\(\int \frac {(e+f x)^3 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx\) [253]

Optimal result

Integrand size = 29, antiderivative size = 139 \[ \int \frac {(e+f x)^3 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i (e+f x)^4}{4 a f}-\frac {2 i (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{a d}-\frac {6 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}+\frac {12 i f^2 (e+f x) \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac {12 i f^3 \operatorname {PolyLog}\left (4,-i e^{c+d x}\right )}{a d^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {5678, 2221, 2611, 6744, 2320, 6724} \[ \int \frac {(e+f x)^3 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {12 i f^3 \operatorname {PolyLog}\left (4,-i e^{c+d x}\right )}{a d^4}+\frac {12 i f^2 (e+f x) \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac {6 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}-\frac {2 i (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{a d}+\frac {i (e+f x)^4}{4 a f} \]

[In]

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

[Out]

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

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 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 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 5678

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :
> Simp[-(e + f*x)^(m + 1)/(b*f*(m + 1)), x] + Dist[2, Int[(e + f*x)^m*(E^(c + d*x)/(a + b*E^(c + d*x))), x], x
] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[a^2 + b^2, 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]

Rule 6744

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Dist[f*(m/(b*c*p*Log[F])), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {i (e+f x)^4}{4 a f}+2 \int \frac {e^{c+d x} (e+f x)^3}{a+i a e^{c+d x}} \, dx \\ & = \frac {i (e+f x)^4}{4 a f}-\frac {2 i (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{a d}+\frac {(6 i f) \int (e+f x)^2 \log \left (1+i e^{c+d x}\right ) \, dx}{a d} \\ & = \frac {i (e+f x)^4}{4 a f}-\frac {2 i (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{a d}-\frac {6 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}+\frac {\left (12 i f^2\right ) \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right ) \, dx}{a d^2} \\ & = \frac {i (e+f x)^4}{4 a f}-\frac {2 i (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{a d}-\frac {6 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}+\frac {12 i f^2 (e+f x) \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac {\left (12 i f^3\right ) \int \operatorname {PolyLog}\left (3,-i e^{c+d x}\right ) \, dx}{a d^3} \\ & = \frac {i (e+f x)^4}{4 a f}-\frac {2 i (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{a d}-\frac {6 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}+\frac {12 i f^2 (e+f x) \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac {\left (12 i f^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,-i x)}{x} \, dx,x,e^{c+d x}\right )}{a d^4} \\ & = \frac {i (e+f x)^4}{4 a f}-\frac {2 i (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{a d}-\frac {6 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}+\frac {12 i f^2 (e+f x) \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac {12 i f^3 \operatorname {PolyLog}\left (4,-i e^{c+d x}\right )}{a d^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.85 \[ \int \frac {(e+f x)^3 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i \left (\frac {(e+f x)^4}{f}-\frac {8 (e+f x)^3 \log \left (1+i e^{c+d x}\right )}{d}-\frac {24 f \left (d^2 (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )-2 d f (e+f x) \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )+2 f^2 \operatorname {PolyLog}\left (4,-i e^{c+d x}\right )\right )}{d^4}\right )}{4 a} \]

[In]

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

[Out]

((I/4)*((e + f*x)^4/f - (8*(e + f*x)^3*Log[1 + I*E^(c + d*x)])/d - (24*f*(d^2*(e + f*x)^2*PolyLog[2, (-I)*E^(c
+ d*x)] - 2*d*f*(e + f*x)*PolyLog[3, (-I)*E^(c + d*x)] + 2*f^2*PolyLog[4, (-I)*E^(c + d*x)]))/d^4))/a

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 646 vs. \(2 (124 ) = 248\).

Time = 6.17 (sec) , antiderivative size = 647, normalized size of antiderivative = 4.65

method	result	size
risch	\(\frac {i f^{3} x^{4}}{4 a}-\frac {i e^{4}}{4 a f}-\frac {i e^{3} x}{a}+\frac {i f^{2} e \,x^{3}}{a}+\frac {3 i f \,e^{2} x^{2}}{2 a}+\frac {6 i e^{2} f c \ln \left ({\mathrm e}^{d x +c}-i\right )}{d^{2} a}-\frac {6 i e^{2} f c \ln \left ({\mathrm e}^{d x +c}\right )}{d^{2} a}-\frac {6 i f^{2} e x \,c^{2}}{d^{2} a}-\frac {6 i f^{2} e \ln \left (1+i {\mathrm e}^{d x +c}\right ) x^{2}}{d a}+\frac {6 i f^{2} e \ln \left (1+i {\mathrm e}^{d x +c}\right ) c^{2}}{d^{3} a}-\frac {12 i f^{2} e \operatorname {polylog}\left (2, -i {\mathrm e}^{d x +c}\right ) x}{d^{2} a}+\frac {6 i f \,e^{2} c x}{d a}-\frac {6 i f \,e^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{d a}-\frac {6 i f \,e^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{d^{2} a}-\frac {6 i e \,f^{2} c^{2} \ln \left ({\mathrm e}^{d x +c}-i\right )}{d^{3} a}+\frac {6 i e \,f^{2} c^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{d^{3} a}+\frac {3 i f^{3} c^{4}}{2 d^{4} a}+\frac {2 i \ln \left ({\mathrm e}^{d x +c}\right ) e^{3}}{d a}-\frac {2 i \ln \left ({\mathrm e}^{d x +c}-i\right ) e^{3}}{d a}-\frac {12 i f^{3} \operatorname {polylog}\left (4, -i {\mathrm e}^{d x +c}\right )}{a \,d^{4}}-\frac {6 i f^{3} \operatorname {polylog}\left (2, -i {\mathrm e}^{d x +c}\right ) x^{2}}{d^{2} a}+\frac {3 i f \,e^{2} c^{2}}{d^{2} a}-\frac {6 i f \,e^{2} \operatorname {polylog}\left (2, -i {\mathrm e}^{d x +c}\right )}{d^{2} a}+\frac {2 i f^{3} c^{3} \ln \left ({\mathrm e}^{d x +c}-i\right )}{d^{4} a}-\frac {2 i f^{3} c^{3} \ln \left ({\mathrm e}^{d x +c}\right )}{d^{4} a}-\frac {4 i f^{2} e \,c^{3}}{d^{3} a}+\frac {12 i f^{3} \operatorname {polylog}\left (3, -i {\mathrm e}^{d x +c}\right ) x}{d^{3} a}+\frac {2 i f^{3} c^{3} x}{d^{3} a}-\frac {2 i f^{3} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x^{3}}{d a}+\frac {12 i f^{2} e \operatorname {polylog}\left (3, -i {\mathrm e}^{d x +c}\right )}{d^{3} a}-\frac {2 i f^{3} \ln \left (1+i {\mathrm e}^{d x +c}\right ) c^{3}}{d^{4} a}\)	\(647\)

[In]

int((f*x+e)^3*cosh(d*x+c)/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (114) = 228\).

Time = 0.25 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.15 \[ \int \frac {(e+f x)^3 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i \, d^{4} f^{3} x^{4} + 4 i \, d^{4} e f^{2} x^{3} + 6 i \, d^{4} e^{2} f x^{2} + 4 i \, d^{4} e^{3} x + 8 i \, c d^{3} e^{3} - 12 i \, c^{2} d^{2} e^{2} f + 8 i \, c^{3} d e f^{2} - 2 i \, c^{4} f^{3} - 48 i \, f^{3} {\rm polylog}\left (4, -i \, e^{\left (d x + c\right )}\right ) - 24 \, {\left (i \, d^{2} f^{3} x^{2} + 2 i \, d^{2} e f^{2} x + i \, d^{2} e^{2} f\right )} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 8 \, {\left (i \, d^{3} e^{3} - 3 i \, c d^{2} e^{2} f + 3 i \, c^{2} d e f^{2} - i \, c^{3} f^{3}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - 8 \, {\left (i \, d^{3} f^{3} x^{3} + 3 i \, d^{3} e f^{2} x^{2} + 3 i \, d^{3} e^{2} f x + 3 i \, c d^{2} e^{2} f - 3 i \, c^{2} d e f^{2} + i \, c^{3} f^{3}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) - 48 \, {\left (-i \, d f^{3} x - i \, d e f^{2}\right )} {\rm polylog}\left (3, -i \, e^{\left (d x + c\right )}\right )}{4 \, a d^{4}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {(e+f x)^3 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=- \frac {i \left (\int \frac {e^{3} \cosh {\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {f^{3} x^{3} \cosh {\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {3 e f^{2} x^{2} \cosh {\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {3 e^{2} f x \cosh {\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx\right )}{a} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (114) = 228\).

Time = 0.29 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.90 \[ \int \frac {(e+f x)^3 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {i \, e^{3} \log \left (i \, a \sinh \left (d x + c\right ) + a\right )}{a d} - \frac {6 i \, {\left (d x \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right )\right )} e^{2} f}{a d^{2}} - \frac {i \, {\left (f^{3} x^{4} + 4 \, e f^{2} x^{3} + 6 \, e^{2} f x^{2}\right )}}{4 \, a} - \frac {6 i \, {\left (d^{2} x^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(-i \, e^{\left (d x + c\right )})\right )} e f^{2}}{a d^{3}} - \frac {2 i \, {\left (d^{3} x^{3} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 3 \, d^{2} x^{2} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 6 \, d x {\rm Li}_{3}(-i \, e^{\left (d x + c\right )}) + 6 \, {\rm Li}_{4}(-i \, e^{\left (d x + c\right )})\right )} f^{3}}{a d^{4}} + \frac {i \, d^{4} f^{3} x^{4} + 4 i \, d^{4} e f^{2} x^{3} + 6 i \, d^{4} e^{2} f x^{2}}{2 \, a d^{4}} \]

[In]

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

[Out]

-I*e^3*log(I*a*sinh(d*x + c) + a)/(a*d) - 6*I*(d*x*log(I*e^(d*x + c) + 1) + dilog(-I*e^(d*x + c)))*e^2*f/(a*d^
2) - 1/4*I*(f^3*x^4 + 4*e*f^2*x^3 + 6*e^2*f*x^2)/a - 6*I*(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)))*e*f^2/(a*d^3) - 2*I*(d^3*x^3*log(I*e^(d*x + c) + 1) + 3*d^2*x^2*dilog
(-I*e^(d*x + c)) - 6*d*x*polylog(3, -I*e^(d*x + c)) + 6*polylog(4, -I*e^(d*x + c)))*f^3/(a*d^4) + 1/2*(I*d^4*f
^3*x^4 + 4*I*d^4*e*f^2*x^3 + 6*I*d^4*e^2*f*x^2)/(a*d^4)

Giac [F]

\[ \int \frac {(e+f x)^3 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{3} \cosh \left (d x + c\right )}{i \, a \sinh \left (d x + c\right ) + a} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\int \frac {\mathrm {cosh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^3}{a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

[In]

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

[Out]

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