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

Optimal result

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

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)*pol 
ylog(2,-I*exp(d*x+c))/a/d^2+4*I*f^2*polylog(3,-I*exp(d*x+c))/a/d^3
 

Mathematica [A] (verified)

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

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

Output:

((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)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {6093, 2620, 3011, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 6093

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

\(\Big \downarrow \) 2620

\(\displaystyle 2 \left (\frac {2 i f \int (e+f x) \log \left (1+i e^{c+d x}\right )dx}{a d}-\frac {i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d}\right )+\frac {i (e+f x)^3}{3 a f}\)

\(\Big \downarrow \) 3011

\(\displaystyle 2 \left (\frac {2 i f \left (\frac {f \int \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )dx}{d}-\frac {(e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{a d}-\frac {i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d}\right )+\frac {i (e+f x)^3}{3 a f}\)

\(\Big \downarrow \) 2720

\(\displaystyle 2 \left (\frac {2 i f \left (\frac {f \int e^{-c-d x} \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )de^{c+d x}}{d^2}-\frac {(e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{a d}-\frac {i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d}\right )+\frac {i (e+f x)^3}{3 a f}\)

\(\Big \downarrow \) 7143

\(\displaystyle 2 \left (\frac {2 i f \left (\frac {f \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{d^2}-\frac {(e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{a d}-\frac {i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d}\right )+\frac {i (e+f x)^3}{3 a f}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2620

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] - Si 
mp[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 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 3011

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] + Simp[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 6093

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin 
h[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[-(e + f*x)^(m + 1)/(b*f*(m + 1)), 
 x] + Simp[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 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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]
 

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 = 1.74 (sec) , antiderivative size = 405, normalized size of antiderivative = 3.82

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

Input:

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

Output:

2*I/d^3/a*f^2*c^2*ln(exp(d*x+c))-4*I/d^2/a*e*f*ln(1+I*exp(d*x+c))*c+1/3*I/ 
a*f^2*x^3+2*I/d^3/a*f^2*ln(1+I*exp(d*x+c))*c^2-I/a*e^2*x-2*I/d^2/a*f^2*c^2 
*x-4*I/d^2/a*e*f*c*ln(exp(d*x+c))-2*I/d/a*f^2*ln(1+I*exp(d*x+c))*x^2+4*I/d 
/a*e*f*c*x+4*I*f^2*polylog(3,-I*exp(d*x+c))/a/d^3+2*I/d^2/a*e*f*c^2-4/3*I/ 
d^3/a*f^2*c^3+2*I/d/a*ln(exp(d*x+c))*e^2+I/a*f*e*x^2-1/3*I/a/f*e^3+4*I/d^2 
/a*e*f*c*ln(exp(d*x+c)-I)-4*I/d^2/a*f^2*polylog(2,-I*exp(d*x+c))*x-4*I/d/a 
*e*f*ln(1+I*exp(d*x+c))*x-2*I/d/a*ln(exp(d*x+c)-I)*e^2-2*I/d^3/a*f^2*c^2*l 
n(exp(d*x+c)-I)-4*I/d^2/a*e*f*polylog(2,-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. 184 vs. \(2 (86) = 172\).

Time = 0.09 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.74 \[ \int \frac {(e+f x)^2 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i \, d^{3} f^{2} x^{3} + 3 i \, d^{3} e f x^{2} + 3 i \, d^{3} e^{2} x + 6 i \, c d^{2} e^{2} - 6 i \, c^{2} d e f + 2 i \, c^{3} f^{2} + 12 i \, f^{2} {\rm polylog}\left (3, -i \, e^{\left (d x + c\right )}\right ) - 12 \, {\left (i \, d f^{2} x + i \, d e f\right )} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 6 \, {\left (i \, d^{2} e^{2} - 2 i \, c d e f + i \, c^{2} f^{2}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - 6 \, {\left (i \, d^{2} f^{2} x^{2} + 2 i \, d^{2} e f x + 2 i \, c d e f - i \, c^{2} f^{2}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{3 \, a d^{3}} \] Input:

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

Output:

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]

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

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

Output:

-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)

Time = 0.13 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.55 \[ \int \frac {(e+f x)^2 \cosh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {i \, e^{2} \log \left (i \, a \sinh \left (d x + c\right ) + a\right )}{a d} - \frac {i \, f^{2} x^{3} + 3 i \, e f x^{2}}{3 \, a} - \frac {4 i \, {\left (d x \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right )\right )} e f}{a d^{2}} - \frac {2 i \, {\left (d^{2} x^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(-i \, e^{\left (d x + c\right )})\right )} f^{2}}{a d^{3}} - \frac {2 \, {\left (-i \, d^{3} f^{2} x^{3} - 3 i \, d^{3} e f x^{2}\right )}}{3 \, a d^{3}} \] Input:

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

Output:

-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*polyl 
og(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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