Integrand size = 27, antiderivative size = 90 \[ \int \frac {(e+f x) \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {i e x}{a}-\frac {i f x^2}{2 a}-\frac {2 i f \log \left (\cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )\right )}{a d^2}+\frac {i (e+f x) \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d} \]
-I*e*x/a-1/2*I*f*x^2/a-2*I*f*ln(cosh(1/2*c+1/4*I*Pi+1/2*d*x))/a/d^2+I*(f*x +e)*tanh(1/2*c+1/4*I*Pi+1/2*d*x)/a/d
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(239\) vs. \(2(90)=180\).
Time = 0.75 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.66 \[ \int \frac {(e+f x) \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {-2 d f x \cosh \left (c+\frac {d x}{2}\right )-i \cosh \left (\frac {d x}{2}\right ) \left (d^2 x (2 e+f x)+4 i f \arctan \left (\text {sech}\left (c+\frac {d x}{2}\right ) \sinh \left (\frac {d x}{2}\right )\right )+2 f \log (\cosh (c+d x))\right )+4 i d e \sinh \left (\frac {d x}{2}\right )+2 i d f x \sinh \left (\frac {d x}{2}\right )+2 d^2 e x \sinh \left (c+\frac {d x}{2}\right )+d^2 f x^2 \sinh \left (c+\frac {d x}{2}\right )+4 i f \arctan \left (\text {sech}\left (c+\frac {d x}{2}\right ) \sinh \left (\frac {d x}{2}\right )\right ) \sinh \left (c+\frac {d x}{2}\right )+2 f \log (\cosh (c+d x)) \sinh \left (c+\frac {d x}{2}\right )}{2 a d^2 \left (\cosh \left (\frac {c}{2}\right )+i \sinh \left (\frac {c}{2}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )} \]
(-2*d*f*x*Cosh[c + (d*x)/2] - I*Cosh[(d*x)/2]*(d^2*x*(2*e + f*x) + (4*I)*f *ArcTan[Sech[c + (d*x)/2]*Sinh[(d*x)/2]] + 2*f*Log[Cosh[c + d*x]]) + (4*I) *d*e*Sinh[(d*x)/2] + (2*I)*d*f*x*Sinh[(d*x)/2] + 2*d^2*e*x*Sinh[c + (d*x)/ 2] + d^2*f*x^2*Sinh[c + (d*x)/2] + (4*I)*f*ArcTan[Sech[c + (d*x)/2]*Sinh[( d*x)/2]]*Sinh[c + (d*x)/2] + 2*f*Log[Cosh[c + d*x]]*Sinh[c + (d*x)/2])/(2* a*d^2*(Cosh[c/2] + I*Sinh[c/2])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]))
Time = 0.47 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.97, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {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.
\(\displaystyle \int \frac {(e+f x) \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx\) |
\(\Big \downarrow \) 6091 |
\(\displaystyle i \int \frac {e+f x}{i \sinh (c+d x) a+a}dx-\frac {i \int (e+f x)dx}{a}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle i \int \frac {e+f x}{i \sinh (c+d x) a+a}dx-\frac {i (e+f x)^2}{2 a f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle i \int \frac {e+f x}{\sin (i c+i d x) a+a}dx-\frac {i (e+f x)^2}{2 a f}\) |
\(\Big \downarrow \) 3799 |
\(\displaystyle \frac {i \int -\left ((e+f x) \text {csch}^2\left (\frac {c}{2}+\frac {d x}{2}-\frac {i \pi }{4}\right )\right )dx}{2 a}-\frac {i (e+f x)^2}{2 a f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {i \int -\left ((e+f x) \text {sech}^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )dx}{2 a}-\frac {i (e+f x)^2}{2 a f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {i \int (e+f x) \text {sech}^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{2 a}-\frac {i (e+f x)^2}{2 a f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {i \int (e+f x) \csc \left (\frac {i c}{2}+\frac {i d x}{2}+\frac {\pi }{4}\right )^2dx}{2 a}-\frac {i (e+f x)^2}{2 a f}\) |
\(\Big \downarrow \) 4672 |
\(\displaystyle \frac {i \left (\frac {2 (e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}-\frac {2 i f \int -i \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{d}\right )}{2 a}-\frac {i (e+f x)^2}{2 a f}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {i \left (\frac {2 (e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}-\frac {2 f \int \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )dx}{d}\right )}{2 a}-\frac {i (e+f x)^2}{2 a f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {i \left (\frac {2 (e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}-\frac {2 f \int -i \tan \left (\frac {i c}{2}+\frac {i d x}{2}-\frac {\pi }{4}\right )dx}{d}\right )}{2 a}-\frac {i (e+f x)^2}{2 a f}\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {i \left (\frac {2 i f \int \tan \left (\frac {i c}{2}+\frac {i d x}{2}-\frac {\pi }{4}\right )dx}{d}+\frac {2 (e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}\right )}{2 a}-\frac {i (e+f x)^2}{2 a f}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {i \left (\frac {2 (e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{d}-\frac {4 f \log \left (\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{d^2}\right )}{2 a}-\frac {i (e+f x)^2}{2 a f}\) |
((-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
3.2.89.3.1 Defintions of rubi rules used
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[((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 = 1.89 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.96
method | result | size |
risch | \(-\frac {i f \,x^{2}}{2 a}-\frac {i e x}{a}+\frac {2 i f x}{a d}+\frac {2 i f c}{a \,d^{2}}-\frac {2 \left (f x +e \right )}{d a \left ({\mathrm e}^{d x +c}-i\right )}-\frac {2 i f \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{2}}\) | \(86\) |
parallelrisch | \(\frac {2 f \left (-1-i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \ln \left (1-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 f \left (i \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \ln \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+d \left (x \left (\left (\frac {1}{2} i x f +i e \right ) d +\left (-1-i\right ) f \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+x \left (\frac {f x}{2}+e \right ) d +\left (-1-i\right ) x f -2 i e \right )}{d^{2} a \left (i-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\) | \(137\) |
-1/2*I*f*x^2/a-I*e*x/a+2*I*f/a/d*x+2*I*f/a/d^2*c-2*(f*x+e)/d/a/(exp(d*x+c) -I)-2*I*f/a/d^2*ln(exp(d*x+c)-I)
Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.07 \[ \int \frac {(e+f x) \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {d^{2} f x^{2} + 2 \, d^{2} e x + 4 \, d e - {\left (-i \, d^{2} f x^{2} - 2 \, {\left (i \, d^{2} e - 2 i \, d f\right )} x\right )} e^{\left (d x + c\right )} + 4 \, {\left (i \, f e^{\left (d x + c\right )} + f\right )} \log \left (e^{\left (d x + c\right )} - i\right )}{2 \, {\left (a d^{2} e^{\left (d x + c\right )} - i \, a d^{2}\right )}} \]
-1/2*(d^2*f*x^2 + 2*d^2*e*x + 4*d*e - (-I*d^2*f*x^2 - 2*(I*d^2*e - 2*I*d*f )*x)*e^(d*x + c) + 4*(I*f*e^(d*x + c) + f)*log(e^(d*x + c) - I))/(a*d^2*e^ (d*x + c) - I*a*d^2)
Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.81 \[ \int \frac {(e+f x) \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {- 2 e - 2 f x}{a d e^{c} e^{d x} - i a d} - \frac {i f x^{2}}{2 a} + \frac {x \left (- i d e + 2 i f\right )}{a d} - \frac {2 i f \log {\left (e^{d x} - i e^{- c} \right )}}{a d^{2}} \]
(-2*e - 2*f*x)/(a*d*exp(c)*exp(d*x) - I*a*d) - I*f*x**2/(2*a) + x*(-I*d*e + 2*I*f)/(a*d) - 2*I*f*log(exp(d*x) - I*exp(-c))/(a*d**2)
Time = 0.22 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.20 \[ \int \frac {(e+f x) \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {1}{2} \, f {\left (\frac {-i \, d x^{2} + {\left (d x^{2} e^{c} - 4 \, x e^{c}\right )} e^{\left (d x\right )}}{i \, a d e^{\left (d x + c\right )} + a d} - \frac {4 i \, \log \left ({\left (e^{\left (d x + c\right )} - i\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} - e {\left (\frac {i \, {\left (d x + c\right )}}{a d} + \frac {2}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d}\right )} \]
1/2*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*(I*(d*x + c)/(a*d) + 2/ ((a*e^(-d*x - c) + I*a)*d))
Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.23 \[ \int \frac {(e+f x) \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {i \, d^{2} f x^{2} e^{\left (d x + c\right )} + d^{2} f x^{2} + 2 i \, d^{2} e x e^{\left (d x + c\right )} + 2 \, d^{2} e x - 4 i \, d f x e^{\left (d x + c\right )} + 4 i \, f e^{\left (d x + c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 4 \, d e + 4 \, f \log \left (e^{\left (d x + c\right )} - i\right )}{2 \, {\left (a d^{2} e^{\left (d x + c\right )} - i \, a d^{2}\right )}} \]
-1/2*(I*d^2*f*x^2*e^(d*x + c) + d^2*f*x^2 + 2*I*d^2*e*x*e^(d*x + c) + 2*d^ 2*e*x - 4*I*d*f*x*e^(d*x + c) + 4*I*f*e^(d*x + c)*log(e^(d*x + c) - I) + 4 *d*e + 4*f*log(e^(d*x + c) - I))/(a*d^2*e^(d*x + c) - I*a*d^2)
Time = 1.30 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.82 \[ \int \frac {(e+f x) \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {f\,x^2\,1{}\mathrm {i}}{2\,a}-\frac {2\,\left (e+f\,x\right )}{a\,d\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )}+\frac {x\,\left (2\,f-d\,e\right )\,1{}\mathrm {i}}{a\,d}-\frac {f\,\ln \left ({\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-\mathrm {i}\right )\,2{}\mathrm {i}}{a\,d^2} \]