\(\int \frac {(e+f x)^2 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx\) [254]

Optimal result

Integrand size = 29, antiderivative size = 106 \[ \int \frac {(e+f x)^2 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i (e+f x)^3}{3 a f}-\frac {2 i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d}-\frac {4 i f (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}+\frac {4 i f^2 \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {5678, 2221, 2611, 2320, 6724} \[ \int \frac {(e+f x)^2 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {4 i f^2 \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac {4 i f (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}-\frac {2 i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d}+\frac {i (e+f x)^3}{3 a f} \]

[In]

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

[Out]

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

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]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.89 \[ \int \frac {(e+f x)^2 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i \left (d^2 (e+f x)^2 \left (d (e+f x)-6 f \log \left (1+i e^{c+d x}\right )\right )-12 d f^2 (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )+12 f^3 \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )\right )}{3 a d^3 f} \]

[In]

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

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 404 vs. \(2 (94 ) = 188\).

Time = 2.90 (sec) , antiderivative size = 405, normalized size of antiderivative = 3.82

method	result	size
risch	\(-\frac {4 i f^{2} c^{3}}{3 d^{3} a}+\frac {2 i f^{2} c^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{d^{3} a}-\frac {4 i e f c \ln \left ({\mathrm e}^{d x +c}\right )}{d^{2} a}-\frac {2 i f^{2} c^{2} \ln \left ({\mathrm e}^{d x +c}-i\right )}{d^{3} a}+\frac {2 i e f \,c^{2}}{d^{2} a}+\frac {i f e \,x^{2}}{a}-\frac {i e^{3}}{3 a f}-\frac {4 i e f \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{d^{2} a}+\frac {2 i f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) c^{2}}{d^{3} a}+\frac {2 i \ln \left ({\mathrm e}^{d x +c}\right ) e^{2}}{d a}+\frac {4 i e f c x}{d a}-\frac {2 i f^{2} x \,c^{2}}{d^{2} a}+\frac {4 i e f c \ln \left ({\mathrm e}^{d x +c}-i\right )}{d^{2} a}-\frac {4 i e f \operatorname {polylog}\left (2, -i {\mathrm e}^{d x +c}\right )}{d^{2} a}-\frac {i e^{2} x}{a}-\frac {4 i f^{2} \operatorname {polylog}\left (2, -i {\mathrm e}^{d x +c}\right ) x}{d^{2} a}+\frac {4 i f^{2} \operatorname {polylog}\left (3, -i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {2 i \ln \left ({\mathrm e}^{d x +c}-i\right ) e^{2}}{d a}+\frac {i f^{2} x^{3}}{3 a}-\frac {2 i f^{2} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x^{2}}{d a}-\frac {4 i e f \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{d a}\)	\(405\)

[In]

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

[Out]

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

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. 184 vs. \(2 (86) = 172\).

Time = 0.26 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.74 \[ \int \frac {(e+f x)^2 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i \, d^{3} f^{2} x^{3} + 3 i \, d^{3} e f x^{2} + 3 i \, d^{3} e^{2} x + 6 i \, c d^{2} e^{2} - 6 i \, c^{2} d e f + 2 i \, c^{3} f^{2} + 12 i \, f^{2} {\rm polylog}\left (3, -i \, e^{\left (d x + c\right )}\right ) - 12 \, {\left (i \, d f^{2} x + i \, d e f\right )} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 6 \, {\left (i \, d^{2} e^{2} - 2 i \, c d e f + i \, c^{2} f^{2}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - 6 \, {\left (i \, d^{2} f^{2} x^{2} + 2 i \, d^{2} e f x + 2 i \, c d e f - i \, c^{2} f^{2}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{3 \, a d^{3}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.55 \[ \int \frac {(e+f x)^2 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {i \, e^{2} \log \left (i \, a \sinh \left (d x + c\right ) + a\right )}{a d} - \frac {i \, f^{2} x^{3} + 3 i \, e f x^{2}}{3 \, a} - \frac {4 i \, {\left (d x \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right )\right )} e f}{a d^{2}} - \frac {2 i \, {\left (d^{2} x^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(-i \, e^{\left (d x + c\right )})\right )} f^{2}}{a d^{3}} - \frac {2 \, {\left (-i \, d^{3} f^{2} x^{3} - 3 i \, d^{3} e f x^{2}\right )}}{3 \, a d^{3}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^2 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\int \frac {\mathrm {cosh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^2}{a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

[In]

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

[Out]

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