3.3.1 \(\int \frac {(e+f x) \sinh ^3(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [201]

Optimal. Leaf size=175 \[ \frac {3 i e x}{2 a}+\frac {3 i f x^2}{4 a}+\frac {(e+f x) \cosh (c+d x)}{a d}+\frac {2 i f \log \left (\cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )\right )}{a d^2}-\frac {f \sinh (c+d x)}{a d^2}-\frac {i (e+f x) \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac {i f \sinh ^2(c+d x)}{4 a d^2}-\frac {i (e+f x) \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d} \]

[Out]

3/2*I*e*x/a+3/4*I*f*x^2/a+(f*x+e)*cosh(d*x+c)/a/d+2*I*f*ln(cosh(1/2*c+1/4*I*Pi+1/2*d*x))/a/d^2-f*sinh(d*x+c)/a
/d^2-1/2*I*(f*x+e)*cosh(d*x+c)*sinh(d*x+c)/a/d+1/4*I*f*sinh(d*x+c)^2/a/d^2-I*(f*x+e)*tanh(1/2*c+1/4*I*Pi+1/2*d
*x)/a/d

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Rubi [A]
time = 0.19, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {5676, 3391, 3377, 2717, 3399, 4269, 3556} \begin {gather*} \frac {i f \sinh ^2(c+d x)}{4 a d^2}-\frac {f \sinh (c+d x)}{a d^2}+\frac {2 i f \log \left (\cosh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )\right )}{a d^2}+\frac {(e+f x) \cosh (c+d x)}{a d}-\frac {i (e+f x) \tanh \left (\frac {c}{2}+\frac {d x}{2}+\frac {i \pi }{4}\right )}{a d}-\frac {i (e+f x) \sinh (c+d x) \cosh (c+d x)}{2 a d}+\frac {3 i e x}{2 a}+\frac {3 i f x^2}{4 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e + f*x)*Sinh[c + d*x]^3)/(a + I*a*Sinh[c + d*x]),x]

[Out]

(((3*I)/2)*e*x)/a + (((3*I)/4)*f*x^2)/a + ((e + f*x)*Cosh[c + d*x])/(a*d) + ((2*I)*f*Log[Cosh[c/2 + (I/4)*Pi +
(d*x)/2]])/(a*d^2) - (f*Sinh[c + d*x])/(a*d^2) - ((I/2)*(e + f*x)*Cosh[c + d*x]*Sinh[c + d*x])/(a*d) + ((I/4)
*f*Sinh[c + d*x]^2)/(a*d^2) - (I*(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 3391

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^
2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[b*(c + d*x)*Cos[e + f*x
]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

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*} \int \frac {(e+f x) \sinh ^3(c+d x)}{a+i a \sinh (c+d x)} \, dx &=i \int \frac {(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx-\frac {i \int (e+f x) \sinh ^2(c+d x) \, dx}{a}\\ &=-\frac {i (e+f x) \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac {i f \sinh ^2(c+d x)}{4 a d^2}+\frac {i \int (e+f x) \, dx}{2 a}+\frac {\int (e+f x) \sinh (c+d x) \, dx}{a}-\int \frac {(e+f x) \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx\\ &=\frac {i e x}{2 a}+\frac {i f x^2}{4 a}+\frac {(e+f x) \cosh (c+d x)}{a d}-\frac {i (e+f x) \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac {i f \sinh ^2(c+d x)}{4 a d^2}-i \int \frac {e+f x}{a+i a \sinh (c+d x)} \, dx+\frac {i \int (e+f x) \, dx}{a}-\frac {f \int \cosh (c+d x) \, dx}{a d}\\ &=\frac {3 i e x}{2 a}+\frac {3 i f x^2}{4 a}+\frac {(e+f x) \cosh (c+d x)}{a d}-\frac {f \sinh (c+d x)}{a d^2}-\frac {i (e+f x) \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac {i f \sinh ^2(c+d x)}{4 a d^2}-\frac {i \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 {3 i e x}{2 a}+\frac {3 i f x^2}{4 a}+\frac {(e+f x) \cosh (c+d x)}{a d}-\frac {f \sinh (c+d x)}{a d^2}-\frac {i (e+f x) \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac {i f \sinh ^2(c+d x)}{4 a d^2}-\frac {i (e+f x) \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}+\frac {(i f) \int \coth \left (\frac {c}{2}-\frac {i \pi }{4}+\frac {d x}{2}\right ) \, dx}{a d}\\ &=\frac {3 i e x}{2 a}+\frac {3 i f x^2}{4 a}+\frac {(e+f x) \cosh (c+d x)}{a d}+\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 {f \sinh (c+d x)}{a d^2}-\frac {i (e+f x) \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac {i f \sinh ^2(c+d x)}{4 a d^2}-\frac {i (e+f x) \tanh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )}{a d}\\ \end {align*}

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Mathematica [A]
time = 1.13, size = 325, normalized size = 1.86 \begin {gather*} \frac {\left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right ) \left (-8 i d (e+f x) \cosh (c+d x)+f \cosh (2 (c+d x))+2 \left (6 c d e-4 i c f-3 c^2 f+6 d^2 e x-4 i d f x+3 d^2 f x^2+8 i f \text {ArcTan}\left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+4 f \log (\cosh (c+d x))+4 i f \sinh (c+d x)-d (e+f x) \sinh (2 (c+d x))\right )\right )+\sinh \left (\frac {1}{2} (c+d x)\right ) \left (8 d (e+f x) \cosh (c+d x)+i \left (f \cosh (2 (c+d x))+2 \left (8 i d e+6 c d e-4 i c f-3 c^2 f+6 d^2 e x+4 i d f x+3 d^2 f x^2+8 i f \text {ArcTan}\left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+4 f \log (\cosh (c+d x))+4 i f \sinh (c+d x)-d (e+f x) \sinh (2 (c+d x))\right )\right )\right )\right )}{8 a d^2 (-i+\sinh (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)*Sinh[c + d*x]^3)/(a + I*a*Sinh[c + d*x]),x]

[Out]

((Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])*(Cosh[(c + d*x)/2]*((-8*I)*d*(e + f*x)*Cosh[c + d*x] + f*Cosh[2*(c
+ d*x)] + 2*(6*c*d*e - (4*I)*c*f - 3*c^2*f + 6*d^2*e*x - (4*I)*d*f*x + 3*d^2*f*x^2 + (8*I)*f*ArcTan[Tanh[(c +
d*x)/2]] + 4*f*Log[Cosh[c + d*x]] + (4*I)*f*Sinh[c + d*x] - d*(e + f*x)*Sinh[2*(c + d*x)])) + Sinh[(c + d*x)/2
]*(8*d*(e + f*x)*Cosh[c + d*x] + I*(f*Cosh[2*(c + d*x)] + 2*((8*I)*d*e + 6*c*d*e - (4*I)*c*f - 3*c^2*f + 6*d^2
*e*x + (4*I)*d*f*x + 3*d^2*f*x^2 + (8*I)*f*ArcTan[Tanh[(c + d*x)/2]] + 4*f*Log[Cosh[c + d*x]] + (4*I)*f*Sinh[c
+ d*x] - d*(e + f*x)*Sinh[2*(c + d*x)])))))/(8*a*d^2*(-I + Sinh[c + d*x]))

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Maple [A]
time = 2.60, size = 197, normalized size = 1.13

method	result	size
risch	\(\frac {3 i f \,x^{2}}{4 a}+\frac {3 i e x}{2 a}-\frac {i \left (2 d x f +2 d e -f \right ) {\mathrm e}^{2 d x +2 c}}{16 a \,d^{2}}+\frac {\left (d x f +d e -f \right ) {\mathrm e}^{d x +c}}{2 a \,d^{2}}+\frac {\left (d x f +d e +f \right ) {\mathrm e}^{-d x -c}}{2 a \,d^{2}}+\frac {i \left (2 d x f +2 d e +f \right ) {\mathrm e}^{-2 d x -2 c}}{16 a \,d^{2}}-\frac {2 i f x}{a d}-\frac {2 i f c}{a \,d^{2}}+\frac {2 f x +2 e}{d a \left ({\mathrm e}^{d x +c}-i\right )}+\frac {2 i f \ln \left ({\mathrm e}^{d x +c}-i\right )}{a \,d^{2}}\)	\(197\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*sinh(d*x+c)^3/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

3/4*I*f*x^2/a+3/2*I*e*x/a-1/16*I*(2*d*f*x+2*d*e-f)/a/d^2*exp(2*d*x+2*c)+1/2*(d*f*x+d*e-f)/a/d^2*exp(d*x+c)+1/2
*(d*f*x+d*e+f)/a/d^2*exp(-d*x-c)+1/16*I*(2*d*f*x+2*d*e+f)/a/d^2*exp(-2*d*x-2*c)-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)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sinh(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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Fricas [A]
time = 0.41, size = 237, normalized size = 1.35 \begin {gather*} \frac {2 \, d f x + 2 \, d e + {\left (-2 i \, d f x - 2 i \, d e + i \, f\right )} e^{\left (5 \, d x + 5 \, c\right )} + {\left (6 \, d f x + 6 \, d e - 7 \, f\right )} e^{\left (4 \, d x + 4 \, c\right )} - 4 \, {\left (-3 i \, d^{2} f x^{2} + 10 i \, d f x + 2 \, {\left (-3 i \, d^{2} x + i \, d\right )} e - 2 i \, f\right )} e^{\left (3 \, d x + 3 \, c\right )} + 4 \, {\left (3 \, d^{2} f x^{2} + 2 \, d f x + 2 \, {\left (3 \, d^{2} x + 5 \, d\right )} e + 2 \, f\right )} e^{\left (2 \, d x + 2 \, c\right )} + {\left (-6 i \, d f x - 6 i \, d e - 7 i \, f\right )} e^{\left (d x + c\right )} - 32 \, {\left (-i \, f e^{\left (3 \, d x + 3 \, c\right )} - f e^{\left (2 \, d x + 2 \, c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + f}{16 \, {\left (a d^{2} e^{\left (3 \, d x + 3 \, c\right )} - i \, a d^{2} e^{\left (2 \, d x + 2 \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sinh(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(2*d*f*x + 2*d*e + (-2*I*d*f*x - 2*I*d*e + I*f)*e^(5*d*x + 5*c) + (6*d*f*x + 6*d*e - 7*f)*e^(4*d*x + 4*c)
- 4*(-3*I*d^2*f*x^2 + 10*I*d*f*x + 2*(-3*I*d^2*x + I*d)*e - 2*I*f)*e^(3*d*x + 3*c) + 4*(3*d^2*f*x^2 + 2*d*f*x
+ 2*(3*d^2*x + 5*d)*e + 2*f)*e^(2*d*x + 2*c) + (-6*I*d*f*x - 6*I*d*e - 7*I*f)*e^(d*x + c) - 32*(-I*f*e^(3*d*x
+ 3*c) - f*e^(2*d*x + 2*c))*log(e^(d*x + c) - I) + f)/(a*d^2*e^(3*d*x + 3*c) - I*a*d^2*e^(2*d*x + 2*c))

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Sympy [A]
time = 0.50, size = 396, normalized size = 2.26 \begin {gather*} \frac {2 e + 2 f x}{a d e^{c} e^{d x} - i a d} + \begin {cases} \frac {\left (\left (512 a^{3} d^{7} e e^{2 c} + 512 a^{3} d^{7} f x e^{2 c} + 512 a^{3} d^{6} f e^{2 c}\right ) e^{- d x} + \left (512 a^{3} d^{7} e e^{4 c} + 512 a^{3} d^{7} f x e^{4 c} - 512 a^{3} d^{6} f e^{4 c}\right ) e^{d x} + \left (128 i a^{3} d^{7} e e^{c} + 128 i a^{3} d^{7} f x e^{c} + 64 i a^{3} d^{6} f e^{c}\right ) e^{- 2 d x} + \left (- 128 i a^{3} d^{7} e e^{5 c} - 128 i a^{3} d^{7} f x e^{5 c} + 64 i a^{3} d^{6} f e^{5 c}\right ) e^{2 d x}\right ) e^{- 3 c}}{1024 a^{4} d^{8}} & \text {for}\: a^{4} d^{8} e^{3 c} \neq 0 \\\frac {x^{2} \left (- i f e^{4 c} + 2 f e^{3 c} - 2 f e^{c} - i f\right ) e^{- 2 c}}{8 a} + \frac {x \left (- i e e^{4 c} + 2 e e^{3 c} - 2 e e^{c} - i e\right ) e^{- 2 c}}{4 a} & \text {otherwise} \end {cases} + \frac {3 i f x^{2}}{4 a} + \frac {x \left (3 i d e - 4 i f\right )}{2 a d} + \frac {2 i f \log {\left (e^{d x} - i e^{- c} \right )}}{a d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sinh(d*x+c)**3/(a+I*a*sinh(d*x+c)),x)

[Out]

(2*e + 2*f*x)/(a*d*exp(c)*exp(d*x) - I*a*d) + Piecewise((((512*a**3*d**7*e*exp(2*c) + 512*a**3*d**7*f*x*exp(2*
c) + 512*a**3*d**6*f*exp(2*c))*exp(-d*x) + (512*a**3*d**7*e*exp(4*c) + 512*a**3*d**7*f*x*exp(4*c) - 512*a**3*d
**6*f*exp(4*c))*exp(d*x) + (128*I*a**3*d**7*e*exp(c) + 128*I*a**3*d**7*f*x*exp(c) + 64*I*a**3*d**6*f*exp(c))*e
xp(-2*d*x) + (-128*I*a**3*d**7*e*exp(5*c) - 128*I*a**3*d**7*f*x*exp(5*c) + 64*I*a**3*d**6*f*exp(5*c))*exp(2*d*
x))*exp(-3*c)/(1024*a**4*d**8), Ne(a**4*d**8*exp(3*c), 0)), (x**2*(-I*f*exp(4*c) + 2*f*exp(3*c) - 2*f*exp(c) -
I*f)*exp(-2*c)/(8*a) + x*(-I*e*exp(4*c) + 2*e*exp(3*c) - 2*e*exp(c) - I*e)*exp(-2*c)/(4*a), True)) + 3*I*f*x*
*2/(4*a) + x*(3*I*d*e - 4*I*f)/(2*a*d) + 2*I*f*log(exp(d*x) - I*exp(-c))/(a*d**2)

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (145) = 290\).
time = 0.45, size = 341, normalized size = 1.95 \begin {gather*} \frac {12 i \, d^{2} f x^{2} e^{\left (3 \, d x + 3 \, c\right )} + 12 \, d^{2} f x^{2} e^{\left (2 \, d x + 2 \, c\right )} + 24 i \, d^{2} e x e^{\left (3 \, d x + 3 \, c\right )} + 24 \, d^{2} e x e^{\left (2 \, d x + 2 \, c\right )} - 2 i \, d f x e^{\left (5 \, d x + 5 \, c\right )} + 6 \, d f x e^{\left (4 \, d x + 4 \, c\right )} - 40 i \, d f x e^{\left (3 \, d x + 3 \, c\right )} + 8 \, d f x e^{\left (2 \, d x + 2 \, c\right )} - 6 i \, d f x e^{\left (d x + c\right )} + 2 \, d f x - 2 i \, d e e^{\left (5 \, d x + 5 \, c\right )} + 6 \, d e e^{\left (4 \, d x + 4 \, c\right )} - 8 i \, d e e^{\left (3 \, d x + 3 \, c\right )} + 40 \, d e e^{\left (2 \, d x + 2 \, c\right )} - 6 i \, d e e^{\left (d x + c\right )} + 32 i \, f e^{\left (3 \, d x + 3 \, c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 32 \, f e^{\left (2 \, d x + 2 \, c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 2 \, d e + i \, f e^{\left (5 \, d x + 5 \, c\right )} - 7 \, f e^{\left (4 \, d x + 4 \, c\right )} + 8 i \, f e^{\left (3 \, d x + 3 \, c\right )} + 8 \, f e^{\left (2 \, d x + 2 \, c\right )} - 7 i \, f e^{\left (d x + c\right )} + f}{16 \, {\left (a d^{2} e^{\left (3 \, d x + 3 \, c\right )} - i \, a d^{2} e^{\left (2 \, d x + 2 \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sinh(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

1/16*(12*I*d^2*f*x^2*e^(3*d*x + 3*c) + 12*d^2*f*x^2*e^(2*d*x + 2*c) + 24*I*d^2*e*x*e^(3*d*x + 3*c) + 24*d^2*e*
x*e^(2*d*x + 2*c) - 2*I*d*f*x*e^(5*d*x + 5*c) + 6*d*f*x*e^(4*d*x + 4*c) - 40*I*d*f*x*e^(3*d*x + 3*c) + 8*d*f*x
*e^(2*d*x + 2*c) - 6*I*d*f*x*e^(d*x + c) + 2*d*f*x - 2*I*d*e*e^(5*d*x + 5*c) + 6*d*e*e^(4*d*x + 4*c) - 8*I*d*e
*e^(3*d*x + 3*c) + 40*d*e*e^(2*d*x + 2*c) - 6*I*d*e*e^(d*x + c) + 32*I*f*e^(3*d*x + 3*c)*log(e^(d*x + c) - I)
+ 32*f*e^(2*d*x + 2*c)*log(e^(d*x + c) - I) + 2*d*e + I*f*e^(5*d*x + 5*c) - 7*f*e^(4*d*x + 4*c) + 8*I*f*e^(3*d
*x + 3*c) + 8*f*e^(2*d*x + 2*c) - 7*I*f*e^(d*x + c) + f)/(a*d^2*e^(3*d*x + 3*c) - I*a*d^2*e^(2*d*x + 2*c))

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Mupad [B]
time = 0.71, size = 215, normalized size = 1.23 \begin {gather*} {\mathrm {e}}^{-c-d\,x}\,\left (\frac {f+d\,e}{2\,a\,d^2}+\frac {f\,x}{2\,a\,d}\right )+{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (\frac {\left (f+2\,d\,e\right )\,1{}\mathrm {i}}{16\,a\,d^2}+\frac {f\,x\,1{}\mathrm {i}}{8\,a\,d}\right )+{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (\frac {\left (f-2\,d\,e\right )\,1{}\mathrm {i}}{16\,a\,d^2}-\frac {f\,x\,1{}\mathrm {i}}{8\,a\,d}\right )-{\mathrm {e}}^{c+d\,x}\,\left (\frac {f-d\,e}{2\,a\,d^2}-\frac {f\,x}{2\,a\,d}\right )+\frac {f\,x^2\,3{}\mathrm {i}}{4\,a}+\frac {2\,\left (e+f\,x\right )}{a\,d\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )}-\frac {x\,\left (4\,f-3\,d\,e\right )\,1{}\mathrm {i}}{2\,a\,d}+\frac {f\,\ln \left ({\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-\mathrm {i}\right )\,2{}\mathrm {i}}{a\,d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sinh(c + d*x)^3*(e + f*x))/(a + a*sinh(c + d*x)*1i),x)

[Out]

exp(- c - d*x)*((f + d*e)/(2*a*d^2) + (f*x)/(2*a*d)) + exp(- 2*c - 2*d*x)*(((f + 2*d*e)*1i)/(16*a*d^2) + (f*x*
1i)/(8*a*d)) + exp(2*c + 2*d*x)*(((f - 2*d*e)*1i)/(16*a*d^2) - (f*x*1i)/(8*a*d)) - exp(c + d*x)*((f - d*e)/(2*
a*d^2) - (f*x)/(2*a*d)) + (f*x^2*3i)/(4*a) + (2*(e + f*x))/(a*d*(exp(c + d*x) - 1i)) - (x*(4*f - 3*d*e)*1i)/(2
*a*d) + (f*log(exp(d*x)*exp(c) - 1i)*2i)/(a*d^2)

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