3.196 \(\int \frac{\sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.08879, antiderivative size = 52, normalized size of antiderivative = 1., 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 = {2746, 12, 2735, 2648} \[ -\frac{i \cosh (c+d x)}{a d}-\frac{i \cosh (c+d x)}{a d (1+i \sinh (c+d x))}+\frac{x}{a} \]

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 2746

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]

Rule 12

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

Rule 2735

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 2648

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]

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*}

Mathematica [A]  time = 0.220904, size = 59, normalized size = 1.13 \[ \frac{\cosh (c+d x) \left (\frac{\sinh ^{-1}(\sinh (c+d x))}{\sqrt{\cosh ^2(c+d x)}}+\frac{-2-i \sinh (c+d x)}{\sinh (c+d x)-i}\right )}{a d} \]

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 [B]  time = 0.036, size = 107, normalized size = 2.1 \begin{align*} -2\,{\frac{1}{da \left ( -i+\tanh \left ( 1/2\,dx+c/2 \right ) \right ) }}-{\frac{i}{da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{i}{da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

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)

[Out]

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

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Maxima [A]  time = 1.00892, size = 100, normalized size = 1.92 \begin{align*} \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{align*}

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 = 2.46326, size = 178, normalized size = 3.42 \begin{align*} \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 \, a d e^{\left (2 \, d x + 2 \, c\right )} - 2 i \, a d e^{\left (d x + c\right )}} \end{align*}

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]

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

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Sympy [A]  time = 0.747901, size = 102, normalized size = 1.96 \begin{align*} \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}\: 4 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{x}{a} - \frac{2 i e^{- c}}{a d \left (e^{d x} - i e^{- c}\right )} \end{align*}

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

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Giac [A]  time = 1.41128, size = 89, normalized size = 1.71 \begin{align*} \frac{d x + c}{a d} - \frac{i \, e^{\left (d x + c\right )}}{2 \, a d} + \frac{{\left (5 \, e^{\left (d x + c\right )} - i\right )} e^{\left (-d x - c\right )}}{2 \, a d{\left (i \, e^{\left (d x + c\right )} + 1\right )}} \end{align*}

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]

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