Integrand size = 29, antiderivative size = 119 \[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {(e+f x)^2}{2 a f}-\frac {i (e+f x) \cosh (c+d x)}{a d}+\frac {2 f \log \left (\cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )\right )}{a d^2}+\frac {i f \sinh (c+d x)}{a d^2}-\frac {(e+f x) \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d} \] Output:
1/2*(f*x+e)^2/a/f-I*(f*x+e)*cosh(d*x+c)/a/d+2*f*ln(cosh(1/2*c+1/4*I*Pi+1/2 *d*x))/a/d^2+I*f*sinh(d*x+c)/a/d^2-(f*x+e)*tanh(1/2*c+1/4*I*Pi+1/2*d*x)/a/ d
Time = 2.83 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.00 \[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {\left (-i \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sinh \left (\frac {1}{2} (c+d x)\right ) \left (i (2 i+c+d x) (2 d e-c f+d f x)-4 f \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+2 d (e+f x) \cosh (c+d x)+2 i f \log (\cosh (c+d x))-2 f \sinh (c+d x)\right )+\cosh \left (\frac {1}{2} (c+d x)\right ) \left (2 c d e-2 i c f-c^2 f+2 d^2 e x-2 i d f x+d^2 f x^2+4 i f \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )-2 i d (e+f x) \cosh (c+d x)+2 f \log (\cosh (c+d x))+2 i f \sinh (c+d x)\right )\right )}{2 a d^2 (-i+\sinh (c+d x))} \] Input:
Integrate[((e + f*x)*Sinh[c + d*x]^2)/(a + I*a*Sinh[c + d*x]),x]
Output:
(((-I)*Cosh[(c + d*x)/2] + Sinh[(c + d*x)/2])*(Sinh[(c + d*x)/2]*(I*(2*I + c + d*x)*(2*d*e - c*f + d*f*x) - 4*f*ArcTan[Tanh[(c + d*x)/2]] + 2*d*(e + f*x)*Cosh[c + d*x] + (2*I)*f*Log[Cosh[c + d*x]] - 2*f*Sinh[c + d*x]) + Co sh[(c + d*x)/2]*(2*c*d*e - (2*I)*c*f - c^2*f + 2*d^2*e*x - (2*I)*d*f*x + d ^2*f*x^2 + (4*I)*f*ArcTan[Tanh[(c + d*x)/2]] - (2*I)*d*(e + f*x)*Cosh[c + d*x] + 2*f*Log[Cosh[c + d*x]] + (2*I)*f*Sinh[c + d*x])))/(2*a*d^2*(-I + Si nh[c + d*x]))
Time = 0.90 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.09, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.621, Rules used = {6091, 3042, 26, 3777, 3042, 3117, 6091, 17, 3042, 3799, 25, 25, 3042, 4672, 26, 3042, 26, 3956}
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.
| \(\displaystyle \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx\) |
| \(\Big \downarrow \) 6091 |
| \(\displaystyle i \int \frac {(e+f x) \sinh (c+d x)}{i \sinh (c+d x) a+a}dx-\frac {i \int (e+f x) \sinh (c+d x)dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle i \int \frac {(e+f x) \sinh (c+d x)}{i \sinh (c+d x) a+a}dx-\frac {i \int -i (e+f x) \sin (i c+i d x)dx}{a}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \int \frac {(e+f x) \sinh (c+d x)}{i \sinh (c+d x) a+a}dx-\frac {\int (e+f x) \sin (i c+i d x)dx}{a}\) |
| \(\Big \downarrow \) 3777 |
| \(\displaystyle i \int \frac {(e+f x) \sinh (c+d x)}{i \sinh (c+d x) a+a}dx-\frac {\frac {i (e+f x) \cosh (c+d x)}{d}-\frac {i f \int \cosh (c+d x)dx}{d}}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle i \int \frac {(e+f x) \sinh (c+d x)}{i \sinh (c+d x) a+a}dx-\frac {\frac {i (e+f x) \cosh (c+d x)}{d}-\frac {i f \int \sin \left (i c+i d x+\frac {\pi }{2}\right )dx}{d}}{a}\) |
| \(\Big \downarrow \) 3117 |
| \(\displaystyle i \int \frac {(e+f x) \sinh (c+d x)}{i \sinh (c+d x) a+a}dx-\frac {\frac {i (e+f x) \cosh (c+d x)}{d}-\frac {i f \sinh (c+d x)}{d^2}}{a}\) |
| \(\Big \downarrow \) 6091 |
| \(\displaystyle i \left (i \int \frac {e+f x}{i \sinh (c+d x) a+a}dx-\frac {i \int (e+f x)dx}{a}\right )-\frac {\frac {i (e+f x) \cosh (c+d x)}{d}-\frac {i f \sinh (c+d x)}{d^2}}{a}\) |
| \(\Big \downarrow \) 17 |
| \(\displaystyle i \left (i \int \frac {e+f x}{i \sinh (c+d x) a+a}dx-\frac {i (e+f x)^2}{2 a f}\right )-\frac {\frac {i (e+f x) \cosh (c+d x)}{d}-\frac {i f \sinh (c+d x)}{d^2}}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle i \left (i \int \frac {e+f x}{\sin (i c+i d x) a+a}dx-\frac {i (e+f x)^2}{2 a f}\right )-\frac {\frac {i (e+f x) \cosh (c+d x)}{d}-\frac {i f \sinh (c+d x)}{d^2}}{a}\) |
| \(\Big \downarrow \) 3799 |
| \(\displaystyle i \left (\frac {i \int -\left ((e+f x) \text {csch}^2\left (\frac {c}{2}+\frac {d x}{2}-\frac {i \pi }{4}\right )\right )dx}{2 a}-\frac {i (e+f x)^2}{2 a f}\right )-\frac {\frac {i (e+f x) \cosh (c+d x)}{d}-\frac {i f \sinh (c+d x)}{d^2}}{a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle i \left (-\frac {i \int -\left ((e+f x) \text {sech}^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )dx}{2 a}-\frac {i (e+f x)^2}{2 a f}\right )-\frac {\frac {i (e+f x) \cosh (c+d x)}{d}-\frac {i f \sinh (c+d x)}{d^2}}{a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle i \left (\frac {i \int (e+f x) \text {sech}^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{2 a}-\frac {i (e+f x)^2}{2 a f}\right )-\frac {\frac {i (e+f x) \cosh (c+d x)}{d}-\frac {i f \sinh (c+d x)}{d^2}}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle i \left (\frac {i \int (e+f x) \csc \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )^2dx}{2 a}-\frac {i (e+f x)^2}{2 a f}\right )-\frac {\frac {i (e+f x) \cosh (c+d x)}{d}-\frac {i f \sinh (c+d x)}{d^2}}{a}\) |
| \(\Big \downarrow \) 4672 |
| \(\displaystyle i \left (\frac {i \left (\frac {2 (e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}-\frac {2 i f \int -i \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{d}\right )}{2 a}-\frac {i (e+f x)^2}{2 a f}\right )-\frac {\frac {i (e+f x) \cosh (c+d x)}{d}-\frac {i f \sinh (c+d x)}{d^2}}{a}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \left (\frac {i \left (\frac {2 (e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}-\frac {2 f \int \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{d}\right )}{2 a}-\frac {i (e+f x)^2}{2 a f}\right )-\frac {\frac {i (e+f x) \cosh (c+d x)}{d}-\frac {i f \sinh (c+d x)}{d^2}}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle i \left (\frac {i \left (\frac {2 (e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}-\frac {2 f \int -i \tan \left (\frac {i c}{2}+\frac {i d x}{2}-\frac {\pi }{4}\right )dx}{d}\right )}{2 a}-\frac {i (e+f x)^2}{2 a f}\right )-\frac {\frac {i (e+f x) \cosh (c+d x)}{d}-\frac {i f \sinh (c+d x)}{d^2}}{a}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \left (\frac {i \left (\frac {2 i f \int \tan \left (\frac {i c}{2}+\frac {i d x}{2}-\frac {\pi }{4}\right )dx}{d}+\frac {2 (e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{2 a}-\frac {i (e+f x)^2}{2 a f}\right )-\frac {\frac {i (e+f x) \cosh (c+d x)}{d}-\frac {i f \sinh (c+d x)}{d^2}}{a}\) |
| \(\Big \downarrow \) 3956 |
| \(\displaystyle i \left (\frac {i \left (\frac {2 (e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}-\frac {4 f \log \left (\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{d^2}\right )}{2 a}-\frac {i (e+f x)^2}{2 a f}\right )-\frac {\frac {i (e+f x) \cosh (c+d x)}{d}-\frac {i f \sinh (c+d x)}{d^2}}{a}\) |
Input:
Int[((e + f*x)*Sinh[c + d*x]^2)/(a + I*a*Sinh[c + d*x]),x]
Output:
-(((I*(e + f*x)*Cosh[c + d*x])/d - (I*f*Sinh[c + d*x])/d^2)/a) + I*(((-1/2 *I)*(e + f*x)^2)/(a*f) + ((I/2)*((-4*f*Log[Cosh[c/2 + (I/4)*Pi + (d*x)/2]] )/d^2 + (2*(e + f*x)*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[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, 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[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
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]
Time = 0.90 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.13
| method | result | size |
| risch | \(\frac {f \,x^{2}}{2 a}+\frac {e x}{a}-\frac {i \left (d x f +d e -f \right ) {\mathrm e}^{d x +c}}{2 a \,d^{2}}-\frac {i \left (d x f +d e +f \right ) {\mathrm e}^{-d x -c}}{2 a \,d^{2}}-\frac {2 f x}{a d}-\frac {2 f c}{a \,d^{2}}-\frac {2 i \left (f x +e \right )}{d a \left ({\mathrm e}^{d x +c}-i\right )}+\frac {2 f \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{2}}\) | \(134\) |
| parallelrisch | \(\frac {4 f \left (i \cosh \left (\frac {d x}{2}+\frac {c}{2}\right )-\sinh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \ln \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 f \left (i \cosh \left (\frac {d x}{2}+\frac {c}{2}\right )-\sinh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\left (i+i x^{2} d^{2}+\left (3-2 i\right ) x d \right ) f +2 i d^{2} e x +7 d e \right ) \cosh \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (\left (-1-x^{2} d^{2}+\left (2-3 i\right ) x d \right ) f +e d \left (-2 d x +i\right )\right ) \sinh \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (\left (d x -i\right ) f +d e \right ) \cosh \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+\sinh \left (\frac {3 d x}{2}+\frac {3 c}{2}\right ) \left (i d f x +i e d -f \right )}{2 d^{2} a \left (i \cosh \left (\frac {d x}{2}+\frac {c}{2}\right )-\sinh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\) | \(244\) |
Input:
int((f*x+e)*sinh(d*x+c)^2/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/2/a*f*x^2+1/a*e*x-1/2*I*(d*f*x+d*e-f)/a/d^2*exp(d*x+c)-1/2*I*(d*f*x+d*e+ f)/a/d^2*exp(-d*x-c)-2*f*x/a/d-2*f/a/d^2*c-2*I*(f*x+e)/d/a/(exp(d*x+c)-I)+ 2*f/a/d^2*ln(exp(d*x+c)-I)
Time = 0.10 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.45 \[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {d f x + d e - {\left (-i \, d f x - i \, d e + i \, f\right )} e^{\left (3 \, d x + 3 \, c\right )} - {\left (d^{2} f x^{2} - d e + {\left (2 \, d^{2} e - 5 \, d f\right )} x + f\right )} e^{\left (2 \, d x + 2 \, c\right )} - {\left (-i \, d^{2} f x^{2} - 5 i \, d e + {\left (-2 i \, d^{2} e - i \, d f\right )} x - i \, f\right )} e^{\left (d x + c\right )} - 4 \, {\left (f e^{\left (2 \, d x + 2 \, c\right )} - i \, f e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + f}{2 \, {\left (a d^{2} e^{\left (2 \, d x + 2 \, c\right )} - i \, a d^{2} e^{\left (d x + c\right )}\right )}} \] Input:
integrate((f*x+e)*sinh(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")
Output:
-1/2*(d*f*x + d*e - (-I*d*f*x - I*d*e + I*f)*e^(3*d*x + 3*c) - (d^2*f*x^2 - d*e + (2*d^2*e - 5*d*f)*x + f)*e^(2*d*x + 2*c) - (-I*d^2*f*x^2 - 5*I*d*e + (-2*I*d^2*e - I*d*f)*x - I*f)*e^(d*x + c) - 4*(f*e^(2*d*x + 2*c) - I*f* e^(d*x + c))*log(e^(d*x + c) - I) + f)/(a*d^2*e^(2*d*x + 2*c) - I*a*d^2*e^ (d*x + c))
Time = 0.27 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.88 \[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {- 2 i e - 2 i f x}{a d e^{c} e^{d x} - i a d} + \begin {cases} \frac {\left (\left (- 2 i a d^{3} e - 2 i a d^{3} f x - 2 i a d^{2} f\right ) e^{- d x} + \left (- 2 i a d^{3} e e^{2 c} - 2 i a d^{3} f x e^{2 c} + 2 i a d^{2} f e^{2 c}\right ) e^{d x}\right ) e^{- c}}{4 a^{2} d^{4}} & \text {for}\: a^{2} d^{4} e^{c} \neq 0 \\\frac {x^{2} \left (- i f e^{2 c} + i f\right ) e^{- c}}{4 a} + \frac {x \left (- i e e^{2 c} + i e\right ) e^{- c}}{2 a} & \text {otherwise} \end {cases} + \frac {f x^{2}}{2 a} + \frac {x \left (d e - 2 f\right )}{a d} + \frac {2 f \log {\left (e^{d x} - i e^{- c} \right )}}{a d^{2}} \] Input:
integrate((f*x+e)*sinh(d*x+c)**2/(a+I*a*sinh(d*x+c)),x)
Output:
(-2*I*e - 2*I*f*x)/(a*d*exp(c)*exp(d*x) - I*a*d) + Piecewise((((-2*I*a*d** 3*e - 2*I*a*d**3*f*x - 2*I*a*d**2*f)*exp(-d*x) + (-2*I*a*d**3*e*exp(2*c) - 2*I*a*d**3*f*x*exp(2*c) + 2*I*a*d**2*f*exp(2*c))*exp(d*x))*exp(-c)/(4*a** 2*d**4), Ne(a**2*d**4*exp(c), 0)), (x**2*(-I*f*exp(2*c) + I*f)*exp(-c)/(4* a) + x*(-I*e*exp(2*c) + I*e)*exp(-c)/(2*a), True)) + f*x**2/(2*a) + x*(d*e - 2*f)/(a*d) + 2*f*log(exp(d*x) - I*exp(-c))/(a*d**2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (100) = 200\).
Time = 0.12 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.00 \[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {1}{2} \, f {\left (\frac {2 \, x e^{\left (d x + c\right )}}{a d e^{\left (d x + c\right )} - i \, a d} + \frac {i \, d^{2} x^{2} e^{c} + i \, d x e^{c} - {\left (-i \, d x e^{\left (3 \, c\right )} + i \, e^{\left (3 \, c\right )}\right )} e^{\left (2 \, d x\right )} - {\left (d^{2} x^{2} e^{\left (2 \, c\right )} - 3 \, d x e^{\left (2 \, c\right )} + e^{\left (2 \, c\right )}\right )} e^{\left (d x\right )} + {\left (d x + 1\right )} e^{\left (-d x\right )} + i \, e^{c}}{a d^{2} e^{\left (d x + 2 \, c\right )} - i \, a d^{2} e^{c}} - \frac {4 \, \log \left ({\left (e^{\left (d x + c\right )} - i\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} + \frac {1}{2} \, e {\left (\frac {2 \, {\left (d x + c\right )}}{a d} + \frac {-5 i \, e^{\left (-d x - c\right )} + 1}{{\left (i \, a e^{\left (-d x - c\right )} + a e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} - \frac {i \, e^{\left (-d x - c\right )}}{a d}\right )} \] Input:
integrate((f*x+e)*sinh(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")
Output:
-1/2*f*(2*x*e^(d*x + c)/(a*d*e^(d*x + c) - I*a*d) + (I*d^2*x^2*e^c + I*d*x *e^c - (-I*d*x*e^(3*c) + I*e^(3*c))*e^(2*d*x) - (d^2*x^2*e^(2*c) - 3*d*x*e ^(2*c) + e^(2*c))*e^(d*x) + (d*x + 1)*e^(-d*x) + I*e^c)/(a*d^2*e^(d*x + 2* c) - I*a*d^2*e^c) - 4*log((e^(d*x + c) - I)*e^(-c))/(a*d^2)) + 1/2*e*(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))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (100) = 200\).
Time = 0.12 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.11 \[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {d^{2} f x^{2} e^{\left (2 \, d x + 2 \, c\right )} - i \, d^{2} f x^{2} e^{\left (d x + c\right )} + 2 \, d^{2} e x e^{\left (2 \, d x + 2 \, c\right )} - 2 i \, d^{2} e x e^{\left (d x + c\right )} - i \, d f x e^{\left (3 \, d x + 3 \, c\right )} - 5 \, d f x e^{\left (2 \, d x + 2 \, c\right )} - i \, d f x e^{\left (d x + c\right )} - d f x - i \, d e e^{\left (3 \, d x + 3 \, c\right )} - d e e^{\left (2 \, d x + 2 \, c\right )} - 5 i \, d e e^{\left (d x + c\right )} + 4 \, f e^{\left (2 \, d x + 2 \, c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - 4 i \, f e^{\left (d x + c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - d e + i \, f e^{\left (3 \, d x + 3 \, c\right )} + f e^{\left (2 \, d x + 2 \, c\right )} - i \, f e^{\left (d x + c\right )} - f}{2 \, {\left (a d^{2} e^{\left (2 \, d x + 2 \, c\right )} - i \, a d^{2} e^{\left (d x + c\right )}\right )}} \] Input:
integrate((f*x+e)*sinh(d*x+c)^2/(a+I*a*sinh(d*x+c)),x, algorithm="giac")
Output:
1/2*(d^2*f*x^2*e^(2*d*x + 2*c) - I*d^2*f*x^2*e^(d*x + c) + 2*d^2*e*x*e^(2* d*x + 2*c) - 2*I*d^2*e*x*e^(d*x + c) - I*d*f*x*e^(3*d*x + 3*c) - 5*d*f*x*e ^(2*d*x + 2*c) - I*d*f*x*e^(d*x + c) - d*f*x - I*d*e*e^(3*d*x + 3*c) - d*e *e^(2*d*x + 2*c) - 5*I*d*e*e^(d*x + c) + 4*f*e^(2*d*x + 2*c)*log(e^(d*x + c) - I) - 4*I*f*e^(d*x + c)*log(e^(d*x + c) - I) - d*e + I*f*e^(3*d*x + 3* c) + f*e^(2*d*x + 2*c) - I*f*e^(d*x + c) - f)/(a*d^2*e^(2*d*x + 2*c) - I*a *d^2*e^(d*x + c))
Time = 1.45 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.20 \[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {f\,x^2}{2\,a}+{\mathrm {e}}^{c+d\,x}\,\left (\frac {\left (f-d\,e\right )\,1{}\mathrm {i}}{2\,a\,d^2}-\frac {f\,x\,1{}\mathrm {i}}{2\,a\,d}\right )-{\mathrm {e}}^{-c-d\,x}\,\left (\frac {\left (f+d\,e\right )\,1{}\mathrm {i}}{2\,a\,d^2}+\frac {f\,x\,1{}\mathrm {i}}{2\,a\,d}\right )-\frac {\left (e+f\,x\right )\,2{}\mathrm {i}}{a\,d\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )}-\frac {x\,\left (2\,f-d\,e\right )}{a\,d}+\frac {2\,f\,\ln \left ({\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-\mathrm {i}\right )}{a\,d^2} \] Input:
int((sinh(c + d*x)^2*(e + f*x))/(a + a*sinh(c + d*x)*1i),x)
Output:
exp(c + d*x)*(((f - d*e)*1i)/(2*a*d^2) - (f*x*1i)/(2*a*d)) - exp(- c - d*x )*(((f + d*e)*1i)/(2*a*d^2) + (f*x*1i)/(2*a*d)) + (f*x^2)/(2*a) - ((e + f* x)*2i)/(a*d*(exp(c + d*x) - 1i)) - (x*(2*f - d*e))/(a*d) + (2*f*log(exp(d* x)*exp(c) - 1i))/(a*d^2)
\[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {-2 \cosh \left (d x +c \right ) d e i -2 \cosh \left (d x +c \right ) d f i x -2 \left (\int \frac {x}{\sinh \left (d x +c \right ) i +1}d x \right ) d^{2} f -2 \left (\int \frac {1}{\sinh \left (d x +c \right ) i +1}d x \right ) d^{2} e +2 \sinh \left (d x +c \right ) f i +2 d^{2} e x +d^{2} f \,x^{2}}{2 a \,d^{2}} \] Input:
int((f*x+e)*sinh(d*x+c)^2/(a+I*a*sinh(d*x+c)),x)
Output:
( - 2*cosh(c + d*x)*d*e*i - 2*cosh(c + d*x)*d*f*i*x - 2*int(x/(sinh(c + d* x)*i + 1),x)*d**2*f - 2*int(1/(sinh(c + d*x)*i + 1),x)*d**2*e + 2*sinh(c + d*x)*f*i + 2*d**2*e*x + d**2*f*x**2)/(2*a*d**2)