3.2.7 \(\int x^3 (A+B x^2) (b x^2+c x^4)^{3/2} \, dx\) [107]

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

[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

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Rubi [A]
time = 0.17, 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 = {2059, 793, 626, 634, 212} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/1024*(b^3*(7*b*B - 12*A*c)*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 626

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 634

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 793

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 2059

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} \text {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)) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.30, size = 191, normalized size = 1.14 \begin {gather*} \frac {x \sqrt {b+c x^2} \left (\sqrt {c} x \sqrt {b+c x^2} \left (-105 b^5 B+48 b^2 c^3 x^4 \left (2 A+B x^2\right )+256 c^5 x^8 \left (6 A+5 B x^2\right )-8 b^3 c^2 x^2 \left (15 A+7 B x^2\right )+10 b^4 c \left (18 A+7 B x^2\right )+64 b c^4 x^6 \left (33 A+26 B x^2\right )\right )-15 b^5 (7 b B-12 A c) \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )\right )}{15360 c^{9/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[b + c*x^2]*(Sqrt[c]*x*Sqrt[b + c*x^2]*(-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
^5*(7*b*B - 12*A*c)*Log[-(Sqrt[c]*x) + Sqrt[b + c*x^2]]))/(15360*c^(9/2)*Sqrt[x^2*(b + c*x^2)])

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Maple [A]
time = 0.39, size = 286, normalized size = 1.71

method	result	size
risch	\(\frac {\left (1280 B \,c^{5} x^{10}+1536 A \,c^{5} x^{8}+1664 B b \,c^{4} x^{8}+2112 A b \,c^{4} x^{6}+48 B \,b^{2} c^{3} x^{6}+96 A \,b^{2} c^{3} x^{4}-56 B \,b^{3} c^{2} x^{4}-120 A \,b^{3} c^{2} x^{2}+70 B \,b^{4} c \,x^{2}+180 A \,b^{4} c -105 b^{5} B \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{15360 c^{4}}+\frac {\left (-\frac {3 b^{5} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) A}{256 c^{\frac {7}{2}}}+\frac {7 b^{6} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) B}{1024 c^{\frac {9}{2}}}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{x \sqrt {c \,x^{2}+b}}\)	\(207\)
default	\(\frac {\left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}} \left (1280 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {7}{2}} x^{7}+1536 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {7}{2}} x^{5}-896 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {5}{2}} b \,x^{5}-960 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {5}{2}} b \,x^{3}+560 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {3}{2}} b^{2} x^{3}+480 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{\frac {3}{2}} b^{2} x -280 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} \sqrt {c}\, b^{3} x -120 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}} b^{3} x +70 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}\, b^{4} x -180 A \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} b^{4} x +105 B \sqrt {c \,x^{2}+b}\, \sqrt {c}\, b^{5} x -180 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{5} c +105 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{6}\right )}{15360 x^{3} \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {9}{2}}}\)	\(286\)

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,method=_RETURNVERBOSE)

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 315 vs. \(2 (147) = 294\).
time = 0.28, size = 315, normalized size = 1.89 \begin {gather*} \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 \end {gather*}

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

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Fricas [A]
time = 2.23, size = 369, normalized size = 2.21 \begin {gather*} \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 ] \end {gather*}

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]

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

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)

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Giac [A]
time = 0.87, size = 246, normalized size = 1.47 \begin {gather*} \frac {1}{15360} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, B c x^{2} \mathrm {sgn}\left (x\right ) + \frac {13 \, B b c^{10} \mathrm {sgn}\left (x\right ) + 12 \, A c^{11} \mathrm {sgn}\left (x\right )}{c^{10}}\right )} x^{2} + \frac {3 \, {\left (B b^{2} c^{9} \mathrm {sgn}\left (x\right ) + 44 \, A b c^{10} \mathrm {sgn}\left (x\right )\right )}}{c^{10}}\right )} x^{2} - \frac {7 \, B b^{3} c^{8} \mathrm {sgn}\left (x\right ) - 12 \, A b^{2} c^{9} \mathrm {sgn}\left (x\right )}{c^{10}}\right )} x^{2} + \frac {5 \, {\left (7 \, B b^{4} c^{7} \mathrm {sgn}\left (x\right ) - 12 \, A b^{3} c^{8} \mathrm {sgn}\left (x\right )\right )}}{c^{10}}\right )} x^{2} - \frac {15 \, {\left (7 \, B b^{5} c^{6} \mathrm {sgn}\left (x\right ) - 12 \, A b^{4} c^{7} \mathrm {sgn}\left (x\right )\right )}}{c^{10}}\right )} \sqrt {c x^{2} + b} x - \frac {{\left (7 \, B b^{6} \mathrm {sgn}\left (x\right ) - 12 \, A b^{5} c \mathrm {sgn}\left (x\right )\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}\left (x\right )}{2048 \, c^{\frac {9}{2}}} \end {gather*}

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)

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

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)

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