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

Optimal. Leaf size=131 \[ -\frac{45 i \cosh (c+d x)}{512 d (3+5 i \sinh (c+d x))}+\frac{5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}+\frac{43 i \log \left (3 \cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}-\frac{43 i \log \left (\cosh \left (\frac{1}{2} (c+d x)\right )+3 i \sinh \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d} \]

[Out]

(((43*I)/2048)*Log[3*Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]])/d - (((43*I)/2048)*Log[Cosh[(c + d*x)/2] + (3*I
)*Sinh[(c + d*x)/2]])/d + (((5*I)/32)*Cosh[c + d*x])/(d*(3 + (5*I)*Sinh[c + d*x])^2) - (((45*I)/512)*Cosh[c +
d*x])/(d*(3 + (5*I)*Sinh[c + d*x]))

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Rubi [A]  time = 0.0851339, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {2664, 2754, 12, 2660, 616, 31} \[ -\frac{45 i \cosh (c+d x)}{512 d (3+5 i \sinh (c+d x))}+\frac{5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}+\frac{43 i \log \left (3 \cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}-\frac{43 i \log \left (\cosh \left (\frac{1}{2} (c+d x)\right )+3 i \sinh \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d} \]

Antiderivative was successfully verified.

[In]

Int[(3 + (5*I)*Sinh[c + d*x])^(-3),x]

[Out]

(((43*I)/2048)*Log[3*Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]])/d - (((43*I)/2048)*Log[Cosh[(c + d*x)/2] + (3*I
)*Sinh[(c + d*x)/2]])/d + (((5*I)/32)*Cosh[c + d*x])/(d*(3 + (5*I)*Sinh[c + d*x])^2) - (((45*I)/512)*Cosh[c +
d*x])/(d*(3 + (5*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 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 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 31

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

Rubi steps

\begin{align*} \int \frac{1}{(3+5 i \sinh (c+d x))^3} \, dx &=\frac{5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}+\frac{1}{32} \int \frac{-6+5 i \sinh (c+d x)}{(3+5 i \sinh (c+d x))^2} \, dx\\ &=\frac{5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}-\frac{45 i \cosh (c+d x)}{512 d (3+5 i \sinh (c+d x))}+\frac{1}{512} \int \frac{43}{3+5 i \sinh (c+d x)} \, dx\\ &=\frac{5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}-\frac{45 i \cosh (c+d x)}{512 d (3+5 i \sinh (c+d x))}+\frac{43}{512} \int \frac{1}{3+5 i \sinh (c+d x)} \, dx\\ &=\frac{5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}-\frac{45 i \cosh (c+d x)}{512 d (3+5 i \sinh (c+d x))}-\frac{(43 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 )}{256 d}\\ &=\frac{5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}-\frac{45 i \cosh (c+d x)}{512 d (3+5 i \sinh (c+d x))}-\frac{(129 i) \operatorname{Subst}\left (\int \frac{1}{1+3 x} \, dx,x,\tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{2048 d}+\frac{(129 i) \operatorname{Subst}\left (\int \frac{1}{9+3 x} \, dx,x,\tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{2048 d}\\ &=\frac{43 i \log \left (3+i \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}-\frac{43 i \log \left (1+3 i \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}+\frac{5 i \cosh (c+d x)}{32 d (3+5 i \sinh (c+d x))^2}-\frac{45 i \cosh (c+d x)}{512 d (3+5 i \sinh (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.484808, size = 204, normalized size = 1.56 \[ \frac{86 \tan ^{-1}\left (3 \tanh \left (\frac{1}{2} (c+d x)\right )\right )-43 i \log (4-5 \cosh (c+d x))+43 i \log (5 \cosh (c+d x)+4)+\sinh \left (\frac{1}{2} (c+d x)\right ) \left (-\frac{360}{\cosh \left (\frac{1}{2} (c+d x)\right )+3 i \sinh \left (\frac{1}{2} (c+d x)\right )}-\frac{120}{3 \cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )}\right )-\frac{80 i}{\left (3 \cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{80 i}{\left (\cosh \left (\frac{1}{2} (c+d x)\right )+3 i \sinh \left (\frac{1}{2} (c+d x)\right )\right )^2}+86 \tan ^{-1}\left (3 \coth \left (\frac{1}{2} (c+d x)\right )\right )}{4096 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(86*ArcTan[3*Coth[(c + d*x)/2]] + 86*ArcTan[3*Tanh[(c + d*x)/2]] - (43*I)*Log[4 - 5*Cosh[c + d*x]] + (43*I)*Lo
g[4 + 5*Cosh[c + d*x]] - (80*I)/(3*Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2 + (80*I)/(Cosh[(c + d*x)/2] + (3
*I)*Sinh[(c + d*x)/2])^2 + (-120/(3*Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2]) - 360/(Cosh[(c + d*x)/2] + (3*I)*
Sinh[(c + d*x)/2]))*Sinh[(c + d*x)/2])/(4096*d)

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Maple [A]  time = 0.051, size = 124, normalized size = 1. \begin{align*}{\frac{-{\frac{43\,i}{2048}}}{d}\ln \left ( 3\,\tanh \left ( 1/2\,dx+c/2 \right ) -i \right ) }-{\frac{{\frac{25\,i}{1152}}}{d} \left ( 3\,\tanh \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-2}}-{\frac{155}{4608\,d} \left ( 3\,\tanh \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-1}}+{\frac{{\frac{25\,i}{128}}}{d} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -3\,i \right ) ^{-2}}+{\frac{{\frac{43\,i}{2048}}}{d}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -3\,i \right ) }+{\frac{15}{512\,d} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -3\,i \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-43/2048*I/d*ln(3*tanh(1/2*d*x+1/2*c)-I)-25/1152*I/d/(3*tanh(1/2*d*x+1/2*c)-I)^2-155/4608/d/(3*tanh(1/2*d*x+1/
2*c)-I)+25/128*I/d/(tanh(1/2*d*x+1/2*c)-3*I)^2+43/2048*I/d*ln(tanh(1/2*d*x+1/2*c)-3*I)+15/512/d/(tanh(1/2*d*x+
1/2*c)-3*I)

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Maxima [A]  time = 1.67748, size = 167, normalized size = 1.27 \begin{align*} -\frac{43 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 )}{2048 \, d} - \frac{-325 i \, e^{\left (-d x - c\right )} - 387 \, e^{\left (-2 \, d x - 2 \, c\right )} + 215 i \, e^{\left (-3 \, d x - 3 \, c\right )} + 225}{d{\left (-15360 i \, e^{\left (-d x - c\right )} - 22016 \, e^{\left (-2 \, d x - 2 \, c\right )} + 15360 i \, e^{\left (-3 \, d x - 3 \, c\right )} + 6400 \, e^{\left (-4 \, d x - 4 \, c\right )} + 6400\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-43/2048*I*log((10*e^(-d*x - c) + 6*I - 8)/(10*e^(-d*x - c) + 6*I + 8))/d - (-325*I*e^(-d*x - c) - 387*e^(-2*d
*x - 2*c) + 215*I*e^(-3*d*x - 3*c) + 225)/(d*(-15360*I*e^(-d*x - c) - 22016*e^(-2*d*x - 2*c) + 15360*I*e^(-3*d
*x - 3*c) + 6400*e^(-4*d*x - 4*c) + 6400))

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Fricas [A]  time = 2.19338, size = 625, normalized size = 4.77 \begin{align*} \frac{{\left (1075 i \, e^{\left (4 \, d x + 4 \, c\right )} + 2580 \, e^{\left (3 \, d x + 3 \, c\right )} - 3698 i \, e^{\left (2 \, d x + 2 \, c\right )} - 2580 \, e^{\left (d x + c\right )} + 1075 i\right )} \log \left (e^{\left (d x + c\right )} - \frac{3}{5} i + \frac{4}{5}\right ) +{\left (-1075 i \, e^{\left (4 \, d x + 4 \, c\right )} - 2580 \, e^{\left (3 \, d x + 3 \, c\right )} + 3698 i \, e^{\left (2 \, d x + 2 \, c\right )} + 2580 \, e^{\left (d x + c\right )} - 1075 i\right )} \log \left (e^{\left (d x + c\right )} - \frac{3}{5} i - \frac{4}{5}\right ) - 1720 i \, e^{\left (3 \, d x + 3 \, c\right )} - 3096 \, e^{\left (2 \, d x + 2 \, c\right )} + 2600 i \, e^{\left (d x + c\right )} + 1800}{51200 \, d e^{\left (4 \, d x + 4 \, c\right )} - 122880 i \, d e^{\left (3 \, d x + 3 \, c\right )} - 176128 \, d e^{\left (2 \, d x + 2 \, c\right )} + 122880 i \, d e^{\left (d x + c\right )} + 51200 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((1075*I*e^(4*d*x + 4*c) + 2580*e^(3*d*x + 3*c) - 3698*I*e^(2*d*x + 2*c) - 2580*e^(d*x + c) + 1075*I)*log(e^(d
*x + c) - 3/5*I + 4/5) + (-1075*I*e^(4*d*x + 4*c) - 2580*e^(3*d*x + 3*c) + 3698*I*e^(2*d*x + 2*c) + 2580*e^(d*
x + c) - 1075*I)*log(e^(d*x + c) - 3/5*I - 4/5) - 1720*I*e^(3*d*x + 3*c) - 3096*e^(2*d*x + 2*c) + 2600*I*e^(d*
x + c) + 1800)/(51200*d*e^(4*d*x + 4*c) - 122880*I*d*e^(3*d*x + 3*c) - 176128*d*e^(2*d*x + 2*c) + 122880*I*d*e
^(d*x + c) + 51200*d)

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Sympy [A]  time = 2.55733, size = 156, normalized size = 1.19 \begin{align*} \frac{- \frac{9 e^{4 c}}{256 d} + \frac{13 i e^{3 c} e^{- d x}}{256 d} + \frac{387 e^{2 c} e^{- 2 d x}}{6400 d} - \frac{43 i e^{c} e^{- 3 d x}}{1280 d}}{e^{4 c} - \frac{12 i e^{3 c} e^{- d x}}{5} - \frac{86 e^{2 c} e^{- 2 d x}}{25} + \frac{12 i e^{c} e^{- 3 d x}}{5} + e^{- 4 d x}} + \frac{\operatorname{RootSum}{\left (4194304 z^{2} + 1849, \left ( i \mapsto i \log{\left (- \frac{8192 i i e^{c}}{215} + \frac{3 i e^{c}}{5} + e^{- d x} \right )} \right )\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-9*exp(4*c)/(256*d) + 13*I*exp(3*c)*exp(-d*x)/(256*d) + 387*exp(2*c)*exp(-2*d*x)/(6400*d) - 43*I*exp(c)*exp(-
3*d*x)/(1280*d))/(exp(4*c) - 12*I*exp(3*c)*exp(-d*x)/5 - 86*exp(2*c)*exp(-2*d*x)/25 + 12*I*exp(c)*exp(-3*d*x)/
5 + exp(-4*d*x)) + RootSum(4194304*_z**2 + 1849, Lambda(_i, _i*log(-8192*_i*I*exp(c)/215 + 3*I*exp(c)/5 + exp(
-d*x))))/d

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Giac [A]  time = 1.30706, size = 126, normalized size = 0.96 \begin{align*} \frac{43 i \, \log \left (-\left (i - 2\right ) \, e^{\left (d x + c\right )} - 2 i + 1\right )}{2048 \, d} - \frac{43 i \, \log \left (-\left (2 i - 1\right ) \, e^{\left (d x + c\right )} + i - 2\right )}{2048 \, d} - \frac{-215 i \, e^{\left (3 \, d x + 3 \, c\right )} - 387 \, e^{\left (2 \, d x + 2 \, c\right )} + 325 i \, e^{\left (d x + c\right )} + 225}{256 \, d{\left (-5 i \, e^{\left (2 \, d x + 2 \, c\right )} - 6 \, e^{\left (d x + c\right )} + 5 i\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

43/2048*I*log(-(I - 2)*e^(d*x + c) - 2*I + 1)/d - 43/2048*I*log(-(2*I - 1)*e^(d*x + c) + I - 2)/d - 1/256*(-21
5*I*e^(3*d*x + 3*c) - 387*e^(2*d*x + 2*c) + 325*I*e^(d*x + c) + 225)/(d*(-5*I*e^(2*d*x + 2*c) - 6*e^(d*x + c)
+ 5*I)^2)