3.89 \(\int \frac{1}{(3+5 i \sinh (c+d x))^2} \, dx\)

Optimal. Leaf size=102 \[ \frac{5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}-\frac{3 i \log \left (3 \cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}+\frac{3 i \log \left (\cosh \left (\frac{1}{2} (c+d x)\right )+3 i \sinh \left (\frac{1}{2} (c+d x)\right )\right )}{64 d} \]

[Out]

(((-3*I)/64)*Log[3*Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]])/d + (((3*I)/64)*Log[Cosh[(c + d*x)/2] + (3*I)*Sin
h[(c + d*x)/2]])/d + (((5*I)/16)*Cosh[c + d*x])/(d*(3 + (5*I)*Sinh[c + d*x]))

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Rubi [A]  time = 0.0508041, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {2664, 12, 2660, 616, 31} \[ \frac{5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}-\frac{3 i \log \left (3 \cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}+\frac{3 i \log \left (\cosh \left (\frac{1}{2} (c+d x)\right )+3 i \sinh \left (\frac{1}{2} (c+d x)\right )\right )}{64 d} \]

Antiderivative was successfully verified.

[In]

Int[(3 + (5*I)*Sinh[c + d*x])^(-2),x]

[Out]

(((-3*I)/64)*Log[3*Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]])/d + (((3*I)/64)*Log[Cosh[(c + d*x)/2] + (3*I)*Sin
h[(c + d*x)/2]])/d + (((5*I)/16)*Cosh[c + d*x])/(d*(3 + (5*I)*Sinh[c + d*x]))

Rule 2664

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

Rule 12

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{1}{(3+5 i \sinh (c+d x))^2} \, dx &=\frac{5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}+\frac{1}{16} \int -\frac{3}{3+5 i \sinh (c+d x)} \, dx\\ &=\frac{5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}-\frac{3}{16} \int \frac{1}{3+5 i \sinh (c+d x)} \, dx\\ &=\frac{5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}+\frac{(3 i) \operatorname{Subst}\left (\int \frac{1}{3+10 x+3 x^2} \, dx,x,\tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{8 d}\\ &=\frac{5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}+\frac{(9 i) \operatorname{Subst}\left (\int \frac{1}{1+3 x} \, dx,x,\tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{64 d}-\frac{(9 i) \operatorname{Subst}\left (\int \frac{1}{9+3 x} \, dx,x,\tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{64 d}\\ &=-\frac{3 i \log \left (3+i \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}+\frac{3 i \log \left (1+3 i \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}+\frac{5 i \cosh (c+d x)}{16 d (3+5 i \sinh (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.245723, size = 142, normalized size = 1.39 \[ \frac{40 \sinh \left (\frac{1}{2} (c+d x)\right ) \left (\frac{3}{\cosh \left (\frac{1}{2} (c+d x)\right )+3 i \sinh \left (\frac{1}{2} (c+d x)\right )}+\frac{1}{3 \cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )}\right )-9 \left (2 \tan ^{-1}\left (3 \tanh \left (\frac{1}{2} (c+d x)\right )\right )-i \log (4-5 \cosh (c+d x))+i \log (5 \cosh (c+d x)+4)+2 \tan ^{-1}\left (3 \coth \left (\frac{1}{2} (c+d x)\right )\right )\right )}{384 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + (5*I)*Sinh[c + d*x])^(-2),x]

[Out]

(-9*(2*ArcTan[3*Coth[(c + d*x)/2]] + 2*ArcTan[3*Tanh[(c + d*x)/2]] - I*Log[4 - 5*Cosh[c + d*x]] + I*Log[4 + 5*
Cosh[c + d*x]]) + 40*((3*Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^(-1) + 3/(Cosh[(c + d*x)/2] + (3*I)*Sinh[(c
+ d*x)/2]))*Sinh[(c + d*x)/2])/(384*d)

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Maple [A]  time = 0.043, size = 82, normalized size = 0.8 \begin{align*}{\frac{{\frac{3\,i}{64}}}{d}\ln \left ( 3\,\tanh \left ( 1/2\,dx+c/2 \right ) -i \right ) }+{\frac{5}{48\,d} \left ( 3\,\tanh \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-1}}-{\frac{{\frac{3\,i}{64}}}{d}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -3\,i \right ) }+{\frac{5}{16\,d} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -3\,i \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3+5*I*sinh(d*x+c))^2,x)

[Out]

3/64*I/d*ln(3*tanh(1/2*d*x+1/2*c)-I)+5/48/d/(3*tanh(1/2*d*x+1/2*c)-I)-3/64*I/d*ln(tanh(1/2*d*x+1/2*c)-3*I)+5/1
6/d/(tanh(1/2*d*x+1/2*c)-3*I)

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Maxima [A]  time = 1.60047, size = 107, normalized size = 1.05 \begin{align*} \frac{3 i \, \log \left (\frac{10 \, e^{\left (-d x - c\right )} + 6 i - 8}{10 \, e^{\left (-d x - c\right )} + 6 i + 8}\right )}{64 \, d} + \frac{3 i \, e^{\left (-d x - c\right )} - 5}{-8 \, d{\left (-6 i \, e^{\left (-d x - c\right )} - 5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 5\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*I*sinh(d*x+c))^2,x, algorithm="maxima")

[Out]

3/64*I*log((10*e^(-d*x - c) + 6*I - 8)/(10*e^(-d*x - c) + 6*I + 8))/d + (3*I*e^(-d*x - c) - 5)/(d*(48*I*e^(-d*
x - c) + 40*e^(-2*d*x - 2*c) - 40))

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Fricas [A]  time = 2.1833, size = 323, normalized size = 3.17 \begin{align*} \frac{{\left (-15 i \, e^{\left (2 \, d x + 2 \, c\right )} - 18 \, e^{\left (d x + c\right )} + 15 i\right )} \log \left (e^{\left (d x + c\right )} - \frac{3}{5} i + \frac{4}{5}\right ) +{\left (15 i \, e^{\left (2 \, d x + 2 \, c\right )} + 18 \, e^{\left (d x + c\right )} - 15 i\right )} \log \left (e^{\left (d x + c\right )} - \frac{3}{5} i - \frac{4}{5}\right ) + 24 i \, e^{\left (d x + c\right )} + 40}{64 \,{\left (5 \, d e^{\left (2 \, d x + 2 \, c\right )} - 6 i \, d e^{\left (d x + c\right )} - 5 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*I*sinh(d*x+c))^2,x, algorithm="fricas")

[Out]

1/64*((-15*I*e^(2*d*x + 2*c) - 18*e^(d*x + c) + 15*I)*log(e^(d*x + c) - 3/5*I + 4/5) + (15*I*e^(2*d*x + 2*c) +
18*e^(d*x + c) - 15*I)*log(e^(d*x + c) - 3/5*I - 4/5) + 24*I*e^(d*x + c) + 40)/(5*d*e^(2*d*x + 2*c) - 6*I*d*e
^(d*x + c) - 5*d)

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Sympy [A]  time = 1.17428, size = 87, normalized size = 0.85 \begin{align*} \frac{\frac{3 i e^{- c} e^{d x}}{40 d} + \frac{e^{- 2 c}}{8 d}}{e^{2 d x} - \frac{6 i e^{- c} e^{d x}}{5} - e^{- 2 c}} + \frac{\operatorname{RootSum}{\left (4096 z^{2} + 9, \left ( i \mapsto i \log{\left (\frac{256 i i e^{- c}}{15} + e^{d x} - \frac{3 i e^{- c}}{5} \right )} \right )\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*I*sinh(d*x+c))**2,x)

[Out]

(3*I*exp(-c)*exp(d*x)/(40*d) + exp(-2*c)/(8*d))/(exp(2*d*x) - 6*I*exp(-c)*exp(d*x)/5 - exp(-2*c)) + RootSum(40
96*_z**2 + 9, Lambda(_i, _i*log(256*_i*I*exp(-c)/15 + exp(d*x) - 3*I*exp(-c)/5)))/d

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Giac [A]  time = 1.60727, size = 96, normalized size = 0.94 \begin{align*} -\frac{3 i \, \log \left (-\left (i - 2\right ) \, e^{\left (d x + c\right )} - 2 i + 1\right )}{64 \, d} + \frac{3 i \, \log \left (-\left (2 i - 1\right ) \, e^{\left (d x + c\right )} + i - 2\right )}{64 \, d} + \frac{3 i \, e^{\left (d x + c\right )} + 5}{8 \, d{\left (5 \, e^{\left (2 \, d x + 2 \, c\right )} - 6 i \, e^{\left (d x + c\right )} - 5\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*I*sinh(d*x+c))^2,x, algorithm="giac")

[Out]

-3/64*I*log(-(I - 2)*e^(d*x + c) - 2*I + 1)/d + 3/64*I*log(-(2*I - 1)*e^(d*x + c) + I - 2)/d + 1/8*(3*I*e^(d*x
+ c) + 5)/(d*(5*e^(2*d*x + 2*c) - 6*I*e^(d*x + c) - 5))