3.16 \(\int \frac{x^5}{a+b \sec (c+d x^2)} \, dx\)

Optimal. Leaf size=382 \[ \frac{b x^2 \text{PolyLog}\left (2,-\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{b^2-a^2}}\right )}{a d^2 \sqrt{b^2-a^2}}-\frac{b x^2 \text{PolyLog}\left (2,-\frac{a e^{i \left (c+d x^2\right )}}{\sqrt{b^2-a^2}+b}\right )}{a d^2 \sqrt{b^2-a^2}}+\frac{i b \text{PolyLog}\left (3,-\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{b^2-a^2}}\right )}{a d^3 \sqrt{b^2-a^2}}-\frac{i b \text{PolyLog}\left (3,-\frac{a e^{i \left (c+d x^2\right )}}{\sqrt{b^2-a^2}+b}\right )}{a d^3 \sqrt{b^2-a^2}}+\frac{i b x^4 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{b^2-a^2}}\right )}{2 a d \sqrt{b^2-a^2}}-\frac{i b x^4 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{\sqrt{b^2-a^2}+b}\right )}{2 a d \sqrt{b^2-a^2}}+\frac{x^6}{6 a} \]

[Out]

x^6/(6*a) + ((I/2)*b*x^4*Log[1 + (a*E^(I*(c + d*x^2)))/(b - Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) - ((I/2
)*b*x^4*Log[1 + (a*E^(I*(c + d*x^2)))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) + (b*x^2*PolyLog[2, -((a
*E^(I*(c + d*x^2)))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) - (b*x^2*PolyLog[2, -((a*E^(I*(c + d*x^
2)))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) + (I*b*PolyLog[3, -((a*E^(I*(c + d*x^2)))/(b - Sqrt[-a
^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^3) - (I*b*PolyLog[3, -((a*E^(I*(c + d*x^2)))/(b + Sqrt[-a^2 + b^2]))])/(a*
Sqrt[-a^2 + b^2]*d^3)

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Rubi [A]  time = 0.877774, antiderivative size = 382, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {4204, 4191, 3321, 2264, 2190, 2531, 2282, 6589} \[ \frac{b x^2 \text{PolyLog}\left (2,-\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{b^2-a^2}}\right )}{a d^2 \sqrt{b^2-a^2}}-\frac{b x^2 \text{PolyLog}\left (2,-\frac{a e^{i \left (c+d x^2\right )}}{\sqrt{b^2-a^2}+b}\right )}{a d^2 \sqrt{b^2-a^2}}+\frac{i b \text{PolyLog}\left (3,-\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{b^2-a^2}}\right )}{a d^3 \sqrt{b^2-a^2}}-\frac{i b \text{PolyLog}\left (3,-\frac{a e^{i \left (c+d x^2\right )}}{\sqrt{b^2-a^2}+b}\right )}{a d^3 \sqrt{b^2-a^2}}+\frac{i b x^4 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{b^2-a^2}}\right )}{2 a d \sqrt{b^2-a^2}}-\frac{i b x^4 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{\sqrt{b^2-a^2}+b}\right )}{2 a d \sqrt{b^2-a^2}}+\frac{x^6}{6 a} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*Sec[c + d*x^2]),x]

[Out]

x^6/(6*a) + ((I/2)*b*x^4*Log[1 + (a*E^(I*(c + d*x^2)))/(b - Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) - ((I/2
)*b*x^4*Log[1 + (a*E^(I*(c + d*x^2)))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) + (b*x^2*PolyLog[2, -((a
*E^(I*(c + d*x^2)))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) - (b*x^2*PolyLog[2, -((a*E^(I*(c + d*x^
2)))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) + (I*b*PolyLog[3, -((a*E^(I*(c + d*x^2)))/(b - Sqrt[-a
^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^3) - (I*b*PolyLog[3, -((a*E^(I*(c + d*x^2)))/(b + Sqrt[-a^2 + b^2]))])/(a*
Sqrt[-a^2 + b^2]*d^3)

Rule 4204

Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sec[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[
(m + 1)/n], 0] && IntegerQ[p]

Rule 4191

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] &
& IGtQ[m, 0]

Rule 3321

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c
+ d*x)^m*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)))/(b + 2*a*E^(I*Pi*(k - 1/2))*E^(I*(e + f*x)) - b*E^(2*I*k*Pi)*E^(
2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

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

Rubi steps

\begin{align*} \int \frac{x^5}{a+b \sec \left (c+d x^2\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{a+b \sec (c+d x)} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{x^2}{a}-\frac{b x^2}{a (b+a \cos (c+d x))}\right ) \, dx,x,x^2\right )\\ &=\frac{x^6}{6 a}-\frac{b \operatorname{Subst}\left (\int \frac{x^2}{b+a \cos (c+d x)} \, dx,x,x^2\right )}{2 a}\\ &=\frac{x^6}{6 a}-\frac{b \operatorname{Subst}\left (\int \frac{e^{i (c+d x)} x^2}{a+2 b e^{i (c+d x)}+a e^{2 i (c+d x)}} \, dx,x,x^2\right )}{a}\\ &=\frac{x^6}{6 a}-\frac{b \operatorname{Subst}\left (\int \frac{e^{i (c+d x)} x^2}{2 b-2 \sqrt{-a^2+b^2}+2 a e^{i (c+d x)}} \, dx,x,x^2\right )}{\sqrt{-a^2+b^2}}+\frac{b \operatorname{Subst}\left (\int \frac{e^{i (c+d x)} x^2}{2 b+2 \sqrt{-a^2+b^2}+2 a e^{i (c+d x)}} \, dx,x,x^2\right )}{\sqrt{-a^2+b^2}}\\ &=\frac{x^6}{6 a}+\frac{i b x^4 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{-a^2+b^2}}\right )}{2 a \sqrt{-a^2+b^2} d}-\frac{i b x^4 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{b+\sqrt{-a^2+b^2}}\right )}{2 a \sqrt{-a^2+b^2} d}-\frac{(i b) \operatorname{Subst}\left (\int x \log \left (1+\frac{2 a e^{i (c+d x)}}{2 b-2 \sqrt{-a^2+b^2}}\right ) \, dx,x,x^2\right )}{a \sqrt{-a^2+b^2} d}+\frac{(i b) \operatorname{Subst}\left (\int x \log \left (1+\frac{2 a e^{i (c+d x)}}{2 b+2 \sqrt{-a^2+b^2}}\right ) \, dx,x,x^2\right )}{a \sqrt{-a^2+b^2} d}\\ &=\frac{x^6}{6 a}+\frac{i b x^4 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{-a^2+b^2}}\right )}{2 a \sqrt{-a^2+b^2} d}-\frac{i b x^4 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{b+\sqrt{-a^2+b^2}}\right )}{2 a \sqrt{-a^2+b^2} d}+\frac{b x^2 \text{Li}_2\left (-\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^2}-\frac{b x^2 \text{Li}_2\left (-\frac{a e^{i \left (c+d x^2\right )}}{b+\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^2}-\frac{b \operatorname{Subst}\left (\int \text{Li}_2\left (-\frac{2 a e^{i (c+d x)}}{2 b-2 \sqrt{-a^2+b^2}}\right ) \, dx,x,x^2\right )}{a \sqrt{-a^2+b^2} d^2}+\frac{b \operatorname{Subst}\left (\int \text{Li}_2\left (-\frac{2 a e^{i (c+d x)}}{2 b+2 \sqrt{-a^2+b^2}}\right ) \, dx,x,x^2\right )}{a \sqrt{-a^2+b^2} d^2}\\ &=\frac{x^6}{6 a}+\frac{i b x^4 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{-a^2+b^2}}\right )}{2 a \sqrt{-a^2+b^2} d}-\frac{i b x^4 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{b+\sqrt{-a^2+b^2}}\right )}{2 a \sqrt{-a^2+b^2} d}+\frac{b x^2 \text{Li}_2\left (-\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^2}-\frac{b x^2 \text{Li}_2\left (-\frac{a e^{i \left (c+d x^2\right )}}{b+\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^2}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{a x}{-b+\sqrt{-a^2+b^2}}\right )}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{a \sqrt{-a^2+b^2} d^3}-\frac{(i b) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{a x}{b+\sqrt{-a^2+b^2}}\right )}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{a \sqrt{-a^2+b^2} d^3}\\ &=\frac{x^6}{6 a}+\frac{i b x^4 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{-a^2+b^2}}\right )}{2 a \sqrt{-a^2+b^2} d}-\frac{i b x^4 \log \left (1+\frac{a e^{i \left (c+d x^2\right )}}{b+\sqrt{-a^2+b^2}}\right )}{2 a \sqrt{-a^2+b^2} d}+\frac{b x^2 \text{Li}_2\left (-\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^2}-\frac{b x^2 \text{Li}_2\left (-\frac{a e^{i \left (c+d x^2\right )}}{b+\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^2}+\frac{i b \text{Li}_3\left (-\frac{a e^{i \left (c+d x^2\right )}}{b-\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^3}-\frac{i b \text{Li}_3\left (-\frac{a e^{i \left (c+d x^2\right )}}{b+\sqrt{-a^2+b^2}}\right )}{a \sqrt{-a^2+b^2} d^3}\\ \end{align*}

Mathematica [A]  time = 1.09081, size = 472, normalized size = 1.24 \[ \frac{6 b e^{i c} d x^2 \text{PolyLog}\left (2,-\frac{a e^{i \left (2 c+d x^2\right )}}{b e^{i c}-\sqrt{e^{2 i c} \left (b^2-a^2\right )}}\right )-6 b e^{i c} d x^2 \text{PolyLog}\left (2,-\frac{a e^{i \left (2 c+d x^2\right )}}{\sqrt{e^{2 i c} \left (b^2-a^2\right )}+b e^{i c}}\right )+6 i b e^{i c} \text{PolyLog}\left (3,-\frac{a e^{i \left (2 c+d x^2\right )}}{b e^{i c}-\sqrt{e^{2 i c} \left (b^2-a^2\right )}}\right )-6 i b e^{i c} \text{PolyLog}\left (3,-\frac{a e^{i \left (2 c+d x^2\right )}}{\sqrt{e^{2 i c} \left (b^2-a^2\right )}+b e^{i c}}\right )+d^3 x^6 \sqrt{e^{2 i c} \left (b^2-a^2\right )}+3 i b e^{i c} d^2 x^4 \log \left (1+\frac{a e^{i \left (2 c+d x^2\right )}}{b e^{i c}-\sqrt{e^{2 i c} \left (b^2-a^2\right )}}\right )-3 i b e^{i c} d^2 x^4 \log \left (1+\frac{a e^{i \left (2 c+d x^2\right )}}{\sqrt{e^{2 i c} \left (b^2-a^2\right )}+b e^{i c}}\right )}{6 a d^3 \sqrt{e^{2 i c} \left (b^2-a^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*Sec[c + d*x^2]),x]

[Out]

(d^3*Sqrt[(-a^2 + b^2)*E^((2*I)*c)]*x^6 + (3*I)*b*d^2*E^(I*c)*x^4*Log[1 + (a*E^(I*(2*c + d*x^2)))/(b*E^(I*c) -
Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - (3*I)*b*d^2*E^(I*c)*x^4*Log[1 + (a*E^(I*(2*c + d*x^2)))/(b*E^(I*c) + Sqrt[
(-a^2 + b^2)*E^((2*I)*c)])] + 6*b*d*E^(I*c)*x^2*PolyLog[2, -((a*E^(I*(2*c + d*x^2)))/(b*E^(I*c) - Sqrt[(-a^2 +
b^2)*E^((2*I)*c)]))] - 6*b*d*E^(I*c)*x^2*PolyLog[2, -((a*E^(I*(2*c + d*x^2)))/(b*E^(I*c) + Sqrt[(-a^2 + b^2)*
E^((2*I)*c)]))] + (6*I)*b*E^(I*c)*PolyLog[3, -((a*E^(I*(2*c + d*x^2)))/(b*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)
*c)]))] - (6*I)*b*E^(I*c)*PolyLog[3, -((a*E^(I*(2*c + d*x^2)))/(b*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))])
/(6*a*d^3*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])

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Maple [F]  time = 0.151, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{5}}{a+b\sec \left ( d{x}^{2}+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a+b*sec(d*x^2+c)),x)

[Out]

int(x^5/(a+b*sec(d*x^2+c)),x)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(a+b*sec(d*x^2+c)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [C]  time = 2.61519, size = 3491, normalized size = 9.14 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(a+b*sec(d*x^2+c)),x, algorithm="fricas")

[Out]

1/24*(4*(a^2 - b^2)*d^3*x^6 - 12*a*b*d*x^2*sqrt(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*b*cos(d*x^2 + c) + 2*I*b*sin(d
*x^2 + c) + 2*(a*cos(d*x^2 + c) + I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a + 1) + 12*a*b*d*x^2*sqrt
(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*b*cos(d*x^2 + c) + 2*I*b*sin(d*x^2 + c) - 2*(a*cos(d*x^2 + c) + I*a*sin(d*x^2
+ c))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a + 1) - 12*a*b*d*x^2*sqrt(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*b*cos(d*x^2 +
c) - 2*I*b*sin(d*x^2 + c) + 2*(a*cos(d*x^2 + c) - I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a + 1) + 1
2*a*b*d*x^2*sqrt(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*b*cos(d*x^2 + c) - 2*I*b*sin(d*x^2 + c) - 2*(a*cos(d*x^2 + c)
- I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a + 1) + 6*I*a*b*c^2*sqrt(-(a^2 - b^2)/a^2)*log(2*a*cos(d
*x^2 + c) + 2*I*a*sin(d*x^2 + c) + 2*a*sqrt(-(a^2 - b^2)/a^2) + 2*b) - 6*I*a*b*c^2*sqrt(-(a^2 - b^2)/a^2)*log(
2*a*cos(d*x^2 + c) - 2*I*a*sin(d*x^2 + c) + 2*a*sqrt(-(a^2 - b^2)/a^2) + 2*b) + 6*I*a*b*c^2*sqrt(-(a^2 - b^2)/
a^2)*log(-2*a*cos(d*x^2 + c) + 2*I*a*sin(d*x^2 + c) + 2*a*sqrt(-(a^2 - b^2)/a^2) - 2*b) - 6*I*a*b*c^2*sqrt(-(a
^2 - b^2)/a^2)*log(-2*a*cos(d*x^2 + c) - 2*I*a*sin(d*x^2 + c) + 2*a*sqrt(-(a^2 - b^2)/a^2) - 2*b) - 12*I*a*b*s
qrt(-(a^2 - b^2)/a^2)*polylog(3, -(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c) + (a*cos(d*x^2 + c) + I*a*sin(d*x^2 +
c))*sqrt(-(a^2 - b^2)/a^2))/a) + 12*I*a*b*sqrt(-(a^2 - b^2)/a^2)*polylog(3, -(b*cos(d*x^2 + c) + I*b*sin(d*x^
2 + c) - (a*cos(d*x^2 + c) + I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2))/a) + 12*I*a*b*sqrt(-(a^2 - b^2)/a^2)*
polylog(3, -(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c) + (a*cos(d*x^2 + c) - I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)
/a^2))/a) - 12*I*a*b*sqrt(-(a^2 - b^2)/a^2)*polylog(3, -(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c) - (a*cos(d*x^2
+ c) - I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2))/a) - 2*(3*I*a*b*d^2*x^4 - 3*I*a*b*c^2)*sqrt(-(a^2 - b^2)/a^
2)*log(1/2*(2*b*cos(d*x^2 + c) + 2*I*b*sin(d*x^2 + c) + 2*(a*cos(d*x^2 + c) + I*a*sin(d*x^2 + c))*sqrt(-(a^2 -
b^2)/a^2) + 2*a)/a) - 2*(-3*I*a*b*d^2*x^4 + 3*I*a*b*c^2)*sqrt(-(a^2 - b^2)/a^2)*log(1/2*(2*b*cos(d*x^2 + c) +
2*I*b*sin(d*x^2 + c) - 2*(a*cos(d*x^2 + c) + I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a) - 2*(-3*I*a
*b*d^2*x^4 + 3*I*a*b*c^2)*sqrt(-(a^2 - b^2)/a^2)*log(1/2*(2*b*cos(d*x^2 + c) - 2*I*b*sin(d*x^2 + c) + 2*(a*cos
(d*x^2 + c) - I*a*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a) - 2*(3*I*a*b*d^2*x^4 - 3*I*a*b*c^2)*sqrt(-(
a^2 - b^2)/a^2)*log(1/2*(2*b*cos(d*x^2 + c) - 2*I*b*sin(d*x^2 + c) - 2*(a*cos(d*x^2 + c) - I*a*sin(d*x^2 + c))
*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a))/((a^3 - a*b^2)*d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(a+b*sec(d*x**2+c)),x)

[Out]

Integral(x**5/(a + b*sec(c + d*x**2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{b \sec \left (d x^{2} + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(a+b*sec(d*x^2+c)),x, algorithm="giac")

[Out]

integrate(x^5/(b*sec(d*x^2 + c) + a), x)