3.73 \(\int (e x)^{-1+2 n} (a+b \text {csch}(c+d x^n)) \, dx\)

Optimal. Leaf size=124 \[ \frac {a (e x)^{2 n}}{2 e n}-\frac {b x^{-2 n} (e x)^{2 n} \text {Li}_2\left (-e^{d x^n+c}\right )}{d^2 e n}+\frac {b x^{-2 n} (e x)^{2 n} \text {Li}_2\left (e^{d x^n+c}\right )}{d^2 e n}-\frac {2 b x^{-n} (e x)^{2 n} \tanh ^{-1}\left (e^{c+d x^n}\right )}{d e n} \]

[Out]

1/2*a*(e*x)^(2*n)/e/n-2*b*(e*x)^(2*n)*arctanh(exp(c+d*x^n))/d/e/n/(x^n)-b*(e*x)^(2*n)*polylog(2,-exp(c+d*x^n))
/d^2/e/n/(x^(2*n))+b*(e*x)^(2*n)*polylog(2,exp(c+d*x^n))/d^2/e/n/(x^(2*n))

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Rubi [A]  time = 0.11, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {14, 5441, 5437, 4182, 2279, 2391} \[ -\frac {b x^{-2 n} (e x)^{2 n} \text {PolyLog}\left (2,-e^{c+d x^n}\right )}{d^2 e n}+\frac {b x^{-2 n} (e x)^{2 n} \text {PolyLog}\left (2,e^{c+d x^n}\right )}{d^2 e n}+\frac {a (e x)^{2 n}}{2 e n}-\frac {2 b x^{-n} (e x)^{2 n} \tanh ^{-1}\left (e^{c+d x^n}\right )}{d e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(e*x)^(2*n))/(2*e*n) - (2*b*(e*x)^(2*n)*ArcTanh[E^(c + d*x^n)])/(d*e*n*x^n) - (b*(e*x)^(2*n)*PolyLog[2, -E^
(c + d*x^n)])/(d^2*e*n*x^(2*n)) + (b*(e*x)^(2*n)*PolyLog[2, E^(c + d*x^n)])/(d^2*e*n*x^(2*n))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5437

Int[((a_.) + Csch[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Csch[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplif
y[(m + 1)/n], 0] && IntegerQ[p]

Rule 5441

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

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Mathematica [A]  time = 0.15, size = 175, normalized size = 1.41 \[ \frac {x^{-2 n} (e x)^{2 n} \left (a d^2 x^{2 n}+2 b \text {Li}_2\left (-e^{-d x^n-c}\right )-2 b \text {Li}_2\left (e^{-d x^n-c}\right )+2 b d x^n \log \left (1-e^{-c-d x^n}\right )-2 b d x^n \log \left (e^{-c-d x^n}+1\right )+2 b c \log \left (1-e^{-c-d x^n}\right )-2 b c \log \left (e^{-c-d x^n}+1\right )-2 b c \log \left (\tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )\right )}{2 d^2 e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^(2*n)*(a*d^2*x^(2*n) + 2*b*c*Log[1 - E^(-c - d*x^n)] + 2*b*d*x^n*Log[1 - E^(-c - d*x^n)] - 2*b*c*Log[1
+ E^(-c - d*x^n)] - 2*b*d*x^n*Log[1 + E^(-c - d*x^n)] - 2*b*c*Log[Tanh[(c + d*x^n)/2]] + 2*b*PolyLog[2, -E^(-c
- d*x^n)] - 2*b*PolyLog[2, E^(-c - d*x^n)]))/(2*d^2*e*n*x^(2*n))

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fricas [B]  time = 0.44, size = 555, normalized size = 4.48 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*d^2*cosh((2*n - 1)*log(e))*cosh(n*log(x))^2 + a*d^2*cosh(n*log(x))^2*sinh((2*n - 1)*log(e)) + (a*d^2*co
sh((2*n - 1)*log(e)) + a*d^2*sinh((2*n - 1)*log(e)))*sinh(n*log(x))^2 + 2*(b*cosh((2*n - 1)*log(e)) + b*sinh((
2*n - 1)*log(e)))*dilog(cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + sinh(d*cosh(n*log(x)) + d*sinh(n*log(x
)) + c)) - 2*(b*cosh((2*n - 1)*log(e)) + b*sinh((2*n - 1)*log(e)))*dilog(-cosh(d*cosh(n*log(x)) + d*sinh(n*log
(x)) + c) - sinh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c)) - 2*(b*d*cosh((2*n - 1)*log(e))*cosh(n*log(x)) + b*
d*cosh(n*log(x))*sinh((2*n - 1)*log(e)) + (b*d*cosh((2*n - 1)*log(e)) + b*d*sinh((2*n - 1)*log(e)))*sinh(n*log
(x)))*log(cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + sinh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + 1) -
2*(b*c*cosh((2*n - 1)*log(e)) + b*c*sinh((2*n - 1)*log(e)))*log(cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c)
+ sinh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) - 1) + 2*(b*d*cosh((2*n - 1)*log(e))*cosh(n*log(x)) + b*c*cos
h((2*n - 1)*log(e)) + (b*d*cosh(n*log(x)) + b*c)*sinh((2*n - 1)*log(e)) + (b*d*cosh((2*n - 1)*log(e)) + b*d*si
nh((2*n - 1)*log(e)))*sinh(n*log(x)))*log(-cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) - sinh(d*cosh(n*log(x
)) + d*sinh(n*log(x)) + c) + 1) + 2*(a*d^2*cosh((2*n - 1)*log(e))*cosh(n*log(x)) + a*d^2*cosh(n*log(x))*sinh((
2*n - 1)*log(e)))*sinh(n*log(x)))/(d^2*n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.92, size = 326, normalized size = 2.63 \[ \frac {a x \,{\mathrm e}^{\frac {\left (-1+2 n \right ) \left (-i \pi \,\mathrm {csgn}\left (i e \right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i e x \right )+i \pi \,\mathrm {csgn}\left (i e \right ) \mathrm {csgn}\left (i e x \right )^{2}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i e x \right )^{2}-i \pi \mathrm {csgn}\left (i e x \right )^{3}+2 \ln \relax (x )+2 \ln \relax (e )\right )}{2}}}{2 n}+\frac {2 b \,{\mathrm e}^{-i \pi n \,\mathrm {csgn}\left (i e \right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i e x \right )} {\mathrm e}^{i \pi n \,\mathrm {csgn}\left (i e \right ) \mathrm {csgn}\left (i e x \right )^{2}} {\mathrm e}^{i \pi n \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i e x \right )^{2}} {\mathrm e}^{-i \pi n \mathrm {csgn}\left (i e x \right )^{3}} {\mathrm e}^{\frac {i \pi \,\mathrm {csgn}\left (i e \right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i e x \right )}{2}} {\mathrm e}^{-\frac {i \pi \,\mathrm {csgn}\left (i e \right ) \mathrm {csgn}\left (i e x \right )^{2}}{2}} {\mathrm e}^{-\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i e x \right )^{2}}{2}} {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (i e x \right )^{3}}{2}} e^{2 n} {\mathrm e}^{c} \left (\frac {\left (\ln \left (1-{\mathrm e}^{c +d \,x^{n}}\right )-\ln \left ({\mathrm e}^{c +d \,x^{n}}+1\right )\right ) d \,x^{n} {\mathrm e}^{-c}}{2}+\frac {\left (\dilog \left (1-{\mathrm e}^{c +d \,x^{n}}\right )-\dilog \left ({\mathrm e}^{c +d \,x^{n}}+1\right )\right ) {\mathrm e}^{-c}}{2}\right )}{e n \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (a+\frac {b}{\mathrm {sinh}\left (c+d\,x^n\right )}\right )\,{\left (e\,x\right )}^{2\,n-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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