3.539 \(\int x^4 (a+b x^2)^{5/2} (A+B x^2) \, dx\)

Optimal. Leaf size=221 \[ -\frac{a^4 x \sqrt{a+b x^2} (12 A b-5 a B)}{1024 b^3}+\frac{a^3 x^3 \sqrt{a+b x^2} (12 A b-5 a B)}{1536 b^2}+\frac{a^5 (12 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{1024 b^{7/2}}+\frac{a^2 x^5 \sqrt{a+b x^2} (12 A b-5 a B)}{384 b}+\frac{a x^5 \left (a+b x^2\right )^{3/2} (12 A b-5 a B)}{192 b}+\frac{x^5 \left (a+b x^2\right )^{5/2} (12 A b-5 a B)}{120 b}+\frac{B x^5 \left (a+b x^2\right )^{7/2}}{12 b} \]

[Out]

-(a^4*(12*A*b - 5*a*B)*x*Sqrt[a + b*x^2])/(1024*b^3) + (a^3*(12*A*b - 5*a*B)*x^3*Sqrt[a + b*x^2])/(1536*b^2) +
(a^2*(12*A*b - 5*a*B)*x^5*Sqrt[a + b*x^2])/(384*b) + (a*(12*A*b - 5*a*B)*x^5*(a + b*x^2)^(3/2))/(192*b) + ((1
2*A*b - 5*a*B)*x^5*(a + b*x^2)^(5/2))/(120*b) + (B*x^5*(a + b*x^2)^(7/2))/(12*b) + (a^5*(12*A*b - 5*a*B)*ArcTa
nh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(1024*b^(7/2))

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Rubi [A]  time = 0.10279, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {459, 279, 321, 217, 206} \[ -\frac{a^4 x \sqrt{a+b x^2} (12 A b-5 a B)}{1024 b^3}+\frac{a^3 x^3 \sqrt{a+b x^2} (12 A b-5 a B)}{1536 b^2}+\frac{a^5 (12 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{1024 b^{7/2}}+\frac{a^2 x^5 \sqrt{a+b x^2} (12 A b-5 a B)}{384 b}+\frac{a x^5 \left (a+b x^2\right )^{3/2} (12 A b-5 a B)}{192 b}+\frac{x^5 \left (a+b x^2\right )^{5/2} (12 A b-5 a B)}{120 b}+\frac{B x^5 \left (a+b x^2\right )^{7/2}}{12 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^4*(12*A*b - 5*a*B)*x*Sqrt[a + b*x^2])/(1024*b^3) + (a^3*(12*A*b - 5*a*B)*x^3*Sqrt[a + b*x^2])/(1536*b^2) +
(a^2*(12*A*b - 5*a*B)*x^5*Sqrt[a + b*x^2])/(384*b) + (a*(12*A*b - 5*a*B)*x^5*(a + b*x^2)^(3/2))/(192*b) + ((1
2*A*b - 5*a*B)*x^5*(a + b*x^2)^(5/2))/(120*b) + (B*x^5*(a + b*x^2)^(7/2))/(12*b) + (a^5*(12*A*b - 5*a*B)*ArcTa
nh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(1024*b^(7/2))

Rule 459

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

Rule 279

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

Rule 321

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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^4 \left (a+b x^2\right )^{5/2} \left (A+B x^2\right ) \, dx &=\frac{B x^5 \left (a+b x^2\right )^{7/2}}{12 b}-\frac{(-12 A b+5 a B) \int x^4 \left (a+b x^2\right )^{5/2} \, dx}{12 b}\\ &=\frac{(12 A b-5 a B) x^5 \left (a+b x^2\right )^{5/2}}{120 b}+\frac{B x^5 \left (a+b x^2\right )^{7/2}}{12 b}+\frac{(a (12 A b-5 a B)) \int x^4 \left (a+b x^2\right )^{3/2} \, dx}{24 b}\\ &=\frac{a (12 A b-5 a B) x^5 \left (a+b x^2\right )^{3/2}}{192 b}+\frac{(12 A b-5 a B) x^5 \left (a+b x^2\right )^{5/2}}{120 b}+\frac{B x^5 \left (a+b x^2\right )^{7/2}}{12 b}+\frac{\left (a^2 (12 A b-5 a B)\right ) \int x^4 \sqrt{a+b x^2} \, dx}{64 b}\\ &=\frac{a^2 (12 A b-5 a B) x^5 \sqrt{a+b x^2}}{384 b}+\frac{a (12 A b-5 a B) x^5 \left (a+b x^2\right )^{3/2}}{192 b}+\frac{(12 A b-5 a B) x^5 \left (a+b x^2\right )^{5/2}}{120 b}+\frac{B x^5 \left (a+b x^2\right )^{7/2}}{12 b}+\frac{\left (a^3 (12 A b-5 a B)\right ) \int \frac{x^4}{\sqrt{a+b x^2}} \, dx}{384 b}\\ &=\frac{a^3 (12 A b-5 a B) x^3 \sqrt{a+b x^2}}{1536 b^2}+\frac{a^2 (12 A b-5 a B) x^5 \sqrt{a+b x^2}}{384 b}+\frac{a (12 A b-5 a B) x^5 \left (a+b x^2\right )^{3/2}}{192 b}+\frac{(12 A b-5 a B) x^5 \left (a+b x^2\right )^{5/2}}{120 b}+\frac{B x^5 \left (a+b x^2\right )^{7/2}}{12 b}-\frac{\left (a^4 (12 A b-5 a B)\right ) \int \frac{x^2}{\sqrt{a+b x^2}} \, dx}{512 b^2}\\ &=-\frac{a^4 (12 A b-5 a B) x \sqrt{a+b x^2}}{1024 b^3}+\frac{a^3 (12 A b-5 a B) x^3 \sqrt{a+b x^2}}{1536 b^2}+\frac{a^2 (12 A b-5 a B) x^5 \sqrt{a+b x^2}}{384 b}+\frac{a (12 A b-5 a B) x^5 \left (a+b x^2\right )^{3/2}}{192 b}+\frac{(12 A b-5 a B) x^5 \left (a+b x^2\right )^{5/2}}{120 b}+\frac{B x^5 \left (a+b x^2\right )^{7/2}}{12 b}+\frac{\left (a^5 (12 A b-5 a B)\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{1024 b^3}\\ &=-\frac{a^4 (12 A b-5 a B) x \sqrt{a+b x^2}}{1024 b^3}+\frac{a^3 (12 A b-5 a B) x^3 \sqrt{a+b x^2}}{1536 b^2}+\frac{a^2 (12 A b-5 a B) x^5 \sqrt{a+b x^2}}{384 b}+\frac{a (12 A b-5 a B) x^5 \left (a+b x^2\right )^{3/2}}{192 b}+\frac{(12 A b-5 a B) x^5 \left (a+b x^2\right )^{5/2}}{120 b}+\frac{B x^5 \left (a+b x^2\right )^{7/2}}{12 b}+\frac{\left (a^5 (12 A b-5 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{1024 b^3}\\ &=-\frac{a^4 (12 A b-5 a B) x \sqrt{a+b x^2}}{1024 b^3}+\frac{a^3 (12 A b-5 a B) x^3 \sqrt{a+b x^2}}{1536 b^2}+\frac{a^2 (12 A b-5 a B) x^5 \sqrt{a+b x^2}}{384 b}+\frac{a (12 A b-5 a B) x^5 \left (a+b x^2\right )^{3/2}}{192 b}+\frac{(12 A b-5 a B) x^5 \left (a+b x^2\right )^{5/2}}{120 b}+\frac{B x^5 \left (a+b x^2\right )^{7/2}}{12 b}+\frac{a^5 (12 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{1024 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.346228, size = 172, normalized size = 0.78 \[ \frac{\sqrt{a+b x^2} \left (\sqrt{b} x \left (48 a^2 b^3 x^4 \left (62 A+45 B x^2\right )+40 a^3 b^2 x^2 \left (3 A+B x^2\right )-10 a^4 b \left (18 A+5 B x^2\right )+75 a^5 B+64 a b^4 x^6 \left (63 A+50 B x^2\right )+256 b^5 x^8 \left (6 A+5 B x^2\right )\right )-\frac{15 a^{9/2} (5 a B-12 A b) \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{\frac{b x^2}{a}+1}}\right )}{15360 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*x^2)^(5/2)*(A + B*x^2),x]

[Out]

(Sqrt[a + b*x^2]*(Sqrt[b]*x*(75*a^5*B + 40*a^3*b^2*x^2*(3*A + B*x^2) + 256*b^5*x^8*(6*A + 5*B*x^2) - 10*a^4*b*
(18*A + 5*B*x^2) + 48*a^2*b^3*x^4*(62*A + 45*B*x^2) + 64*a*b^4*x^6*(63*A + 50*B*x^2)) - (15*a^(9/2)*(-12*A*b +
5*a*B)*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[1 + (b*x^2)/a]))/(15360*b^(7/2))

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Maple [A]  time = 0.012, size = 257, normalized size = 1.2 \begin{align*}{\frac{B{x}^{5}}{12\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{Ba{x}^{3}}{24\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{{a}^{2}Bx}{64\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{B{a}^{3}x}{384\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{5\,B{a}^{4}x}{1536\,{b}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,B{a}^{5}x}{1024\,{b}^{3}}\sqrt{b{x}^{2}+a}}-{\frac{5\,B{a}^{6}}{1024}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{7}{2}}}}+{\frac{A{x}^{3}}{10\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{3\,aAx}{80\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{{a}^{2}Ax}{160\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{A{a}^{3}x}{128\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,A{a}^{4}x}{256\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{3\,A{a}^{5}}{256}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*B*x^5*(b*x^2+a)^(7/2)/b-1/24*B/b^2*a*x^3*(b*x^2+a)^(7/2)+1/64*B/b^3*a^2*x*(b*x^2+a)^(7/2)-1/384*B/b^3*a^3
*x*(b*x^2+a)^(5/2)-5/1536*B/b^3*a^4*x*(b*x^2+a)^(3/2)-5/1024*B/b^3*a^5*x*(b*x^2+a)^(1/2)-5/1024*B/b^(7/2)*a^6*
ln(x*b^(1/2)+(b*x^2+a)^(1/2))+1/10*A*x^3*(b*x^2+a)^(7/2)/b-3/80*A/b^2*a*x*(b*x^2+a)^(7/2)+1/160*A/b^2*a^2*x*(b
*x^2+a)^(5/2)+1/128*A/b^2*a^3*x*(b*x^2+a)^(3/2)+3/256*A/b^2*a^4*x*(b*x^2+a)^(1/2)+3/256*A/b^(5/2)*a^5*ln(x*b^(
1/2)+(b*x^2+a)^(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^4*(b*x^2+a)^(5/2)*(B*x^2+A),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.452, size = 844, normalized size = 3.82 \begin{align*} \left [-\frac{15 \,{\left (5 \, B a^{6} - 12 \, A a^{5} b\right )} \sqrt{b} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) - 2 \,{\left (1280 \, B b^{6} x^{11} + 128 \,{\left (25 \, B a b^{5} + 12 \, A b^{6}\right )} x^{9} + 144 \,{\left (15 \, B a^{2} b^{4} + 28 \, A a b^{5}\right )} x^{7} + 8 \,{\left (5 \, B a^{3} b^{3} + 372 \, A a^{2} b^{4}\right )} x^{5} - 10 \,{\left (5 \, B a^{4} b^{2} - 12 \, A a^{3} b^{3}\right )} x^{3} + 15 \,{\left (5 \, B a^{5} b - 12 \, A a^{4} b^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{30720 \, b^{4}}, \frac{15 \,{\left (5 \, B a^{6} - 12 \, A a^{5} b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) +{\left (1280 \, B b^{6} x^{11} + 128 \,{\left (25 \, B a b^{5} + 12 \, A b^{6}\right )} x^{9} + 144 \,{\left (15 \, B a^{2} b^{4} + 28 \, A a b^{5}\right )} x^{7} + 8 \,{\left (5 \, B a^{3} b^{3} + 372 \, A a^{2} b^{4}\right )} x^{5} - 10 \,{\left (5 \, B a^{4} b^{2} - 12 \, A a^{3} b^{3}\right )} x^{3} + 15 \,{\left (5 \, B a^{5} b - 12 \, A a^{4} b^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{15360 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/30720*(15*(5*B*a^6 - 12*A*a^5*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(1280*B*b^6*x
^11 + 128*(25*B*a*b^5 + 12*A*b^6)*x^9 + 144*(15*B*a^2*b^4 + 28*A*a*b^5)*x^7 + 8*(5*B*a^3*b^3 + 372*A*a^2*b^4)*
x^5 - 10*(5*B*a^4*b^2 - 12*A*a^3*b^3)*x^3 + 15*(5*B*a^5*b - 12*A*a^4*b^2)*x)*sqrt(b*x^2 + a))/b^4, 1/15360*(15
*(5*B*a^6 - 12*A*a^5*b)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (1280*B*b^6*x^11 + 128*(25*B*a*b^5 + 12*
A*b^6)*x^9 + 144*(15*B*a^2*b^4 + 28*A*a*b^5)*x^7 + 8*(5*B*a^3*b^3 + 372*A*a^2*b^4)*x^5 - 10*(5*B*a^4*b^2 - 12*
A*a^3*b^3)*x^3 + 15*(5*B*a^5*b - 12*A*a^4*b^2)*x)*sqrt(b*x^2 + a))/b^4]

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Sympy [A]  time = 61.4179, size = 405, normalized size = 1.83 \begin{align*} - \frac{3 A a^{\frac{9}{2}} x}{256 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{A a^{\frac{7}{2}} x^{3}}{256 b \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{129 A a^{\frac{5}{2}} x^{5}}{640 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{73 A a^{\frac{3}{2}} b x^{7}}{160 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{29 A \sqrt{a} b^{2} x^{9}}{80 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 A a^{5} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{256 b^{\frac{5}{2}}} + \frac{A b^{3} x^{11}}{10 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{5 B a^{\frac{11}{2}} x}{1024 b^{3} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{5 B a^{\frac{9}{2}} x^{3}}{3072 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{B a^{\frac{7}{2}} x^{5}}{1536 b \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{55 B a^{\frac{5}{2}} x^{7}}{384 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{67 B a^{\frac{3}{2}} b x^{9}}{192 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{7 B \sqrt{a} b^{2} x^{11}}{24 \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{5 B a^{6} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{1024 b^{\frac{7}{2}}} + \frac{B b^{3} x^{13}}{12 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*A*a**(9/2)*x/(256*b**2*sqrt(1 + b*x**2/a)) - A*a**(7/2)*x**3/(256*b*sqrt(1 + b*x**2/a)) + 129*A*a**(5/2)*x*
*5/(640*sqrt(1 + b*x**2/a)) + 73*A*a**(3/2)*b*x**7/(160*sqrt(1 + b*x**2/a)) + 29*A*sqrt(a)*b**2*x**9/(80*sqrt(
1 + b*x**2/a)) + 3*A*a**5*asinh(sqrt(b)*x/sqrt(a))/(256*b**(5/2)) + A*b**3*x**11/(10*sqrt(a)*sqrt(1 + b*x**2/a
)) + 5*B*a**(11/2)*x/(1024*b**3*sqrt(1 + b*x**2/a)) + 5*B*a**(9/2)*x**3/(3072*b**2*sqrt(1 + b*x**2/a)) - B*a**
(7/2)*x**5/(1536*b*sqrt(1 + b*x**2/a)) + 55*B*a**(5/2)*x**7/(384*sqrt(1 + b*x**2/a)) + 67*B*a**(3/2)*b*x**9/(1
92*sqrt(1 + b*x**2/a)) + 7*B*sqrt(a)*b**2*x**11/(24*sqrt(1 + b*x**2/a)) - 5*B*a**6*asinh(sqrt(b)*x/sqrt(a))/(1
024*b**(7/2)) + B*b**3*x**13/(12*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]  time = 1.62696, size = 263, normalized size = 1.19 \begin{align*} \frac{1}{15360} \,{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (10 \, B b^{2} x^{2} + \frac{25 \, B a b^{11} + 12 \, A b^{12}}{b^{10}}\right )} x^{2} + \frac{9 \,{\left (15 \, B a^{2} b^{10} + 28 \, A a b^{11}\right )}}{b^{10}}\right )} x^{2} + \frac{5 \, B a^{3} b^{9} + 372 \, A a^{2} b^{10}}{b^{10}}\right )} x^{2} - \frac{5 \,{\left (5 \, B a^{4} b^{8} - 12 \, A a^{3} b^{9}\right )}}{b^{10}}\right )} x^{2} + \frac{15 \,{\left (5 \, B a^{5} b^{7} - 12 \, A a^{4} b^{8}\right )}}{b^{10}}\right )} \sqrt{b x^{2} + a} x + \frac{{\left (5 \, B a^{6} - 12 \, A a^{5} b\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{1024 \, b^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15360*(2*(4*(2*(8*(10*B*b^2*x^2 + (25*B*a*b^11 + 12*A*b^12)/b^10)*x^2 + 9*(15*B*a^2*b^10 + 28*A*a*b^11)/b^10
)*x^2 + (5*B*a^3*b^9 + 372*A*a^2*b^10)/b^10)*x^2 - 5*(5*B*a^4*b^8 - 12*A*a^3*b^9)/b^10)*x^2 + 15*(5*B*a^5*b^7
- 12*A*a^4*b^8)/b^10)*sqrt(b*x^2 + a)*x + 1/1024*(5*B*a^6 - 12*A*a^5*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))
/b^(7/2)