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

Optimal. Leaf size=66 \[ -\frac{3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}-\frac{5 i \tan ^{-1}\left (\frac{\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{32 d}+\frac{5 x}{64} \]

[Out]

(5*x)/64 - (((5*I)/32)*ArcTan[Cosh[c + d*x]/(3 + I*Sinh[c + d*x])])/d - (((3*I)/16)*Cosh[c + d*x])/(d*(5 + (3*
I)*Sinh[c + d*x]))

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Rubi [A]  time = 0.0346976, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2664, 12, 2657} \[ -\frac{3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}-\frac{5 i \tan ^{-1}\left (\frac{\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{32 d}+\frac{5 x}{64} \]

Antiderivative was successfully verified.

[In]

Int[(5 + (3*I)*Sinh[c + d*x])^(-2),x]

[Out]

(5*x)/64 - (((5*I)/32)*ArcTan[Cosh[c + d*x]/(3 + I*Sinh[c + d*x])])/d - (((3*I)/16)*Cosh[c + d*x])/(d*(5 + (3*
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 2657

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

Rubi steps

\begin{align*} \int \frac{1}{(5+3 i \sinh (c+d x))^2} \, dx &=-\frac{3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}-\frac{1}{16} \int -\frac{5}{5+3 i \sinh (c+d x)} \, dx\\ &=-\frac{3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}+\frac{5}{16} \int \frac{1}{5+3 i \sinh (c+d x)} \, dx\\ &=\frac{5 x}{64}-\frac{5 i \tan ^{-1}\left (\frac{\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{32 d}-\frac{3 i \cosh (c+d x)}{16 d (5+3 i \sinh (c+d x))}\\ \end{align*}

Mathematica [B]  time = 0.228439, size = 183, normalized size = 2.77 \[ \frac{-\frac{120 \cosh (c+d x)}{3 \sinh (c+d x)-5 i}-25 \log (5 \cosh (c+d x)-4 \sinh (c+d x))+25 \log (4 \sinh (c+d x)+5 \cosh (c+d x))-50 i \tan ^{-1}\left (\frac{2 \cosh \left (\frac{1}{2} (c+d x)\right )-\sinh \left (\frac{1}{2} (c+d x)\right )}{\cosh \left (\frac{1}{2} (c+d x)\right )-2 \sinh \left (\frac{1}{2} (c+d x)\right )}\right )+50 i \tan ^{-1}\left (\frac{2 \sinh \left (\frac{1}{2} (c+d x)\right )+\cosh \left (\frac{1}{2} (c+d x)\right )}{\sinh \left (\frac{1}{2} (c+d x)\right )+2 \cosh \left (\frac{1}{2} (c+d x)\right )}\right )+24 i}{640 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*I - (50*I)*ArcTan[(2*Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2])/(Cosh[(c + d*x)/2] - 2*Sinh[(c + d*x)/2])] + (
50*I)*ArcTan[(Cosh[(c + d*x)/2] + 2*Sinh[(c + d*x)/2])/(2*Cosh[(c + d*x)/2] + Sinh[(c + d*x)/2])] - 25*Log[5*C
osh[c + d*x] - 4*Sinh[c + d*x]] + 25*Log[5*Cosh[c + d*x] + 4*Sinh[c + d*x]] - (120*Cosh[c + d*x])/(-5*I + 3*Si
nh[c + d*x]))/(640*d)

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Maple [B]  time = 0.043, size = 134, normalized size = 2. \begin{align*} -{\frac{9}{80\,d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) +4-3\,i \right ) ^{-1}}-{\frac{{\frac{3\,i}{20}}}{d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) +4-3\,i \right ) ^{-1}}+{\frac{5}{64\,d}\ln \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) +4-3\,i \right ) }-{\frac{9}{80\,d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) -4-3\,i \right ) ^{-1}}+{\frac{{\frac{3\,i}{20}}}{d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) -4-3\,i \right ) ^{-1}}-{\frac{5}{64\,d}\ln \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) -4-3\,i \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/80/d/(5*tanh(1/2*d*x+1/2*c)+4-3*I)-3/20*I/d/(5*tanh(1/2*d*x+1/2*c)+4-3*I)+5/64/d*ln(5*tanh(1/2*d*x+1/2*c)+4
-3*I)-9/80/d/(5*tanh(1/2*d*x+1/2*c)-4-3*I)+3/20*I/d/(5*tanh(1/2*d*x+1/2*c)-4-3*I)-5/64/d*ln(5*tanh(1/2*d*x+1/2
*c)-4-3*I)

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Maxima [A]  time = 1.78205, size = 86, normalized size = 1.3 \begin{align*} -\frac{5 i \, \arctan \left (\frac{3}{4} \, e^{\left (-d x - c\right )} + \frac{5}{4} i\right )}{32 \, d} - \frac{5 i \, e^{\left (-d x - c\right )} - 3}{-8 \, d{\left (-10 i \, e^{\left (-d x - c\right )} - 3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/32*I*arctan(3/4*e^(-d*x - c) + 5/4*I)/d - (5*I*e^(-d*x - c) - 3)/(d*(80*I*e^(-d*x - c) + 24*e^(-2*d*x - 2*c
) - 24))

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Fricas [A]  time = 2.08619, size = 298, normalized size = 4.52 \begin{align*} \frac{5 \,{\left (3 \, e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, e^{\left (d x + c\right )} - 3\right )} \log \left (e^{\left (d x + c\right )} - \frac{1}{3} i\right ) - 5 \,{\left (3 \, e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, e^{\left (d x + c\right )} - 3\right )} \log \left (e^{\left (d x + c\right )} - 3 i\right ) - 40 i \, e^{\left (d x + c\right )} - 24}{64 \,{\left (3 \, d e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, d e^{\left (d x + c\right )} - 3 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(5*(3*e^(2*d*x + 2*c) - 10*I*e^(d*x + c) - 3)*log(e^(d*x + c) - 1/3*I) - 5*(3*e^(2*d*x + 2*c) - 10*I*e^(d
*x + c) - 3)*log(e^(d*x + c) - 3*I) - 40*I*e^(d*x + c) - 24)/(3*d*e^(2*d*x + 2*c) - 10*I*d*e^(d*x + c) - 3*d)

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Sympy [A]  time = 1.11671, size = 88, normalized size = 1.33 \begin{align*} \frac{- \frac{5 i e^{- c} e^{d x}}{24 d} - \frac{e^{- 2 c}}{8 d}}{e^{2 d x} - \frac{10 i e^{- c} e^{d x}}{3} - e^{- 2 c}} + \frac{- \frac{5 \log{\left (e^{d x} - 3 i e^{- c} \right )}}{64} + \frac{5 \log{\left (e^{d x} - \frac{i e^{- c}}{3} \right )}}{64}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-5*I*exp(-c)*exp(d*x)/(24*d) - exp(-2*c)/(8*d))/(exp(2*d*x) - 10*I*exp(-c)*exp(d*x)/3 - exp(-2*c)) + (-5*log(
exp(d*x) - 3*I*exp(-c))/64 + 5*log(exp(d*x) - I*exp(-c)/3)/64)/d

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Giac [A]  time = 1.24642, size = 93, normalized size = 1.41 \begin{align*} \frac{5 \, \log \left (3 \, e^{\left (d x + c\right )} - i\right )}{64 \, d} - \frac{5 \, \log \left (e^{\left (d x + c\right )} - 3 i\right )}{64 \, d} + \frac{-5 i \, e^{\left (d x + c\right )} - 3}{8 \, d{\left (3 \, e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, e^{\left (d x + c\right )} - 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/64*log(3*e^(d*x + c) - I)/d - 5/64*log(e^(d*x + c) - 3*I)/d + 1/8*(-5*I*e^(d*x + c) - 3)/(d*(3*e^(2*d*x + 2*
c) - 10*I*e^(d*x + c) - 3))