3.2.96 \(\int \frac {\sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [196]

Optimal. Leaf size=52 \[ \frac {x}{a}-\frac {i \cosh (c+d x)}{a d}-\frac {i \cosh (c+d x)}{a d (1+i \sinh (c+d x))} \]

[Out]

x/a-I*cosh(d*x+c)/a/d-I*cosh(d*x+c)/a/d/(1+I*sinh(d*x+c))

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Rubi [A]
time = 0.07, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2825, 12, 2814, 2727} \begin {gather*} -\frac {i \cosh (c+d x)}{a d}-\frac {i \cosh (c+d x)}{a d (1+i \sinh (c+d x))}+\frac {x}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/a - (I*Cosh[c + d*x])/(a*d) - (I*Cosh[c + d*x])/(a*d*(1 + I*Sinh[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2814

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

Rule 2825

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

Rubi steps

\begin {align*} \int \frac {\sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx &=-\frac {i \cosh (c+d x)}{a d}+\frac {i \int \frac {a \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx}{a}\\ &=-\frac {i \cosh (c+d x)}{a d}+i \int \frac {\sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx\\ &=\frac {x}{a}-\frac {i \cosh (c+d x)}{a d}-\int \frac {1}{a+i a \sinh (c+d x)} \, dx\\ &=\frac {x}{a}-\frac {i \cosh (c+d x)}{a d}-\frac {i \cosh (c+d x)}{d (a+i a \sinh (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 59, normalized size = 1.13 \begin {gather*} \frac {\cosh (c+d x) \left (\frac {\sinh ^{-1}(\sinh (c+d x))}{\sqrt {\cosh ^2(c+d x)}}+\frac {-2-i \sinh (c+d x)}{-i+\sinh (c+d x)}\right )}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.91, size = 86, normalized size = 1.65

method	result	size
risch	\(\frac {x}{a}-\frac {i {\mathrm e}^{d x +c}}{2 a d}-\frac {i {\mathrm e}^{-d x -c}}{2 a d}-\frac {2 i}{d a \left ({\mathrm e}^{d x +c}-i\right )}\)	\(60\)
derivativedivides	\(\frac {\frac {8 i}{8 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-8}-\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {i}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2}{-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{a d}\)	\(86\)
default	\(\frac {\frac {8 i}{8 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-8}-\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {i}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2}{-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{a d}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/d/a*(1/8*I/(tanh(1/2*d*x+1/2*c)-1)-1/8*ln(tanh(1/2*d*x+1/2*c)-1)-1/8*I/(tanh(1/2*d*x+1/2*c)+1)+1/8*ln(tanh(1
/2*d*x+1/2*c)+1)-1/4/(-I+tanh(1/2*d*x+1/2*c)))

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Maxima [A]
time = 0.27, size = 74, normalized size = 1.42 \begin {gather*} \frac {d x + c}{a d} + \frac {-5 i \, e^{\left (-d x - c\right )} + 1}{2 \, {\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 )}}{2 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.39, size = 69, normalized size = 1.33 \begin {gather*} \frac {{\left (2 \, d x - 1\right )} e^{\left (2 \, d x + 2 \, c\right )} + {\left (-2 i \, d x - 5 i\right )} e^{\left (d x + c\right )} - i \, e^{\left (3 \, d x + 3 \, c\right )} - 1}{2 \, {\left (a d e^{\left (2 \, d x + 2 \, c\right )} - i \, a d e^{\left (d x + c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.16, size = 99, normalized size = 1.90 \begin {gather*} \begin {cases} \frac {\left (- 2 i a d e^{2 c} e^{d x} - 2 i a d e^{- d x}\right ) e^{- c}}{4 a^{2} d^{2}} & \text {for}\: a^{2} d^{2} e^{c} \neq 0 \\x \left (\frac {\left (- i e^{2 c} + 2 e^{c} + i\right ) e^{- c}}{2 a} - \frac {1}{a}\right ) & \text {otherwise} \end {cases} - \frac {2 i}{a d e^{c} e^{d x} - i a d} + \frac {x}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-2*I*a*d*exp(2*c)*exp(d*x) - 2*I*a*d*exp(-d*x))*exp(-c)/(4*a**2*d**2), Ne(a**2*d**2*exp(c), 0)), (
x*((-I*exp(2*c) + 2*exp(c) + I)*exp(-c)/(2*a) - 1/a), True)) - 2*I/(a*d*exp(c)*exp(d*x) - I*a*d) + x/a

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Giac [A]
time = 0.44, size = 63, normalized size = 1.21 \begin {gather*} \frac {\frac {2 \, {\left (d x + c\right )}}{a} - \frac {i \, e^{\left (d x + c\right )}}{a} - \frac {{\left (5 \, e^{\left (d x + c\right )} - i\right )} e^{\left (-d x - c\right )}}{a {\left (-i \, e^{\left (d x + c\right )} - 1\right )}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.30, size = 59, normalized size = 1.13 \begin {gather*} \frac {x}{a}-\frac {2{}\mathrm {i}}{a\,d\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )}-\frac {{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}}{2\,a\,d}-\frac {{\mathrm {e}}^{-c-d\,x}\,1{}\mathrm {i}}{2\,a\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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