3.107 \(\int x^3 (A+B x^2) (b x^2+c x^4)^{3/2} \, dx\)

Optimal. Leaf size=167 \[ \frac {b^5 (7 b B-12 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{1024 c^{9/2}}-\frac {b^3 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4} (7 b B-12 A c)}{1024 c^4}+\frac {b \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2} (7 b B-12 A c)}{384 c^3}-\frac {\left (b x^2+c x^4\right )^{5/2} \left (-12 A c+7 b B-10 B c x^2\right )}{120 c^2} \]

[Out]

1/384*b*(-12*A*c+7*B*b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(3/2)/c^3-1/120*(-10*B*c*x^2-12*A*c+7*B*b)*(c*x^4+b*x^2)^(5/
2)/c^2+1/1024*b^5*(-12*A*c+7*B*b)*arctanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/2))/c^(9/2)-1/1024*b^3*(-12*A*c+7*B*b)*
(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c^4

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2034, 779, 612, 620, 206} \[ -\frac {b^3 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4} (7 b B-12 A c)}{1024 c^4}+\frac {b^5 (7 b B-12 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{1024 c^{9/2}}+\frac {b \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2} (7 b B-12 A c)}{384 c^3}-\frac {\left (b x^2+c x^4\right )^{5/2} \left (-12 A c+7 b B-10 B c x^2\right )}{120 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(A + B*x^2)*(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(b^3*(7*b*B - 12*A*c)*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(1024*c^4) + (b*(7*b*B - 12*A*c)*(b + 2*c*x^2)*(b*x^
2 + c*x^4)^(3/2))/(384*c^3) - ((7*b*B - 12*A*c - 10*B*c*x^2)*(b*x^2 + c*x^4)^(5/2))/(120*c^2) + (b^5*(7*b*B -
12*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(1024*c^(9/2))

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])

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 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 2034

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

\begin {align*} \int x^3 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x (A+B x) \left (b x+c x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=-\frac {\left (7 b B-12 A c-10 B c x^2\right ) \left (b x^2+c x^4\right )^{5/2}}{120 c^2}+\frac {(b (7 b B-12 A c)) \operatorname {Subst}\left (\int \left (b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{48 c^2}\\ &=\frac {b (7 b B-12 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{384 c^3}-\frac {\left (7 b B-12 A c-10 B c x^2\right ) \left (b x^2+c x^4\right )^{5/2}}{120 c^2}-\frac {\left (b^3 (7 b B-12 A c)\right ) \operatorname {Subst}\left (\int \sqrt {b x+c x^2} \, dx,x,x^2\right )}{256 c^3}\\ &=-\frac {b^3 (7 b B-12 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{1024 c^4}+\frac {b (7 b B-12 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{384 c^3}-\frac {\left (7 b B-12 A c-10 B c x^2\right ) \left (b x^2+c x^4\right )^{5/2}}{120 c^2}+\frac {\left (b^5 (7 b B-12 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{2048 c^4}\\ &=-\frac {b^3 (7 b B-12 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{1024 c^4}+\frac {b (7 b B-12 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{384 c^3}-\frac {\left (7 b B-12 A c-10 B c x^2\right ) \left (b x^2+c x^4\right )^{5/2}}{120 c^2}+\frac {\left (b^5 (7 b B-12 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right )}{1024 c^4}\\ &=-\frac {b^3 (7 b B-12 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{1024 c^4}+\frac {b (7 b B-12 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{384 c^3}-\frac {\left (7 b B-12 A c-10 B c x^2\right ) \left (b x^2+c x^4\right )^{5/2}}{120 c^2}+\frac {b^5 (7 b B-12 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{1024 c^{9/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.32, size = 193, normalized size = 1.16 \[ \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (15 b^{9/2} (7 b B-12 A c) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )+\sqrt {c} x \sqrt {\frac {c x^2}{b}+1} \left (10 b^4 c \left (18 A+7 B x^2\right )-8 b^3 c^2 x^2 \left (15 A+7 B x^2\right )+48 b^2 c^3 x^4 \left (2 A+B x^2\right )+64 b c^4 x^6 \left (33 A+26 B x^2\right )+256 c^5 x^8 \left (6 A+5 B x^2\right )-105 b^5 B\right )\right )}{15360 c^{9/2} x \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(A + B*x^2)*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(Sqrt[x^2*(b + c*x^2)]*(Sqrt[c]*x*Sqrt[1 + (c*x^2)/b]*(-105*b^5*B + 48*b^2*c^3*x^4*(2*A + B*x^2) + 256*c^5*x^8
*(6*A + 5*B*x^2) - 8*b^3*c^2*x^2*(15*A + 7*B*x^2) + 10*b^4*c*(18*A + 7*B*x^2) + 64*b*c^4*x^6*(33*A + 26*B*x^2)
) + 15*b^(9/2)*(7*b*B - 12*A*c)*ArcSinh[(Sqrt[c]*x)/Sqrt[b]]))/(15360*c^(9/2)*x*Sqrt[1 + (c*x^2)/b])

________________________________________________________________________________________

fricas [A]  time = 1.29, size = 369, normalized size = 2.21 \[ \left [-\frac {15 \, {\left (7 \, B b^{6} - 12 \, A b^{5} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (1280 \, B c^{6} x^{10} + 128 \, {\left (13 \, B b c^{5} + 12 \, A c^{6}\right )} x^{8} - 105 \, B b^{5} c + 180 \, A b^{4} c^{2} + 48 \, {\left (B b^{2} c^{4} + 44 \, A b c^{5}\right )} x^{6} - 8 \, {\left (7 \, B b^{3} c^{3} - 12 \, A b^{2} c^{4}\right )} x^{4} + 10 \, {\left (7 \, B b^{4} c^{2} - 12 \, A b^{3} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{30720 \, c^{5}}, -\frac {15 \, {\left (7 \, B b^{6} - 12 \, A b^{5} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (1280 \, B c^{6} x^{10} + 128 \, {\left (13 \, B b c^{5} + 12 \, A c^{6}\right )} x^{8} - 105 \, B b^{5} c + 180 \, A b^{4} c^{2} + 48 \, {\left (B b^{2} c^{4} + 44 \, A b c^{5}\right )} x^{6} - 8 \, {\left (7 \, B b^{3} c^{3} - 12 \, A b^{2} c^{4}\right )} x^{4} + 10 \, {\left (7 \, B b^{4} c^{2} - 12 \, A b^{3} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15360 \, c^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/30720*(15*(7*B*b^6 - 12*A*b^5*c)*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(1280*B*c^6
*x^10 + 128*(13*B*b*c^5 + 12*A*c^6)*x^8 - 105*B*b^5*c + 180*A*b^4*c^2 + 48*(B*b^2*c^4 + 44*A*b*c^5)*x^6 - 8*(7
*B*b^3*c^3 - 12*A*b^2*c^4)*x^4 + 10*(7*B*b^4*c^2 - 12*A*b^3*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^5, -1/15360*(15*(
7*B*b^6 - 12*A*b^5*c)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - (1280*B*c^6*x^10 + 128*(13*B
*b*c^5 + 12*A*c^6)*x^8 - 105*B*b^5*c + 180*A*b^4*c^2 + 48*(B*b^2*c^4 + 44*A*b*c^5)*x^6 - 8*(7*B*b^3*c^3 - 12*A
*b^2*c^4)*x^4 + 10*(7*B*b^4*c^2 - 12*A*b^3*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^5]

________________________________________________________________________________________

giac [A]  time = 0.23, size = 246, normalized size = 1.47 \[ \frac {1}{15360} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, B c x^{2} \mathrm {sgn}\relax (x) + \frac {13 \, B b c^{10} \mathrm {sgn}\relax (x) + 12 \, A c^{11} \mathrm {sgn}\relax (x)}{c^{10}}\right )} x^{2} + \frac {3 \, {\left (B b^{2} c^{9} \mathrm {sgn}\relax (x) + 44 \, A b c^{10} \mathrm {sgn}\relax (x)\right )}}{c^{10}}\right )} x^{2} - \frac {7 \, B b^{3} c^{8} \mathrm {sgn}\relax (x) - 12 \, A b^{2} c^{9} \mathrm {sgn}\relax (x)}{c^{10}}\right )} x^{2} + \frac {5 \, {\left (7 \, B b^{4} c^{7} \mathrm {sgn}\relax (x) - 12 \, A b^{3} c^{8} \mathrm {sgn}\relax (x)\right )}}{c^{10}}\right )} x^{2} - \frac {15 \, {\left (7 \, B b^{5} c^{6} \mathrm {sgn}\relax (x) - 12 \, A b^{4} c^{7} \mathrm {sgn}\relax (x)\right )}}{c^{10}}\right )} \sqrt {c x^{2} + b} x - \frac {{\left (7 \, B b^{6} \mathrm {sgn}\relax (x) - 12 \, A b^{5} c \mathrm {sgn}\relax (x)\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{1024 \, c^{\frac {9}{2}}} + \frac {{\left (7 \, B b^{6} \log \left ({\left | b \right |}\right ) - 12 \, A b^{5} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\relax (x)}{2048 \, c^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

1/15360*(2*(4*(2*(8*(10*B*c*x^2*sgn(x) + (13*B*b*c^10*sgn(x) + 12*A*c^11*sgn(x))/c^10)*x^2 + 3*(B*b^2*c^9*sgn(
x) + 44*A*b*c^10*sgn(x))/c^10)*x^2 - (7*B*b^3*c^8*sgn(x) - 12*A*b^2*c^9*sgn(x))/c^10)*x^2 + 5*(7*B*b^4*c^7*sgn
(x) - 12*A*b^3*c^8*sgn(x))/c^10)*x^2 - 15*(7*B*b^5*c^6*sgn(x) - 12*A*b^4*c^7*sgn(x))/c^10)*sqrt(c*x^2 + b)*x -
1/1024*(7*B*b^6*sgn(x) - 12*A*b^5*c*sgn(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^(9/2) + 1/2048*(7*B*b^6*
log(abs(b)) - 12*A*b^5*c*log(abs(b)))*sgn(x)/c^(9/2)

________________________________________________________________________________________

maple [A]  time = 0.07, size = 286, normalized size = 1.71 \[ \frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (1280 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B \,c^{\frac {7}{2}} x^{7}+1536 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A \,c^{\frac {7}{2}} x^{5}-896 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B b \,c^{\frac {5}{2}} x^{5}-180 A \,b^{5} c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+105 B \,b^{6} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-180 \sqrt {c \,x^{2}+b}\, A \,b^{4} c^{\frac {3}{2}} x -960 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A b \,c^{\frac {5}{2}} x^{3}+105 \sqrt {c \,x^{2}+b}\, B \,b^{5} \sqrt {c}\, x +560 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B \,b^{2} c^{\frac {3}{2}} x^{3}-120 \left (c \,x^{2}+b \right )^{\frac {3}{2}} A \,b^{3} c^{\frac {3}{2}} x +70 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B \,b^{4} \sqrt {c}\, x +480 \left (c \,x^{2}+b \right )^{\frac {5}{2}} A \,b^{2} c^{\frac {3}{2}} x -280 \left (c \,x^{2}+b \right )^{\frac {5}{2}} B \,b^{3} \sqrt {c}\, x \right )}{15360 \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {9}{2}} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x)

[Out]

1/15360*(c*x^4+b*x^2)^(3/2)*(1280*B*(c*x^2+b)^(5/2)*c^(7/2)*x^7+1536*A*(c*x^2+b)^(5/2)*c^(7/2)*x^5-896*B*(c*x^
2+b)^(5/2)*c^(5/2)*x^5*b-960*A*(c*x^2+b)^(5/2)*c^(5/2)*x^3*b+560*B*(c*x^2+b)^(5/2)*c^(3/2)*x^3*b^2+480*A*(c*x^
2+b)^(5/2)*c^(3/2)*x*b^2-280*B*(c*x^2+b)^(5/2)*c^(1/2)*x*b^3-120*(c*x^2+b)^(3/2)*A*b^3*c^(3/2)*x+70*(c*x^2+b)^
(3/2)*B*b^4*c^(1/2)*x-180*(c*x^2+b)^(1/2)*A*b^4*c^(3/2)*x+105*(c*x^2+b)^(1/2)*B*b^5*c^(1/2)*x-180*A*b^5*c*ln(c
^(1/2)*x+(c*x^2+b)^(1/2))+105*B*b^6*ln(c^(1/2)*x+(c*x^2+b)^(1/2)))/x^3/(c*x^2+b)^(3/2)/c^(9/2)

________________________________________________________________________________________

maxima [B]  time = 1.56, size = 315, normalized size = 1.89 \[ \frac {1}{2560} \, {\left (\frac {60 \, \sqrt {c x^{4} + b x^{2}} b^{3} x^{2}}{c^{2}} - \frac {160 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b x^{2}}{c} - \frac {15 \, b^{5} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}} + \frac {30 \, \sqrt {c x^{4} + b x^{2}} b^{4}}{c^{3}} - \frac {80 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2}}{c^{2}} + \frac {256 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {5}{2}}}{c}\right )} A - \frac {1}{30720} \, {\left (\frac {420 \, \sqrt {c x^{4} + b x^{2}} b^{4} x^{2}}{c^{3}} - \frac {1120 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2} x^{2}}{c^{2}} - \frac {2560 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {5}{2}} x^{2}}{c} - \frac {105 \, b^{6} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {9}{2}}} + \frac {210 \, \sqrt {c x^{4} + b x^{2}} b^{5}}{c^{4}} - \frac {560 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{3}}{c^{3}} + \frac {1792 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {5}{2}} b}{c^{2}}\right )} B \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

1/2560*(60*sqrt(c*x^4 + b*x^2)*b^3*x^2/c^2 - 160*(c*x^4 + b*x^2)^(3/2)*b*x^2/c - 15*b^5*log(2*c*x^2 + b + 2*sq
rt(c*x^4 + b*x^2)*sqrt(c))/c^(7/2) + 30*sqrt(c*x^4 + b*x^2)*b^4/c^3 - 80*(c*x^4 + b*x^2)^(3/2)*b^2/c^2 + 256*(
c*x^4 + b*x^2)^(5/2)/c)*A - 1/30720*(420*sqrt(c*x^4 + b*x^2)*b^4*x^2/c^3 - 1120*(c*x^4 + b*x^2)^(3/2)*b^2*x^2/
c^2 - 2560*(c*x^4 + b*x^2)^(5/2)*x^2/c - 105*b^6*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(9/2) + 21
0*sqrt(c*x^4 + b*x^2)*b^5/c^4 - 560*(c*x^4 + b*x^2)^(3/2)*b^3/c^3 + 1792*(c*x^4 + b*x^2)^(5/2)*b/c^2)*B

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(A + B*x^2)*(b*x^2 + c*x^4)^(3/2),x)

[Out]

int(x^3*(A + B*x^2)*(b*x^2 + c*x^4)^(3/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(B*x**2+A)*(c*x**4+b*x**2)**(3/2),x)

[Out]

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

________________________________________________________________________________________