Integrand size = 29, antiderivative size = 163 \[ \int \frac {(e+f x)^3 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i (e+f x)^3}{a d}-\frac {i (e+f x)^4}{4 a f}-\frac {6 i f (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac {12 i f^2 (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac {12 i f^3 \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^4}+\frac {i (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d} \] Output:
I*(f*x+e)^3/a/d-1/4*I*(f*x+e)^4/a/f-6*I*f*(f*x+e)^2*ln(1+I*exp(d*x+c))/a/d ^2-12*I*f^2*(f*x+e)*polylog(2,-I*exp(d*x+c))/a/d^3+12*I*f^3*polylog(3,-I*e xp(d*x+c))/a/d^4+I*(f*x+e)^3*tanh(1/2*c+1/4*I*Pi+1/2*d*x)/a/d
Time = 1.48 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.31 \[ \int \frac {(e+f x)^3 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {-\frac {8 (e+f x)^3}{d \left (-i+e^c\right )}-i x \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right )-\frac {24 i f (e+f x)^2 \log \left (1-i e^{-c-d x}\right )}{d^2}+\frac {48 i f^2 \left (d (e+f x) \operatorname {PolyLog}\left (2,i e^{-c-d x}\right )+f \operatorname {PolyLog}\left (3,i e^{-c-d x}\right )\right )}{d^4}+\frac {8 i (e+f x)^3 \sinh \left (\frac {d x}{2}\right )}{d \left (\cosh \left (\frac {c}{2}\right )+i \sinh \left (\frac {c}{2}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}}{4 a} \] Input:
Integrate[((e + f*x)^3*Sinh[c + d*x])/(a + I*a*Sinh[c + d*x]),x]
Output:
((-8*(e + f*x)^3)/(d*(-I + E^c)) - I*x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3) - ((24*I)*f*(e + f*x)^2*Log[1 - I*E^(-c - d*x)])/d^2 + ((48*I)*f^ 2*(d*(e + f*x)*PolyLog[2, I*E^(-c - d*x)] + f*PolyLog[3, I*E^(-c - d*x)])) /d^4 + ((8*I)*(e + f*x)^3*Sinh[(d*x)/2])/(d*(Cosh[c/2] + I*Sinh[c/2])*(Cos h[(c + d*x)/2] + I*Sinh[(c + d*x)/2])))/(4*a)
Time = 0.98 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.03, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.586, Rules used = {6091, 17, 3042, 3799, 25, 25, 3042, 4672, 26, 3042, 26, 4199, 26, 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.
| \(\displaystyle \int \frac {(e+f x)^3 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx\) |
| \(\Big \downarrow \) 6091 |
| \(\displaystyle i \int \frac {(e+f x)^3}{i \sinh (c+d x) a+a}dx-\frac {i \int (e+f x)^3dx}{a}\) |
| \(\Big \downarrow \) 17 |
| \(\displaystyle i \int \frac {(e+f x)^3}{i \sinh (c+d x) a+a}dx-\frac {i (e+f x)^4}{4 a f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle i \int \frac {(e+f x)^3}{\sin (i c+i d x) a+a}dx-\frac {i (e+f x)^4}{4 a f}\) |
| \(\Big \downarrow \) 3799 |
| \(\displaystyle \frac {i \int -(e+f x)^3 \text {csch}^2\left (\frac {c}{2}+\frac {d x}{2}-\frac {i \pi }{4}\right )dx}{2 a}-\frac {i (e+f x)^4}{4 a f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {i \int -(e+f x)^3 \text {sech}^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{2 a}-\frac {i (e+f x)^4}{4 a f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {i \int (e+f x)^3 \text {sech}^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{2 a}-\frac {i (e+f x)^4}{4 a f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {i \int (e+f x)^3 \csc \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )^2dx}{2 a}-\frac {i (e+f x)^4}{4 a f}\) |
| \(\Big \downarrow \) 4672 |
| \(\displaystyle \frac {i \left (\frac {2 (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}-\frac {6 i f \int -i (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{d}\right )}{2 a}-\frac {i (e+f x)^4}{4 a f}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {i \left (\frac {2 (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}-\frac {6 f \int (e+f x)^2 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{d}\right )}{2 a}-\frac {i (e+f x)^4}{4 a f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {i \left (\frac {2 (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}-\frac {6 f \int -i (e+f x)^2 \tan \left (\frac {i c}{2}+\frac {i d x}{2}-\frac {\pi }{4}\right )dx}{d}\right )}{2 a}-\frac {i (e+f x)^4}{4 a f}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {i \left (\frac {6 i f \int (e+f x)^2 \tan \left (\frac {i c}{2}+\frac {i d x}{2}-\frac {\pi }{4}\right )dx}{d}+\frac {2 (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{2 a}-\frac {i (e+f x)^4}{4 a f}\) |
| \(\Big \downarrow \) 4199 |
| \(\displaystyle \frac {i \left (\frac {6 i f \left (2 i \int \frac {i e^{c+d x} (e+f x)^2}{1+i e^{c+d x}}dx-\frac {i (e+f x)^3}{3 f}\right )}{d}+\frac {2 (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{2 a}-\frac {i (e+f x)^4}{4 a f}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {i \left (\frac {6 i f \left (-2 \int \frac {e^{c+d x} (e+f x)^2}{1+i e^{c+d x}}dx-\frac {i (e+f x)^3}{3 f}\right )}{d}+\frac {2 (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{2 a}-\frac {i (e+f x)^4}{4 a f}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {i \left (\frac {6 i f \left (-2 \left (\frac {2 i f \int (e+f x) \log \left (1+i e^{c+d x}\right )dx}{d}-\frac {i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{d}\right )-\frac {i (e+f x)^3}{3 f}\right )}{d}+\frac {2 (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{2 a}-\frac {i (e+f x)^4}{4 a f}\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {i \left (\frac {6 i f \left (-2 \left (\frac {2 i f \left (\frac {f \int \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )dx}{d}-\frac {(e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{d}\right )-\frac {i (e+f x)^3}{3 f}\right )}{d}+\frac {2 (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{2 a}-\frac {i (e+f x)^4}{4 a f}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {i \left (\frac {6 i f \left (-2 \left (\frac {2 i f \left (\frac {f \int e^{-c-d x} \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )de^{c+d x}}{d^2}-\frac {(e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{d}\right )-\frac {i (e+f x)^3}{3 f}\right )}{d}+\frac {2 (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{2 a}-\frac {i (e+f x)^4}{4 a f}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {i \left (\frac {6 i f \left (-2 \left (\frac {2 i f \left (\frac {f \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{d^2}-\frac {(e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+i e^{c+d x}\right )}{d}\right )-\frac {i (e+f x)^3}{3 f}\right )}{d}+\frac {2 (e+f x)^3 \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{2 a}-\frac {i (e+f x)^4}{4 a f}\) |
Input:
Int[((e + f*x)^3*Sinh[c + d*x])/(a + I*a*Sinh[c + d*x]),x]
Output:
((-1/4*I)*(e + f*x)^4)/(a*f) + ((I/2)*(((6*I)*f*(((-1/3*I)*(e + f*x)^3)/f - 2*(((-I)*(e + f*x)^2*Log[1 + I*E^(c + d*x)])/d + ((2*I)*f*(-(((e + f*x)* PolyLog[2, (-I)*E^(c + d*x)])/d) + (f*PolyLog[3, (-I)*E^(c + d*x)])/d^2))/ d)))/d + (2*(e + f*x)^3*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/d))/a
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
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]
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]]]
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]
Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) , x_Symbol] :> Simp[(2*a)^n Int[(c + d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) + 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])
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] + Simp [2*I Int[((c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x ))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && In tegerQ[4*k] && IGtQ[m, 0]
Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp [(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1) *Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Int[(((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_ .)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[1/b Int[(e + f*x)^m*Sinh[ c + d*x]^(n - 1), x], x] - Simp[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]
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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 578 vs. \(2 (142 ) = 284\).
Time = 0.42 (sec) , antiderivative size = 579, normalized size of antiderivative = 3.55
| method | result | size |
| risch | \(\frac {6 i f^{2} e \,c^{2}}{a \,d^{3}}-\frac {12 c \,f^{2} e \arctan \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {12 i c \,f^{2} e \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {12 i f^{2} e c x}{a \,d^{2}}+\frac {6 i c \,f^{2} e \ln \left (1+{\mathrm e}^{2 d x +2 c}\right )}{a \,d^{3}}-\frac {12 i f^{2} e \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{a \,d^{3}}-\frac {12 i f^{2} e \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{a \,d^{2}}-\frac {i f^{2} e \,x^{3}}{a}-\frac {3 i f \,e^{2} x^{2}}{2 a}+\frac {2 i f^{3} x^{3}}{a d}-\frac {4 i f^{3} c^{3}}{a \,d^{4}}+\frac {12 i f^{3} \operatorname {polylog}\left (3, -i {\mathrm e}^{d x +c}\right )}{a \,d^{4}}+\frac {6 c^{2} f^{3} \arctan \left ({\mathrm e}^{d x +c}\right )}{a \,d^{4}}+\frac {6 e^{2} f \arctan \left ({\mathrm e}^{d x +c}\right )}{a \,d^{2}}-\frac {6 i f^{3} \ln \left (1+i {\mathrm e}^{d x +c}\right ) x^{2}}{a \,d^{2}}+\frac {6 i f^{2} e \,x^{2}}{a d}+\frac {6 i e^{2} f \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{2}}+\frac {6 i f^{3} \ln \left (1+i {\mathrm e}^{d x +c}\right ) c^{2}}{a \,d^{4}}-\frac {12 i f^{3} \operatorname {polylog}\left (2, -i {\mathrm e}^{d x +c}\right ) x}{a \,d^{3}}-\frac {6 i f^{3} c^{2} x}{a \,d^{3}}-\frac {12 i f^{2} e \operatorname {polylog}\left (2, -i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {3 i e^{2} f \ln \left (1+{\mathrm e}^{2 d x +2 c}\right )}{a \,d^{2}}+\frac {6 i c^{2} f^{3} \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{4}}-\frac {3 i c^{2} f^{3} \ln \left (1+{\mathrm e}^{2 d x +2 c}\right )}{a \,d^{4}}-\frac {i f^{3} x^{4}}{4 a}-\frac {i e^{3} x}{a}-\frac {2 \left (x^{3} f^{3}+3 e \,f^{2} x^{2}+3 e^{2} f x +e^{3}\right )}{d a \left ({\mathrm e}^{d x +c}-i\right )}-\frac {i e^{4}}{4 a f}\) | \(579\) |
Input:
int((f*x+e)^3*sinh(d*x+c)/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)
Output:
-I/a*f^2*e*x^3-3/2*I/a*f*e^2*x^2-1/4*I/a*f^3*x^4-I/a*e^3*x-12/a/d^3*c*f^2* e*arctan(exp(d*x+c))+6*I/a/d^3*f^2*e*c^2-6*I/a/d^2*f^3*ln(1+I*exp(d*x+c))* x^2+6*I/a/d*f^2*e*x^2+6*I/a/d^2*e^2*f*ln(exp(d*x+c))+6*I/a/d^4*f^3*ln(1+I* exp(d*x+c))*c^2-12*I/a/d^3*f^3*polylog(2,-I*exp(d*x+c))*x+6*I/a/d^3*c*f^2* e*ln(1+exp(2*d*x+2*c))-12*I/a/d^3*f^2*e*ln(1+I*exp(d*x+c))*c-12*I/a/d^2*f^ 2*e*ln(1+I*exp(d*x+c))*x-12*I/a/d^3*c*f^2*e*ln(exp(d*x+c))+12*I/a/d^2*f^2* e*c*x+2*I/a/d*f^3*x^3-2*(f^3*x^3+3*e*f^2*x^2+3*e^2*f*x+e^3)/d/a/(exp(d*x+c )-I)+6/a/d^4*c^2*f^3*arctan(exp(d*x+c))+6/a/d^2*e^2*f*arctan(exp(d*x+c))-4 *I/a/d^4*f^3*c^3-6*I/a/d^3*f^3*c^2*x-12*I/a/d^3*f^2*e*polylog(2,-I*exp(d*x +c))-3*I/a/d^2*e^2*f*ln(1+exp(2*d*x+2*c))+6*I/a/d^4*c^2*f^3*ln(exp(d*x+c)) -3*I/a/d^4*c^2*f^3*ln(1+exp(2*d*x+2*c))+12*I*f^3*polylog(3,-I*exp(d*x+c))/ a/d^4-1/4*I/a/f*e^4
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 456 vs. \(2 (131) = 262\).
Time = 0.11 (sec) , antiderivative size = 456, normalized size of antiderivative = 2.80 \[ \int \frac {(e+f x)^3 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {d^{4} f^{3} x^{4} + 4 \, d^{4} e f^{2} x^{3} + 6 \, d^{4} e^{2} f x^{2} + 4 \, d^{4} e^{3} x + 8 \, d^{3} e^{3} - 24 \, c d^{2} e^{2} f + 24 \, c^{2} d e f^{2} - 8 \, c^{3} f^{3} + 48 \, {\left (d f^{3} x + d e f^{2} + {\left (i \, d f^{3} x + i \, d e f^{2}\right )} e^{\left (d x + c\right )}\right )} {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - {\left (-i \, d^{4} f^{3} x^{4} + 24 i \, c d^{2} e^{2} f - 24 i \, c^{2} d e f^{2} + 8 i \, c^{3} f^{3} - 4 \, {\left (i \, d^{4} e f^{2} - 2 i \, d^{3} f^{3}\right )} x^{3} - 6 \, {\left (i \, d^{4} e^{2} f - 4 i \, d^{3} e f^{2}\right )} x^{2} - 4 \, {\left (i \, d^{4} e^{3} - 6 i \, d^{3} e^{2} f\right )} x\right )} e^{\left (d x + c\right )} + 24 \, {\left (d^{2} e^{2} f - 2 \, c d e f^{2} + c^{2} f^{3} + {\left (i \, d^{2} e^{2} f - 2 i \, c d e f^{2} + i \, c^{2} f^{3}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 24 \, {\left (d^{2} f^{3} x^{2} + 2 \, d^{2} e f^{2} x + 2 \, c d e f^{2} - c^{2} f^{3} + {\left (i \, d^{2} f^{3} x^{2} + 2 i \, d^{2} e f^{2} x + 2 i \, c d e f^{2} - i \, c^{2} f^{3}\right )} e^{\left (d x + c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 48 \, {\left (-i \, f^{3} e^{\left (d x + c\right )} - f^{3}\right )} {\rm polylog}\left (3, -i \, e^{\left (d x + c\right )}\right )}{4 \, {\left (a d^{4} e^{\left (d x + c\right )} - i \, a d^{4}\right )}} \] Input:
integrate((f*x+e)^3*sinh(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")
Output:
-1/4*(d^4*f^3*x^4 + 4*d^4*e*f^2*x^3 + 6*d^4*e^2*f*x^2 + 4*d^4*e^3*x + 8*d^ 3*e^3 - 24*c*d^2*e^2*f + 24*c^2*d*e*f^2 - 8*c^3*f^3 + 48*(d*f^3*x + d*e*f^ 2 + (I*d*f^3*x + I*d*e*f^2)*e^(d*x + c))*dilog(-I*e^(d*x + c)) - (-I*d^4*f ^3*x^4 + 24*I*c*d^2*e^2*f - 24*I*c^2*d*e*f^2 + 8*I*c^3*f^3 - 4*(I*d^4*e*f^ 2 - 2*I*d^3*f^3)*x^3 - 6*(I*d^4*e^2*f - 4*I*d^3*e*f^2)*x^2 - 4*(I*d^4*e^3 - 6*I*d^3*e^2*f)*x)*e^(d*x + c) + 24*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3 + (I*d^2*e^2*f - 2*I*c*d*e*f^2 + 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 + (I*d^2*f^3*x ^2 + 2*I*d^2*e*f^2*x + 2*I*c*d*e*f^2 - I*c^2*f^3)*e^(d*x + c))*log(I*e^(d* x + c) + 1) + 48*(-I*f^3*e^(d*x + c) - f^3)*polylog(3, -I*e^(d*x + c)))/(a *d^4*e^(d*x + c) - I*a*d^4)
\[ \int \frac {(e+f x)^3 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {- 2 e^{3} - 6 e^{2} f x - 6 e f^{2} x^{2} - 2 f^{3} x^{3}}{a d e^{c} e^{d x} - i a d} - \frac {i \left (\int \left (- \frac {i d e^{3}}{e^{c} e^{d x} - i}\right )\, dx + \int \frac {6 i e^{2} f}{e^{c} e^{d x} - i}\, dx + \int \frac {6 i f^{3} x^{2}}{e^{c} e^{d x} - i}\, dx + \int \left (- \frac {i d f^{3} x^{3}}{e^{c} e^{d x} - i}\right )\, dx + \int \frac {12 i e f^{2} x}{e^{c} e^{d x} - i}\, dx + \int \frac {d e^{3} e^{c} e^{d x}}{e^{c} e^{d x} - i}\, dx + \int \left (- \frac {3 i d e f^{2} x^{2}}{e^{c} e^{d x} - i}\right )\, dx + \int \left (- \frac {3 i d e^{2} f x}{e^{c} e^{d x} - i}\right )\, dx + \int \frac {d f^{3} x^{3} e^{c} e^{d x}}{e^{c} e^{d x} - i}\, dx + \int \frac {3 d e f^{2} x^{2} e^{c} e^{d x}}{e^{c} e^{d x} - i}\, dx + \int \frac {3 d e^{2} f x e^{c} e^{d x}}{e^{c} e^{d x} - i}\, dx\right )}{a d} \] Input:
integrate((f*x+e)**3*sinh(d*x+c)/(a+I*a*sinh(d*x+c)),x)
Output:
(-2*e**3 - 6*e**2*f*x - 6*e*f**2*x**2 - 2*f**3*x**3)/(a*d*exp(c)*exp(d*x) - I*a*d) - I*(Integral(-I*d*e**3/(exp(c)*exp(d*x) - I), x) + Integral(6*I* e**2*f/(exp(c)*exp(d*x) - I), x) + Integral(6*I*f**3*x**2/(exp(c)*exp(d*x) - I), x) + Integral(-I*d*f**3*x**3/(exp(c)*exp(d*x) - I), x) + Integral(1 2*I*e*f**2*x/(exp(c)*exp(d*x) - I), x) + Integral(d*e**3*exp(c)*exp(d*x)/( exp(c)*exp(d*x) - I), x) + Integral(-3*I*d*e*f**2*x**2/(exp(c)*exp(d*x) - I), x) + Integral(-3*I*d*e**2*f*x/(exp(c)*exp(d*x) - I), x) + Integral(d*f **3*x**3*exp(c)*exp(d*x)/(exp(c)*exp(d*x) - I), x) + Integral(3*d*e*f**2*x **2*exp(c)*exp(d*x)/(exp(c)*exp(d*x) - I), x) + Integral(3*d*e**2*f*x*exp( c)*exp(d*x)/(exp(c)*exp(d*x) - I), x))/(a*d)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (131) = 262\).
Time = 0.16 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.95 \[ \int \frac {(e+f x)^3 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {3}{2} \, e^{2} f {\left (\frac {-i \, d x^{2} + {\left (d x^{2} e^{c} - 4 \, x e^{c}\right )} e^{\left (d x\right )}}{i \, a d e^{\left (d x + c\right )} + a d} - \frac {4 i \, \log \left ({\left (e^{\left (d x + c\right )} - i\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} - e^{3} {\left (\frac {i \, {\left (d x + c\right )}}{a d} + \frac {2}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d}\right )} - \frac {d f^{3} x^{4} + 24 \, e f^{2} x^{2} + 4 \, {\left (d e f^{2} + 2 \, f^{3}\right )} x^{3} + {\left (i \, d f^{3} x^{4} e^{c} + 4 i \, d e f^{2} x^{3} e^{c}\right )} e^{\left (d x\right )}}{4 \, {\left (a d e^{\left (d x + c\right )} - i \, a d\right )}} - \frac {12 i \, {\left (d x \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right )\right )} e f^{2}}{a d^{3}} - \frac {6 i \, {\left (d^{2} x^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(-i \, e^{\left (d x + c\right )})\right )} f^{3}}{a d^{4}} - \frac {2 \, {\left (-i \, d^{3} f^{3} x^{3} - 3 i \, d^{3} e f^{2} x^{2}\right )}}{a d^{4}} \] Input:
integrate((f*x+e)^3*sinh(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")
Output:
3/2*e^2*f*((-I*d*x^2 + (d*x^2*e^c - 4*x*e^c)*e^(d*x))/(I*a*d*e^(d*x + c) + a*d) - 4*I*log((e^(d*x + c) - I)*e^(-c))/(a*d^2)) - e^3*(I*(d*x + c)/(a*d ) + 2/((a*e^(-d*x - c) + I*a)*d)) - 1/4*(d*f^3*x^4 + 24*e*f^2*x^2 + 4*(d*e *f^2 + 2*f^3)*x^3 + (I*d*f^3*x^4*e^c + 4*I*d*e*f^2*x^3*e^c)*e^(d*x))/(a*d* e^(d*x + c) - I*a*d) - 12*I*(d*x*log(I*e^(d*x + c) + 1) + dilog(-I*e^(d*x + c)))*e*f^2/(a*d^3) - 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)))*f^3/(a*d^4) - 2*(-I*d^3*f^3 *x^3 - 3*I*d^3*e*f^2*x^2)/(a*d^4)
\[ \int \frac {(e+f x)^3 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{3} \sinh \left (d x + c\right )}{i \, a \sinh \left (d x + c\right ) + a} \,d x } \] Input:
integrate((f*x+e)^3*sinh(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")
Output:
integrate((f*x + e)^3*sinh(d*x + c)/(I*a*sinh(d*x + c) + a), x)
Timed out. \[ \int \frac {(e+f x)^3 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\int \frac {\mathrm {sinh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^3}{a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \] Input:
int((sinh(c + d*x)*(e + f*x)^3)/(a + a*sinh(c + d*x)*1i),x)
Output:
int((sinh(c + d*x)*(e + f*x)^3)/(a + a*sinh(c + d*x)*1i), x)
\[ \int \frac {(e+f x)^3 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {4 \left (\int \frac {x^{3}}{\sinh \left (d x +c \right )-i}d x \right ) f^{3}+12 \left (\int \frac {x^{2}}{\sinh \left (d x +c \right )-i}d x \right ) e \,f^{2}+12 \left (\int \frac {x}{\sinh \left (d x +c \right )-i}d x \right ) e^{2} f +4 \left (\int \frac {1}{\sinh \left (d x +c \right )-i}d x \right ) e^{3}-4 e^{3} i x -6 e^{2} f i \,x^{2}-4 e \,f^{2} i \,x^{3}-f^{3} i \,x^{4}}{4 a} \] Input:
int((f*x+e)^3*sinh(d*x+c)/(a+I*a*sinh(d*x+c)),x)
Output:
(4*int(x**3/(sinh(c + d*x) - i),x)*f**3 + 12*int(x**2/(sinh(c + d*x) - i), x)*e*f**2 + 12*int(x/(sinh(c + d*x) - i),x)*e**2*f + 4*int(1/(sinh(c + d*x ) - i),x)*e**3 - 4*e**3*i*x - 6*e**2*f*i*x**2 - 4*e*f**2*i*x**3 - f**3*i*x **4)/(4*a)