4128768 \, c^{7}}, -\frac{315 \,{\left (11 \, B b^{9} - 18 \, A b^{8} c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (229376 \, B c^{9} x^{8} - 3465 \, B b^{8} c + 5670 \, A b^{7} c^{2} + 14336 \,{\left (37 \, B b c^{8} + 18 \, A c^{9}\right )} x^{7} + 3072 \,{\left (103 \, B b^{2} c^{7} + 198 \, A b c^{8}\right )} x^{6} + 256 \,{\left (5 \, B b^{3} c^{6} + 1458 \, A b^{2} c^{7}\right )} x^{5} - 128 \,{\left (11 \, B b^{4} c^{5} - 18 \, A b^{3} c^{6}\right )} x^{4} + 144 \,{\left (11 \, B b^{5} c^{4} - 18 \, A b^{4} c^{5}\right )} x^{3} - 168 \,{\left (11 \, B b^{6} c^{3} - 18 \, A b^{5} c^{4}\right )} x^{2} + 210 \,{\left (11 \, B b^{7} c^{2} - 18 \, A b^{6} c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{2064384 \, c^{7}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4128768*(315*(11*B*b^9 - 18*A*b^8*c)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(229376*B*c^
9*x^8 - 3465*B*b^8*c + 5670*A*b^7*c^2 + 14336*(37*B*b*c^8 + 18*A*c^9)*x^7 + 3072*(103*B*b^2*c^7 + 198*A*b*c^8)
*x^6 + 256*(5*B*b^3*c^6 + 1458*A*b^2*c^7)*x^5 - 128*(11*B*b^4*c^5 - 18*A*b^3*c^6)*x^4 + 144*(11*B*b^5*c^4 - 18
*A*b^4*c^5)*x^3 - 168*(11*B*b^6*c^3 - 18*A*b^5*c^4)*x^2 + 210*(11*B*b^7*c^2 - 18*A*b^6*c^3)*x)*sqrt(c*x^2 + b*
x))/c^7, -1/2064384*(315*(11*B*b^9 - 18*A*b^8*c)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (229376*B
*c^9*x^8 - 3465*B*b^8*c + 5670*A*b^7*c^2 + 14336*(37*B*b*c^8 + 18*A*c^9)*x^7 + 3072*(103*B*b^2*c^7 + 198*A*b*c
^8)*x^6 + 256*(5*B*b^3*c^6 + 1458*A*b^2*c^7)*x^5 - 128*(11*B*b^4*c^5 - 18*A*b^3*c^6)*x^4 + 144*(11*B*b^5*c^4 -
18*A*b^4*c^5)*x^3 - 168*(11*B*b^6*c^3 - 18*A*b^5*c^4)*x^2 + 210*(11*B*b^7*c^2 - 18*A*b^6*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^{2} \left (x \left (b + c x\right )\right )^{\frac{5}{2}} \left (A + B x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1708, size = 381, normalized size = 1.58 \begin{align*} \frac{1}{2064384} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (2 \,{\left (4 \,{\left (14 \,{\left (16 \, B c^{2} x + \frac{37 \, B b c^{9} + 18 \, A c^{10}}{c^{8}}\right )} x + \frac{3 \,{\left (103 \, B b^{2} c^{8} + 198 \, A b c^{9}\right )}}{c^{8}}\right )} x + \frac{5 \, B b^{3} c^{7} + 1458 \, A b^{2} c^{8}}{c^{8}}\right )} x - \frac{11 \, B b^{4} c^{6} - 18 \, A b^{3} c^{7}}{c^{8}}\right )} x + \frac{9 \,{\left (11 \, B b^{5} c^{5} - 18 \, A b^{4} c^{6}\right )}}{c^{8}}\right )} x - \frac{21 \,{\left (11 \, B b^{6} c^{4} - 18 \, A b^{5} c^{5}\right )}}{c^{8}}\right )} x + \frac{105 \,{\left (11 \, B b^{7} c^{3} - 18 \, A b^{6} c^{4}\right )}}{c^{8}}\right )} x - \frac{315 \,{\left (11 \, B b^{8} c^{2} - 18 \, A b^{7} c^{3}\right )}}{c^{8}}\right )} - \frac{5 \,{\left (11 \, B b^{9} - 18 \, A b^{8} c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{65536 \, c^{\frac{13}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2064384*sqrt(c*x^2 + b*x)*(2*(4*(2*(8*(2*(4*(14*(16*B*c^2*x + (37*B*b*c^9 + 18*A*c^10)/c^8)*x + 3*(103*B*b^2
*c^8 + 198*A*b*c^9)/c^8)*x + (5*B*b^3*c^7 + 1458*A*b^2*c^8)/c^8)*x - (11*B*b^4*c^6 - 18*A*b^3*c^7)/c^8)*x + 9*
(11*B*b^5*c^5 - 18*A*b^4*c^6)/c^8)*x - 21*(11*B*b^6*c^4 - 18*A*b^5*c^5)/c^8)*x + 105*(11*B*b^7*c^3 - 18*A*b^6*
c^4)/c^8)*x - 315*(11*B*b^8*c^2 - 18*A*b^7*c^3)/c^8) - 5/65536*(11*B*b^9 - 18*A*b^8*c)*log(abs(-2*(sqrt(c)*x -
sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(13/2)