3.1.15 \(\int x^3 (A+B x) (a+b x^2)^{5/2} \, dx\)

Optimal. Leaf size=173 \[ \frac {3 a^5 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{5/2}}+\frac {3 a^4 B x \sqrt {a+b x^2}}{256 b^2}+\frac {a^3 B x \left (a+b x^2\right )^{3/2}}{128 b^2}+\frac {a^2 B x \left (a+b x^2\right )^{5/2}}{160 b^2}-\frac {a \left (a+b x^2\right )^{7/2} (160 A+189 B x)}{5040 b^2}+\frac {A x^2 \left (a+b x^2\right )^{7/2}}{9 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b} \]

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Rubi [A]  time = 0.10, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {833, 780, 195, 217, 206} \begin {gather*} \frac {3 a^4 B x \sqrt {a+b x^2}}{256 b^2}+\frac {a^3 B x \left (a+b x^2\right )^{3/2}}{128 b^2}+\frac {a^2 B x \left (a+b x^2\right )^{5/2}}{160 b^2}+\frac {3 a^5 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{5/2}}-\frac {a \left (a+b x^2\right )^{7/2} (160 A+189 B x)}{5040 b^2}+\frac {A x^2 \left (a+b x^2\right )^{7/2}}{9 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^4*B*x*Sqrt[a + b*x^2])/(256*b^2) + (a^3*B*x*(a + b*x^2)^(3/2))/(128*b^2) + (a^2*B*x*(a + b*x^2)^(5/2))/(1
60*b^2) + (A*x^2*(a + b*x^2)^(7/2))/(9*b) + (B*x^3*(a + b*x^2)^(7/2))/(10*b) - (a*(160*A + 189*B*x)*(a + b*x^2
)^(7/2))/(5040*b^2) + (3*a^5*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(256*b^(5/2))

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])

Rubi steps

\begin {align*} \int x^3 (A+B x) \left (a+b x^2\right )^{5/2} \, dx &=\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}+\frac {\int x^2 (-3 a B+10 A b x) \left (a+b x^2\right )^{5/2} \, dx}{10 b}\\ &=\frac {A x^2 \left (a+b x^2\right )^{7/2}}{9 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}+\frac {\int x (-20 a A b-27 a b B x) \left (a+b x^2\right )^{5/2} \, dx}{90 b^2}\\ &=\frac {A x^2 \left (a+b x^2\right )^{7/2}}{9 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {a (160 A+189 B x) \left (a+b x^2\right )^{7/2}}{5040 b^2}+\frac {\left (3 a^2 B\right ) \int \left (a+b x^2\right )^{5/2} \, dx}{80 b^2}\\ &=\frac {a^2 B x \left (a+b x^2\right )^{5/2}}{160 b^2}+\frac {A x^2 \left (a+b x^2\right )^{7/2}}{9 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {a (160 A+189 B x) \left (a+b x^2\right )^{7/2}}{5040 b^2}+\frac {\left (a^3 B\right ) \int \left (a+b x^2\right )^{3/2} \, dx}{32 b^2}\\ &=\frac {a^3 B x \left (a+b x^2\right )^{3/2}}{128 b^2}+\frac {a^2 B x \left (a+b x^2\right )^{5/2}}{160 b^2}+\frac {A x^2 \left (a+b x^2\right )^{7/2}}{9 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {a (160 A+189 B x) \left (a+b x^2\right )^{7/2}}{5040 b^2}+\frac {\left (3 a^4 B\right ) \int \sqrt {a+b x^2} \, dx}{128 b^2}\\ &=\frac {3 a^4 B x \sqrt {a+b x^2}}{256 b^2}+\frac {a^3 B x \left (a+b x^2\right )^{3/2}}{128 b^2}+\frac {a^2 B x \left (a+b x^2\right )^{5/2}}{160 b^2}+\frac {A x^2 \left (a+b x^2\right )^{7/2}}{9 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {a (160 A+189 B x) \left (a+b x^2\right )^{7/2}}{5040 b^2}+\frac {\left (3 a^5 B\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{256 b^2}\\ &=\frac {3 a^4 B x \sqrt {a+b x^2}}{256 b^2}+\frac {a^3 B x \left (a+b x^2\right )^{3/2}}{128 b^2}+\frac {a^2 B x \left (a+b x^2\right )^{5/2}}{160 b^2}+\frac {A x^2 \left (a+b x^2\right )^{7/2}}{9 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {a (160 A+189 B x) \left (a+b x^2\right )^{7/2}}{5040 b^2}+\frac {\left (3 a^5 B\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{256 b^2}\\ &=\frac {3 a^4 B x \sqrt {a+b x^2}}{256 b^2}+\frac {a^3 B x \left (a+b x^2\right )^{3/2}}{128 b^2}+\frac {a^2 B x \left (a+b x^2\right )^{5/2}}{160 b^2}+\frac {A x^2 \left (a+b x^2\right )^{7/2}}{9 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {a (160 A+189 B x) \left (a+b x^2\right )^{7/2}}{5040 b^2}+\frac {3 a^5 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 145, normalized size = 0.84 \begin {gather*} \frac {\sqrt {a+b x^2} \left (\frac {945 a^{9/2} B \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}+\sqrt {b} \left (-5 a^4 (512 A+189 B x)+10 a^3 b x^2 (128 A+63 B x)+24 a^2 b^2 x^4 (800 A+651 B x)+16 a b^3 x^6 (1520 A+1323 B x)+896 b^4 x^8 (10 A+9 B x)\right )\right )}{80640 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(A + B*x)*(a + b*x^2)^(5/2),x]

[Out]

(Sqrt[a + b*x^2]*(Sqrt[b]*(896*b^4*x^8*(10*A + 9*B*x) + 10*a^3*b*x^2*(128*A + 63*B*x) - 5*a^4*(512*A + 189*B*x
) + 24*a^2*b^2*x^4*(800*A + 651*B*x) + 16*a*b^3*x^6*(1520*A + 1323*B*x)) + (945*a^(9/2)*B*ArcSinh[(Sqrt[b]*x)/
Sqrt[a]])/Sqrt[1 + (b*x^2)/a]))/(80640*b^(5/2))

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IntegrateAlgebraic [A]  time = 0.46, size = 149, normalized size = 0.86 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-2560 a^4 A-945 a^4 B x+1280 a^3 A b x^2+630 a^3 b B x^3+19200 a^2 A b^2 x^4+15624 a^2 b^2 B x^5+24320 a A b^3 x^6+21168 a b^3 B x^7+8960 A b^4 x^8+8064 b^4 B x^9\right )}{80640 b^2}-\frac {3 a^5 B \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{256 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^3*(A + B*x)*(a + b*x^2)^(5/2),x]

[Out]

(Sqrt[a + b*x^2]*(-2560*a^4*A - 945*a^4*B*x + 1280*a^3*A*b*x^2 + 630*a^3*b*B*x^3 + 19200*a^2*A*b^2*x^4 + 15624
*a^2*b^2*B*x^5 + 24320*a*A*b^3*x^6 + 21168*a*b^3*B*x^7 + 8960*A*b^4*x^8 + 8064*b^4*B*x^9))/(80640*b^2) - (3*a^
5*B*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(256*b^(5/2))

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fricas [A]  time = 1.02, size = 302, normalized size = 1.75 \begin {gather*} \left [\frac {945 \, B a^{5} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (8064 \, B b^{5} x^{9} + 8960 \, A b^{5} x^{8} + 21168 \, B a b^{4} x^{7} + 24320 \, A a b^{4} x^{6} + 15624 \, B a^{2} b^{3} x^{5} + 19200 \, A a^{2} b^{3} x^{4} + 630 \, B a^{3} b^{2} x^{3} + 1280 \, A a^{3} b^{2} x^{2} - 945 \, B a^{4} b x - 2560 \, A a^{4} b\right )} \sqrt {b x^{2} + a}}{161280 \, b^{3}}, -\frac {945 \, B a^{5} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8064 \, B b^{5} x^{9} + 8960 \, A b^{5} x^{8} + 21168 \, B a b^{4} x^{7} + 24320 \, A a b^{4} x^{6} + 15624 \, B a^{2} b^{3} x^{5} + 19200 \, A a^{2} b^{3} x^{4} + 630 \, B a^{3} b^{2} x^{3} + 1280 \, A a^{3} b^{2} x^{2} - 945 \, B a^{4} b x - 2560 \, A a^{4} b\right )} \sqrt {b x^{2} + a}}{80640 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/161280*(945*B*a^5*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(8064*B*b^5*x^9 + 8960*A*b^5*
x^8 + 21168*B*a*b^4*x^7 + 24320*A*a*b^4*x^6 + 15624*B*a^2*b^3*x^5 + 19200*A*a^2*b^3*x^4 + 630*B*a^3*b^2*x^3 +
1280*A*a^3*b^2*x^2 - 945*B*a^4*b*x - 2560*A*a^4*b)*sqrt(b*x^2 + a))/b^3, -1/80640*(945*B*a^5*sqrt(-b)*arctan(s
qrt(-b)*x/sqrt(b*x^2 + a)) - (8064*B*b^5*x^9 + 8960*A*b^5*x^8 + 21168*B*a*b^4*x^7 + 24320*A*a*b^4*x^6 + 15624*
B*a^2*b^3*x^5 + 19200*A*a^2*b^3*x^4 + 630*B*a^3*b^2*x^3 + 1280*A*a^3*b^2*x^2 - 945*B*a^4*b*x - 2560*A*a^4*b)*s
qrt(b*x^2 + a))/b^3]

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giac [A]  time = 0.53, size = 140, normalized size = 0.81 \begin {gather*} -\frac {3 \, B a^{5} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{256 \, b^{\frac {5}{2}}} - \frac {1}{80640} \, {\left (\frac {2560 \, A a^{4}}{b^{2}} + {\left (\frac {945 \, B a^{4}}{b^{2}} - 2 \, {\left (\frac {640 \, A a^{3}}{b} + {\left (\frac {315 \, B a^{3}}{b} + 4 \, {\left (2400 \, A a^{2} + {\left (1953 \, B a^{2} + 2 \, {\left (1520 \, A a b + 7 \, {\left (189 \, B a b + 8 \, {\left (9 \, B b^{2} x + 10 \, A b^{2}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {b x^{2} + a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/256*B*a^5*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2) - 1/80640*(2560*A*a^4/b^2 + (945*B*a^4/b^2 - 2*(64
0*A*a^3/b + (315*B*a^3/b + 4*(2400*A*a^2 + (1953*B*a^2 + 2*(1520*A*a*b + 7*(189*B*a*b + 8*(9*B*b^2*x + 10*A*b^
2)*x)*x)*x)*x)*x)*x)*x)*x)*sqrt(b*x^2 + a)

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maple [A]  time = 0.01, size = 153, normalized size = 0.88 \begin {gather*} \frac {3 B \,a^{5} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{256 b^{\frac {5}{2}}}+\frac {3 \sqrt {b \,x^{2}+a}\, B \,a^{4} x}{256 b^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,a^{3} x}{128 b^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} B \,x^{3}}{10 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} A \,x^{2}}{9 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B \,a^{2} x}{160 b^{2}}-\frac {3 \left (b \,x^{2}+a \right )^{\frac {7}{2}} B a x}{80 b^{2}}-\frac {2 \left (b \,x^{2}+a \right )^{\frac {7}{2}} A a}{63 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*B*x^3*(b*x^2+a)^(7/2)/b-3/80*B*a/b^2*x*(b*x^2+a)^(7/2)+1/160*a^2*B*x*(b*x^2+a)^(5/2)/b^2+1/128*a^3*B*x*(b
*x^2+a)^(3/2)/b^2+3/256*a^4*B*x*(b*x^2+a)^(1/2)/b^2+3/256*B*a^5/b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))+1/9*A*x^
2*(b*x^2+a)^(7/2)/b-2/63*A*a/b^2*(b*x^2+a)^(7/2)

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maxima [A]  time = 1.36, size = 145, normalized size = 0.84 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B x^{3}}{10 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A x^{2}}{9 \, b} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a x}{80 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{2} x}{160 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{3} x}{128 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} B a^{4} x}{256 \, b^{2}} + \frac {3 \, B a^{5} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{256 \, b^{\frac {5}{2}}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A a}{63 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(b*x^2 + a)^(7/2)*B*x^3/b + 1/9*(b*x^2 + a)^(7/2)*A*x^2/b - 3/80*(b*x^2 + a)^(7/2)*B*a*x/b^2 + 1/160*(b*x
^2 + a)^(5/2)*B*a^2*x/b^2 + 1/128*(b*x^2 + a)^(3/2)*B*a^3*x/b^2 + 3/256*sqrt(b*x^2 + a)*B*a^4*x/b^2 + 3/256*B*
a^5*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 2/63*(b*x^2 + a)^(7/2)*A*a/b^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 )}^{5/2}\,\left (A+B\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 37.44, size = 469, normalized size = 2.71 \begin {gather*} A a^{2} \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {a x^{2} \sqrt {a + b x^{2}}}{15 b} + \frac {x^{4} \sqrt {a + b x^{2}}}{5} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + 2 A a b \left (\begin {cases} \frac {8 a^{3} \sqrt {a + b x^{2}}}{105 b^{3}} - \frac {4 a^{2} x^{2} \sqrt {a + b x^{2}}}{105 b^{2}} + \frac {a x^{4} \sqrt {a + b x^{2}}}{35 b} + \frac {x^{6} \sqrt {a + b x^{2}}}{7} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{6}}{6} & \text {otherwise} \end {cases}\right ) + A b^{2} \left (\begin {cases} - \frac {16 a^{4} \sqrt {a + b x^{2}}}{315 b^{4}} + \frac {8 a^{3} x^{2} \sqrt {a + b x^{2}}}{315 b^{3}} - \frac {2 a^{2} x^{4} \sqrt {a + b x^{2}}}{105 b^{2}} + \frac {a x^{6} \sqrt {a + b x^{2}}}{63 b} + \frac {x^{8} \sqrt {a + b x^{2}}}{9} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{8}}{8} & \text {otherwise} \end {cases}\right ) - \frac {3 B a^{\frac {9}{2}} x}{256 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{\frac {7}{2}} x^{3}}{256 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {129 B a^{\frac {5}{2}} x^{5}}{640 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {73 B a^{\frac {3}{2}} b x^{7}}{160 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {29 B \sqrt {a} b^{2} x^{9}}{80 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 B a^{5} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{256 b^{\frac {5}{2}}} + \frac {B b^{3} x^{11}}{10 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**2*Piecewise((-2*a**2*sqrt(a + b*x**2)/(15*b**2) + a*x**2*sqrt(a + b*x**2)/(15*b) + x**4*sqrt(a + b*x**2)/
5, Ne(b, 0)), (sqrt(a)*x**4/4, True)) + 2*A*a*b*Piecewise((8*a**3*sqrt(a + b*x**2)/(105*b**3) - 4*a**2*x**2*sq
rt(a + b*x**2)/(105*b**2) + a*x**4*sqrt(a + b*x**2)/(35*b) + x**6*sqrt(a + b*x**2)/7, Ne(b, 0)), (sqrt(a)*x**6
/6, True)) + A*b**2*Piecewise((-16*a**4*sqrt(a + b*x**2)/(315*b**4) + 8*a**3*x**2*sqrt(a + b*x**2)/(315*b**3)
- 2*a**2*x**4*sqrt(a + b*x**2)/(105*b**2) + a*x**6*sqrt(a + b*x**2)/(63*b) + x**8*sqrt(a + b*x**2)/9, Ne(b, 0)
), (sqrt(a)*x**8/8, True)) - 3*B*a**(9/2)*x/(256*b**2*sqrt(1 + b*x**2/a)) - B*a**(7/2)*x**3/(256*b*sqrt(1 + b*
x**2/a)) + 129*B*a**(5/2)*x**5/(640*sqrt(1 + b*x**2/a)) + 73*B*a**(3/2)*b*x**7/(160*sqrt(1 + b*x**2/a)) + 29*B
*sqrt(a)*b**2*x**9/(80*sqrt(1 + b*x**2/a)) + 3*B*a**5*asinh(sqrt(b)*x/sqrt(a))/(256*b**(5/2)) + B*b**3*x**11/(
10*sqrt(a)*sqrt(1 + b*x**2/a))

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