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

Optimal. Leaf size=224 \[ \frac {4 i f^2 \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac {2 f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac {4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}-\frac {i (e+f x)^2}{a d} \]

[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 = 0.35, 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 = {5575, 4182, 2531, 2282, 6589, 3318, 4184, 3716, 2190, 2279, 2391} \[ -\frac {2 f (e+f x) \text {PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {PolyLog}\left (2,e^{c+d x}\right )}{a d^2}+\frac {4 i f^2 \text {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac {2 f^2 \text {PolyLog}\left (3,-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {PolyLog}\left (3,e^{c+d x}\right )}{a d^3}+\frac {4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}-\frac {i (e+f x)^2}{a d} \]

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 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 3318

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*(e + (Pi*a)/(2*b)))/2 + (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 3716

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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

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 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 5575

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 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 \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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 = 4.13, size = 275, normalized size = 1.23 \[ \frac {d^2 (e+f x)^2 \log \left (1-e^{c+d x}\right )-d^2 (e+f x)^2 \log \left (e^{c+d x}+1\right )-\frac {2 i d^2 \sinh \left (\frac {d x}{2}\right ) (e+f x)^2}{\left (\cosh \left (\frac {c}{2}\right )+i \sinh \left (\frac {c}{2}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 d (e+f x) \left (2 \left (e^c-i\right ) f \log \left (1-i e^{-c-d x}\right )-i d (e+f x)\right )-4 \left (e^c-i\right ) f^2 \text {Li}_2\left (i e^{-c-d x}\right )}{-1-i e^c}-2 d f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )+2 d f (e+f x) \text {Li}_2\left (e^{c+d x}\right )+2 f^2 \text {Li}_3\left (-e^{c+d x}\right )-2 f^2 \text {Li}_3\left (e^{c+d x}\right )}{a d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(e + f*x)^2*Log[1 - E^(c + d*x)] - d^2*(e + f*x)^2*Log[1 + E^(c + d*x)] + (2*d*(e + f*x)*((-I)*d*(e + f*x
) + 2*(-I + E^c)*f*Log[1 - I*E^(-c - d*x)]) - 4*(-I + E^c)*f^2*PolyLog[2, I*E^(-c - d*x)])/(-1 - I*E^c) - 2*d*
f*(e + f*x)*PolyLog[2, -E^(c + d*x)] + 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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fricas [C]  time = 0.45, size = 557, normalized size = 2.49 \[ \frac {2 \, d^{2} e^{2} - 4 \, c d e f + 2 \, c^{2} f^{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 ) + {\left (2 i \, d f^{2} x + 2 i \, d e f - 2 \, {\left (d f^{2} x + d e f\right )} e^{\left (d x + c\right )}\right )} {\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) + {\left (-2 i \, d f^{2} x - 2 i \, d e f + 2 \, {\left (d f^{2} x + d e f\right )} e^{\left (d x + c\right )}\right )} {\rm Li}_2\left (e^{\left (d x + c\right )}\right ) + {\left (-2 i \, d^{2} f^{2} x^{2} - 4 i \, d^{2} e f x - 4 i \, c d e f + 2 i \, c^{2} f^{2}\right )} e^{\left (d x + c\right )} + {\left (i \, d^{2} f^{2} x^{2} + 2 i \, d^{2} e f x + i \, d^{2} e^{2} - {\left (d^{2} f^{2} x^{2} + 2 \, d^{2} e f x + d^{2} e^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} + 1\right ) + {\left (4 \, d e f - 4 \, c f^{2} + {\left (4 i \, d e f - 4 i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + {\left (-i \, d^{2} e^{2} + 2 i \, c d e f - i \, c^{2} f^{2} + {\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - 1\right ) + {\left (4 \, d f^{2} x + 4 \, c f^{2} + {\left (4 i \, d f^{2} x + 4 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} - 2 i \, d^{2} e f x - 2 i \, c d e f + i \, c^{2} f^{2} + {\left (d^{2} f^{2} x^{2} + 2 \, d^{2} e f x + 2 \, c d e f - c^{2} f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (-e^{\left (d x + c\right )} + 1\right ) + {\left (2 \, f^{2} e^{\left (d x + c\right )} - 2 i \, f^{2}\right )} {\rm polylog}\left (3, -e^{\left (d x + c\right )}\right ) - {\left (2 \, f^{2} e^{\left (d x + c\right )} - 2 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}} \]

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

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

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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maple [B]  time = 0.21, size = 573, normalized size = 2.56 \[ \frac {4 i \ln \left ({\mathrm e}^{d x +c}-i\right ) e f}{a \,d^{2}}-\frac {4 i \ln \left ({\mathrm e}^{d x +c}\right ) e f}{a \,d^{2}}-\frac {4 i f^{2} c x}{a \,d^{2}}+\frac {4 i f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{a \,d^{2}}+\frac {4 i f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{a \,d^{3}}-\frac {4 i f^{2} c \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{3}}+\frac {4 i f^{2} c \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {2 \ln \left (1-{\mathrm e}^{d x +c}\right ) e f x}{a d}+\frac {2 \ln \left (1-{\mathrm e}^{d x +c}\right ) c e f}{a \,d^{2}}-\frac {2 e f c \ln \left ({\mathrm e}^{d x +c}-1\right )}{a \,d^{2}}-\frac {2 \ln \left ({\mathrm e}^{d x +c}+1\right ) e f x}{a d}+\frac {2 f^{2} \polylog \left (3, -{\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {2 f^{2} \polylog \left (3, {\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {f^{2} c^{2} \ln \left (1-{\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {f^{2} \ln \left ({\mathrm e}^{d x +c}+1\right ) x^{2}}{a d}+\frac {f^{2} \ln \left (1-{\mathrm e}^{d x +c}\right ) x^{2}}{a d}-\frac {e^{2} \ln \left ({\mathrm e}^{d x +c}+1\right )}{a d}+\frac {e^{2} \ln \left ({\mathrm e}^{d x +c}-1\right )}{a d}+\frac {2 e f \polylog \left (2, {\mathrm e}^{d x +c}\right )}{a \,d^{2}}+\frac {f^{2} c^{2} \ln \left ({\mathrm e}^{d x +c}-1\right )}{a \,d^{3}}-\frac {2 f^{2} \polylog \left (2, -{\mathrm e}^{d x +c}\right ) x}{a \,d^{2}}+\frac {2 f^{2} \polylog \left (2, {\mathrm e}^{d x +c}\right ) x}{a \,d^{2}}-\frac {2 e f \polylog \left (2, -{\mathrm e}^{d x +c}\right )}{a \,d^{2}}+\frac {2 x^{2} f^{2}+4 e f x +2 e^{2}}{d a \left ({\mathrm e}^{d x +c}-i\right )}-\frac {2 i f^{2} x^{2}}{a d}-\frac {2 i f^{2} c^{2}}{a \,d^{3}}+\frac {4 i f^{2} \polylog \left (2, -i {\mathrm e}^{d x +c}\right )}{a \,d^{3}} \]

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)

[Out]

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

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maxima [A]  time = 0.76, size = 347, normalized size = 1.55 \[ -e^{2} {\left (\frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} - \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} - \frac {2}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d}\right )} - \frac {2 i \, f^{2} x^{2}}{a d} - \frac {4 i \, e f x}{a d} + \frac {2 \, {\left (f^{2} x^{2} + 2 \, e f x\right )}}{a d e^{\left (d x + c\right )} - i \, a d} - \frac {2 \, {\left (d x \log \left (e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (d x + c\right )}\right )\right )} e f}{a d^{2}} + \frac {2 \, {\left (d x \log \left (-e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (d x + c\right )}\right )\right )} e f}{a d^{2}} + \frac {4 i \, e f \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{a d^{2}} - \frac {{\left (d^{2} x^{2} \log \left (e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (d x + c\right )})\right )} f^{2}}{a d^{3}} + \frac {{\left (d^{2} x^{2} \log \left (-e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (d x + c\right )})\right )} f^{2}}{a d^{3}} + \frac {4 i \, {\left (d x \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right )\right )} f^{2}}{a d^{3}} \]

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]

-e^2*(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)) - 2*I*f^2*x^2/
(a*d) - 4*I*e*f*x/(a*d) + 2*(f^2*x^2 + 2*e*f*x)/(a*d*e^(d*x + c) - I*a*d) - 2*(d*x*log(e^(d*x + c) + 1) + dilo
g(-e^(d*x + c)))*e*f/(a*d^2) + 2*(d*x*log(-e^(d*x + c) + 1) + dilog(e^(d*x + c)))*e*f/(a*d^2) + 4*I*e*f*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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (e+f\,x\right )}^2}{\mathrm {sinh}\left (c+d\,x\right )\,\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \]

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

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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