3.3.6 \(\int \frac {(e+f x)^2 \text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [206]

Optimal. Leaf size=224 \[ -\frac {i (e+f x)^2}{a d}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac {4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {2 f (e+f x) \text {PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}+\frac {4 i f^2 \text {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac {2 f (e+f x) \text {PolyLog}\left (2,e^{c+d x}\right )}{a d^2}+\frac {2 f^2 \text {PolyLog}\left (3,-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {PolyLog}\left (3,e^{c+d x}\right )}{a d^3}-\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d} \]

[Out]

-I*(f*x+e)^2/a/d-2*(f*x+e)^2*arctanh(exp(d*x+c))/a/d+4*I*f*(f*x+e)*ln(1+I*exp(d*x+c))/a/d^2-2*f*(f*x+e)*polylo
g(2,-exp(d*x+c))/a/d^2+4*I*f^2*polylog(2,-I*exp(d*x+c))/a/d^3+2*f*(f*x+e)*polylog(2,exp(d*x+c))/a/d^2+2*f^2*po
lylog(3,-exp(d*x+c))/a/d^3-2*f^2*polylog(3,exp(d*x+c))/a/d^3-I*(f*x+e)^2*tanh(1/2*c+1/4*I*Pi+1/2*d*x)/a/d

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Rubi [A]
time = 0.25, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {5694, 4267, 2611, 2320, 6724, 3399, 4269, 3797, 2221, 2317, 2438} \begin {gather*} \frac {4 i f^2 \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac {2 f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac {4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}-\frac {i (e+f x)^2}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^2*Csch[c + d*x])/(a + I*a*Sinh[c + d*x]),x]

[Out]

((-I)*(e + f*x)^2)/(a*d) - (2*(e + f*x)^2*ArcTanh[E^(c + d*x)])/(a*d) + ((4*I)*f*(e + f*x)*Log[1 + I*E^(c + d*
x)])/(a*d^2) - (2*f*(e + f*x)*PolyLog[2, -E^(c + d*x)])/(a*d^2) + ((4*I)*f^2*PolyLog[2, (-I)*E^(c + d*x)])/(a*
d^3) + (2*f*(e + f*x)*PolyLog[2, E^(c + d*x)])/(a*d^2) + (2*f^2*PolyLog[3, -E^(c + d*x)])/(a*d^3) - (2*f^2*Pol
yLog[3, E^(c + d*x)])/(a*d^3) - (I*(e + f*x)^2*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(a*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 3399

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) + f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

Rule 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(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*fz*x))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4267

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 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 5694

Int[(Csch[(c_.) + (d_.)*(x_)]^(n_.)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/a, Int[(e + f*x)^m*Csch[c + d*x]^n, x], x] - Dist[b/a, Int[(e + f*x)^m*(Csch[c + d*x]^(n - 1)/(
a + b*Sinh[c + d*x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && IGtQ[n, 0]

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 \frac {(e+f x)^2 \text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\left (i \int \frac {(e+f x)^2}{a+i a \sinh (c+d x)} \, dx\right )+\frac {\int (e+f x)^2 \text {csch}(c+d x) \, dx}{a}\\ &=-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {i \int (e+f x)^2 \csc ^2\left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {i d x}{2}\right ) \, dx}{2 a}-\frac {(2 f) \int (e+f x) \log \left (1-e^{c+d x}\right ) \, dx}{a d}+\frac {(2 f) \int (e+f x) \log \left (1+e^{c+d x}\right ) \, dx}{a d}\\ &=-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}-\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}+\frac {(2 i f) \int (e+f x) \coth \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx}{a d}+\frac {\left (2 f^2\right ) \int \text {Li}_2\left (-e^{c+d x}\right ) \, dx}{a d^2}-\frac {\left (2 f^2\right ) \int \text {Li}_2\left (e^{c+d x}\right ) \, dx}{a d^2}\\ &=-\frac {i (e+f x)^2}{a d}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}-\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {(4 f) \int \frac {e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )} (e+f x)}{1+i e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )}} \, dx}{a d}+\frac {\left (2 f^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{c+d x}\right )}{a d^3}-\frac {\left (2 f^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{c+d x}\right )}{a d^3}\\ &=-\frac {i (e+f x)^2}{a d}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac {4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac {2 f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {\left (4 i f^2\right ) \int \log \left (1+i e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \, dx}{a d^2}\\ &=-\frac {i (e+f x)^2}{a d}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac {4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac {2 f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {\left (4 i f^2\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )}\right )}{a d^3}\\ &=-\frac {i (e+f x)^2}{a d}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}+\frac {4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {4 i f^2 \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac {2 f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}\\ \end {align*}

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Mathematica [A]
time = 3.76, size = 327, normalized size = 1.46 \begin {gather*} \frac {-2 d^2 e^2 \tanh ^{-1}\left (e^{c+d x}\right )+2 d^2 e f x \log \left (1-e^{c+d x}\right )+d^2 f^2 x^2 \log \left (1-e^{c+d x}\right )-2 d^2 e f x \log \left (1+e^{c+d x}\right )-d^2 f^2 x^2 \log \left (1+e^{c+d x}\right )-2 d f (e+f x) \text {PolyLog}\left (2,-e^{c+d x}\right )+\frac {2 f \left (d \left (-i d e^c x (2 e+f x)+2 \left (1+i e^c\right ) (e+f x) \log \left (1+i e^{c+d x}\right )\right )+2 \left (1+i e^c\right ) f \text {PolyLog}\left (2,-i e^{c+d x}\right )\right )}{-i+e^c}+2 d f (e+f x) \text {PolyLog}\left (2,e^{c+d x}\right )+2 f^2 \text {PolyLog}\left (3,-e^{c+d x}\right )-2 f^2 \text {PolyLog}\left (3,e^{c+d x}\right )-\frac {2 i d^2 (e+f x)^2 \sinh \left (\frac {d x}{2}\right )}{\left (\cosh \left (\frac {c}{2}\right )+i \sinh \left (\frac {c}{2}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}}{a d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^2*Csch[c + d*x])/(a + I*a*Sinh[c + d*x]),x]

[Out]

(-2*d^2*e^2*ArcTanh[E^(c + d*x)] + 2*d^2*e*f*x*Log[1 - E^(c + d*x)] + d^2*f^2*x^2*Log[1 - E^(c + d*x)] - 2*d^2
*e*f*x*Log[1 + E^(c + d*x)] - d^2*f^2*x^2*Log[1 + E^(c + d*x)] - 2*d*f*(e + f*x)*PolyLog[2, -E^(c + d*x)] + (2
*f*(d*((-I)*d*E^c*x*(2*e + f*x) + 2*(1 + I*E^c)*(e + f*x)*Log[1 + I*E^(c + d*x)]) + 2*(1 + I*E^c)*f*PolyLog[2,
(-I)*E^(c + d*x)]))/(-I + E^c) + 2*d*f*(e + f*x)*PolyLog[2, E^(c + d*x)] + 2*f^2*PolyLog[3, -E^(c + d*x)] - 2
*f^2*PolyLog[3, E^(c + d*x)] - ((2*I)*d^2*(e + f*x)^2*Sinh[(d*x)/2])/((Cosh[c/2] + I*Sinh[c/2])*(Cosh[(c + d*x
)/2] + I*Sinh[(c + d*x)/2])))/(a*d^3)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 572 vs. \(2 (204 ) = 408\).
time = 3.06, size = 573, normalized size = 2.56

method	result	size
risch	\(-\frac {e^{2} \ln \left ({\mathrm e}^{d x +c}+1\right )}{a d}+\frac {e^{2} \ln \left ({\mathrm e}^{d x +c}-1\right )}{a d}+\frac {2 f^{2} \polylog \left (3, -{\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {2 f^{2} \polylog \left (3, {\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {4 i f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{a \,d^{2}}+\frac {4 i f^{2} c \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {4 i f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{a \,d^{3}}-\frac {2 i f^{2} x^{2}}{a d}-\frac {2 e f \polylog \left (2, -{\mathrm e}^{d x +c}\right )}{a \,d^{2}}+\frac {2 e f \polylog \left (2, {\mathrm e}^{d x +c}\right )}{a \,d^{2}}-\frac {4 i f^{2} c \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{3}}-\frac {4 i f^{2} c x}{a \,d^{2}}-\frac {2 e f c \ln \left ({\mathrm e}^{d x +c}-1\right )}{a \,d^{2}}+\frac {2 \ln \left (1-{\mathrm e}^{d x +c}\right ) c e f}{a \,d^{2}}+\frac {4 i f^{2} \polylog \left (2, -i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {2 f^{2} \polylog \left (2, {\mathrm e}^{d x +c}\right ) x}{a \,d^{2}}-\frac {2 \ln \left ({\mathrm e}^{d x +c}+1\right ) e f x}{a d}+\frac {2 \ln \left (1-{\mathrm e}^{d x +c}\right ) e f x}{a d}-\frac {4 i e f \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{2}}+\frac {4 i e f \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{2}}-\frac {2 f^{2} \polylog \left (2, -{\mathrm e}^{d x +c}\right ) x}{a \,d^{2}}-\frac {2 i f^{2} c^{2}}{a \,d^{3}}+\frac {f^{2} c^{2} \ln \left ({\mathrm e}^{d x +c}-1\right )}{a \,d^{3}}-\frac {f^{2} \ln \left ({\mathrm e}^{d x +c}+1\right ) x^{2}}{a d}+\frac {f^{2} \ln \left (1-{\mathrm e}^{d x +c}\right ) x^{2}}{a d}-\frac {f^{2} \ln \left (1-{\mathrm e}^{d x +c}\right ) c^{2}}{a \,d^{3}}+\frac {2 x^{2} f^{2}+4 e f x +2 e^{2}}{d a \left ({\mathrm e}^{d x +c}-i\right )}\)	\(573\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^2*csch(d*x+c)/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2*f^2*polylog(3,-exp(d*x+c))/a/d^3-2*f^2*polylog(3,exp(d*x+c))/a/d^3-1/a/d*e^2*ln(exp(d*x+c)+1)+1/a/d*e^2*ln(e
xp(d*x+c)-1)+4*I*f^2*polylog(2,-I*exp(d*x+c))/a/d^3-2/a/d^2*e*f*c*ln(exp(d*x+c)-1)+2/a/d^2*ln(1-exp(d*x+c))*c*
e*f-2/a/d*ln(exp(d*x+c)+1)*e*f*x+2/a/d*ln(1-exp(d*x+c))*e*f*x-4*I/a/d^2*e*f*ln(exp(d*x+c))+4*I/a/d^2*e*f*ln(ex
p(d*x+c)-I)-4*I/a/d^2*f^2*c*x+4*I/a/d^3*f^2*c*ln(exp(d*x+c))-4*I/a/d^3*f^2*c*ln(exp(d*x+c)-I)+4*I/a/d^2*f^2*ln
(1+I*exp(d*x+c))*x+4*I/a/d^3*f^2*ln(1+I*exp(d*x+c))*c+2*(f^2*x^2+2*e*f*x+e^2)/d/a/(exp(d*x+c)-I)+2/a/d^2*f^2*p
olylog(2,exp(d*x+c))*x-2/a/d^2*e*f*polylog(2,-exp(d*x+c))+2/a/d^2*e*f*polylog(2,exp(d*x+c))+1/a/d^3*f^2*c^2*ln
(exp(d*x+c)-1)-1/a/d*f^2*ln(exp(d*x+c)+1)*x^2-2/a/d^2*f^2*polylog(2,-exp(d*x+c))*x+1/a/d*f^2*ln(1-exp(d*x+c))*
x^2-1/a/d^3*f^2*ln(1-exp(d*x+c))*c^2-2*I/a/d*f^2*x^2-2*I/a/d^3*f^2*c^2

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Maxima [A]
time = 0.43, size = 351, normalized size = 1.57 \begin {gather*} -\frac {2 i \, f^{2} x^{2}}{a d} - {\left (\frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} - \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} - \frac {2}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d}\right )} e^{2} - \frac {4 i \, f x e}{a d} + \frac {2 \, {\left (f^{2} x^{2} + 2 \, f x e\right )}}{a d e^{\left (d x + c\right )} - i \, a d} - \frac {2 \, {\left (d x \log \left (e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (d x + c\right )}\right )\right )} f e}{a d^{2}} + \frac {2 \, {\left (d x \log \left (-e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (d x + c\right )}\right )\right )} f e}{a d^{2}} + \frac {4 i \, f e \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{a d^{2}} - \frac {{\left (d^{2} x^{2} \log \left (e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (d x + c\right )})\right )} f^{2}}{a d^{3}} + \frac {{\left (d^{2} x^{2} \log \left (-e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (d x + c\right )})\right )} f^{2}}{a d^{3}} + \frac {4 i \, {\left (d x \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right )\right )} f^{2}}{a d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-2*I*f^2*x^2/(a*d) - (log(e^(-d*x - c) + 1)/(a*d) - log(e^(-d*x - c) - 1)/(a*d) - 2/((a*e^(-d*x - c) + I*a)*d)
)*e^2 - 4*I*f*x*e/(a*d) + 2*(f^2*x^2 + 2*f*x*e)/(a*d*e^(d*x + c) - I*a*d) - 2*(d*x*log(e^(d*x + c) + 1) + dilo
g(-e^(d*x + c)))*f*e/(a*d^2) + 2*(d*x*log(-e^(d*x + c) + 1) + dilog(e^(d*x + c)))*f*e/(a*d^2) + 4*I*f*e*log(I*
e^(d*x + c) + 1)/(a*d^2) - (d^2*x^2*log(e^(d*x + c) + 1) + 2*d*x*dilog(-e^(d*x + c)) - 2*polylog(3, -e^(d*x +
c)))*f^2/(a*d^3) + (d^2*x^2*log(-e^(d*x + c) + 1) + 2*d*x*dilog(e^(d*x + c)) - 2*polylog(3, e^(d*x + c)))*f^2/
(a*d^3) + 4*I*(d*x*log(I*e^(d*x + c) + 1) + dilog(-I*e^(d*x + c)))*f^2/(a*d^3)

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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. 571 vs. \(2 (200) = 400\).
time = 0.49, size = 571, normalized size = 2.55 \begin {gather*} \frac {2 \, c^{2} f^{2} - 4 \, c d f e + 2 \, d^{2} e^{2} - 4 \, {\left (-i \, f^{2} e^{\left (d x + c\right )} - f^{2}\right )} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 2 \, {\left (-i \, d f^{2} x - i \, d f e + {\left (d f^{2} x + d f e\right )} e^{\left (d x + c\right )}\right )} {\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) - 2 \, {\left (i \, d f^{2} x + i \, d f e - {\left (d f^{2} x + d f e\right )} e^{\left (d x + c\right )}\right )} {\rm Li}_2\left (e^{\left (d x + c\right )}\right ) - 2 \, {\left (i \, d^{2} f^{2} x^{2} - i \, c^{2} f^{2} + 2 \, {\left (i \, d^{2} f x + i \, c d f\right )} e\right )} e^{\left (d x + c\right )} + {\left (i \, d^{2} f^{2} x^{2} + 2 i \, d^{2} f x e + i \, d^{2} e^{2} - {\left (d^{2} f^{2} x^{2} + 2 \, d^{2} f x e + d^{2} e^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} + 1\right ) - 4 \, {\left (c f^{2} - d f e + {\left (i \, c f^{2} - i \, d f e\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + {\left (-i \, c^{2} f^{2} + 2 i \, c d f e - i \, d^{2} e^{2} + {\left (c^{2} f^{2} - 2 \, c d f e + d^{2} e^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - 1\right ) + 4 \, {\left (d f^{2} x + c f^{2} - {\left (-i \, d f^{2} x - i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\left (-i \, d^{2} f^{2} x^{2} + i \, c^{2} f^{2} - 2 \, {\left (i \, d^{2} f x + i \, c d f\right )} e + {\left (d^{2} f^{2} x^{2} - c^{2} f^{2} + 2 \, {\left (d^{2} f x + c d f\right )} e\right )} e^{\left (d x + c\right )}\right )} \log \left (-e^{\left (d x + c\right )} + 1\right ) + 2 \, {\left (f^{2} e^{\left (d x + c\right )} - i \, f^{2}\right )} {\rm polylog}\left (3, -e^{\left (d x + c\right )}\right ) - 2 \, {\left (f^{2} e^{\left (d x + c\right )} - i \, f^{2}\right )} {\rm polylog}\left (3, e^{\left (d x + c\right )}\right )}{a d^{3} e^{\left (d x + c\right )} - i \, a d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

(2*c^2*f^2 - 4*c*d*f*e + 2*d^2*e^2 - 4*(-I*f^2*e^(d*x + c) - f^2)*dilog(-I*e^(d*x + c)) - 2*(-I*d*f^2*x - I*d*
f*e + (d*f^2*x + d*f*e)*e^(d*x + c))*dilog(-e^(d*x + c)) - 2*(I*d*f^2*x + I*d*f*e - (d*f^2*x + d*f*e)*e^(d*x +
c))*dilog(e^(d*x + c)) - 2*(I*d^2*f^2*x^2 - I*c^2*f^2 + 2*(I*d^2*f*x + I*c*d*f)*e)*e^(d*x + c) + (I*d^2*f^2*x
^2 + 2*I*d^2*f*x*e + I*d^2*e^2 - (d^2*f^2*x^2 + 2*d^2*f*x*e + d^2*e^2)*e^(d*x + c))*log(e^(d*x + c) + 1) - 4*(
c*f^2 - d*f*e + (I*c*f^2 - I*d*f*e)*e^(d*x + c))*log(e^(d*x + c) - I) + (-I*c^2*f^2 + 2*I*c*d*f*e - I*d^2*e^2
+ (c^2*f^2 - 2*c*d*f*e + d^2*e^2)*e^(d*x + c))*log(e^(d*x + c) - 1) + 4*(d*f^2*x + c*f^2 - (-I*d*f^2*x - I*c*f
^2)*e^(d*x + c))*log(I*e^(d*x + c) + 1) + (-I*d^2*f^2*x^2 + I*c^2*f^2 - 2*(I*d^2*f*x + I*c*d*f)*e + (d^2*f^2*x
^2 - c^2*f^2 + 2*(d^2*f*x + c*d*f)*e)*e^(d*x + c))*log(-e^(d*x + c) + 1) + 2*(f^2*e^(d*x + c) - I*f^2)*polylog
(3, -e^(d*x + c)) - 2*(f^2*e^(d*x + c) - I*f^2)*polylog(3, e^(d*x + c)))/(a*d^3*e^(d*x + c) - I*a*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \left (\int \frac {e^{2} \operatorname {csch}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {f^{2} x^{2} \operatorname {csch}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {2 e f x \operatorname {csch}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**2*csch(d*x+c)/(a+I*a*sinh(d*x+c)),x)

[Out]

-I*(Integral(e**2*csch(c + d*x)/(sinh(c + d*x) - I), x) + Integral(f**2*x**2*csch(c + d*x)/(sinh(c + d*x) - I)
, x) + Integral(2*e*f*x*csch(c + d*x)/(sinh(c + d*x) - I), x))/a

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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((f*x+e)^2*csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)^2*csch(d*x + c)/(I*a*sinh(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)^2/(sinh(c + d*x)*(a + a*sinh(c + d*x)*1i)),x)

[Out]

int((e + f*x)^2/(sinh(c + d*x)*(a + a*sinh(c + d*x)*1i)), x)

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