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

Optimal. Leaf size=73 \[ \frac {i \log \left (3 \cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}-\frac {i \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )+3 i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \]

[Out]

1/4*I*ln(3*cosh(1/2*d*x+1/2*c)+I*sinh(1/2*d*x+1/2*c))/d-1/4*I*ln(cosh(1/2*d*x+1/2*c)+3*I*sinh(1/2*d*x+1/2*c))/
d

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Rubi [A]  time = 0.03, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2660, 616, 31} \[ \frac {i \log \left (3 \cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}-\frac {i \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )+3 i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(3 + (5*I)*Sinh[c + d*x])^(-1),x]

[Out]

((I/4)*Log[3*Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]])/d - ((I/4)*Log[Cosh[(c + d*x)/2] + (3*I)*Sinh[(c + d*x)
/2]])/d

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]

Rubi steps

\begin {align*} \int \frac {1}{3+5 i \sinh (c+d x)} \, dx &=-\frac {(2 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 )}{d}\\ &=-\frac {(3 i) \operatorname {Subst}\left (\int \frac {1}{1+3 x} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{4 d}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {1}{9+3 x} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{4 d}\\ &=\frac {i \log \left (3+i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}-\frac {i \log \left (1+3 i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 81, normalized size = 1.11 \[ \frac {\tan ^{-1}\left (3 \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{4 d}-\frac {i \log (4-5 \cosh (c+d x))}{8 d}+\frac {i \log (5 \cosh (c+d x)+4)}{8 d}+\frac {\tan ^{-1}\left (3 \coth \left (\frac {1}{2} (c+d x)\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + (5*I)*Sinh[c + d*x])^(-1),x]

[Out]

ArcTan[3*Coth[(c + d*x)/2]]/(4*d) + ArcTan[3*Tanh[(c + d*x)/2]]/(4*d) - ((I/8)*Log[4 - 5*Cosh[c + d*x]])/d + (
(I/8)*Log[4 + 5*Cosh[c + d*x]])/d

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fricas [A]  time = 0.51, size = 28, normalized size = 0.38 \[ \frac {i \, \log \left (e^{\left (d x + c\right )} - \frac {3}{5} i + \frac {4}{5}\right ) - i \, \log \left (e^{\left (d x + c\right )} - \frac {3}{5} i - \frac {4}{5}\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(I*log(e^(d*x + c) - 3/5*I + 4/5) - I*log(e^(d*x + c) - 3/5*I - 4/5))/d

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giac [A]  time = 0.58, size = 32, normalized size = 0.44 \[ -\frac {-i \, \log \left (-\left (i - 2\right ) \, e^{\left (d x + c\right )} - 2 i + 1\right ) + i \, \log \left (-\left (2 i - 1\right ) \, e^{\left (d x + c\right )} + i - 2\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-I*log(-(I - 2)*e^(d*x + c) - 2*I + 1) + I*log(-(2*I - 1)*e^(d*x + c) + I - 2))/d

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maple [A]  time = 0.05, size = 42, normalized size = 0.58 \[ \frac {i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}{4 d}-\frac {i \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I/d*ln(tanh(1/2*d*x+1/2*c)-3*I)-1/4*I/d*ln(3*tanh(1/2*d*x+1/2*c)-I)

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maxima [A]  time = 0.42, size = 19, normalized size = 0.26 \[ \frac {\arctan \left (\frac {5}{4} i \, e^{\left (-d x - c\right )} - \frac {3}{4}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(5/4*I*e^(-d*x - c) - 3/4)/d

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mupad [B]  time = 0.35, size = 39, normalized size = 0.53 \[ -\frac {\ln \left (-\frac {5}{2}+{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (2-\frac {3}{2}{}\mathrm {i}\right )\right )\,1{}\mathrm {i}}{4\,d}+\frac {\ln \left (\frac {5}{2}+{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (2+\frac {3}{2}{}\mathrm {i}\right )\right )\,1{}\mathrm {i}}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(c + d*x)*5i + 3),x)

[Out]

(log(exp(d*x)*exp(c)*(2 + 3i/2) + 5/2)*1i)/(4*d) - (log(exp(d*x)*exp(c)*(2 - 3i/2) - 5/2)*1i)/(4*d)

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sympy [A]  time = 0.31, size = 34, normalized size = 0.47 \[ \frac {\operatorname {RootSum} {\left (400 z^{2} + 1, \left (i \mapsto i \log {\left (16 i i e^{c} + \frac {3 i e^{c}}{5} + e^{- d x} \right )} \right )\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(400*_z**2 + 1, Lambda(_i, _i*log(16*_i*I*exp(c) + 3*I*exp(c)/5 + exp(-d*x))))/d

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