3.1.8 \(\int x^5 (a+b \sec (c+d x^2))^2 \, dx\) [8]

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

[Out]

-1/2*I*b^2*x^4/d+1/6*a^2*x^6-2*I*a*b*x^4*arctan(exp(I*(d*x^2+c)))/d+b^2*x^2*ln(1+exp(2*I*(d*x^2+c)))/d^2+2*I*a
*b*x^2*polylog(2,-I*exp(I*(d*x^2+c)))/d^2-2*I*a*b*x^2*polylog(2,I*exp(I*(d*x^2+c)))/d^2-1/2*I*b^2*polylog(2,-e
xp(2*I*(d*x^2+c)))/d^3-2*a*b*polylog(3,-I*exp(I*(d*x^2+c)))/d^3+2*a*b*polylog(3,I*exp(I*(d*x^2+c)))/d^3+1/2*b^
2*x^4*tan(d*x^2+c)/d

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Rubi [A]
time = 0.28, antiderivative size = 242, normalized size of antiderivative = 1.00, 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 = {4289, 4275, 4266, 2611, 2320, 6724, 4269, 3800, 2221, 2317, 2438} \begin {gather*} \frac {a^2 x^6}{6}-\frac {2 i a b x^4 \text {ArcTan}\left (e^{i \left (c+d x^2\right )}\right )}{d}-\frac {2 a b \text {Li}_3\left (-i e^{i \left (d x^2+c\right )}\right )}{d^3}+\frac {2 a b \text {Li}_3\left (i e^{i \left (d x^2+c\right )}\right )}{d^3}+\frac {2 i a b x^2 \text {Li}_2\left (-i e^{i \left (d x^2+c\right )}\right )}{d^2}-\frac {2 i a b x^2 \text {Li}_2\left (i e^{i \left (d x^2+c\right )}\right )}{d^2}-\frac {i b^2 \text {Li}_2\left (-e^{2 i \left (d x^2+c\right )}\right )}{2 d^3}+\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*Sec[c + d*x^2])^2,x]

[Out]

((-1/2*I)*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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2320

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 2438

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]

Rule 2611

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 3800

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 4266

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 4269

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 4275

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 4289

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 6724

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 x^5 \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx &=\frac {1}{2} \text {Subst}\left (\int x^2 (a+b \sec (c+d x))^2 \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {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) \text {Subst}\left (\int x^2 \sec (c+d x) \, dx,x,x^2\right )+\frac {1}{2} b^2 \text {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) \text {Subst}\left (\int x \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d}+\frac {(2 a b) \text {Subst}\left (\int x \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d}-\frac {b^2 \text {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) \text {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) \text {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 ) \text {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) \text {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) \text {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 \text {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 ) \text {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*}

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Mathematica [A]
time = 0.63, size = 229, normalized size = 0.95 \begin {gather*} \frac {-3 i b^2 d^2 x^4+a^2 d^3 x^6-12 i a b d^2 x^4 \text {ArcTan}\left (e^{i \left (c+d x^2\right )}\right )+6 b^2 d x^2 \log \left (1+e^{2 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 i a b d x^2 \text {PolyLog}\left (2,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 )-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 b^2 d^2 x^4 \tan \left (c+d x^2\right )}{6 d^3} \end {gather*}

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 = 0.30, size = 0, normalized size = 0.00 \[\int x^{5} \left (a +b \sec \left (d \,x^{2}+c \right )\right )^{2}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 799 vs. \(2 (199) = 398\).
time = 2.93, size = 799, normalized size = 3.30 \begin {gather*} \frac {a^{2} d^{3} x^{6} \cos \left (d x^{2} + c\right ) + 3 \, b^{2} d^{2} x^{4} \sin \left (d x^{2} + c\right ) - 6 \, a b \cos \left (d x^{2} + c\right ) {\rm polylog}\left (3, i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + 6 \, a b \cos \left (d x^{2} + c\right ) {\rm polylog}\left (3, i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) - 6 \, a b \cos \left (d x^{2} + c\right ) {\rm polylog}\left (3, -i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + 6 \, a b \cos \left (d x^{2} + c\right ) {\rm polylog}\left (3, -i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) - 3 \, {\left (2 i \, a b d x^{2} - i \, b^{2}\right )} \cos \left (d x^{2} + c\right ) {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) - 3 \, {\left (2 i \, a b d x^{2} + i \, b^{2}\right )} \cos \left (d x^{2} + c\right ) {\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) - 3 \, {\left (-2 i \, a b d x^{2} + i \, b^{2}\right )} \cos \left (d x^{2} + c\right ) {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) - 3 \, {\left (-2 i \, a b d x^{2} - i \, b^{2}\right )} \cos \left (d x^{2} + c\right ) {\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + 3 \, {\left (a b c^{2} - b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) - 3 \, {\left (a b c^{2} + b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right ) + 3 \, {\left (a b d^{2} x^{4} + b^{2} d x^{2} - a b c^{2} + b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - 3 \, {\left (a b d^{2} x^{4} - b^{2} d x^{2} - a b c^{2} - b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right ) + 3 \, {\left (a b d^{2} x^{4} + b^{2} d x^{2} - a b c^{2} + b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - 3 \, {\left (a b d^{2} x^{4} - b^{2} d x^{2} - a b c^{2} - b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right ) + 3 \, {\left (a b c^{2} - b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (-\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) - 3 \, {\left (a b c^{2} + b^{2} c\right )} \cos \left (d x^{2} + c\right ) \log \left (-\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right )}{6 \, d^{3} \cos \left (d x^{2} + c\right )} \end {gather*}

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)) - 3*(2*I*a*b*d*x^2 - I*b^2)*cos(d*x^2 + c)*dilog(I*cos(d*x^2 + c) + sin(d*x^2 + c)) - 3*(2*I*a*b*d*
x^2 + I*b^2)*cos(d*x^2 + c)*dilog(I*cos(d*x^2 + c) - sin(d*x^2 + c)) - 3*(-2*I*a*b*d*x^2 + I*b^2)*cos(d*x^2 +
c)*dilog(-I*cos(d*x^2 + c) + sin(d*x^2 + c)) - 3*(-2*I*a*b*d*x^2 - 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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{5} \left (a + b \sec {\left (c + d x^{2} \right )}\right )^{2}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^5\,{\left (a+\frac {b}{\cos \left (d\,x^2+c\right )}\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b/cos(c + d*x^2))^2,x)

[Out]

int(x^5*(a + b/cos(c + d*x^2))^2, x)

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