\(\int (e x)^{-1+n} (a+b \sec (c+d x^n))^2 \, dx\) [75]

Optimal result

Integrand size = 22, antiderivative size = 79 \[ \int (e x)^{-1+n} \left (a+b \sec \left (c+d x^n\right )\right )^2 \, dx=\frac {a^2 (e x)^n}{e n}+\frac {2 a b x^{-n} (e x)^n \text {arctanh}\left (\sin \left (c+d x^n\right )\right )}{d e n}+\frac {b^2 x^{-n} (e x)^n \tan \left (c+d x^n\right )}{d e n} \]

[Out]

a^2*(e*x)^n/e/n+2*a*b*(e*x)^n*arctanh(sin(c+d*x^n))/d/e/n/(x^n)+b^2*(e*x)^n*tan(c+d*x^n)/d/e/n/(x^n)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4293, 4289, 3858, 3855, 3852, 8} \[ \int (e x)^{-1+n} \left (a+b \sec \left (c+d x^n\right )\right )^2 \, dx=\frac {a^2 (e x)^n}{e n}+\frac {2 a b x^{-n} (e x)^n \text {arctanh}\left (\sin \left (c+d x^n\right )\right )}{d e n}+\frac {b^2 x^{-n} (e x)^n \tan \left (c+d x^n\right )}{d e n} \]

[In]

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

[Out]

(a^2*(e*x)^n)/(e*n) + (2*a*b*(e*x)^n*ArcTanh[Sin[c + d*x^n]])/(d*e*n*x^n) + (b^2*(e*x)^n*Tan[c + d*x^n])/(d*e*
n*x^n)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

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

Rule 3858

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[a^2*x, x] + (Dist[2*a*b, Int[Csc[c + d*x], x],
x] + Dist[b^2, Int[Csc[c + d*x]^2, x], x]) /; FreeQ[{a, b, c, d}, x]

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*} \text {integral}& = \frac {\left (x^{-n} (e x)^n\right ) \int x^{-1+n} \left (a+b \sec \left (c+d x^n\right )\right )^2 \, dx}{e} \\ & = \frac {\left (x^{-n} (e x)^n\right ) \text {Subst}\left (\int (a+b \sec (c+d x))^2 \, dx,x,x^n\right )}{e n} \\ & = \frac {a^2 (e x)^n}{e n}+\frac {\left (2 a b x^{-n} (e x)^n\right ) \text {Subst}\left (\int \sec (c+d x) \, dx,x,x^n\right )}{e n}+\frac {\left (b^2 x^{-n} (e x)^n\right ) \text {Subst}\left (\int \sec ^2(c+d x) \, dx,x,x^n\right )}{e n} \\ & = \frac {a^2 (e x)^n}{e n}+\frac {2 a b x^{-n} (e x)^n \text {arctanh}\left (\sin \left (c+d x^n\right )\right )}{d e n}-\frac {\left (b^2 x^{-n} (e x)^n\right ) \text {Subst}\left (\int 1 \, dx,x,-\tan \left (c+d x^n\right )\right )}{d e n} \\ & = \frac {a^2 (e x)^n}{e n}+\frac {2 a b x^{-n} (e x)^n \text {arctanh}\left (\sin \left (c+d x^n\right )\right )}{d e n}+\frac {b^2 x^{-n} (e x)^n \tan \left (c+d x^n\right )}{d e n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.68 \[ \int (e x)^{-1+n} \left (a+b \sec \left (c+d x^n\right )\right )^2 \, dx=\frac {x^{-n} (e x)^n \left (a^2 d x^n+2 a b \text {arctanh}\left (\sin \left (c+d x^n\right )\right )+b^2 \tan \left (c+d x^n\right )\right )}{d e n} \]

[In]

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

[Out]

((e*x)^n*(a^2*d*x^n + 2*a*b*ArcTanh[Sin[c + d*x^n]] + b^2*Tan[c + d*x^n]))/(d*e*n*x^n)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 4.02 (sec) , antiderivative size = 276, normalized size of antiderivative = 3.49

method	result	size
risch	\(\frac {a^{2} x \,{\mathrm e}^{\frac {\left (-1+n \right ) \left (-i \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )+i \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}-i \pi \operatorname {csgn}\left (i e x \right )^{3}+2 \ln \left (x \right )+2 \ln \left (e \right )\right )}{2}}}{n}+\frac {2 i x \,{\mathrm e}^{\frac {\left (-1+n \right ) \left (-i \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )+i \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}-i \pi \operatorname {csgn}\left (i e x \right )^{3}+2 \ln \left (x \right )+2 \ln \left (e \right )\right )}{2}} b^{2} x^{-n}}{d n \left (1+{\mathrm e}^{2 i \left (c +d \,x^{n}\right )}\right )}-\frac {4 i \arctan \left ({\mathrm e}^{i \left (c +d \,x^{n}\right )}\right ) e^{n} a b \,{\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i e x \right ) \left (-1+n \right ) \left (\operatorname {csgn}\left (i e x \right )-\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i e x \right )+\operatorname {csgn}\left (i e \right )\right )}{2}}}{d e n}\)	\(276\)

[In]

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

[Out]

a^2/n*x*exp(1/2*(-1+n)*(-I*Pi*csgn(I*e)*csgn(I*x)*csgn(I*e*x)+I*Pi*csgn(I*e)*csgn(I*e*x)^2+I*Pi*csgn(I*x)*csgn
(I*e*x)^2-I*Pi*csgn(I*e*x)^3+2*ln(x)+2*ln(e)))+2*I*x*exp(1/2*(-1+n)*(-I*Pi*csgn(I*e)*csgn(I*x)*csgn(I*e*x)+I*P
i*csgn(I*e)*csgn(I*e*x)^2+I*Pi*csgn(I*x)*csgn(I*e*x)^2-I*Pi*csgn(I*e*x)^3+2*ln(x)+2*ln(e)))*b^2/d/n/(x^n)/(1+e
xp(2*I*(c+d*x^n)))-4*I*arctan(exp(I*(c+d*x^n)))/d/e*e^n/n*a*b*exp(1/2*I*Pi*csgn(I*e*x)*(-1+n)*(csgn(I*e*x)-csg
n(I*x))*(-csgn(I*e*x)+csgn(I*e)))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.43 \[ \int (e x)^{-1+n} \left (a+b \sec \left (c+d x^n\right )\right )^2 \, dx=\frac {a^{2} d e^{n - 1} x^{n} \cos \left (d x^{n} + c\right ) + a b e^{n - 1} \cos \left (d x^{n} + c\right ) \log \left (\sin \left (d x^{n} + c\right ) + 1\right ) - a b e^{n - 1} \cos \left (d x^{n} + c\right ) \log \left (-\sin \left (d x^{n} + c\right ) + 1\right ) + b^{2} e^{n - 1} \sin \left (d x^{n} + c\right )}{d n \cos \left (d x^{n} + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

(e*x)^n*a^2/(e*n) + 2*(b^2*e^n*sin(2*d*x^n + 2*c) + 2*(a*b*d*e^(n + 1)*n*cos(2*d*x^n + 2*c)^2 + a*b*d*e^(n + 1
)*n*sin(2*d*x^n + 2*c)^2 + 2*a*b*d*e^(n + 1)*n*cos(2*d*x^n + 2*c) + a*b*d*e^(n + 1)*n)*integrate((x^n*cos(2*d*
x^n + 2*c)*cos(d*x^n + c) + x^n*sin(2*d*x^n + 2*c)*sin(d*x^n + c) + x^n*cos(d*x^n + c))/(e*x*cos(2*d*x^n + 2*c
)^2 + e*x*sin(2*d*x^n + 2*c)^2 + 2*e*x*cos(2*d*x^n + 2*c) + 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)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 15.51 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.28 \[ \int (e x)^{-1+n} \left (a+b \sec \left (c+d x^n\right )\right )^2 \, dx=\frac {a^2\,x\,{\left (e\,x\right )}^{n-1}}{n}+\frac {b^2\,x\,{\left (e\,x\right )}^{n-1}\,2{}\mathrm {i}}{d\,n\,x^n\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x^n\,2{}\mathrm {i}}+1\right )}+\frac {2\,a\,b\,x\,\ln \left (-a\,b\,{\left (e\,x\right )}^{n-1}\,4{}\mathrm {i}-4\,a\,b\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x^n\,1{}\mathrm {i}}\,{\left (e\,x\right )}^{n-1}\right )\,{\left (e\,x\right )}^{n-1}}{d\,n\,x^n}-\frac {2\,a\,b\,x\,\ln \left (a\,b\,{\left (e\,x\right )}^{n-1}\,4{}\mathrm {i}-4\,a\,b\,{\mathrm {e}}^{c\,1{}\mathrm {i}}\,{\mathrm {e}}^{d\,x^n\,1{}\mathrm {i}}\,{\left (e\,x\right )}^{n-1}\right )\,{\left (e\,x\right )}^{n-1}}{d\,n\,x^n} \]

[In]

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

[Out]

(a^2*x*(e*x)^(n - 1))/n + (b^2*x*(e*x)^(n - 1)*2i)/(d*n*x^n*(exp(c*2i + d*x^n*2i) + 1)) + (2*a*b*x*log(- a*b*(
e*x)^(n - 1)*4i - 4*a*b*exp(c*1i)*exp(d*x^n*1i)*(e*x)^(n - 1))*(e*x)^(n - 1))/(d*n*x^n) - (2*a*b*x*log(a*b*(e*
x)^(n - 1)*4i - 4*a*b*exp(c*1i)*exp(d*x^n*1i)*(e*x)^(n - 1))*(e*x)^(n - 1))/(d*n*x^n)