3.77 \(\int (e x)^{-1+3 n} (a+b \sec (c+d x^n))^2 \, dx\)

Optimal. Leaf size=390 \[ \frac {a^2 (e x)^{3 n}}{3 e n}-\frac {4 a b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (-i e^{i \left (d x^n+c\right )}\right )}{d^3 e n}+\frac {4 a b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (i e^{i \left (d x^n+c\right )}\right )}{d^3 e n}+\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (-i e^{i \left (d x^n+c\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (i e^{i \left (d x^n+c\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}-\frac {i b^2 x^{-3 n} (e x)^{3 n} \text {Li}_2\left (-e^{2 i \left (d x^n+c\right )}\right )}{d^3 e n}+\frac {2 b^2 x^{-2 n} (e x)^{3 n} \log \left (1+e^{2 i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tan \left (c+d x^n\right )}{d e n}-\frac {i b^2 x^{-n} (e x)^{3 n}}{d e n} \]

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 390, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4208, 4204, 4190, 4181, 2531, 2282, 6589, 4184, 3719, 2190, 2279, 2391} \[ -\frac {4 a b x^{-3 n} (e x)^{3 n} \text {PolyLog}\left (3,-i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {4 a b x^{-3 n} (e x)^{3 n} \text {PolyLog}\left (3,i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {PolyLog}\left (2,-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {PolyLog}\left (2,i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {i b^2 x^{-3 n} (e x)^{3 n} \text {PolyLog}\left (2,-e^{2 i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {a^2 (e x)^{3 n}}{3 e n}-\frac {4 i a b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {2 b^2 x^{-2 n} (e x)^{3 n} \log \left (1+e^{2 i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tan \left (c+d x^n\right )}{d e n}-\frac {i b^2 x^{-n} (e x)^{3 n}}{d e n} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^(-1 + 3*n)*(a + b*Sec[c + d*x^n])^2,x]

[Out]

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

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 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 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]

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 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 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 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 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 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 4208

Int[((e_)*(x_))^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[(e^IntPart[m]*(e*x
)^FracPart[m])/x^FracPart[m], Int[x^m*(a + b*Sec[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, 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 (e x)^{-1+3 n} \left (a+b \sec \left (c+d x^n\right )\right )^2 \, dx &=\frac {\left (x^{-3 n} (e x)^{3 n}\right ) \int x^{-1+3 n} \left (a+b \sec \left (c+d x^n\right )\right )^2 \, dx}{e}\\ &=\frac {\left (x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int x^2 (a+b \sec (c+d x))^2 \, dx,x,x^n\right )}{e n}\\ &=\frac {\left (x^{-3 n} (e x)^{3 n}\right ) \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^n\right )}{e n}\\ &=\frac {a^2 (e x)^{3 n}}{3 e n}+\frac {\left (2 a b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int x^2 \sec (c+d x) \, dx,x,x^n\right )}{e n}+\frac {\left (b^2 x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int x^2 \sec ^2(c+d x) \, dx,x,x^n\right )}{e n}\\ &=\frac {a^2 (e x)^{3 n}}{3 e n}-\frac {4 i a b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tan \left (c+d x^n\right )}{d e n}-\frac {\left (4 a b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int x \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,x^n\right )}{d e n}+\frac {\left (4 a b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int x \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,x^n\right )}{d e n}-\frac {\left (2 b^2 x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int x \tan (c+d x) \, dx,x,x^n\right )}{d e n}\\ &=\frac {a^2 (e x)^{3 n}}{3 e n}-\frac {i b^2 x^{-n} (e x)^{3 n}}{d e n}-\frac {4 i a b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tan \left (c+d x^n\right )}{d e n}-\frac {\left (4 i a b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,x^n\right )}{d^2 e n}+\frac {\left (4 i a b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,x^n\right )}{d^2 e n}+\frac {\left (4 i b^2 x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x}{1+e^{2 i (c+d x)}} \, dx,x,x^n\right )}{d e n}\\ &=\frac {a^2 (e x)^{3 n}}{3 e n}-\frac {i b^2 x^{-n} (e x)^{3 n}}{d e n}-\frac {4 i a b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {2 b^2 x^{-2 n} (e x)^{3 n} \log \left (1+e^{2 i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tan \left (c+d x^n\right )}{d e n}-\frac {\left (4 a b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {\left (4 a b x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \left (c+d x^n\right )}\right )}{d^3 e n}-\frac {\left (2 b^2 x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,x^n\right )}{d^2 e n}\\ &=\frac {a^2 (e x)^{3 n}}{3 e n}-\frac {i b^2 x^{-n} (e x)^{3 n}}{d e n}-\frac {4 i a b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {2 b^2 x^{-2 n} (e x)^{3 n} \log \left (1+e^{2 i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {4 a b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (-i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {4 a b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tan \left (c+d x^n\right )}{d e n}+\frac {\left (i b^2 x^{-3 n} (e x)^{3 n}\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \left (c+d x^n\right )}\right )}{d^3 e n}\\ &=\frac {a^2 (e x)^{3 n}}{3 e n}-\frac {i b^2 x^{-n} (e x)^{3 n}}{d e n}-\frac {4 i a b x^{-n} (e x)^{3 n} \tan ^{-1}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {2 b^2 x^{-2 n} (e x)^{3 n} \log \left (1+e^{2 i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {4 i a b x^{-2 n} (e x)^{3 n} \text {Li}_2\left (i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {i b^2 x^{-3 n} (e x)^{3 n} \text {Li}_2\left (-e^{2 i \left (c+d x^n\right )}\right )}{d^3 e n}-\frac {4 a b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (-i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {4 a b x^{-3 n} (e x)^{3 n} \text {Li}_3\left (i e^{i \left (c+d x^n\right )}\right )}{d^3 e n}+\frac {b^2 x^{-n} (e x)^{3 n} \tan \left (c+d x^n\right )}{d e n}\\ \end {align*}

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Mathematica [F]  time = 11.66, size = 0, normalized size = 0.00 \[ \int (e x)^{-1+3 n} \left (a+b \sec \left (c+d x^n\right )\right )^2 \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(e*x)^(-1 + 3*n)*(a + b*Sec[c + d*x^n])^2,x]

[Out]

Integrate[(e*x)^(-1 + 3*n)*(a + b*Sec[c + d*x^n])^2, x]

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fricas [C]  time = 1.28, size = 1028, normalized size = 2.64 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(-1+3*n)*(a+b*sec(c+d*x^n))^2,x, algorithm="fricas")

[Out]

1/3*(a^2*d^3*e^(3*n - 1)*x^(3*n)*cos(d*x^n + c) + 3*b^2*d^2*e^(3*n - 1)*x^(2*n)*sin(d*x^n + c) - 6*a*b*e^(3*n
- 1)*cos(d*x^n + c)*polylog(3, I*cos(d*x^n + c) + sin(d*x^n + c)) + 6*a*b*e^(3*n - 1)*cos(d*x^n + c)*polylog(3
, I*cos(d*x^n + c) - sin(d*x^n + c)) - 6*a*b*e^(3*n - 1)*cos(d*x^n + c)*polylog(3, -I*cos(d*x^n + c) + sin(d*x
^n + c)) + 6*a*b*e^(3*n - 1)*cos(d*x^n + c)*polylog(3, -I*cos(d*x^n + c) - sin(d*x^n + c)) + 3*(a*b*c^2 - b^2*
c)*e^(3*n - 1)*cos(d*x^n + c)*log(cos(d*x^n + c) + I*sin(d*x^n + c) + I) - 3*(a*b*c^2 + b^2*c)*e^(3*n - 1)*cos
(d*x^n + c)*log(cos(d*x^n + c) - I*sin(d*x^n + c) + I) + 3*(a*b*c^2 - b^2*c)*e^(3*n - 1)*cos(d*x^n + c)*log(-c
os(d*x^n + c) + I*sin(d*x^n + c) + I) - 3*(a*b*c^2 + b^2*c)*e^(3*n - 1)*cos(d*x^n + c)*log(-cos(d*x^n + c) - I
*sin(d*x^n + c) + I) + (-6*I*a*b*d*e^(3*n - 1)*x^n + 3*I*b^2*e^(3*n - 1))*cos(d*x^n + c)*dilog(I*cos(d*x^n + c
) + sin(d*x^n + c)) + (-6*I*a*b*d*e^(3*n - 1)*x^n - 3*I*b^2*e^(3*n - 1))*cos(d*x^n + c)*dilog(I*cos(d*x^n + c)
- sin(d*x^n + c)) + (6*I*a*b*d*e^(3*n - 1)*x^n - 3*I*b^2*e^(3*n - 1))*cos(d*x^n + c)*dilog(-I*cos(d*x^n + c)
+ sin(d*x^n + c)) + (6*I*a*b*d*e^(3*n - 1)*x^n + 3*I*b^2*e^(3*n - 1))*cos(d*x^n + c)*dilog(-I*cos(d*x^n + c) -
sin(d*x^n + c)) + 3*(a*b*d^2*e^(3*n - 1)*x^(2*n) + b^2*d*e^(3*n - 1)*x^n - (a*b*c^2 - b^2*c)*e^(3*n - 1))*cos
(d*x^n + c)*log(I*cos(d*x^n + c) + sin(d*x^n + c) + 1) - 3*(a*b*d^2*e^(3*n - 1)*x^(2*n) - b^2*d*e^(3*n - 1)*x^
n - (a*b*c^2 + b^2*c)*e^(3*n - 1))*cos(d*x^n + c)*log(I*cos(d*x^n + c) - sin(d*x^n + c) + 1) + 3*(a*b*d^2*e^(3
*n - 1)*x^(2*n) + b^2*d*e^(3*n - 1)*x^n - (a*b*c^2 - b^2*c)*e^(3*n - 1))*cos(d*x^n + c)*log(-I*cos(d*x^n + c)
+ sin(d*x^n + c) + 1) - 3*(a*b*d^2*e^(3*n - 1)*x^(2*n) - b^2*d*e^(3*n - 1)*x^n - (a*b*c^2 + b^2*c)*e^(3*n - 1)
)*cos(d*x^n + c)*log(-I*cos(d*x^n + c) - sin(d*x^n + c) + 1))/(d^3*n*cos(d*x^n + c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x^{n} + c\right ) + a\right )}^{2} \left (e x\right )^{3 \, n - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(-1+3*n)*(a+b*sec(c+d*x^n))^2,x, algorithm="giac")

[Out]

integrate((b*sec(d*x^n + c) + a)^2*(e*x)^(3*n - 1), x)

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maple [F]  time = 3.02, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{-1+3 n} \left (a +b \sec \left (c +d \,x^{n}\right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(-1+3*n)*(a+b*sec(c+d*x^n))^2,x)

[Out]

int((e*x)^(-1+3*n)*(a+b*sec(c+d*x^n))^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\left (e x\right )^{3 \, n} a^{2}}{3 \, e n} + \frac {2 \, b^{2} e^{3 \, n} x^{2 \, n} \sin \left (2 \, d x^{n} + 2 \, c\right ) + 4 \, {\left (d e n \cos \left (2 \, d x^{n} + 2 \, c\right )^{2} + d e n \sin \left (2 \, d x^{n} + 2 \, c\right )^{2} + 2 \, d e n \cos \left (2 \, d x^{n} + 2 \, c\right ) + d e n\right )} \int \frac {a b d e^{3 \, n} x^{3 \, n} \cos \left (2 \, d x^{n} + 2 \, c\right ) \cos \left (d x^{n} + c\right ) + a b d e^{3 \, n} x^{3 \, n} \cos \left (d x^{n} + c\right ) + {\left (a b d e^{3 \, n} x^{3 \, n} \sin \left (d x^{n} + c\right ) - b^{2} e^{3 \, n} x^{2 \, n}\right )} \sin \left (2 \, d x^{n} + 2 \, c\right )}{d e x \cos \left (2 \, d x^{n} + 2 \, c\right )^{2} + d e x \sin \left (2 \, d x^{n} + 2 \, c\right )^{2} + 2 \, d e x \cos \left (2 \, d x^{n} + 2 \, c\right ) + d e x}\,{d x}}{d e n \cos \left (2 \, d x^{n} + 2 \, c\right )^{2} + d e n \sin \left (2 \, d x^{n} + 2 \, c\right )^{2} + 2 \, d e n \cos \left (2 \, d x^{n} + 2 \, c\right ) + d e n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(-1+3*n)*(a+b*sec(c+d*x^n))^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+\frac {b}{\cos \left (c+d\,x^n\right )}\right )}^2\,{\left (e\,x\right )}^{3\,n-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x^n))^2*(e*x)^(3*n - 1),x)

[Out]

int((a + b/cos(c + d*x^n))^2*(e*x)^(3*n - 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{3 n - 1} \left (a + b \sec {\left (c + d x^{n} \right )}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(-1+3*n)*(a+b*sec(c+d*x**n))**2,x)

[Out]

Integral((e*x)**(3*n - 1)*(a + b*sec(c + d*x**n))**2, x)

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