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

Optimal. Leaf size=95 \[ -\frac{45 i \cosh (c+d x)}{512 d (5+3 i \sinh (c+d x))}-\frac{3 i \cosh (c+d x)}{32 d (5+3 i \sinh (c+d x))^2}-\frac{59 i \tan ^{-1}\left (\frac{\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{1024 d}+\frac{59 x}{2048} \]

[Out]

(59*x)/2048 - (((59*I)/1024)*ArcTan[Cosh[c + d*x]/(3 + I*Sinh[c + d*x])])/d - (((3*I)/32)*Cosh[c + d*x])/(d*(5
+ (3*I)*Sinh[c + d*x])^2) - (((45*I)/512)*Cosh[c + d*x])/(d*(5 + (3*I)*Sinh[c + d*x]))

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Rubi [A]  time = 0.0674412, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2664, 2754, 12, 2657} \[ -\frac{45 i \cosh (c+d x)}{512 d (5+3 i \sinh (c+d x))}-\frac{3 i \cosh (c+d x)}{32 d (5+3 i \sinh (c+d x))^2}-\frac{59 i \tan ^{-1}\left (\frac{\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{1024 d}+\frac{59 x}{2048} \]

Antiderivative was successfully verified.

[In]

Int[(5 + (3*I)*Sinh[c + d*x])^(-3),x]

[Out]

(59*x)/2048 - (((59*I)/1024)*ArcTan[Cosh[c + d*x]/(3 + I*Sinh[c + d*x])])/d - (((3*I)/32)*Cosh[c + d*x])/(d*(5
+ (3*I)*Sinh[c + d*x])^2) - (((45*I)/512)*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 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

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))^3} \, dx &=-\frac{3 i \cosh (c+d x)}{32 d (5+3 i \sinh (c+d x))^2}-\frac{1}{32} \int \frac{-10+3 i \sinh (c+d x)}{(5+3 i \sinh (c+d x))^2} \, dx\\ &=-\frac{3 i \cosh (c+d x)}{32 d (5+3 i \sinh (c+d x))^2}-\frac{45 i \cosh (c+d x)}{512 d (5+3 i \sinh (c+d x))}+\frac{1}{512} \int \frac{59}{5+3 i \sinh (c+d x)} \, dx\\ &=-\frac{3 i \cosh (c+d x)}{32 d (5+3 i \sinh (c+d x))^2}-\frac{45 i \cosh (c+d x)}{512 d (5+3 i \sinh (c+d x))}+\frac{59}{512} \int \frac{1}{5+3 i \sinh (c+d x)} \, dx\\ &=\frac{59 x}{2048}-\frac{59 i \tan ^{-1}\left (\frac{\cosh (c+d x)}{3+i \sinh (c+d x)}\right )}{1024 d}-\frac{3 i \cosh (c+d x)}{32 d (5+3 i \sinh (c+d x))^2}-\frac{45 i \cosh (c+d x)}{512 d (5+3 i \sinh (c+d x))}\\ \end{align*}

Mathematica [B]  time = 0.648357, size = 277, normalized size = 2.92 \[ \frac{-\frac{144 \sinh \left (\frac{1}{2} (c+d x)\right ) \left (5 \sinh \left (\frac{1}{2} (c+d x)\right )-3 i \cosh \left (\frac{1}{2} (c+d x)\right )\right )}{3 \sinh (c+d x)-5 i}+\frac{48}{\left ((1+2 i) \cosh \left (\frac{1}{2} (c+d x)\right )-(2+i) \sinh \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{48}{\left ((1+2 i) \sinh \left (\frac{1}{2} (c+d x)\right )+(2+i) \cosh \left (\frac{1}{2} (c+d x)\right )\right )^2}-59 \log (5 \cosh (c+d x)-4 \sinh (c+d x))+59 \log (4 \sinh (c+d x)+5 \cosh (c+d x))-118 i \tan ^{-1}\left (\frac{2 \cosh \left (\frac{1}{2} (c+d x)\right )-\sinh \left (\frac{1}{2} (c+d x)\right )}{\cosh \left (\frac{1}{2} (c+d x)\right )-2 \sinh \left (\frac{1}{2} (c+d x)\right )}\right )+118 i \tan ^{-1}\left (\frac{2 \sinh \left (\frac{1}{2} (c+d x)\right )+\cosh \left (\frac{1}{2} (c+d x)\right )}{\sinh \left (\frac{1}{2} (c+d x)\right )+2 \cosh \left (\frac{1}{2} (c+d x)\right )}\right )}{4096 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-118*I)*ArcTan[(2*Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2])/(Cosh[(c + d*x)/2] - 2*Sinh[(c + d*x)/2])] + (118*I
)*ArcTan[(Cosh[(c + d*x)/2] + 2*Sinh[(c + d*x)/2])/(2*Cosh[(c + d*x)/2] + Sinh[(c + d*x)/2])] - 59*Log[5*Cosh[
c + d*x] - 4*Sinh[c + d*x]] + 59*Log[5*Cosh[c + d*x] + 4*Sinh[c + d*x]] + 48/((1 + 2*I)*Cosh[(c + d*x)/2] - (2
+ I)*Sinh[(c + d*x)/2])^2 + 48/((2 + I)*Cosh[(c + d*x)/2] + (1 + 2*I)*Sinh[(c + d*x)/2])^2 - (144*Sinh[(c + d
*x)/2]*((-3*I)*Cosh[(c + d*x)/2] + 5*Sinh[(c + d*x)/2]))/(-5*I + 3*Sinh[c + d*x]))/(4096*d)

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Maple [B]  time = 0.05, size = 224, normalized size = 2.4 \begin{align*}{\frac{63}{3200\,d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) +4-3\,i \right ) ^{-2}}-{\frac{{\frac{27\,i}{400}}}{d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) +4-3\,i \right ) ^{-2}}-{\frac{963}{12800\,d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) +4-3\,i \right ) ^{-1}}-{\frac{{\frac{123\,i}{1600}}}{d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) +4-3\,i \right ) ^{-1}}+{\frac{59}{2048\,d}\ln \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) +4-3\,i \right ) }-{\frac{63}{3200\,d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) -4-3\,i \right ) ^{-2}}-{\frac{{\frac{27\,i}{400}}}{d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) -4-3\,i \right ) ^{-2}}-{\frac{963}{12800\,d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) -4-3\,i \right ) ^{-1}}+{\frac{{\frac{123\,i}{1600}}}{d} \left ( 5\,\tanh \left ( 1/2\,dx+c/2 \right ) -4-3\,i \right ) ^{-1}}-{\frac{59}{2048\,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))^3,x)

[Out]

63/3200/d/(5*tanh(1/2*d*x+1/2*c)+4-3*I)^2-27/400*I/d/(5*tanh(1/2*d*x+1/2*c)+4-3*I)^2-963/12800/d/(5*tanh(1/2*d
*x+1/2*c)+4-3*I)-123/1600*I/d/(5*tanh(1/2*d*x+1/2*c)+4-3*I)+59/2048/d*ln(5*tanh(1/2*d*x+1/2*c)+4-3*I)-63/3200/
d/(5*tanh(1/2*d*x+1/2*c)-4-3*I)^2-27/400*I/d/(5*tanh(1/2*d*x+1/2*c)-4-3*I)^2-963/12800/d/(5*tanh(1/2*d*x+1/2*c
)-4-3*I)+123/1600*I/d/(5*tanh(1/2*d*x+1/2*c)-4-3*I)-59/2048/d*ln(5*tanh(1/2*d*x+1/2*c)-4-3*I)

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Maxima [A]  time = 1.60141, size = 146, normalized size = 1.54 \begin{align*} -\frac{59 i \, \arctan \left (\frac{3}{4} \, e^{\left (-d x - c\right )} + \frac{5}{4} i\right )}{1024 \, d} - \frac{-723 i \, e^{\left (-d x - c\right )} - 885 \, e^{\left (-2 \, d x - 2 \, c\right )} + 177 i \, e^{\left (-3 \, d x - 3 \, c\right )} + 135}{d{\left (-15360 i \, e^{\left (-d x - c\right )} - 30208 \, e^{\left (-2 \, d x - 2 \, c\right )} + 15360 i \, e^{\left (-3 \, d x - 3 \, c\right )} + 2304 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2304\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-59/1024*I*arctan(3/4*e^(-d*x - c) + 5/4*I)/d - (-723*I*e^(-d*x - c) - 885*e^(-2*d*x - 2*c) + 177*I*e^(-3*d*x
- 3*c) + 135)/(d*(-15360*I*e^(-d*x - c) - 30208*e^(-2*d*x - 2*c) + 15360*I*e^(-3*d*x - 3*c) + 2304*e^(-4*d*x -
4*c) + 2304))

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Fricas [B]  time = 2.10321, size = 594, normalized size = 6.25 \begin{align*} \frac{{\left (531 \, e^{\left (4 \, d x + 4 \, c\right )} - 3540 i \, e^{\left (3 \, d x + 3 \, c\right )} - 6962 \, e^{\left (2 \, d x + 2 \, c\right )} + 3540 i \, e^{\left (d x + c\right )} + 531\right )} \log \left (e^{\left (d x + c\right )} - \frac{1}{3} i\right ) -{\left (531 \, e^{\left (4 \, d x + 4 \, c\right )} - 3540 i \, e^{\left (3 \, d x + 3 \, c\right )} - 6962 \, e^{\left (2 \, d x + 2 \, c\right )} + 3540 i \, e^{\left (d x + c\right )} + 531\right )} \log \left (e^{\left (d x + c\right )} - 3 i\right ) - 1416 i \, e^{\left (3 \, d x + 3 \, c\right )} - 7080 \, e^{\left (2 \, d x + 2 \, c\right )} + 5784 i \, e^{\left (d x + c\right )} + 1080}{18432 \, d e^{\left (4 \, d x + 4 \, c\right )} - 122880 i \, d e^{\left (3 \, d x + 3 \, c\right )} - 241664 \, d e^{\left (2 \, d x + 2 \, c\right )} + 122880 i \, d e^{\left (d x + c\right )} + 18432 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((531*e^(4*d*x + 4*c) - 3540*I*e^(3*d*x + 3*c) - 6962*e^(2*d*x + 2*c) + 3540*I*e^(d*x + c) + 531)*log(e^(d*x +
c) - 1/3*I) - (531*e^(4*d*x + 4*c) - 3540*I*e^(3*d*x + 3*c) - 6962*e^(2*d*x + 2*c) + 3540*I*e^(d*x + c) + 531
)*log(e^(d*x + c) - 3*I) - 1416*I*e^(3*d*x + 3*c) - 7080*e^(2*d*x + 2*c) + 5784*I*e^(d*x + c) + 1080)/(18432*d
*e^(4*d*x + 4*c) - 122880*I*d*e^(3*d*x + 3*c) - 241664*d*e^(2*d*x + 2*c) + 122880*I*d*e^(d*x + c) + 18432*d)

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Sympy [A]  time = 2.60271, size = 158, normalized size = 1.66 \begin{align*} \frac{- \frac{15 e^{4 c}}{256 d} + \frac{241 i e^{3 c} e^{- d x}}{768 d} + \frac{295 e^{2 c} e^{- 2 d x}}{768 d} - \frac{59 i e^{c} e^{- 3 d x}}{768 d}}{e^{4 c} - \frac{20 i e^{3 c} e^{- d x}}{3} - \frac{118 e^{2 c} e^{- 2 d x}}{9} + \frac{20 i e^{c} e^{- 3 d x}}{3} + e^{- 4 d x}} + \frac{- \frac{59 \log{\left (\frac{i e^{c}}{3} + e^{- d x} \right )}}{2048} + \frac{59 \log{\left (3 i e^{c} + e^{- d x} \right )}}{2048}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-15*exp(4*c)/(256*d) + 241*I*exp(3*c)*exp(-d*x)/(768*d) + 295*exp(2*c)*exp(-2*d*x)/(768*d) - 59*I*exp(c)*exp(
-3*d*x)/(768*d))/(exp(4*c) - 20*I*exp(3*c)*exp(-d*x)/3 - 118*exp(2*c)*exp(-2*d*x)/9 + 20*I*exp(c)*exp(-3*d*x)/
3 + exp(-4*d*x)) + (-59*log(I*exp(c)/3 + exp(-d*x))/2048 + 59*log(3*I*exp(c) + exp(-d*x))/2048)/d

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Giac [A]  time = 1.27225, size = 123, normalized size = 1.29 \begin{align*} \frac{59 \, \log \left (3 \, e^{\left (d x + c\right )} - i\right )}{2048 \, d} - \frac{59 \, \log \left (e^{\left (d x + c\right )} - 3 i\right )}{2048 \, d} - \frac{-177 i \, e^{\left (3 \, d x + 3 \, c\right )} - 885 \, e^{\left (2 \, d x + 2 \, c\right )} + 723 i \, e^{\left (d x + c\right )} + 135}{256 \, d{\left (-3 i \, e^{\left (2 \, d x + 2 \, c\right )} - 10 \, e^{\left (d x + c\right )} + 3 i\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

59/2048*log(3*e^(d*x + c) - I)/d - 59/2048*log(e^(d*x + c) - 3*I)/d - 1/256*(-177*I*e^(3*d*x + 3*c) - 885*e^(2
*d*x + 2*c) + 723*I*e^(d*x + c) + 135)/(d*(-3*I*e^(2*d*x + 2*c) - 10*e^(d*x + c) + 3*I)^2)