3.1.90 \(\int \frac {1}{(3+5 i \sinh (c+d x))^3} \, dx\) [90]

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

[Out]

43/2048*I*ln(3*cosh(1/2*d*x+1/2*c)+I*sinh(1/2*d*x+1/2*c))/d-43/2048*I*ln(cosh(1/2*d*x+1/2*c)+3*I*sinh(1/2*d*x+
1/2*c))/d+5/32*I*cosh(d*x+c)/d/(3+5*I*sinh(d*x+c))^2-45/512*I*cosh(d*x+c)/d/(3+5*I*sinh(d*x+c))

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Rubi [A]
time = 0.06, antiderivative size = 131, normalized size of antiderivative = 1.00, 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 = {2743, 2833, 12, 2739, 630, 31} \begin {gather*} -\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} \end {gather*}

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 12

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

Rule 31

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

Rule 630

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 2739

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 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]

Rule 2833

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}{(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) \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 0.38, size = 204, normalized size = 1.56 \begin {gather*} \frac {86 \text {ArcTan}\left (3 \coth \left (\frac {1}{2} (c+d x)\right )\right )+86 \text {ArcTan}\left (3 \tanh \left (\frac {1}{2} (c+d x)\right )\right )-43 i \log (4-5 \cosh (c+d x))+43 i \log (4+5 \cosh (c+d x))-\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}+\left (-\frac {120}{3 \cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )}-\frac {360}{\cosh \left (\frac {1}{2} (c+d x)\right )+3 i \sinh \left (\frac {1}{2} (c+d x)\right )}\right ) \sinh \left (\frac {1}{2} (c+d x)\right )}{4096 d} \end {gather*}

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 = 1.40, size = 110, normalized size = 0.84

method	result	size
risch	\(-\frac {i \left (-387 i {\mathrm e}^{2 d x +2 c}+215 \,{\mathrm e}^{3 d x +3 c}+225 i-325 \,{\mathrm e}^{d x +c}\right )}{256 d \left (5 \,{\mathrm e}^{2 d x +2 c}-5-6 i {\mathrm e}^{d x +c}\right )^{2}}-\frac {43 i \ln \left ({\mathrm e}^{d x +c}-\frac {4}{5}-\frac {3 i}{5}\right )}{2048 d}+\frac {43 i \ln \left ({\mathrm e}^{d x +c}+\frac {4}{5}-\frac {3 i}{5}\right )}{2048 d}\)	\(100\)
derivativedivides	\(\frac {-\frac {43 i \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}{2048}-\frac {25 i}{1152 \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}-\frac {155}{4608 \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}+\frac {25 i}{128 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )^{2}}+\frac {43 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}{2048}+\frac {15}{512 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}}{d}\)	\(110\)
default	\(\frac {-\frac {43 i \ln \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}{2048}-\frac {25 i}{1152 \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}-\frac {155}{4608 \left (3 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}+\frac {25 i}{128 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )^{2}}+\frac {43 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}{2048}+\frac {15}{512 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-3 i\right )}}{d}\)	\(110\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 124, normalized size = 0.95 \begin {gather*} -\frac {43 i \, \log \left (\frac {5 \, e^{\left (-d x - c\right )} + 3 i - 4}{5 \, e^{\left (-d x - c\right )} + 3 i + 4}\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}{-256 \, d {\left (60 i \, e^{\left (-d x - c\right )} + 86 \, e^{\left (-2 \, d x - 2 \, c\right )} - 60 i \, e^{\left (-3 \, d x - 3 \, c\right )} - 25 \, e^{\left (-4 \, d x - 4 \, c\right )} - 25\right )}} \end {gather*}

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((5*e^(-d*x - c) + 3*I - 4)/(5*e^(-d*x - c) + 3*I + 4))/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 = 0.63, size = 193, normalized size = 1.47 \begin {gather*} -\frac {43 \, {\left (-25 i \, e^{\left (4 \, d x + 4 \, c\right )} - 60 \, e^{\left (3 \, d x + 3 \, c\right )} + 86 i \, e^{\left (2 \, d x + 2 \, c\right )} + 60 \, e^{\left (d x + c\right )} - 25 i\right )} \log \left (e^{\left (d x + c\right )} - \frac {3}{5} i + \frac {4}{5}\right ) + 43 \, {\left (25 i \, e^{\left (4 \, d x + 4 \, c\right )} + 60 \, e^{\left (3 \, d x + 3 \, c\right )} - 86 i \, e^{\left (2 \, d x + 2 \, c\right )} - 60 \, e^{\left (d x + c\right )} + 25 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}{2048 \, {\left (25 \, d e^{\left (4 \, d x + 4 \, c\right )} - 60 i \, d e^{\left (3 \, d x + 3 \, c\right )} - 86 \, d e^{\left (2 \, d x + 2 \, c\right )} + 60 i \, d e^{\left (d x + c\right )} + 25 \, d\right )}} \end {gather*}

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]

-1/2048*(43*(-25*I*e^(4*d*x + 4*c) - 60*e^(3*d*x + 3*c) + 86*I*e^(2*d*x + 2*c) + 60*e^(d*x + c) - 25*I)*log(e^
(d*x + c) - 3/5*I + 4/5) + 43*(25*I*e^(4*d*x + 4*c) + 60*e^(3*d*x + 3*c) - 86*I*e^(2*d*x + 2*c) - 60*e^(d*x +
c) + 25*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)/(25*d*e^(4*d*x + 4*c) - 60*I*d*e^(3*d*x + 3*c) - 86*d*e^(2*d*x + 2*c) + 60*I*d*e^(d*x + c) + 25*d)

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Sympy [A]
time = 0.29, size = 138, normalized size = 1.05 \begin {gather*} \frac {- 215 i e^{3 c} e^{3 d x} - 387 e^{2 c} e^{2 d x} + 325 i e^{c} e^{d x} + 225}{6400 d e^{4 c} e^{4 d x} - 15360 i d e^{3 c} e^{3 d x} - 22016 d e^{2 c} e^{2 d x} + 15360 i d e^{c} e^{d x} + 6400 d} + \frac {\operatorname {RootSum} {\left (4194304 z^{2} + 1849, \left ( i \mapsto i \log {\left (\frac {\left (- 8192 i i - 129 i\right ) e^{- c}}{215} + e^{d x} \right )} \right )\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-215*I*exp(3*c)*exp(3*d*x) - 387*exp(2*c)*exp(2*d*x) + 325*I*exp(c)*exp(d*x) + 225)/(6400*d*exp(4*c)*exp(4*d*
x) - 15360*I*d*exp(3*c)*exp(3*d*x) - 22016*d*exp(2*c)*exp(2*d*x) + 15360*I*d*exp(c)*exp(d*x) + 6400*d) + RootS
um(4194304*_z**2 + 1849, Lambda(_i, _i*log((-8192*_i*I - 129*I)*exp(-c)/215 + exp(d*x))))/d

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Giac [A]
time = 0.41, size = 89, normalized size = 0.68 \begin {gather*} -\frac {\frac {8 \, {\left (-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\right )}}{{\left (-5 i \, e^{\left (2 \, d x + 2 \, c\right )} - 6 \, e^{\left (d x + c\right )} + 5 i\right )}^{2}} - 43 i \, \log \left (-\left (i - 2\right ) \, e^{\left (d x + c\right )} - 2 i + 1\right ) + 43 i \, \log \left (-\left (2 i - 1\right ) \, e^{\left (d x + c\right )} + i - 2\right )}{2048 \, d} \end {gather*}

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]

-1/2048*(8*(-215*I*e^(3*d*x + 3*c) - 387*e^(2*d*x + 2*c) + 325*I*e^(d*x + c) + 225)/(-5*I*e^(2*d*x + 2*c) - 6*
e^(d*x + c) + 5*I)^2 - 43*I*log(-(I - 2)*e^(d*x + c) - 2*I + 1) + 43*I*log(-(2*I - 1)*e^(d*x + c) + I - 2))/d

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Mupad [B]
time = 1.06, size = 147, normalized size = 1.12 \begin {gather*} \frac {\frac {129}{6400\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,43{}\mathrm {i}}{1280\,d}}{1-{\mathrm {e}}^{2\,c+2\,d\,x}+\frac {{\mathrm {e}}^{c+d\,x}\,6{}\mathrm {i}}{5}}-\frac {\ln \left (-\frac {215}{4}+{\mathrm {e}}^{c+d\,x}\,\left (43-\frac {129}{4}{}\mathrm {i}\right )\right )\,43{}\mathrm {i}}{2048\,d}+\frac {\ln \left (\frac {215}{4}+{\mathrm {e}}^{c+d\,x}\,\left (43+\frac {129}{4}{}\mathrm {i}\right )\right )\,43{}\mathrm {i}}{2048\,d}-\frac {-\frac {3}{200\,d}+\frac {{\mathrm {e}}^{c+d\,x}\,7{}\mathrm {i}}{1000\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-\frac {86\,{\mathrm {e}}^{2\,c+2\,d\,x}}{25}+1+\frac {{\mathrm {e}}^{c+d\,x}\,12{}\mathrm {i}}{5}-\frac {{\mathrm {e}}^{3\,c+3\,d\,x}\,12{}\mathrm {i}}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(c + d*x)*5i + 3)^3,x)

[Out]

((exp(c + d*x)*43i)/(1280*d) + 129/(6400*d))/((exp(c + d*x)*6i)/5 - exp(2*c + 2*d*x) + 1) - (log(exp(c + d*x)*
(43 - 129i/4) - 215/4)*43i)/(2048*d) + (log(exp(c + d*x)*(43 + 129i/4) + 215/4)*43i)/(2048*d) - ((exp(c + d*x)
*7i)/(1000*d) - 3/(200*d))/((exp(c + d*x)*12i)/5 - (86*exp(2*c + 2*d*x))/25 - (exp(3*c + 3*d*x)*12i)/5 + exp(4
*c + 4*d*x) + 1)

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