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]

59/2048*x-59/1024*I*arctan(cosh(d*x+c)/(3+I*sinh(d*x+c)))/d-3/32*I*cosh(d*x+c)/d/(5+3*I*sinh(d*x+c))^2-45/512*
I*cosh(d*x+c)/d/(5+3*I*sinh(d*x+c))

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Rubi [A]  time = 0.07, antiderivative size = 95, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [B]  time = 0.69, 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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fricas [B]  time = 0.94, size = 191, normalized size = 2.01 \[ \frac {{\left (531 \, e^{\left (4 \, d x + 4 \, c\right )} - 3540 i \, e^{\left (3 \, d x + 3 \, c\right )} - 6962 \, e^{\left (2 \, d x + 2 \, c\right )} + 3540 i \, e^{\left (d x + c\right )} + 531\right )} \log \left (e^{\left (d x + c\right )} - \frac {1}{3} i\right ) - {\left (531 \, e^{\left (4 \, d x + 4 \, c\right )} - 3540 i \, e^{\left (3 \, d x + 3 \, c\right )} - 6962 \, e^{\left (2 \, d x + 2 \, c\right )} + 3540 i \, e^{\left (d x + c\right )} + 531\right )} \log \left (e^{\left (d x + c\right )} - 3 i\right ) - 1416 i \, e^{\left (3 \, d x + 3 \, c\right )} - 7080 \, e^{\left (2 \, d x + 2 \, c\right )} + 5784 i \, e^{\left (d x + c\right )} + 1080}{18432 \, d e^{\left (4 \, d x + 4 \, c\right )} - 122880 i \, d e^{\left (3 \, d x + 3 \, c\right )} - 241664 \, d e^{\left (2 \, d x + 2 \, c\right )} + 122880 i \, d e^{\left (d x + c\right )} + 18432 \, d} \]

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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giac [A]  time = 0.27, size = 87, normalized size = 0.92 \[ -\frac {\frac {8 \, {\left (-177 i \, e^{\left (3 \, d x + 3 \, c\right )} - 885 \, e^{\left (2 \, d x + 2 \, c\right )} + 723 i \, e^{\left (d x + c\right )} + 135\right )}}{{\left (-3 i \, e^{\left (2 \, d x + 2 \, c\right )} - 10 \, e^{\left (d x + c\right )} + 3 i\right )}^{2}} - 59 \, \log \left (3 \, e^{\left (d x + c\right )} - i\right ) + 59 \, \log \left (e^{\left (d x + c\right )} - 3 i\right )}{2048 \, d} \]

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]

-1/2048*(8*(-177*I*e^(3*d*x + 3*c) - 885*e^(2*d*x + 2*c) + 723*I*e^(d*x + c) + 135)/(-3*I*e^(2*d*x + 2*c) - 10
*e^(d*x + c) + 3*I)^2 - 59*log(3*e^(d*x + c) - I) + 59*log(e^(d*x + c) - 3*I))/d

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maple [B]  time = 0.10, size = 224, normalized size = 2.36 \[ -\frac {63}{3200 d \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )^{2}}-\frac {27 i}{400 d \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )^{2}}-\frac {963}{12800 d \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}+\frac {123 i}{1600 d \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}-\frac {59 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-4-3 i\right )}{2048 d}+\frac {63}{3200 d \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )^{2}}-\frac {27 i}{400 d \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )^{2}}-\frac {963}{12800 d \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}-\frac {123 i}{1600 d \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}+\frac {59 \ln \left (5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+4-3 i\right )}{2048 d} \]

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 = 0.41, size = 108, normalized size = 1.14 \[ -\frac {59 i \, \arctan \left (\frac {3}{4} \, e^{\left (-d x - c\right )} + \frac {5}{4} i\right )}{1024 \, d} - \frac {-723 i \, e^{\left (-d x - c\right )} - 885 \, e^{\left (-2 \, d x - 2 \, c\right )} + 177 i \, e^{\left (-3 \, d x - 3 \, c\right )} + 135}{d {\left (-15360 i \, e^{\left (-d x - c\right )} - 30208 \, e^{\left (-2 \, d x - 2 \, c\right )} + 15360 i \, e^{\left (-3 \, d x - 3 \, c\right )} + 2304 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2304\right )}} \]

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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mupad [B]  time = 1.62, size = 143, normalized size = 1.51 \[ \frac {\frac {295}{2304\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,59{}\mathrm {i}}{768\,d}}{1-{\mathrm {e}}^{2\,c+2\,d\,x}+\frac {{\mathrm {e}}^{c+d\,x}\,10{}\mathrm {i}}{3}}-\frac {59\,\ln \left (-\frac {59\,{\mathrm {e}}^{c+d\,x}}{4}+\frac {177}{4}{}\mathrm {i}\right )}{2048\,d}+\frac {59\,\ln \left (\frac {531\,{\mathrm {e}}^{c+d\,x}}{4}-\frac {177}{4}{}\mathrm {i}\right )}{2048\,d}-\frac {\frac {5}{72\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,41{}\mathrm {i}}{216\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-\frac {118\,{\mathrm {e}}^{2\,c+2\,d\,x}}{9}+1+\frac {{\mathrm {e}}^{c+d\,x}\,20{}\mathrm {i}}{3}-\frac {{\mathrm {e}}^{3\,c+3\,d\,x}\,20{}\mathrm {i}}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(c + d*x)*3i + 5)^3,x)

[Out]

((exp(c + d*x)*59i)/(768*d) + 295/(2304*d))/((exp(c + d*x)*10i)/3 - exp(2*c + 2*d*x) + 1) - (59*log(177i/4 - (
59*exp(c + d*x))/4))/(2048*d) + (59*log((531*exp(c + d*x))/4 - 177i/4))/(2048*d) - ((exp(c + d*x)*41i)/(216*d)
+ 5/(72*d))/((exp(c + d*x)*20i)/3 - (118*exp(2*c + 2*d*x))/9 - (exp(3*c + 3*d*x)*20i)/3 + exp(4*c + 4*d*x) +
1)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotInvertible} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotInvertible

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