3.193 \(\int \frac {(e+f x)^3 \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=241 \[ -\frac {12 f^3 \text {Li}_3\left (-i e^{c+d x}\right )}{a d^4}+\frac {6 i f^3 \sinh (c+d x)}{a d^4}+\frac {12 f^2 (e+f x) \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}-\frac {6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}+\frac {6 f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {3 i f (e+f x)^2 \sinh (c+d x)}{a d^2}-\frac {i (e+f x)^3 \cosh (c+d x)}{a d}-\frac {(e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}-\frac {(e+f x)^3}{a d}+\frac {(e+f x)^4}{4 a f} \]

[Out]

-(f*x+e)^3/a/d+1/4*(f*x+e)^4/a/f-6*I*f^2*(f*x+e)*cosh(d*x+c)/a/d^3-I*(f*x+e)^3*cosh(d*x+c)/a/d+6*f*(f*x+e)^2*l
n(1+I*exp(d*x+c))/a/d^2+12*f^2*(f*x+e)*polylog(2,-I*exp(d*x+c))/a/d^3-12*f^3*polylog(3,-I*exp(d*x+c))/a/d^4+6*
I*f^3*sinh(d*x+c)/a/d^4+3*I*f*(f*x+e)^2*sinh(d*x+c)/a/d^2-(f*x+e)^3*tanh(1/2*c+1/4*I*Pi+1/2*d*x)/a/d

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Rubi [A]  time = 0.53, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {5557, 3296, 2637, 32, 3318, 4184, 3716, 2190, 2531, 2282, 6589} \[ \frac {12 f^2 (e+f x) \text {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}-\frac {12 f^3 \text {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^4}-\frac {6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}+\frac {6 f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {3 i f (e+f x)^2 \sinh (c+d x)}{a d^2}+\frac {6 i f^3 \sinh (c+d x)}{a d^4}-\frac {i (e+f x)^3 \cosh (c+d x)}{a d}-\frac {(e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}-\frac {(e+f x)^3}{a d}+\frac {(e+f x)^4}{4 a f} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Sinh[c + d*x]^2)/(a + I*a*Sinh[c + d*x]),x]

[Out]

-((e + f*x)^3/(a*d)) + (e + f*x)^4/(4*a*f) - ((6*I)*f^2*(e + f*x)*Cosh[c + d*x])/(a*d^3) - (I*(e + f*x)^3*Cosh
[c + d*x])/(a*d) + (6*f*(e + f*x)^2*Log[1 + I*E^(c + d*x)])/(a*d^2) + (12*f^2*(e + f*x)*PolyLog[2, (-I)*E^(c +
d*x)])/(a*d^3) - (12*f^3*PolyLog[3, (-I)*E^(c + d*x)])/(a*d^4) + ((6*I)*f^3*Sinh[c + d*x])/(a*d^4) + ((3*I)*f
*(e + f*x)^2*Sinh[c + d*x])/(a*d^2) - ((e + f*x)^3*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(a*d)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 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 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, 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 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 5557

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/b, Int[(e + f*x)^m*Sinh[c + d*x]^(n - 1), x], x] - Dist[a/b, Int[((e + f*x)^m*Sinh[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)^3 \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx &=i \int \frac {(e+f x)^3 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx-\frac {i \int (e+f x)^3 \sinh (c+d x) \, dx}{a}\\ &=-\frac {i (e+f x)^3 \cosh (c+d x)}{a d}+\frac {\int (e+f x)^3 \, dx}{a}+\frac {(3 i f) \int (e+f x)^2 \cosh (c+d x) \, dx}{a d}-\int \frac {(e+f x)^3}{a+i a \sinh (c+d x)} \, dx\\ &=\frac {(e+f x)^4}{4 a f}-\frac {i (e+f x)^3 \cosh (c+d x)}{a d}+\frac {3 i f (e+f x)^2 \sinh (c+d x)}{a d^2}-\frac {\int (e+f x)^3 \csc ^2\left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {i d x}{2}\right ) \, dx}{2 a}-\frac {\left (6 i f^2\right ) \int (e+f x) \sinh (c+d x) \, dx}{a d^2}\\ &=\frac {(e+f x)^4}{4 a f}-\frac {6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}-\frac {i (e+f x)^3 \cosh (c+d x)}{a d}+\frac {3 i f (e+f x)^2 \sinh (c+d x)}{a d^2}-\frac {(e+f x)^3 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}+\frac {(3 f) \int (e+f x)^2 \coth \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx}{a d}+\frac {\left (6 i f^3\right ) \int \cosh (c+d x) \, dx}{a d^3}\\ &=-\frac {(e+f x)^3}{a d}+\frac {(e+f x)^4}{4 a f}-\frac {6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}-\frac {i (e+f x)^3 \cosh (c+d x)}{a d}+\frac {6 i f^3 \sinh (c+d x)}{a d^4}+\frac {3 i f (e+f x)^2 \sinh (c+d x)}{a d^2}-\frac {(e+f x)^3 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}+\frac {(6 i f) \int \frac {e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )} (e+f x)^2}{1+i e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )}} \, dx}{a d}\\ &=-\frac {(e+f x)^3}{a d}+\frac {(e+f x)^4}{4 a f}-\frac {6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}-\frac {i (e+f x)^3 \cosh (c+d x)}{a d}+\frac {6 f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {6 i f^3 \sinh (c+d x)}{a d^4}+\frac {3 i f (e+f x)^2 \sinh (c+d x)}{a d^2}-\frac {(e+f x)^3 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {\left (12 f^2\right ) \int (e+f x) \log \left (1+i e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \, dx}{a d^2}\\ &=-\frac {(e+f x)^3}{a d}+\frac {(e+f x)^4}{4 a f}-\frac {6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}-\frac {i (e+f x)^3 \cosh (c+d x)}{a d}+\frac {6 f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {12 f^2 (e+f x) \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac {6 i f^3 \sinh (c+d x)}{a d^4}+\frac {3 i f (e+f x)^2 \sinh (c+d x)}{a d^2}-\frac {(e+f x)^3 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {\left (12 f^3\right ) \int \text {Li}_2\left (-i e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \, dx}{a d^3}\\ &=-\frac {(e+f x)^3}{a d}+\frac {(e+f x)^4}{4 a f}-\frac {6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}-\frac {i (e+f x)^3 \cosh (c+d x)}{a d}+\frac {6 f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {12 f^2 (e+f x) \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac {6 i f^3 \sinh (c+d x)}{a d^4}+\frac {3 i f (e+f x)^2 \sinh (c+d x)}{a d^2}-\frac {(e+f x)^3 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {\left (12 f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{2 \left (\frac {c}{2}+\frac {d x}{2}\right )}\right )}{a d^4}\\ &=-\frac {(e+f x)^3}{a d}+\frac {(e+f x)^4}{4 a f}-\frac {6 i f^2 (e+f x) \cosh (c+d x)}{a d^3}-\frac {i (e+f x)^3 \cosh (c+d x)}{a d}+\frac {6 f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac {12 f^2 (e+f x) \text {Li}_2\left (-i e^{c+d x}\right )}{a d^3}-\frac {12 f^3 \text {Li}_3\left (-i e^{c+d x}\right )}{a d^4}+\frac {6 i f^3 \sinh (c+d x)}{a d^4}+\frac {3 i f (e+f x)^2 \sinh (c+d x)}{a d^2}-\frac {(e+f x)^3 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}\\ \end {align*}

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Mathematica [B]  time = 6.53, size = 857, normalized size = 3.56 \[ \frac {\frac {i f^3 x^4 \sinh \left (c+\frac {d x}{2}\right ) d^4+4 i e f^2 x^3 \sinh \left (c+\frac {d x}{2}\right ) d^4+6 i e^2 f x^2 \sinh \left (c+\frac {d x}{2}\right ) d^4+4 i e^3 x \sinh \left (c+\frac {d x}{2}\right ) d^4-10 e^3 \sinh \left (\frac {d x}{2}\right ) d^3-10 f^3 x^3 \sinh \left (\frac {d x}{2}\right ) d^3-30 e f^2 x^2 \sinh \left (\frac {d x}{2}\right ) d^3-30 e^2 f x \sinh \left (\frac {d x}{2}\right ) d^3+2 e^3 \sinh \left (2 c+\frac {3 d x}{2}\right ) d^3+2 f^3 x^3 \sinh \left (2 c+\frac {3 d x}{2}\right ) d^3+6 e f^2 x^2 \sinh \left (2 c+\frac {3 d x}{2}\right ) d^3+6 e^2 f x \sinh \left (2 c+\frac {3 d x}{2}\right ) d^3-6 f^3 x^2 \cosh \left (2 c+\frac {3 d x}{2}\right ) d^2-6 e^2 f \cosh \left (2 c+\frac {3 d x}{2}\right ) d^2-12 e f^2 x \cosh \left (2 c+\frac {3 d x}{2}\right ) d^2+6 i f^3 x^2 \sinh \left (c+\frac {d x}{2}\right ) d^2+6 i e^2 f \sinh \left (c+\frac {d x}{2}\right ) d^2+12 i e f^2 x \sinh \left (c+\frac {d x}{2}\right ) d^2+6 i f^3 x^2 \sinh \left (c+\frac {3 d x}{2}\right ) d^2+6 i e^2 f \sinh \left (c+\frac {3 d x}{2}\right ) d^2+12 i e f^2 x \sinh \left (c+\frac {3 d x}{2}\right ) d^2-2 i (e+f x) \left (6 f^2+d^2 (e+f x)^2\right ) \cosh \left (c+\frac {d x}{2}\right ) d-2 i (e+f x) \left (6 f^2+d^2 (e+f x)^2\right ) \cosh \left (c+\frac {3 d x}{2}\right ) d-12 e f^2 \sinh \left (\frac {d x}{2}\right ) d-12 f^3 x \sinh \left (\frac {d x}{2}\right ) d+12 e f^2 \sinh \left (2 c+\frac {3 d x}{2}\right ) d+12 f^3 x \sinh \left (2 c+\frac {3 d x}{2}\right ) d+\left (x \left (4 e^3+6 f x e^2+4 f^2 x^2 e+f^3 x^3\right ) d^4+6 f (e+f x)^2 d^2+12 f^3\right ) \cosh \left (\frac {d x}{2}\right )-12 f^3 \cosh \left (2 c+\frac {3 d x}{2}\right )+12 i f^3 \sinh \left (c+\frac {d x}{2}\right )+12 i f^3 \sinh \left (c+\frac {3 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 )}-\frac {8 i \left (d^3 (e+f x)^3+3 d^2 \left (1+i e^c\right ) f \log \left (1-i e^{-c-d x}\right ) (e+f x)^2+6 i \left (i-e^c\right ) f^2 \left (d (e+f x) \text {Li}_2\left (i e^{-c-d x}\right )+f \text {Li}_3\left (i e^{-c-d x}\right )\right )\right )}{-i+e^c}}{4 a d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^3*Sinh[c + d*x]^2)/(a + I*a*Sinh[c + d*x]),x]

[Out]

(((-8*I)*(d^3*(e + f*x)^3 + 3*d^2*(1 + I*E^c)*f*(e + f*x)^2*Log[1 - I*E^(-c - d*x)] + (6*I)*(I - E^c)*f^2*(d*(
e + f*x)*PolyLog[2, I*E^(-c - d*x)] + f*PolyLog[3, I*E^(-c - d*x)])))/(-I + E^c) + ((12*f^3 + 6*d^2*f*(e + f*x
)^2 + d^4*x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3))*Cosh[(d*x)/2] - (2*I)*d*(e + f*x)*(6*f^2 + d^2*(e + f
*x)^2)*Cosh[c + (d*x)/2] - (2*I)*d*(e + f*x)*(6*f^2 + d^2*(e + f*x)^2)*Cosh[c + (3*d*x)/2] - 6*d^2*e^2*f*Cosh[
2*c + (3*d*x)/2] - 12*f^3*Cosh[2*c + (3*d*x)/2] - 12*d^2*e*f^2*x*Cosh[2*c + (3*d*x)/2] - 6*d^2*f^3*x^2*Cosh[2*
c + (3*d*x)/2] - 10*d^3*e^3*Sinh[(d*x)/2] - 12*d*e*f^2*Sinh[(d*x)/2] - 30*d^3*e^2*f*x*Sinh[(d*x)/2] - 12*d*f^3
*x*Sinh[(d*x)/2] - 30*d^3*e*f^2*x^2*Sinh[(d*x)/2] - 10*d^3*f^3*x^3*Sinh[(d*x)/2] + (6*I)*d^2*e^2*f*Sinh[c + (d
*x)/2] + (12*I)*f^3*Sinh[c + (d*x)/2] + (4*I)*d^4*e^3*x*Sinh[c + (d*x)/2] + (12*I)*d^2*e*f^2*x*Sinh[c + (d*x)/
2] + (6*I)*d^4*e^2*f*x^2*Sinh[c + (d*x)/2] + (6*I)*d^2*f^3*x^2*Sinh[c + (d*x)/2] + (4*I)*d^4*e*f^2*x^3*Sinh[c
+ (d*x)/2] + I*d^4*f^3*x^4*Sinh[c + (d*x)/2] + (6*I)*d^2*e^2*f*Sinh[c + (3*d*x)/2] + (12*I)*f^3*Sinh[c + (3*d*
x)/2] + (12*I)*d^2*e*f^2*x*Sinh[c + (3*d*x)/2] + (6*I)*d^2*f^3*x^2*Sinh[c + (3*d*x)/2] + 2*d^3*e^3*Sinh[2*c +
(3*d*x)/2] + 12*d*e*f^2*Sinh[2*c + (3*d*x)/2] + 6*d^3*e^2*f*x*Sinh[2*c + (3*d*x)/2] + 12*d*f^3*x*Sinh[2*c + (3
*d*x)/2] + 6*d^3*e*f^2*x^2*Sinh[2*c + (3*d*x)/2] + 2*d^3*f^3*x^3*Sinh[2*c + (3*d*x)/2])/((Cosh[c/2] + I*Sinh[c
/2])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])))/(4*a*d^4)

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fricas [C]  time = 0.52, size = 816, normalized size = 3.39 \[ -\frac {2 \, d^{3} f^{3} x^{3} + 2 \, d^{3} e^{3} + 6 \, d^{2} e^{2} f + 12 \, d e f^{2} + 12 \, f^{3} + 6 \, {\left (d^{3} e f^{2} + d^{2} f^{3}\right )} x^{2} + 6 \, {\left (d^{3} e^{2} f + 2 \, d^{2} e f^{2} + 2 \, d f^{3}\right )} x - {\left (48 \, {\left (d f^{3} x + d e f^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )} + {\left (-48 i \, d f^{3} x - 48 i \, d e f^{2}\right )} e^{\left (d x + c\right )}\right )} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - {\left (-2 i \, d^{3} f^{3} x^{3} - 2 i \, d^{3} e^{3} + 6 i \, d^{2} e^{2} f - 12 i \, d e f^{2} + 12 i \, f^{3} + {\left (-6 i \, d^{3} e f^{2} + 6 i \, d^{2} f^{3}\right )} x^{2} + {\left (-6 i \, d^{3} e^{2} f + 12 i \, d^{2} e f^{2} - 12 i \, d f^{3}\right )} x\right )} e^{\left (3 \, d x + 3 \, c\right )} - {\left (d^{4} f^{3} x^{4} - 2 \, d^{3} e^{3} - 6 \, {\left (4 \, c - 1\right )} d^{2} e^{2} f + 12 \, {\left (2 \, c^{2} - 1\right )} d e f^{2} - 4 \, {\left (2 \, c^{3} - 3\right )} f^{3} + 2 \, {\left (2 \, d^{4} e f^{2} - 5 \, d^{3} f^{3}\right )} x^{3} + 6 \, {\left (d^{4} e^{2} f - 5 \, d^{3} e f^{2} + d^{2} f^{3}\right )} x^{2} + 2 \, {\left (2 \, d^{4} e^{3} - 15 \, d^{3} e^{2} f + 6 \, d^{2} e f^{2} - 6 \, d f^{3}\right )} x\right )} e^{\left (2 \, d x + 2 \, c\right )} - {\left (-i \, d^{4} f^{3} x^{4} - 10 i \, d^{3} e^{3} + {\left (24 i \, c - 6 i\right )} d^{2} e^{2} f + {\left (-24 i \, c^{2} - 12 i\right )} d e f^{2} + {\left (8 i \, c^{3} - 12 i\right )} f^{3} + {\left (-4 i \, d^{4} e f^{2} - 2 i \, d^{3} f^{3}\right )} x^{3} + {\left (-6 i \, d^{4} e^{2} f - 6 i \, d^{3} e f^{2} - 6 i \, d^{2} f^{3}\right )} x^{2} + {\left (-4 i \, d^{4} e^{3} - 6 i \, d^{3} e^{2} f - 12 i \, d^{2} e f^{2} - 12 i \, d f^{3}\right )} x\right )} e^{\left (d x + c\right )} - {\left (24 \, {\left (d^{2} e^{2} f - 2 \, c d e f^{2} + c^{2} f^{3}\right )} e^{\left (2 \, d x + 2 \, c\right )} + {\left (-24 i \, d^{2} e^{2} f + 48 i \, c d e f^{2} - 24 i \, c^{2} f^{3}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - {\left (24 \, {\left (d^{2} f^{3} x^{2} + 2 \, d^{2} e f^{2} x + 2 \, c d e f^{2} - c^{2} f^{3}\right )} e^{\left (2 \, d x + 2 \, c\right )} + {\left (-24 i \, d^{2} f^{3} x^{2} - 48 i \, d^{2} e f^{2} x - 48 i \, c d e f^{2} + 24 i \, c^{2} f^{3}\right )} e^{\left (d x + c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\left (48 \, f^{3} e^{\left (2 \, d x + 2 \, c\right )} - 48 i \, f^{3} e^{\left (d x + c\right )}\right )} {\rm polylog}\left (3, -i \, e^{\left (d x + c\right )}\right )}{4 \, a d^{4} e^{\left (2 \, d x + 2 \, c\right )} - 4 i \, a d^{4} e^{\left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sinh(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

-(2*d^3*f^3*x^3 + 2*d^3*e^3 + 6*d^2*e^2*f + 12*d*e*f^2 + 12*f^3 + 6*(d^3*e*f^2 + d^2*f^3)*x^2 + 6*(d^3*e^2*f +
2*d^2*e*f^2 + 2*d*f^3)*x - (48*(d*f^3*x + d*e*f^2)*e^(2*d*x + 2*c) + (-48*I*d*f^3*x - 48*I*d*e*f^2)*e^(d*x +
c))*dilog(-I*e^(d*x + c)) - (-2*I*d^3*f^3*x^3 - 2*I*d^3*e^3 + 6*I*d^2*e^2*f - 12*I*d*e*f^2 + 12*I*f^3 + (-6*I*
d^3*e*f^2 + 6*I*d^2*f^3)*x^2 + (-6*I*d^3*e^2*f + 12*I*d^2*e*f^2 - 12*I*d*f^3)*x)*e^(3*d*x + 3*c) - (d^4*f^3*x^
4 - 2*d^3*e^3 - 6*(4*c - 1)*d^2*e^2*f + 12*(2*c^2 - 1)*d*e*f^2 - 4*(2*c^3 - 3)*f^3 + 2*(2*d^4*e*f^2 - 5*d^3*f^
3)*x^3 + 6*(d^4*e^2*f - 5*d^3*e*f^2 + d^2*f^3)*x^2 + 2*(2*d^4*e^3 - 15*d^3*e^2*f + 6*d^2*e*f^2 - 6*d*f^3)*x)*e
^(2*d*x + 2*c) - (-I*d^4*f^3*x^4 - 10*I*d^3*e^3 + (24*I*c - 6*I)*d^2*e^2*f + (-24*I*c^2 - 12*I)*d*e*f^2 + (8*I
*c^3 - 12*I)*f^3 + (-4*I*d^4*e*f^2 - 2*I*d^3*f^3)*x^3 + (-6*I*d^4*e^2*f - 6*I*d^3*e*f^2 - 6*I*d^2*f^3)*x^2 + (
-4*I*d^4*e^3 - 6*I*d^3*e^2*f - 12*I*d^2*e*f^2 - 12*I*d*f^3)*x)*e^(d*x + c) - (24*(d^2*e^2*f - 2*c*d*e*f^2 + c^
2*f^3)*e^(2*d*x + 2*c) + (-24*I*d^2*e^2*f + 48*I*c*d*e*f^2 - 24*I*c^2*f^3)*e^(d*x + c))*log(e^(d*x + c) - I) -
(24*(d^2*f^3*x^2 + 2*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3)*e^(2*d*x + 2*c) + (-24*I*d^2*f^3*x^2 - 48*I*d^2*e*f
^2*x - 48*I*c*d*e*f^2 + 24*I*c^2*f^3)*e^(d*x + c))*log(I*e^(d*x + c) + 1) + (48*f^3*e^(2*d*x + 2*c) - 48*I*f^3
*e^(d*x + c))*polylog(3, -I*e^(d*x + c)))/(4*a*d^4*e^(2*d*x + 2*c) - 4*I*a*d^4*e^(d*x + c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{3} \sinh \left (d x + c\right )^{2}}{i \, a \sinh \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sinh(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)^3*sinh(d*x + c)^2/(I*a*sinh(d*x + c) + a), x)

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maple [B]  time = 0.29, size = 688, normalized size = 2.85 \[ \frac {12 f^{2} e \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{a \,d^{2}}+\frac {12 f^{2} e \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{a \,d^{3}}-\frac {12 f^{2} e c \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{3}}+\frac {12 f^{2} e c \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {12 f^{2} e c x}{a \,d^{2}}-\frac {2 f^{3} x^{3}}{a d}+\frac {4 f^{3} c^{3}}{a \,d^{4}}-\frac {12 f^{3} \polylog \left (3, -i {\mathrm e}^{d x +c}\right )}{a \,d^{4}}+\frac {e \,f^{2} x^{3}}{a}+\frac {3 e^{2} f \,x^{2}}{2 a}-\frac {6 f \ln \left ({\mathrm e}^{d x +c}\right ) e^{2}}{a \,d^{2}}+\frac {6 f^{3} c^{2} x}{a \,d^{3}}-\frac {i \left (f^{3} x^{3} d^{3}+3 d^{3} e \,f^{2} x^{2}+3 d^{3} e^{2} f x -3 d^{2} f^{3} x^{2}+d^{3} e^{3}-6 d^{2} e \,f^{2} x -3 d^{2} e^{2} f +6 d \,f^{3} x +6 d e \,f^{2}-6 f^{3}\right ) {\mathrm e}^{d x +c}}{2 a \,d^{4}}-\frac {2 i \left (x^{3} f^{3}+3 e \,f^{2} x^{2}+3 e^{2} f x +e^{3}\right )}{d a \left ({\mathrm e}^{d x +c}-i\right )}-\frac {i \left (f^{3} x^{3} d^{3}+3 d^{3} e \,f^{2} x^{2}+3 d^{3} e^{2} f x +3 d^{2} f^{3} x^{2}+d^{3} e^{3}+6 d^{2} e \,f^{2} x +3 d^{2} e^{2} f +6 d \,f^{3} x +6 d e \,f^{2}+6 f^{3}\right ) {\mathrm e}^{-d x -c}}{2 a \,d^{4}}+\frac {6 f^{3} c^{2} \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{4}}+\frac {6 f \ln \left ({\mathrm e}^{d x +c}-i\right ) e^{2}}{a \,d^{2}}+\frac {12 f^{3} \polylog \left (2, -i {\mathrm e}^{d x +c}\right ) x}{a \,d^{3}}+\frac {12 f^{2} e \polylog \left (2, -i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {6 f^{3} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x^{2}}{a \,d^{2}}-\frac {6 f^{3} \ln \left (1+i {\mathrm e}^{d x +c}\right ) c^{2}}{a \,d^{4}}-\frac {6 f^{2} e \,c^{2}}{a \,d^{3}}-\frac {6 f^{2} e \,x^{2}}{a d}-\frac {6 f^{3} c^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{4}}+\frac {x^{4} f^{3}}{4 a}+\frac {e^{3} x}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*sinh(d*x+c)^2/(a+I*a*sinh(d*x+c)),x)

[Out]

12/a/d^3*f^2*e*c*ln(exp(d*x+c))+12/a/d^2*f^2*e*ln(1+I*exp(d*x+c))*x+12/a/d^3*f^2*e*ln(1+I*exp(d*x+c))*c-12/a/d
^2*f^2*e*c*x-12/a/d^3*f^2*e*c*ln(exp(d*x+c)-I)-12*f^3*polylog(3,-I*exp(d*x+c))/a/d^4-1/2*I*(d^3*f^3*x^3+3*d^3*
e*f^2*x^2+3*d^3*e^2*f*x-3*d^2*f^3*x^2+d^3*e^3-6*d^2*e*f^2*x-3*d^2*e^2*f+6*d*f^3*x+6*d*e*f^2-6*f^3)/a/d^4*exp(d
*x+c)-2*I*(f^3*x^3+3*e*f^2*x^2+3*e^2*f*x+e^3)/d/a/(exp(d*x+c)-I)-2/a/d*f^3*x^3+4/a/d^4*f^3*c^3-1/2*I*(d^3*f^3*
x^3+3*d^3*e*f^2*x^2+3*d^3*e^2*f*x+3*d^2*f^3*x^2+d^3*e^3+6*d^2*e*f^2*x+3*d^2*e^2*f+6*d*f^3*x+6*d*e*f^2+6*f^3)/a
/d^4*exp(-d*x-c)+1/a*e*f^2*x^3+3/2/a*e^2*f*x^2+12/a/d^3*f^3*polylog(2,-I*exp(d*x+c))*x+6/a/d^4*f^3*c^2*ln(exp(
d*x+c)-I)-6/a/d^2*f*ln(exp(d*x+c))*e^2+6/a/d^3*f^3*c^2*x+6/a/d^2*f*ln(exp(d*x+c)-I)*e^2-6/a/d^3*f^2*e*c^2-6/a/
d*f^2*e*x^2+12/a/d^3*f^2*e*polylog(2,-I*exp(d*x+c))-6/a/d^4*f^3*c^2*ln(exp(d*x+c))+6/a/d^2*f^3*ln(1+I*exp(d*x+
c))*x^2-6/a/d^4*f^3*ln(1+I*exp(d*x+c))*c^2+1/4/a*x^4*f^3+1/a*e^3*x

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maxima [B]  time = 0.71, size = 670, normalized size = 2.78 \[ -\frac {3}{4} \, e^{2} f {\left (\frac {4 \, x e^{\left (d x + c\right )}}{a d e^{\left (d x + c\right )} - i \, a d} - \frac {-2 i \, d^{2} x^{2} e^{c} - 2 i \, d x e^{c} - {\left (2 i \, d x e^{\left (3 \, c\right )} - 2 i \, e^{\left (3 \, c\right )}\right )} e^{\left (2 \, d x\right )} + 2 \, {\left (d^{2} x^{2} e^{\left (2 \, c\right )} - 3 \, d x e^{\left (2 \, c\right )} + e^{\left (2 \, c\right )}\right )} e^{\left (d x\right )} - 2 \, {\left (d x + 1\right )} e^{\left (-d x\right )} - 2 i \, e^{c}}{a d^{2} e^{\left (d x + 2 \, c\right )} - i \, a d^{2} e^{c}} - \frac {8 \, \log \left ({\left (e^{\left (d x + c\right )} - i\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} + \frac {1}{2} \, e^{3} {\left (\frac {2 \, {\left (d x + c\right )}}{a d} + \frac {-5 i \, e^{\left (-d x - c\right )} + 1}{{\left (i \, a e^{\left (-d x - c\right )} + a e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} - \frac {i \, e^{\left (-d x - c\right )}}{a d}\right )} + \frac {-i \, d^{4} f^{3} x^{4} - 12 i \, d e f^{2} - {\left (4 i \, d^{4} e f^{2} + 10 i \, d^{3} f^{3}\right )} x^{3} - 12 i \, f^{3} - {\left (30 i \, d^{3} e f^{2} + 6 i \, d^{2} f^{3}\right )} x^{2} - {\left (12 i \, d^{2} e f^{2} + 12 i \, d f^{3}\right )} x - {\left (2 i \, d^{3} f^{3} x^{3} e^{\left (2 \, c\right )} + {\left (6 i \, d^{3} e f^{2} - 6 i \, d^{2} f^{3}\right )} x^{2} e^{\left (2 \, c\right )} + {\left (-12 i \, d^{2} e f^{2} + 12 i \, d f^{3}\right )} x e^{\left (2 \, c\right )} + {\left (12 i \, d e f^{2} - 12 i \, f^{3}\right )} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} + {\left (d^{4} f^{3} x^{4} e^{c} + 2 \, {\left (2 \, d^{4} e f^{2} - d^{3} f^{3}\right )} x^{3} e^{c} - 6 \, {\left (d^{3} e f^{2} - d^{2} f^{3}\right )} x^{2} e^{c} + 12 \, {\left (d^{2} e f^{2} - d f^{3}\right )} x e^{c} - 12 \, {\left (d e f^{2} - f^{3}\right )} e^{c}\right )} e^{\left (d x\right )}}{4 \, {\left (a d^{4} e^{\left (d x + c\right )} - i \, a d^{4}\right )}} + \frac {12 \, {\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 )} e f^{2}}{a d^{3}} + \frac {6 \, {\left (d^{2} x^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(-i \, e^{\left (d x + c\right )})\right )} f^{3}}{a d^{4}} - \frac {2 \, {\left (d^{3} f^{3} x^{3} + 3 \, d^{3} e f^{2} x^{2}\right )}}{a d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sinh(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-3/4*e^2*f*(4*x*e^(d*x + c)/(a*d*e^(d*x + c) - I*a*d) - (-2*I*d^2*x^2*e^c - 2*I*d*x*e^c - (2*I*d*x*e^(3*c) - 2
*I*e^(3*c))*e^(2*d*x) + 2*(d^2*x^2*e^(2*c) - 3*d*x*e^(2*c) + e^(2*c))*e^(d*x) - 2*(d*x + 1)*e^(-d*x) - 2*I*e^c
)/(a*d^2*e^(d*x + 2*c) - I*a*d^2*e^c) - 8*log((e^(d*x + c) - I)*e^(-c))/(a*d^2)) + 1/2*e^3*(2*(d*x + c)/(a*d)
+ (-5*I*e^(-d*x - c) + 1)/((I*a*e^(-d*x - c) + a*e^(-2*d*x - 2*c))*d) - I*e^(-d*x - c)/(a*d)) + 1/4*(-I*d^4*f^
3*x^4 - 12*I*d*e*f^2 - (4*I*d^4*e*f^2 + 10*I*d^3*f^3)*x^3 - 12*I*f^3 - (30*I*d^3*e*f^2 + 6*I*d^2*f^3)*x^2 - (1
2*I*d^2*e*f^2 + 12*I*d*f^3)*x - (2*I*d^3*f^3*x^3*e^(2*c) + (6*I*d^3*e*f^2 - 6*I*d^2*f^3)*x^2*e^(2*c) + (-12*I*
d^2*e*f^2 + 12*I*d*f^3)*x*e^(2*c) + (12*I*d*e*f^2 - 12*I*f^3)*e^(2*c))*e^(2*d*x) + (d^4*f^3*x^4*e^c + 2*(2*d^4
*e*f^2 - d^3*f^3)*x^3*e^c - 6*(d^3*e*f^2 - d^2*f^3)*x^2*e^c + 12*(d^2*e*f^2 - d*f^3)*x*e^c - 12*(d*e*f^2 - f^3
)*e^c)*e^(d*x))/(a*d^4*e^(d*x + c) - I*a*d^4) + 12*(d*x*log(I*e^(d*x + c) + 1) + dilog(-I*e^(d*x + c)))*e*f^2/
(a*d^3) + 6*(d^2*x^2*log(I*e^(d*x + c) + 1) + 2*d*x*dilog(-I*e^(d*x + c)) - 2*polylog(3, -I*e^(d*x + c)))*f^3/
(a*d^4) - 2*(d^3*f^3*x^3 + 3*d^3*e*f^2*x^2)/(a*d^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {sinh}\left (c+d\,x\right )}^2\,{\left (e+f\,x\right )}^3}{a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sinh(c + d*x)^2*(e + f*x)^3)/(a + a*sinh(c + d*x)*1i),x)

[Out]

int((sinh(c + d*x)^2*(e + f*x)^3)/(a + a*sinh(c + d*x)*1i), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 i e^{3} e^{c} + 6 i e^{2} f x e^{c} + 6 i e f^{2} x^{2} e^{c} + 2 i f^{3} x^{3} e^{c}}{- i a d e^{c} - a d e^{- d x}} - \frac {i \left (\int \frac {i d e^{3}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {i d f^{3} x^{3}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {d e^{3} e^{c} e^{d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {d e^{3} e^{3 c} e^{3 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {3 i d e f^{2} x^{2}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {3 i d e^{2} f x}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {i d e^{3} e^{2 c} e^{2 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {12 i e^{2} f e^{2 c} e^{2 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {12 i f^{3} x^{2} e^{2 c} e^{2 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {d f^{3} x^{3} e^{c} e^{d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {d f^{3} x^{3} e^{3 c} e^{3 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {i d f^{3} x^{3} e^{2 c} e^{2 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {24 i e f^{2} x e^{2 c} e^{2 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {3 d e f^{2} x^{2} e^{c} e^{d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {3 d e f^{2} x^{2} e^{3 c} e^{3 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {3 d e^{2} f x e^{c} e^{d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {3 d e^{2} f x e^{3 c} e^{3 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {3 i d e f^{2} x^{2} e^{2 c} e^{2 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx + \int \frac {3 i d e^{2} f x e^{2 c} e^{2 d x}}{e^{c} e^{2 d x} - i e^{d x}}\, dx\right ) e^{- c}}{2 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**3*sinh(d*x+c)**2/(a+I*a*sinh(d*x+c)),x)

[Out]

(2*I*e**3*exp(c) + 6*I*e**2*f*x*exp(c) + 6*I*e*f**2*x**2*exp(c) + 2*I*f**3*x**3*exp(c))/(-I*a*d*exp(c) - a*d*e
xp(-d*x)) - I*(Integral(I*d*e**3/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(I*d*f**3*x**3/(exp(c)*exp(2*d
*x) - I*exp(d*x)), x) + Integral(d*e**3*exp(c)*exp(d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(d*e**3
*exp(3*c)*exp(3*d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(3*I*d*e*f**2*x**2/(exp(c)*exp(2*d*x) - I*
exp(d*x)), x) + Integral(3*I*d*e**2*f*x/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(I*d*e**3*exp(2*c)*exp(
2*d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(12*I*e**2*f*exp(2*c)*exp(2*d*x)/(exp(c)*exp(2*d*x) - I*
exp(d*x)), x) + Integral(12*I*f**3*x**2*exp(2*c)*exp(2*d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(d*
f**3*x**3*exp(c)*exp(d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(d*f**3*x**3*exp(3*c)*exp(3*d*x)/(exp
(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(I*d*f**3*x**3*exp(2*c)*exp(2*d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x))
, x) + Integral(24*I*e*f**2*x*exp(2*c)*exp(2*d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(3*d*e*f**2*x
**2*exp(c)*exp(d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(3*d*e*f**2*x**2*exp(3*c)*exp(3*d*x)/(exp(c
)*exp(2*d*x) - I*exp(d*x)), x) + Integral(3*d*e**2*f*x*exp(c)*exp(d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) +
Integral(3*d*e**2*f*x*exp(3*c)*exp(3*d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(3*I*d*e*f**2*x**2*ex
p(2*c)*exp(2*d*x)/(exp(c)*exp(2*d*x) - I*exp(d*x)), x) + Integral(3*I*d*e**2*f*x*exp(2*c)*exp(2*d*x)/(exp(c)*e
xp(2*d*x) - I*exp(d*x)), x))*exp(-c)/(2*a*d)

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