3.1.93 \(\int \frac {1}{(5+3 i \sinh (c+d x))^2} \, dx\) [93]

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

[Out]

5/64*x-5/32*I*arctan(cosh(d*x+c)/(3+I*sinh(d*x+c)))/d-3/16*I*cosh(d*x+c)/d/(5+3*I*sinh(d*x+c))

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Rubi [A]
time = 0.03, antiderivative size = 66, normalized size of antiderivative = 1.00, 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 = {2743, 12, 2736} \begin {gather*} -\frac {5 i \text {ArcTan}\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))}+\frac {5 x}{64} \end {gather*}

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 12

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

Rule 2736

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

Rule 2743

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] && Integ
erQ[2*n]

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*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(183\) vs. \(2(66)=132\).
time = 0.19, size = 183, normalized size = 2.77 \begin {gather*} \frac {24 i-50 i \text {ArcTan}\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 \text {ArcTan}\left (\frac {\cosh \left (\frac {1}{2} (c+d x)\right )+2 \sinh \left (\frac {1}{2} (c+d x)\right )}{2 \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )}\right )-25 \log (5 \cosh (c+d x)-4 \sinh (c+d x))+25 \log (5 \cosh (c+d x)+4 \sinh (c+d x))-\frac {120 \cosh (c+d x)}{-5 i+3 \sinh (c+d x)}}{640 d} \end {gather*}

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 [A]
time = 1.17, size = 84, normalized size = 1.27

method	result	size
risch	\(-\frac {i \left (5 \,{\mathrm e}^{d x +c}-3 i\right )}{8 d \left (3 \,{\mathrm e}^{2 d x +2 c}-3-10 i {\mathrm e}^{d x +c}\right )}+\frac {5 \ln \left ({\mathrm e}^{d x +c}-\frac {i}{3}\right )}{64 d}-\frac {5 \ln \left (-3 i+{\mathrm e}^{d x +c}\right )}{64 d}\)	\(73\)
derivativedivides	\(\frac {\frac {-\frac {9}{80}-\frac {3 i}{20}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i}+\frac {5 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{64}+\frac {-\frac {9}{80}+\frac {3 i}{20}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i}-\frac {5 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}{64}}{d}\)	\(84\)
default	\(\frac {\frac {-\frac {9}{80}-\frac {3 i}{20}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i}+\frac {5 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{64}+\frac {-\frac {9}{80}+\frac {3 i}{20}}{5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i}-\frac {5 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}{64}}{d}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 64, normalized size = 0.97 \begin {gather*} -\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 {gather*}

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 = 0.45, size = 103, normalized size = 1.56 \begin {gather*} \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 {gather*}

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 = 0.16, size = 82, normalized size = 1.24 \begin {gather*} \frac {- 5 i e^{c} e^{d x} - 3}{24 d e^{2 c} e^{2 d x} - 80 i d e^{c} e^{d x} - 24 d} + \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 {gather*}

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) - 3)/(24*d*exp(2*c)*exp(2*d*x) - 80*I*d*exp(c)*exp(d*x) - 24*d) + (-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 = 0.41, size = 65, normalized size = 0.98 \begin {gather*} -\frac {\frac {8 \, {\left (5 i \, e^{\left (d x + c\right )} + 3\right )}}{3 \, e^{\left (2 \, d x + 2 \, c\right )} - 10 i \, e^{\left (d x + c\right )} - 3} - 5 \, \log \left (3 \, e^{\left (d x + c\right )} - i\right ) + 5 \, \log \left (e^{\left (d x + c\right )} - 3 i\right )}{64 \, d} \end {gather*}

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]

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

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Mupad [B]
time = 1.07, size = 102, normalized size = 1.55 \begin {gather*} \frac {3}{8\,\left (3\,d-3\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{c+d\,x}\,10{}\mathrm {i}\right )}-\frac {5\,\ln \left (-\frac {5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{4}+\frac {15}{4}{}\mathrm {i}\right )}{64\,d}+\frac {5\,\ln \left (\frac {45\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{4}-\frac {15}{4}{}\mathrm {i}\right )}{64\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,5{}\mathrm {i}}{8\,\left (3\,d-3\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{c+d\,x}\,10{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(c + d*x)*3i + 5)^2,x)

[Out]

3/(8*(3*d + d*exp(c + d*x)*10i - 3*d*exp(2*c + 2*d*x))) - (5*log(15i/4 - (5*exp(d*x)*exp(c))/4))/(64*d) + (5*l
og((45*exp(d*x)*exp(c))/4 - 15i/4))/(64*d) + (exp(c + d*x)*5i)/(8*(3*d + d*exp(c + d*x)*10i - 3*d*exp(2*c + 2*
d*x)))

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