3.8 \(\int x^5 (a+b \sec (c+d x^2))^2 \, dx\)

Optimal. Leaf size=242 \[ \frac{2 i a b x^2 \text{PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{2 i a b x^2 \text{PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{2 a b \text{PolyLog}\left (3,-i e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac{2 a b \text{PolyLog}\left (3,i e^{i \left (c+d x^2\right )}\right )}{d^3}-\frac{i b^2 \text{PolyLog}\left (2,-e^{2 i \left (c+d x^2\right )}\right )}{2 d^3}+\frac{a^2 x^6}{6}-\frac{2 i a b x^4 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac{b^2 x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )}{d^2}+\frac{b^2 x^4 \tan \left (c+d x^2\right )}{2 d}-\frac{i b^2 x^4}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.370284, antiderivative size = 242, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.611, Rules used = {4204, 4190, 4181, 2531, 2282, 6589, 4184, 3719, 2190, 2279, 2391} \[ \frac{2 i a b x^2 \text{PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{2 i a b x^2 \text{PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{2 a b \text{PolyLog}\left (3,-i e^{i \left (c+d x^2\right )}\right )}{d^3}+\frac{2 a b \text{PolyLog}\left (3,i e^{i \left (c+d x^2\right )}\right )}{d^3}-\frac{i b^2 \text{PolyLog}\left (2,-e^{2 i \left (c+d x^2\right )}\right )}{2 d^3}+\frac{a^2 x^6}{6}-\frac{2 i a b x^4 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac{b^2 x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )}{d^2}+\frac{b^2 x^4 \tan \left (c+d x^2\right )}{2 d}-\frac{i b^2 x^4}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*Sec[c + d*x^2])^2,x]

[Out]

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

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 4190

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

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && 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]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

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

Mathematica [A]  time = 0.589636, size = 229, normalized size = 0.95 \[ \frac{12 i a b d x^2 \text{PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )-12 i a b d x^2 \text{PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )-12 a b \text{PolyLog}\left (3,-i e^{i \left (c+d x^2\right )}\right )+12 a b \text{PolyLog}\left (3,i e^{i \left (c+d x^2\right )}\right )-3 i b^2 \text{PolyLog}\left (2,-e^{2 i \left (c+d x^2\right )}\right )+a^2 d^3 x^6-12 i a b d^2 x^4 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )+3 b^2 d^2 x^4 \tan \left (c+d x^2\right )+6 b^2 d x^2 \log \left (1+e^{2 i \left (c+d x^2\right )}\right )-3 i b^2 d^2 x^4}{6 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*I)*b^2*d^2*x^4 + a^2*d^3*x^6 - (12*I)*a*b*d^2*x^4*ArcTan[E^(I*(c + d*x^2))] + 6*b^2*d*x^2*Log[1 + E^((2*I
)*(c + d*x^2))] + (12*I)*a*b*d*x^2*PolyLog[2, (-I)*E^(I*(c + d*x^2))] - (12*I)*a*b*d*x^2*PolyLog[2, I*E^(I*(c
+ d*x^2))] - (3*I)*b^2*PolyLog[2, -E^((2*I)*(c + d*x^2))] - 12*a*b*PolyLog[3, (-I)*E^(I*(c + d*x^2))] + 12*a*b
*PolyLog[3, I*E^(I*(c + d*x^2))] + 3*b^2*d^2*x^4*Tan[c + d*x^2])/(6*d^3)

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Maple [F]  time = 0.206, size = 0, normalized size = 0. \begin{align*} \int{x}^{5} \left ( a+b\sec \left ( d{x}^{2}+c \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{6} \, a^{2} x^{6} + \frac{b^{2} x^{4} \sin \left (2 \, d x^{2} + 2 \, c\right ) + 4 \,{\left (d \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + d \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + 2 \, d \cos \left (2 \, d x^{2} + 2 \, c\right ) + d\right )} \int \frac{a b d x^{5} \cos \left (2 \, d x^{2} + 2 \, c\right ) \cos \left (d x^{2} + c\right ) + a b d x^{5} \cos \left (d x^{2} + c\right ) +{\left (a b d x^{5} \sin \left (d x^{2} + c\right ) - b^{2} x^{3}\right )} \sin \left (2 \, d x^{2} + 2 \, c\right )}{d \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + d \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + 2 \, d \cos \left (2 \, d x^{2} + 2 \, c\right ) + d}\,{d x}}{d \cos \left (2 \, d x^{2} + 2 \, c\right )^{2} + d \sin \left (2 \, d x^{2} + 2 \, c\right )^{2} + 2 \, d \cos \left (2 \, d x^{2} + 2 \, c\right ) + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a^2*x^6 + (b^2*x^4*sin(2*d*x^2 + 2*c) + (d*cos(2*d*x^2 + 2*c)^2 + d*sin(2*d*x^2 + 2*c)^2 + 2*d*cos(2*d*x^2
+ 2*c) + d)*integrate(4*(a*b*d*x^5*cos(2*d*x^2 + 2*c)*cos(d*x^2 + c) + a*b*d*x^5*cos(d*x^2 + c) + (a*b*d*x^5*
sin(d*x^2 + c) - b^2*x^3)*sin(2*d*x^2 + 2*c))/(d*cos(2*d*x^2 + 2*c)^2 + d*sin(2*d*x^2 + 2*c)^2 + 2*d*cos(2*d*x
^2 + 2*c) + d), x))/(d*cos(2*d*x^2 + 2*c)^2 + d*sin(2*d*x^2 + 2*c)^2 + 2*d*cos(2*d*x^2 + 2*c) + d)

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Fricas [C]  time = 2.29686, size = 1960, normalized size = 8.1 \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))^2,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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