3.1.76 \(\int (e x)^{-1+2 n} (a+b \sec (c+d x^n))^2 \, dx\) [76]

Optimal. Leaf size=221 \[ \frac {a^2 (e x)^{2 n}}{2 e n}-\frac {4 i a b x^{-n} (e x)^{2 n} \text {ArcTan}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {b^2 x^{-2 n} (e x)^{2 n} \log \left (\cos \left (c+d x^n\right )\right )}{d^2 e n}+\frac {2 i a b x^{-2 n} (e x)^{2 n} \text {PolyLog}\left (2,-i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}-\frac {2 i a b x^{-2 n} (e x)^{2 n} \text {PolyLog}\left (2,i e^{i \left (c+d x^n\right )}\right )}{d^2 e n}+\frac {b^2 x^{-n} (e x)^{2 n} \tan \left (c+d x^n\right )}{d e n} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4293, 4289, 4275, 4266, 2317, 2438, 4269, 3556} \begin {gather*} \frac {a^2 (e x)^{2 n}}{2 e n}-\frac {4 i a b x^{-n} (e x)^{2 n} \text {ArcTan}\left (e^{i \left (c+d x^n\right )}\right )}{d e n}+\frac {2 i a b x^{-2 n} (e x)^{2 n} \text {Li}_2\left (-i e^{i \left (d x^n+c\right )}\right )}{d^2 e n}-\frac {2 i a b x^{-2 n} (e x)^{2 n} \text {Li}_2\left (i e^{i \left (d x^n+c\right )}\right )}{d^2 e n}+\frac {b^2 x^{-2 n} (e x)^{2 n} \log \left (\cos \left (c+d x^n\right )\right )}{d^2 e n}+\frac {b^2 x^{-n} (e x)^{2 n} \tan \left (c+d x^n\right )}{d e n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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 4293

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]

Rubi steps

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

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Mathematica [A]
time = 4.88, size = 347, normalized size = 1.57 \begin {gather*} \frac {x^{-2 n} (e x)^{2 n} \left (8 a b \text {ArcTan}(\cot (c)) \tanh ^{-1}\left (\sin (c)+\cos (c) \tan \left (\frac {d x^n}{2}\right )\right )-\frac {4 a b \csc (c) \left (\left (d x^n-\text {ArcTan}(\cot (c))\right ) \left (\log \left (1-e^{i \left (d x^n-\text {ArcTan}(\cot (c))\right )}\right )-\log \left (1+e^{i \left (d x^n-\text {ArcTan}(\cot (c))\right )}\right )\right )+i \text {PolyLog}\left (2,-e^{i \left (d x^n-\text {ArcTan}(\cot (c))\right )}\right )-i \text {PolyLog}\left (2,e^{i \left (d x^n-\text {ArcTan}(\cot (c))\right )}\right )\right )}{\sqrt {\csc ^2(c)}}+\frac {2 b^2 d x^n \sin \left (\frac {d x^n}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} \left (c+d x^n\right )\right )-\sin \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}+\frac {2 b^2 d x^n \sin \left (\frac {d x^n}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} \left (c+d x^n\right )\right )+\sin \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}-2 b^2 d x^n \tan (c)+d x^n \left (a^2 d x^n+2 b^2 \tan (c)\right )+2 b^2 \left (\log \left (\cos \left (c+d x^n\right )\right )+d x^n \tan (c)\right )\right )}{2 d^2 e n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^(2*n)*(8*a*b*ArcTan[Cot[c]]*ArcTanh[Sin[c] + Cos[c]*Tan[(d*x^n)/2]] - (4*a*b*Csc[c]*((d*x^n - ArcTan[Co
t[c]])*(Log[1 - E^(I*(d*x^n - ArcTan[Cot[c]]))] - Log[1 + E^(I*(d*x^n - ArcTan[Cot[c]]))]) + I*PolyLog[2, -E^(
I*(d*x^n - ArcTan[Cot[c]]))] - I*PolyLog[2, E^(I*(d*x^n - ArcTan[Cot[c]]))]))/Sqrt[Csc[c]^2] + (2*b^2*d*x^n*Si
n[(d*x^n)/2])/((Cos[c/2] - Sin[c/2])*(Cos[(c + d*x^n)/2] - Sin[(c + d*x^n)/2])) + (2*b^2*d*x^n*Sin[(d*x^n)/2])
/((Cos[c/2] + Sin[c/2])*(Cos[(c + d*x^n)/2] + Sin[(c + d*x^n)/2])) - 2*b^2*d*x^n*Tan[c] + d*x^n*(a^2*d*x^n + 2
*b^2*Tan[c]) + 2*b^2*(Log[Cos[c + d*x^n]] + d*x^n*Tan[c])))/(2*d^2*e*n*x^(2*n))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.74, size = 1096, normalized size = 4.96

method	result	size
risch	\(\text {Expression too large to display}\)	\(1096\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2/n*x*exp(1/2*(-1+2*n)*(-I*csgn(I*e)*csgn(I*x)*csgn(I*e*x)*Pi+I*csgn(I*e)*csgn(I*e*x)^2*Pi+I*csgn(I*x)*c
sgn(I*e*x)^2*Pi-I*csgn(I*e*x)^3*Pi+2*ln(e)+2*ln(x)))+2*I*x*b^2*exp(1/2*(-1+2*n)*(-I*csgn(I*e)*csgn(I*x)*csgn(I
*e*x)*Pi+I*csgn(I*e)*csgn(I*e*x)^2*Pi+I*csgn(I*x)*csgn(I*e*x)^2*Pi-I*csgn(I*e*x)^3*Pi+2*ln(e)+2*ln(x)))/d/n/(x
^n)/(1+exp(2*I*(c+d*x^n)))-2*b^2/d^2*ln(exp(I*x^n*d))/n/e*(e^n)^2*exp(1/2*I*csgn(I*e*x)*Pi*(-1+2*n)*(csgn(I*e*
x)-csgn(I*x))*(-csgn(I*e*x)+csgn(I*e)))+b^2/d^2*ln(1+exp(2*I*(c+d*x^n)))/n/e*(e^n)^2*exp(1/2*I*csgn(I*e*x)*Pi*
(-1+2*n)*(csgn(I*e*x)-csgn(I*x))*(-csgn(I*e*x)+csgn(I*e)))+2*I*b/d*ln(1+exp(I*x^n*d)*(-exp(2*I*c))^(1/2))*x^n*
(-exp(2*I*c))^(1/2)/n*a/e*(e^n)^2*exp(-1/2*I*(2*Pi*n*csgn(I*e)*csgn(I*x)*csgn(I*e*x)-2*Pi*n*csgn(I*e)*csgn(I*e
*x)^2-2*Pi*n*csgn(I*x)*csgn(I*e*x)^2+2*Pi*n*csgn(I*e*x)^3-Pi*csgn(I*e)*csgn(I*x)*csgn(I*e*x)+Pi*csgn(I*e)*csgn
(I*e*x)^2+Pi*csgn(I*x)*csgn(I*e*x)^2-Pi*csgn(I*e*x)^3+2*c))-2*I*b/d*ln(1-exp(I*x^n*d)*(-exp(2*I*c))^(1/2))*x^n
*(-exp(2*I*c))^(1/2)/n*a/e*(e^n)^2*exp(-1/2*I*(2*Pi*n*csgn(I*e)*csgn(I*x)*csgn(I*e*x)-2*Pi*n*csgn(I*e)*csgn(I*
e*x)^2-2*Pi*n*csgn(I*x)*csgn(I*e*x)^2+2*Pi*n*csgn(I*e*x)^3-Pi*csgn(I*e)*csgn(I*x)*csgn(I*e*x)+Pi*csgn(I*e)*csg
n(I*e*x)^2+Pi*csgn(I*x)*csgn(I*e*x)^2-Pi*csgn(I*e*x)^3+2*c))+2*b/d^2*dilog(1+exp(I*x^n*d)*(-exp(2*I*c))^(1/2))
*(-exp(2*I*c))^(1/2)/n*a/e*(e^n)^2*exp(-1/2*I*(2*Pi*n*csgn(I*e)*csgn(I*x)*csgn(I*e*x)-2*Pi*n*csgn(I*e)*csgn(I*
e*x)^2-2*Pi*n*csgn(I*x)*csgn(I*e*x)^2+2*Pi*n*csgn(I*e*x)^3-Pi*csgn(I*e)*csgn(I*x)*csgn(I*e*x)+Pi*csgn(I*e)*csg
n(I*e*x)^2+Pi*csgn(I*x)*csgn(I*e*x)^2-Pi*csgn(I*e*x)^3+2*c))-2*b/d^2*dilog(1-exp(I*x^n*d)*(-exp(2*I*c))^(1/2))
*(-exp(2*I*c))^(1/2)/n*a/e*(e^n)^2*exp(-1/2*I*(2*Pi*n*csgn(I*e)*csgn(I*x)*csgn(I*e*x)-2*Pi*n*csgn(I*e)*csgn(I*
e*x)^2-2*Pi*n*csgn(I*x)*csgn(I*e*x)^2+2*Pi*n*csgn(I*e*x)^3-Pi*csgn(I*e)*csgn(I*x)*csgn(I*e*x)+Pi*csgn(I*e)*csg
n(I*e*x)^2+Pi*csgn(I*x)*csgn(I*e*x)^2-Pi*csgn(I*e*x)^3+2*c))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(2*n-1>0)', see `assume?` for m
ore details)

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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. 638 vs. \(2 (207) = 414\).
time = 3.56, size = 638, normalized size = 2.89 \begin {gather*} \frac {a^{2} d^{2} x^{2 \, n} \cos \left (d x^{n} + c\right ) e^{\left (2 \, n - 1\right )} + 2 \, b^{2} d x^{n} e^{\left (2 \, n - 1\right )} \sin \left (d x^{n} + c\right ) - 2 i \, a b \cos \left (d x^{n} + c\right ) {\rm Li}_2\left (i \, \cos \left (d x^{n} + c\right ) + \sin \left (d x^{n} + c\right )\right ) e^{\left (2 \, n - 1\right )} - 2 i \, a b \cos \left (d x^{n} + c\right ) {\rm Li}_2\left (i \, \cos \left (d x^{n} + c\right ) - \sin \left (d x^{n} + c\right )\right ) e^{\left (2 \, n - 1\right )} + 2 i \, a b \cos \left (d x^{n} + c\right ) {\rm Li}_2\left (-i \, \cos \left (d x^{n} + c\right ) + \sin \left (d x^{n} + c\right )\right ) e^{\left (2 \, n - 1\right )} + 2 i \, a b \cos \left (d x^{n} + c\right ) {\rm Li}_2\left (-i \, \cos \left (d x^{n} + c\right ) - \sin \left (d x^{n} + c\right )\right ) e^{\left (2 \, n - 1\right )} - {\left (2 \, a b c - b^{2}\right )} \cos \left (d x^{n} + c\right ) e^{\left (2 \, n - 1\right )} \log \left (\cos \left (d x^{n} + c\right ) + i \, \sin \left (d x^{n} + c\right ) + i\right ) + {\left (2 \, a b c + b^{2}\right )} \cos \left (d x^{n} + c\right ) e^{\left (2 \, n - 1\right )} \log \left (\cos \left (d x^{n} + c\right ) - i \, \sin \left (d x^{n} + c\right ) + i\right ) - {\left (2 \, a b c - b^{2}\right )} \cos \left (d x^{n} + c\right ) e^{\left (2 \, n - 1\right )} \log \left (-\cos \left (d x^{n} + c\right ) + i \, \sin \left (d x^{n} + c\right ) + i\right ) + {\left (2 \, a b c + b^{2}\right )} \cos \left (d x^{n} + c\right ) e^{\left (2 \, n - 1\right )} \log \left (-\cos \left (d x^{n} + c\right ) - i \, \sin \left (d x^{n} + c\right ) + i\right ) + 2 \, {\left (a b d x^{n} e^{\left (2 \, n - 1\right )} + a b c e^{\left (2 \, n - 1\right )}\right )} \cos \left (d x^{n} + c\right ) \log \left (i \, \cos \left (d x^{n} + c\right ) + \sin \left (d x^{n} + c\right ) + 1\right ) - 2 \, {\left (a b d x^{n} e^{\left (2 \, n - 1\right )} + a b c e^{\left (2 \, n - 1\right )}\right )} \cos \left (d x^{n} + c\right ) \log \left (i \, \cos \left (d x^{n} + c\right ) - \sin \left (d x^{n} + c\right ) + 1\right ) + 2 \, {\left (a b d x^{n} e^{\left (2 \, n - 1\right )} + a b c e^{\left (2 \, n - 1\right )}\right )} \cos \left (d x^{n} + c\right ) \log \left (-i \, \cos \left (d x^{n} + c\right ) + \sin \left (d x^{n} + c\right ) + 1\right ) - 2 \, {\left (a b d x^{n} e^{\left (2 \, n - 1\right )} + a b c e^{\left (2 \, n - 1\right )}\right )} \cos \left (d x^{n} + c\right ) \log \left (-i \, \cos \left (d x^{n} + c\right ) - \sin \left (d x^{n} + c\right ) + 1\right )}{2 \, d^{2} n \cos \left (d x^{n} + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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