3.106 \(\int x^5 (A+B x^2) (b x^2+c x^4)^{3/2} \, dx\)

Optimal. Leaf size=223 \[ \frac{b^4 \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4} (9 b B-14 A c)}{2048 c^5}-\frac{b^2 \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2} (9 b B-14 A c)}{768 c^4}-\frac{b^6 (9 b B-14 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{2048 c^{11/2}}+\frac{b \left (b x^2+c x^4\right )^{5/2} (9 b B-14 A c)}{240 c^3}-\frac{x^2 \left (b x^2+c x^4\right )^{5/2} (9 b B-14 A c)}{168 c^2}+\frac{B x^4 \left (b x^2+c x^4\right )^{5/2}}{14 c} \]

[Out]

(b^4*(9*b*B - 14*A*c)*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(2048*c^5) - (b^2*(9*b*B - 14*A*c)*(b + 2*c*x^2)*(b*x
^2 + c*x^4)^(3/2))/(768*c^4) + (b*(9*b*B - 14*A*c)*(b*x^2 + c*x^4)^(5/2))/(240*c^3) - ((9*b*B - 14*A*c)*x^2*(b
*x^2 + c*x^4)^(5/2))/(168*c^2) + (B*x^4*(b*x^2 + c*x^4)^(5/2))/(14*c) - (b^6*(9*b*B - 14*A*c)*ArcTanh[(Sqrt[c]
*x^2)/Sqrt[b*x^2 + c*x^4]])/(2048*c^(11/2))

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Rubi [A]  time = 0.403646, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {2034, 794, 670, 640, 612, 620, 206} \[ \frac{b^4 \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4} (9 b B-14 A c)}{2048 c^5}-\frac{b^2 \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2} (9 b B-14 A c)}{768 c^4}-\frac{b^6 (9 b B-14 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{2048 c^{11/2}}+\frac{b \left (b x^2+c x^4\right )^{5/2} (9 b B-14 A c)}{240 c^3}-\frac{x^2 \left (b x^2+c x^4\right )^{5/2} (9 b B-14 A c)}{168 c^2}+\frac{B x^4 \left (b x^2+c x^4\right )^{5/2}}{14 c} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(A + B*x^2)*(b*x^2 + c*x^4)^(3/2),x]

[Out]

(b^4*(9*b*B - 14*A*c)*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(2048*c^5) - (b^2*(9*b*B - 14*A*c)*(b + 2*c*x^2)*(b*x
^2 + c*x^4)^(3/2))/(768*c^4) + (b*(9*b*B - 14*A*c)*(b*x^2 + c*x^4)^(5/2))/(240*c^3) - ((9*b*B - 14*A*c)*x^2*(b
*x^2 + c*x^4)^(5/2))/(168*c^2) + (B*x^4*(b*x^2 + c*x^4)^(5/2))/(14*c) - (b^6*(9*b*B - 14*A*c)*ArcTanh[(Sqrt[c]
*x^2)/Sqrt[b*x^2 + c*x^4]])/(2048*c^(11/2))

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]

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^5 \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^2 (A+B x) \left (b x+c x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{B x^4 \left (b x^2+c x^4\right )^{5/2}}{14 c}+\frac{\left (2 (-b B+A c)+\frac{5}{2} (-b B+2 A c)\right ) \operatorname{Subst}\left (\int x^2 \left (b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{14 c}\\ &=-\frac{(9 b B-14 A c) x^2 \left (b x^2+c x^4\right )^{5/2}}{168 c^2}+\frac{B x^4 \left (b x^2+c x^4\right )^{5/2}}{14 c}+\frac{(b (9 b B-14 A c)) \operatorname{Subst}\left (\int x \left (b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{48 c^2}\\ &=\frac{b (9 b B-14 A c) \left (b x^2+c x^4\right )^{5/2}}{240 c^3}-\frac{(9 b B-14 A c) x^2 \left (b x^2+c x^4\right )^{5/2}}{168 c^2}+\frac{B x^4 \left (b x^2+c x^4\right )^{5/2}}{14 c}-\frac{\left (b^2 (9 b B-14 A c)\right ) \operatorname{Subst}\left (\int \left (b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{96 c^3}\\ &=-\frac{b^2 (9 b B-14 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{768 c^4}+\frac{b (9 b B-14 A c) \left (b x^2+c x^4\right )^{5/2}}{240 c^3}-\frac{(9 b B-14 A c) x^2 \left (b x^2+c x^4\right )^{5/2}}{168 c^2}+\frac{B x^4 \left (b x^2+c x^4\right )^{5/2}}{14 c}+\frac{\left (b^4 (9 b B-14 A c)\right ) \operatorname{Subst}\left (\int \sqrt{b x+c x^2} \, dx,x,x^2\right )}{512 c^4}\\ &=\frac{b^4 (9 b B-14 A c) \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{2048 c^5}-\frac{b^2 (9 b B-14 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{768 c^4}+\frac{b (9 b B-14 A c) \left (b x^2+c x^4\right )^{5/2}}{240 c^3}-\frac{(9 b B-14 A c) x^2 \left (b x^2+c x^4\right )^{5/2}}{168 c^2}+\frac{B x^4 \left (b x^2+c x^4\right )^{5/2}}{14 c}-\frac{\left (b^6 (9 b B-14 A c)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{4096 c^5}\\ &=\frac{b^4 (9 b B-14 A c) \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{2048 c^5}-\frac{b^2 (9 b B-14 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{768 c^4}+\frac{b (9 b B-14 A c) \left (b x^2+c x^4\right )^{5/2}}{240 c^3}-\frac{(9 b B-14 A c) x^2 \left (b x^2+c x^4\right )^{5/2}}{168 c^2}+\frac{B x^4 \left (b x^2+c x^4\right )^{5/2}}{14 c}-\frac{\left (b^6 (9 b B-14 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 )}{2048 c^5}\\ &=\frac{b^4 (9 b B-14 A c) \left (b+2 c x^2\right ) \sqrt{b x^2+c x^4}}{2048 c^5}-\frac{b^2 (9 b B-14 A c) \left (b+2 c x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{768 c^4}+\frac{b (9 b B-14 A c) \left (b x^2+c x^4\right )^{5/2}}{240 c^3}-\frac{(9 b B-14 A c) x^2 \left (b x^2+c x^4\right )^{5/2}}{168 c^2}+\frac{B x^4 \left (b x^2+c x^4\right )^{5/2}}{14 c}-\frac{b^6 (9 b B-14 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{2048 c^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.324806, size = 215, normalized size = 0.96 \[ \frac{\sqrt{x^2 \left (b+c x^2\right )} \left (\sqrt{c} x \sqrt{\frac{c x^2}{b}+1} \left (96 b^2 c^4 x^6 \left (7 A+4 B x^2\right )-16 b^3 c^3 x^4 \left (49 A+27 B x^2\right )+28 b^4 c^2 x^2 \left (35 A+18 B x^2\right )-210 b^5 c \left (7 A+3 B x^2\right )+256 b c^5 x^8 \left (91 A+75 B x^2\right )+2560 c^6 x^{10} \left (7 A+6 B x^2\right )+945 b^6 B\right )-105 b^{11/2} (9 b B-14 A c) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )\right )}{215040 c^{11/2} x \sqrt{\frac{c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(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]*(945*b^6*B - 210*b^5*c*(7*A + 3*B*x^2) + 96*b^2*c^4*x^6*
(7*A + 4*B*x^2) + 2560*c^6*x^10*(7*A + 6*B*x^2) + 28*b^4*c^2*x^2*(35*A + 18*B*x^2) - 16*b^3*c^3*x^4*(49*A + 27
*B*x^2) + 256*b*c^5*x^8*(91*A + 75*B*x^2)) - 105*b^(11/2)*(9*b*B - 14*A*c)*ArcSinh[(Sqrt[c]*x)/Sqrt[b]]))/(215
040*c^(11/2)*x*Sqrt[1 + (c*x^2)/b])

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Maple [A]  time = 0.032, size = 328, normalized size = 1.5 \begin{align*}{\frac{1}{215040\,{x}^{3}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 15360\,B \left ( c{x}^{2}+b \right ) ^{5/2}{c}^{9/2}{x}^{9}+17920\,A \left ( c{x}^{2}+b \right ) ^{5/2}{c}^{9/2}{x}^{7}-11520\,B \left ( c{x}^{2}+b \right ) ^{5/2}{c}^{7/2}{x}^{7}b-12544\,A \left ( c{x}^{2}+b \right ) ^{5/2}{c}^{7/2}{x}^{5}b+8064\,B \left ( c{x}^{2}+b \right ) ^{5/2}{c}^{5/2}{x}^{5}{b}^{2}+7840\,A \left ( c{x}^{2}+b \right ) ^{5/2}{c}^{5/2}{x}^{3}{b}^{2}-5040\,B \left ( c{x}^{2}+b \right ) ^{5/2}{c}^{3/2}{x}^{3}{b}^{3}-3920\,A \left ( c{x}^{2}+b \right ) ^{5/2}{c}^{3/2}x{b}^{3}+2520\,B \left ( c{x}^{2}+b \right ) ^{5/2}\sqrt{c}x{b}^{4}+980\,A \left ( c{x}^{2}+b \right ) ^{3/2}{c}^{3/2}x{b}^{4}-630\,B \left ( c{x}^{2}+b \right ) ^{3/2}\sqrt{c}x{b}^{5}+1470\,A\sqrt{c{x}^{2}+b}{c}^{3/2}x{b}^{5}-945\,B\sqrt{c{x}^{2}+b}\sqrt{c}x{b}^{6}+1470\,A\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){b}^{6}c-945\,B\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){b}^{7} \right ) \left ( c{x}^{2}+b \right ) ^{-{\frac{3}{2}}}{c}^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x)

[Out]

1/215040*(c*x^4+b*x^2)^(3/2)*(15360*B*(c*x^2+b)^(5/2)*c^(9/2)*x^9+17920*A*(c*x^2+b)^(5/2)*c^(9/2)*x^7-11520*B*
(c*x^2+b)^(5/2)*c^(7/2)*x^7*b-12544*A*(c*x^2+b)^(5/2)*c^(7/2)*x^5*b+8064*B*(c*x^2+b)^(5/2)*c^(5/2)*x^5*b^2+784
0*A*(c*x^2+b)^(5/2)*c^(5/2)*x^3*b^2-5040*B*(c*x^2+b)^(5/2)*c^(3/2)*x^3*b^3-3920*A*(c*x^2+b)^(5/2)*c^(3/2)*x*b^
3+2520*B*(c*x^2+b)^(5/2)*c^(1/2)*x*b^4+980*A*(c*x^2+b)^(3/2)*c^(3/2)*x*b^4-630*B*(c*x^2+b)^(3/2)*c^(1/2)*x*b^5
+1470*A*(c*x^2+b)^(1/2)*c^(3/2)*x*b^5-945*B*(c*x^2+b)^(1/2)*c^(1/2)*x*b^6+1470*A*ln(x*c^(1/2)+(c*x^2+b)^(1/2))
*b^6*c-945*B*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*b^7)/x^3/(c*x^2+b)^(3/2)/c^(11/2)

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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^5*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.33344, size = 984, normalized size = 4.41 \begin{align*} \left [-\frac{105 \,{\left (9 \, B b^{7} - 14 \, A b^{6} c\right )} \sqrt{c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{c}\right ) - 2 \,{\left (15360 \, B c^{7} x^{12} + 1280 \,{\left (15 \, B b c^{6} + 14 \, A c^{7}\right )} x^{10} + 128 \,{\left (3 \, B b^{2} c^{5} + 182 \, A b c^{6}\right )} x^{8} + 945 \, B b^{6} c - 1470 \, A b^{5} c^{2} - 48 \,{\left (9 \, B b^{3} c^{4} - 14 \, A b^{2} c^{5}\right )} x^{6} + 56 \,{\left (9 \, B b^{4} c^{3} - 14 \, A b^{3} c^{4}\right )} x^{4} - 70 \,{\left (9 \, B b^{5} c^{2} - 14 \, A b^{4} c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{430080 \, c^{6}}, \frac{105 \,{\left (9 \, B b^{7} - 14 \, A b^{6} c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-c}}{c x^{2} + b}\right ) +{\left (15360 \, B c^{7} x^{12} + 1280 \,{\left (15 \, B b c^{6} + 14 \, A c^{7}\right )} x^{10} + 128 \,{\left (3 \, B b^{2} c^{5} + 182 \, A b c^{6}\right )} x^{8} + 945 \, B b^{6} c - 1470 \, A b^{5} c^{2} - 48 \,{\left (9 \, B b^{3} c^{4} - 14 \, A b^{2} c^{5}\right )} x^{6} + 56 \,{\left (9 \, B b^{4} c^{3} - 14 \, A b^{3} c^{4}\right )} x^{4} - 70 \,{\left (9 \, B b^{5} c^{2} - 14 \, A b^{4} c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{215040 \, c^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/430080*(105*(9*B*b^7 - 14*A*b^6*c)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(15360*B*
c^7*x^12 + 1280*(15*B*b*c^6 + 14*A*c^7)*x^10 + 128*(3*B*b^2*c^5 + 182*A*b*c^6)*x^8 + 945*B*b^6*c - 1470*A*b^5*
c^2 - 48*(9*B*b^3*c^4 - 14*A*b^2*c^5)*x^6 + 56*(9*B*b^4*c^3 - 14*A*b^3*c^4)*x^4 - 70*(9*B*b^5*c^2 - 14*A*b^4*c
^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^6, 1/215040*(105*(9*B*b^7 - 14*A*b^6*c)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sq
rt(-c)/(c*x^2 + b)) + (15360*B*c^7*x^12 + 1280*(15*B*b*c^6 + 14*A*c^7)*x^10 + 128*(3*B*b^2*c^5 + 182*A*b*c^6)*
x^8 + 945*B*b^6*c - 1470*A*b^5*c^2 - 48*(9*B*b^3*c^4 - 14*A*b^2*c^5)*x^6 + 56*(9*B*b^4*c^3 - 14*A*b^3*c^4)*x^4
- 70*(9*B*b^5*c^2 - 14*A*b^4*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^6]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{5} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac{3}{2}} \left (A + B x^{2}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(B*x**2+A)*(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**5*(x**2*(b + c*x**2))**(3/2)*(A + B*x**2), x)

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Giac [A]  time = 1.16556, size = 378, normalized size = 1.7 \begin{align*} \frac{1}{215040} \,{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (10 \,{\left (12 \, B c x^{2} \mathrm{sgn}\left (x\right ) + \frac{15 \, B b c^{12} \mathrm{sgn}\left (x\right ) + 14 \, A c^{13} \mathrm{sgn}\left (x\right )}{c^{12}}\right )} x^{2} + \frac{3 \, B b^{2} c^{11} \mathrm{sgn}\left (x\right ) + 182 \, A b c^{12} \mathrm{sgn}\left (x\right )}{c^{12}}\right )} x^{2} - \frac{3 \,{\left (9 \, B b^{3} c^{10} \mathrm{sgn}\left (x\right ) - 14 \, A b^{2} c^{11} \mathrm{sgn}\left (x\right )\right )}}{c^{12}}\right )} x^{2} + \frac{7 \,{\left (9 \, B b^{4} c^{9} \mathrm{sgn}\left (x\right ) - 14 \, A b^{3} c^{10} \mathrm{sgn}\left (x\right )\right )}}{c^{12}}\right )} x^{2} - \frac{35 \,{\left (9 \, B b^{5} c^{8} \mathrm{sgn}\left (x\right ) - 14 \, A b^{4} c^{9} \mathrm{sgn}\left (x\right )\right )}}{c^{12}}\right )} x^{2} + \frac{105 \,{\left (9 \, B b^{6} c^{7} \mathrm{sgn}\left (x\right ) - 14 \, A b^{5} c^{8} \mathrm{sgn}\left (x\right )\right )}}{c^{12}}\right )} \sqrt{c x^{2} + b} x + \frac{{\left (9 \, B b^{7} \mathrm{sgn}\left (x\right ) - 14 \, A b^{6} c \mathrm{sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + b} \right |}\right )}{2048 \, c^{\frac{11}{2}}} - \frac{{\left (9 \, B b^{7} \log \left ({\left | b \right |}\right ) - 14 \, A b^{6} c \log \left ({\left | b \right |}\right )\right )} \mathrm{sgn}\left (x\right )}{4096 \, c^{\frac{11}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)*(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

1/215040*(2*(4*(2*(8*(10*(12*B*c*x^2*sgn(x) + (15*B*b*c^12*sgn(x) + 14*A*c^13*sgn(x))/c^12)*x^2 + (3*B*b^2*c^1
1*sgn(x) + 182*A*b*c^12*sgn(x))/c^12)*x^2 - 3*(9*B*b^3*c^10*sgn(x) - 14*A*b^2*c^11*sgn(x))/c^12)*x^2 + 7*(9*B*
b^4*c^9*sgn(x) - 14*A*b^3*c^10*sgn(x))/c^12)*x^2 - 35*(9*B*b^5*c^8*sgn(x) - 14*A*b^4*c^9*sgn(x))/c^12)*x^2 + 1
05*(9*B*b^6*c^7*sgn(x) - 14*A*b^5*c^8*sgn(x))/c^12)*sqrt(c*x^2 + b)*x + 1/2048*(9*B*b^7*sgn(x) - 14*A*b^6*c*sg
n(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^(11/2) - 1/4096*(9*B*b^7*log(abs(b)) - 14*A*b^6*c*log(abs(b)))*
sgn(x)/c^(11/2)