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]

-3/64*I*ln(3*cosh(1/2*d*x+1/2*c)+I*sinh(1/2*d*x+1/2*c))/d+3/64*I*ln(cosh(1/2*d*x+1/2*c)+3*I*sinh(1/2*d*x+1/2*c
))/d+5/16*I*cosh(d*x+c)/d/(3+5*I*sinh(d*x+c))

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Rubi [A]  time = 0.05, antiderivative size = 102, normalized size of antiderivative = 1.00, 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 12

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

Rule 31

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

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 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 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]

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*}

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Mathematica [A]  time = 0.26, 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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fricas [A]  time = 0.62, size = 100, normalized size = 0.98 \[ \frac {{\left (-15 i \, e^{\left (2 \, d x + 2 \, c\right )} - 18 \, e^{\left (d x + c\right )} + 15 i\right )} \log \left (e^{\left (d x + c\right )} - \frac {3}{5} i + \frac {4}{5}\right ) + {\left (15 i \, e^{\left (2 \, d x + 2 \, c\right )} + 18 \, e^{\left (d x + c\right )} - 15 i\right )} \log \left (e^{\left (d x + c\right )} - \frac {3}{5} i - \frac {4}{5}\right ) + 24 i \, e^{\left (d x + c\right )} + 40}{320 \, d e^{\left (2 \, d x + 2 \, c\right )} - 384 i \, d e^{\left (d x + c\right )} - 320 \, d} \]

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]

((-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)/(320*d*e^(2*d*x + 2*c) - 384*I*d*e^
(d*x + c) - 320*d)

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giac [A]  time = 0.16, size = 67, normalized size = 0.66 \[ -\frac {\frac {8 \, {\left (-3 i \, e^{\left (d x + c\right )} - 5\right )}}{5 \, e^{\left (2 \, d x + 2 \, c\right )} - 6 i \, e^{\left (d x + c\right )} - 5} + 3 i \, \log \left (-\left (i - 2\right ) \, e^{\left (d x + c\right )} - 2 i + 1\right ) - 3 i \, \log \left (-\left (2 i - 1\right ) \, e^{\left (d x + c\right )} + i - 2\right )}{64 \, d} \]

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]

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

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maple [A]  time = 0.08, size = 82, normalized size = 0.80 \[ -\frac {3 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}{64 d}+\frac {5}{16 d \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}+\frac {3 i \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}{64 d}+\frac {5}{48 d \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )} \]

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(tanh(1/2*d*x+1/2*c)-3*I)+5/16/d/(tanh(1/2*d*x+1/2*c)-3*I)+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)

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maxima [A]  time = 0.43, size = 79, normalized size = 0.77 \[ \frac {3 i \, \log \left (\frac {10 \, e^{\left (-d x - c\right )} + 6 i - 8}{10 \, e^{\left (-d x - c\right )} + 6 i + 8}\right )}{64 \, d} + \frac {3 i \, e^{\left (-d x - c\right )} - 5}{-8 \, d {\left (-6 i \, e^{\left (-d x - c\right )} - 5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 5\right )}} \]

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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mupad [B]  time = 0.98, size = 106, normalized size = 1.04 \[ -\frac {5}{8\,\left (5\,d-5\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{c+d\,x}\,6{}\mathrm {i}\right )}-\frac {\ln \left (-\frac {15}{4}+{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (-3-\frac {9}{4}{}\mathrm {i}\right )\right )\,3{}\mathrm {i}}{64\,d}+\frac {\ln \left (\frac {15}{4}+{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (-3+\frac {9}{4}{}\mathrm {i}\right )\right )\,3{}\mathrm {i}}{64\,d}-\frac {{\mathrm {e}}^{c+d\,x}\,3{}\mathrm {i}}{8\,\left (5\,d-5\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{c+d\,x}\,6{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(c + d*x)*5i + 3)^2,x)

[Out]

(log(15/4 - exp(d*x)*exp(c)*(3 - 9i/4))*3i)/(64*d) - (log(- exp(d*x)*exp(c)*(3 + 9i/4) - 15/4)*3i)/(64*d) - 5/
(8*(5*d + d*exp(c + d*x)*6i - 5*d*exp(2*c + 2*d*x))) - (exp(c + d*x)*3i)/(8*(5*d + d*exp(c + d*x)*6i - 5*d*exp
(2*c + 2*d*x)))

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sympy [A]  time = 0.41, size = 87, normalized size = 0.85 \[ \frac {- 5 e^{2 c} + 3 i e^{c} e^{- d x}}{- 40 d e^{2 c} + 48 i d e^{c} e^{- d x} + 40 d e^{- 2 d x}} + \frac {\operatorname {RootSum} {\left (2560000 z^{2} + 9, \left (i \mapsto i \log {\left (- \frac {1280 i i e^{c}}{3} + \frac {3 i e^{c}}{5} + e^{- d x} \right )} \right )\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-5*exp(2*c) + 3*I*exp(c)*exp(-d*x))/(-40*d*exp(2*c) + 48*I*d*exp(c)*exp(-d*x) + 40*d*exp(-2*d*x)) + RootSum(2
560000*_z**2 + 9, Lambda(_i, _i*log(-1280*_i*I*exp(c)/3 + 3*I*exp(c)/5 + exp(-d*x))))/d

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