3.3.2 \(\int \frac {\sinh ^3(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [202]

Optimal. Leaf size=83 \[ \frac {3 i x}{2 a}+\frac {2 \cosh (c+d x)}{a d}-\frac {3 i \cosh (c+d x) \sinh (c+d x)}{2 a d}-\frac {\cosh (c+d x) \sinh ^2(c+d x)}{d (a+i a \sinh (c+d x))} \]

[Out]

3/2*I*x/a+2*cosh(d*x+c)/a/d-3/2*I*cosh(d*x+c)*sinh(d*x+c)/a/d-cosh(d*x+c)*sinh(d*x+c)^2/d/(a+I*a*sinh(d*x+c))

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Rubi [A]
time = 0.06, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2846, 2813} \begin {gather*} \frac {2 \cosh (c+d x)}{a d}-\frac {\sinh ^2(c+d x) \cosh (c+d x)}{d (a+i a \sinh (c+d x))}-\frac {3 i \sinh (c+d x) \cosh (c+d x)}{2 a d}+\frac {3 i x}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)/2)*x)/a + (2*Cosh[c + d*x])/(a*d) - (((3*I)/2)*Cosh[c + d*x]*Sinh[c + d*x])/(a*d) - (Cosh[c + d*x]*Sin
h[c + d*x]^2)/(d*(a + I*a*Sinh[c + d*x]))

Rule 2813

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*a*c +
b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Cos[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2846

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
b*c - a*d))*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(a + b*Sin[e + f*x]))), x] - Dist[d/(a*b), Int[(c
+ d*Sin[e + f*x])^(n - 2)*Simp[b*d*(n - 1) - a*c*n + (b*c*(n - 1) - a*d*n)*Sin[e + f*x], x], x], x] /; FreeQ[{
a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 1] && (IntegerQ
[2*n] || EqQ[c, 0])

Rubi steps

\begin {align*} \int \frac {\sinh ^3(c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\frac {\cosh (c+d x) \sinh ^2(c+d x)}{d (a+i a \sinh (c+d x))}+\frac {\int \sinh (c+d x) (2 a-3 i a \sinh (c+d x)) \, dx}{a^2}\\ &=\frac {3 i x}{2 a}+\frac {2 \cosh (c+d x)}{a d}-\frac {3 i \cosh (c+d x) \sinh (c+d x)}{2 a d}-\frac {\cosh (c+d x) \sinh ^2(c+d x)}{d (a+i a \sinh (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 109, normalized size = 1.31 \begin {gather*} \frac {\cosh (c+d x) \left (3 \sinh ^{-1}(\sinh (c+d x)) \sqrt {1+i \sinh (c+d x)}+\sqrt {1-i \sinh (c+d x)} \left (-4 i+\sinh (c+d x)-i \sinh ^2(c+d x)\right )\right )}{2 a d \sqrt {1-i \sinh (c+d x)} (-i+\sinh (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[c + d*x]*(3*ArcSinh[Sinh[c + d*x]]*Sqrt[1 + I*Sinh[c + d*x]] + Sqrt[1 - I*Sinh[c + d*x]]*(-4*I + Sinh[c
+ d*x] - I*Sinh[c + d*x]^2)))/(2*a*d*Sqrt[1 - I*Sinh[c + d*x]]*(-I + Sinh[c + d*x]))

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Maple [A]
time = 1.06, size = 123, normalized size = 1.48

method	result	size
risch	\(\frac {3 i x}{2 a}-\frac {i {\mathrm e}^{2 d x +2 c}}{8 a d}+\frac {{\mathrm e}^{d x +c}}{2 a d}+\frac {{\mathrm e}^{-d x -c}}{2 a d}+\frac {i {\mathrm e}^{-2 d x -2 c}}{8 a d}+\frac {2}{d a \left ({\mathrm e}^{d x +c}-i\right )}\)	\(95\)
derivativedivides	\(\frac {-\frac {2 i}{-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}-\frac {i}{2 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {16 \left (-\frac {1}{16}-\frac {i}{32}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {i}{2 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {3 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {16 \left (\frac {1}{16}-\frac {i}{32}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{a d}\)	\(123\)
default	\(\frac {-\frac {2 i}{-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {3 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}-\frac {i}{2 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {16 \left (-\frac {1}{16}-\frac {i}{32}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {i}{2 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {3 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {16 \left (\frac {1}{16}-\frac {i}{32}\right )}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{a d}\)	\(123\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/d/a*(-1/8*I/(-I+tanh(1/2*d*x+1/2*c))-3/32*I*ln(tanh(1/2*d*x+1/2*c)-1)-1/32*I/(tanh(1/2*d*x+1/2*c)-1)^2-(1/1
6+1/32*I)/(tanh(1/2*d*x+1/2*c)-1)+1/32*I/(tanh(1/2*d*x+1/2*c)+1)^2+3/32*I*ln(tanh(1/2*d*x+1/2*c)+1)+(1/16-1/32
*I)/(tanh(1/2*d*x+1/2*c)+1))

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Maxima [A]
time = 0.27, size = 98, normalized size = 1.18 \begin {gather*} \frac {3 i \, {\left (d x + c\right )}}{2 \, a d} + \frac {3 i \, e^{\left (-d x - c\right )} + 20 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{8 \, {\left (i \, a e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-3 \, d x - 3 \, c\right )}\right )} d} + \frac {i \, {\left (-4 i \, e^{\left (-d x - c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}}{8 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*I*(d*x + c)/(a*d) + 1/8*(3*I*e^(-d*x - c) + 20*e^(-2*d*x - 2*c) + 1)/((I*a*e^(-2*d*x - 2*c) + a*e^(-3*d*x
- 3*c))*d) + 1/8*I*(-4*I*e^(-d*x - c) + e^(-2*d*x - 2*c))/(a*d)

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Fricas [A]
time = 0.39, size = 96, normalized size = 1.16 \begin {gather*} -\frac {4 \, {\left (-3 i \, d x + i\right )} e^{\left (3 \, d x + 3 \, c\right )} - 4 \, {\left (3 \, d x + 5\right )} e^{\left (2 \, d x + 2 \, c\right )} + i \, e^{\left (5 \, d x + 5 \, c\right )} - 3 \, e^{\left (4 \, d x + 4 \, c\right )} + 3 i \, e^{\left (d x + c\right )} - 1}{8 \, {\left (a d e^{\left (3 \, d x + 3 \, c\right )} - i \, a d e^{\left (2 \, d x + 2 \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*(-3*I*d*x + I)*e^(3*d*x + 3*c) - 4*(3*d*x + 5)*e^(2*d*x + 2*c) + I*e^(5*d*x + 5*c) - 3*e^(4*d*x + 4*c)
+ 3*I*e^(d*x + c) - 1)/(a*d*e^(3*d*x + 3*c) - I*a*d*e^(2*d*x + 2*c))

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Sympy [A]
time = 0.24, size = 175, normalized size = 2.11 \begin {gather*} \begin {cases} \frac {\left (- 32 i a^{3} d^{3} e^{5 c} e^{2 d x} + 128 a^{3} d^{3} e^{4 c} e^{d x} + 128 a^{3} d^{3} e^{2 c} e^{- d x} + 32 i a^{3} d^{3} e^{c} e^{- 2 d x}\right ) e^{- 3 c}}{256 a^{4} d^{4}} & \text {for}\: a^{4} d^{4} e^{3 c} \neq 0 \\x \left (\frac {\left (- i e^{4 c} + 2 e^{3 c} + 6 i e^{2 c} - 2 e^{c} - i\right ) e^{- 2 c}}{4 a} - \frac {3 i}{2 a}\right ) & \text {otherwise} \end {cases} + \frac {2}{a d e^{c} e^{d x} - i a d} + \frac {3 i x}{2 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-32*I*a**3*d**3*exp(5*c)*exp(2*d*x) + 128*a**3*d**3*exp(4*c)*exp(d*x) + 128*a**3*d**3*exp(2*c)*exp
(-d*x) + 32*I*a**3*d**3*exp(c)*exp(-2*d*x))*exp(-3*c)/(256*a**4*d**4), Ne(a**4*d**4*exp(3*c), 0)), (x*((-I*exp
(4*c) + 2*exp(3*c) + 6*I*exp(2*c) - 2*exp(c) - I)*exp(-2*c)/(4*a) - 3*I/(2*a)), True)) + 2/(a*d*exp(c)*exp(d*x
) - I*a*d) + 3*I*x/(2*a)

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Giac [A]
time = 0.46, size = 87, normalized size = 1.05 \begin {gather*} -\frac {-\frac {12 i \, {\left (d x + c\right )}}{a} - \frac {{\left (20 \, e^{\left (2 \, d x + 2 \, c\right )} - 3 i \, e^{\left (d x + c\right )} + 1\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{a {\left (e^{\left (d x + c\right )} - i\right )}} + \frac {i \, a e^{\left (2 \, d x + 2 \, c\right )} - 4 \, a e^{\left (d x + c\right )}}{a^{2}}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(-12*I*(d*x + c)/a - (20*e^(2*d*x + 2*c) - 3*I*e^(d*x + c) + 1)*e^(-2*d*x - 2*c)/(a*(e^(d*x + c) - I)) +
(I*a*e^(2*d*x + 2*c) - 4*a*e^(d*x + c))/a^2)/d

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Mupad [B]
time = 0.35, size = 94, normalized size = 1.13 \begin {gather*} \frac {x\,3{}\mathrm {i}}{2\,a}+\frac {2}{a\,d\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )}+\frac {{\mathrm {e}}^{c+d\,x}}{2\,a\,d}+\frac {{\mathrm {e}}^{-c-d\,x}}{2\,a\,d}+\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,1{}\mathrm {i}}{8\,a\,d}-\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,1{}\mathrm {i}}{8\,a\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*3i)/(2*a) + 2/(a*d*(exp(c + d*x) - 1i)) + exp(c + d*x)/(2*a*d) + exp(- c - d*x)/(2*a*d) + (exp(- 2*c - 2*d*
x)*1i)/(8*a*d) - (exp(2*c + 2*d*x)*1i)/(8*a*d)

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