\(\int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [195]

Optimal result

Integrand size = 29, antiderivative size = 119 \[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {e x}{a}+\frac {f x^2}{2 a}-\frac {i (e+f x) \cosh (c+d x)}{a d}+\frac {2 f \log \left (\cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )\right )}{a d^2}+\frac {i f \sinh (c+d x)}{a d^2}-\frac {(e+f x) \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d} \]

[Out]

e*x/a+1/2*f*x^2/a-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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {5676, 3377, 2717, 3399, 4269, 3556} \[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {i f \sinh (c+d x)}{a d^2}+\frac {2 f \log \left (\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{a d^2}-\frac {i (e+f x) \cosh (c+d x)}{a d}-\frac {(e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}+\frac {e x}{a}+\frac {f x^2}{2 a} \]

[In]

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

[Out]

(e*x)/a + (f*x^2)/(2*a) - (I*(e + f*x)*Cosh[c + d*x])/(a*d) + (2*f*Log[Cosh[c/2 + (I/4)*Pi + (d*x)/2]])/(a*d^2
) + (I*f*Sinh[c + d*x])/(a*d^2) - ((e + f*x)*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(a*d)

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3399

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(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])

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4269

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Cot[e + f*x]/f), x
] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 5676

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/b, Int[(e + f*x)^m*Sinh[c + d*x]^(n - 1), x], x] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = i \int \frac {(e+f x) \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx-\frac {i \int (e+f x) \sinh (c+d x) \, dx}{a} \\ & = -\frac {i (e+f x) \cosh (c+d x)}{a d}+\frac {\int (e+f x) \, dx}{a}+\frac {(i f) \int \cosh (c+d x) \, dx}{a d}-\int \frac {e+f x}{a+i a \sinh (c+d x)} \, dx \\ & = \frac {e x}{a}+\frac {f x^2}{2 a}-\frac {i (e+f x) \cosh (c+d x)}{a d}+\frac {i f \sinh (c+d x)}{a d^2}-\frac {\int (e+f x) \csc ^2\left (\frac {1}{2} \left (i c+\frac {\pi }{2}\right )+\frac {i d x}{2}\right ) \, dx}{2 a} \\ & = \frac {e x}{a}+\frac {f x^2}{2 a}-\frac {i (e+f x) \cosh (c+d x)}{a d}+\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}+\frac {f \int \coth \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx}{a d} \\ & = \frac {e x}{a}+\frac {f x^2}{2 a}-\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.45 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.00 \[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {\left (-i \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sinh \left (\frac {1}{2} (c+d x)\right ) \left (i (2 i+c+d x) (2 d e-c f+d f x)-4 f \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+2 d (e+f x) \cosh (c+d x)+2 i f \log (\cosh (c+d x))-2 f \sinh (c+d x)\right )+\cosh \left (\frac {1}{2} (c+d x)\right ) \left (2 c d e-2 i c f-c^2 f+2 d^2 e x-2 i d f x+d^2 f x^2+4 i f \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )-2 i d (e+f x) \cosh (c+d x)+2 f \log (\cosh (c+d x))+2 i f \sinh (c+d x)\right )\right )}{2 a d^2 (-i+\sinh (c+d x))} \]

[In]

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

[Out]

(((-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]) +
Cosh[(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
+ Sinh[c + d*x]))

Maple [A] (verified)

Time = 2.06 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.13

method	result	size
risch	\(\frac {f \,x^{2}}{2 a}+\frac {e x}{a}-\frac {i \left (d f x +d e -f \right ) {\mathrm e}^{d x +c}}{2 a \,d^{2}}-\frac {i \left (d f x +d e +f \right ) {\mathrm e}^{-d x -c}}{2 a \,d^{2}}-\frac {2 f x}{a d}-\frac {2 f c}{a \,d^{2}}-\frac {2 i \left (f x +e \right )}{d a \left ({\mathrm e}^{d x +c}-i\right )}+\frac {2 f \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{2}}\)	\(134\)
parallelrisch	\(\frac {-4 f \left (i \cosh \left (\frac {d x}{2}+\frac {c}{2}\right )-\sinh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \ln \left (1-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 f \left (i \cosh \left (\frac {d x}{2}+\frac {c}{2}\right )-\sinh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \ln \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\left (i+i d^{2} x^{2}+\left (3-2 i\right ) x d \right ) f +2 i d^{2} e x +7 d e \right ) \cosh \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (\left (-1-d^{2} x^{2}+\left (2-3 i\right ) x d \right ) f +e d \left (-2 d x +i\right )\right ) \sinh \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (\left (d x -i\right ) f +d e \right ) \cosh \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+\sinh \left (\frac {3 d x}{2}+\frac {3 c}{2}\right ) \left (i d f x +i d e -f \right )}{2 d^{2} a \left (i \cosh \left (\frac {d x}{2}+\frac {c}{2}\right )-\sinh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(246\)

[In]

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

[Out]

1/2*f*x^2/a+e*x/a-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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.45 \[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {d f x + d e - {\left (-i \, d f x - i \, d e + i \, f\right )} e^{\left (3 \, d x + 3 \, c\right )} - {\left (d^{2} f x^{2} - d e + {\left (2 \, d^{2} e - 5 \, d f\right )} x + f\right )} e^{\left (2 \, d x + 2 \, c\right )} - {\left (-i \, d^{2} f x^{2} - 5 i \, d e + {\left (-2 i \, d^{2} e - i \, d f\right )} x - i \, f\right )} e^{\left (d x + c\right )} - 4 \, {\left (f e^{\left (2 \, d x + 2 \, c\right )} - i \, f e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + f}{2 \, {\left (a d^{2} e^{\left (2 \, d x + 2 \, c\right )} - i \, a d^{2} e^{\left (d x + c\right )}\right )}} \]

[In]

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

[Out]

-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))

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.88 \[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {- 2 i e - 2 i f x}{a d e^{c} e^{d x} - i a d} + \begin {cases} \frac {\left (\left (- 2 i a d^{3} e - 2 i a d^{3} f x - 2 i a d^{2} f\right ) e^{- d x} + \left (- 2 i a d^{3} e e^{2 c} - 2 i a d^{3} f x e^{2 c} + 2 i a d^{2} f e^{2 c}\right ) e^{d x}\right ) e^{- c}}{4 a^{2} d^{4}} & \text {for}\: a^{2} d^{4} e^{c} \neq 0 \\\frac {x^{2} \left (- i f e^{2 c} + i f\right ) e^{- c}}{4 a} + \frac {x \left (- i e e^{2 c} + i e\right ) e^{- c}}{2 a} & \text {otherwise} \end {cases} + \frac {f x^{2}}{2 a} + \frac {x \left (d e - 2 f\right )}{a d} + \frac {2 f \log {\left (e^{d x} - i e^{- c} \right )}}{a d^{2}} \]

[In]

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

[Out]

(-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)

Maxima [B] (verification not implemented)

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 = 0.25 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.00 \[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {1}{2} \, f {\left (\frac {2 \, x e^{\left (d x + c\right )}}{a d e^{\left (d x + c\right )} - i \, a d} + \frac {i \, d^{2} x^{2} e^{c} + i \, d x e^{c} - {\left (-i \, d x e^{\left (3 \, c\right )} + i \, e^{\left (3 \, c\right )}\right )} e^{\left (2 \, d x\right )} - {\left (d^{2} x^{2} e^{\left (2 \, c\right )} - 3 \, d x e^{\left (2 \, c\right )} + e^{\left (2 \, c\right )}\right )} e^{\left (d x\right )} + {\left (d x + 1\right )} e^{\left (-d x\right )} + i \, e^{c}}{a d^{2} e^{\left (d x + 2 \, c\right )} - i \, a d^{2} e^{c}} - \frac {4 \, \log \left ({\left (e^{\left (d x + c\right )} - i\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} + \frac {1}{2} \, e {\left (\frac {2 \, {\left (d x + c\right )}}{a d} + \frac {-5 i \, e^{\left (-d x - c\right )} + 1}{{\left (i \, a e^{\left (-d x - c\right )} + a e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} - \frac {i \, e^{\left (-d x - c\right )}}{a d}\right )} \]

[In]

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

[Out]

-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))

Giac [B] (verification not implemented)

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 = 0.28 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.11 \[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {d^{2} f x^{2} e^{\left (2 \, d x + 2 \, c\right )} - i \, d^{2} f x^{2} e^{\left (d x + c\right )} + 2 \, d^{2} e x e^{\left (2 \, d x + 2 \, c\right )} - 2 i \, d^{2} e x e^{\left (d x + c\right )} - i \, d f x e^{\left (3 \, d x + 3 \, c\right )} - 5 \, d f x e^{\left (2 \, d x + 2 \, c\right )} - i \, d f x e^{\left (d x + c\right )} - d f x - i \, d e e^{\left (3 \, d x + 3 \, c\right )} - d e e^{\left (2 \, d x + 2 \, c\right )} - 5 i \, d e e^{\left (d x + c\right )} + 4 \, f e^{\left (2 \, d x + 2 \, c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - 4 i \, f e^{\left (d x + c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - d e + i \, f e^{\left (3 \, d x + 3 \, c\right )} + f e^{\left (2 \, d x + 2 \, c\right )} - i \, f e^{\left (d x + c\right )} - f}{2 \, {\left (a d^{2} e^{\left (2 \, d x + 2 \, c\right )} - i \, a d^{2} e^{\left (d x + c\right )}\right )}} \]

[In]

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

[Out]

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))

Mupad [B] (verification not implemented)

Time = 1.31 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.20 \[ \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {f\,x^2}{2\,a}+{\mathrm {e}}^{c+d\,x}\,\left (\frac {\left (f-d\,e\right )\,1{}\mathrm {i}}{2\,a\,d^2}-\frac {f\,x\,1{}\mathrm {i}}{2\,a\,d}\right )-{\mathrm {e}}^{-c-d\,x}\,\left (\frac {\left (f+d\,e\right )\,1{}\mathrm {i}}{2\,a\,d^2}+\frac {f\,x\,1{}\mathrm {i}}{2\,a\,d}\right )-\frac {\left (e+f\,x\right )\,2{}\mathrm {i}}{a\,d\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )}-\frac {x\,\left (2\,f-d\,e\right )}{a\,d}+\frac {2\,f\,\ln \left ({\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-\mathrm {i}\right )}{a\,d^2} \]

[In]

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

[Out]

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*1
i)/(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)