3.117 \(\int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx\)

Optimal. Leaf size=68 \[ \frac {(-3 B+2 i A) \cosh (x)}{15 (\sinh (x)+i)}-\frac {(2 A+3 i B) \cosh (x)}{15 (\sinh (x)+i)^2}-\frac {(B+i A) \cosh (x)}{5 (\sinh (x)+i)^3} \]

[Out]

-1/5*(I*A+B)*cosh(x)/(I+sinh(x))^3-1/15*(2*A+3*I*B)*cosh(x)/(I+sinh(x))^2+1/15*(2*I*A-3*B)*cosh(x)/(I+sinh(x))

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2750, 2650, 2648} \[ \frac {(-3 B+2 i A) \cosh (x)}{15 (\sinh (x)+i)}-\frac {(2 A+3 i B) \cosh (x)}{15 (\sinh (x)+i)^2}-\frac {(B+i A) \cosh (x)}{5 (\sinh (x)+i)^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sinh[x])/(I + Sinh[x])^3,x]

[Out]

-((I*A + B)*Cosh[x])/(5*(I + Sinh[x])^3) - ((2*A + (3*I)*B)*Cosh[x])/(15*(I + Sinh[x])^2) + (((2*I)*A - 3*B)*C
osh[x])/(15*(I + Sinh[x]))

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2650

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

Rule 2750

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)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx &=-\frac {(i A+B) \cosh (x)}{5 (i+\sinh (x))^3}+\frac {1}{5} (-2 i A+3 B) \int \frac {1}{(i+\sinh (x))^2} \, dx\\ &=-\frac {(i A+B) \cosh (x)}{5 (i+\sinh (x))^3}-\frac {(2 A+3 i B) \cosh (x)}{15 (i+\sinh (x))^2}+\frac {1}{15} (-2 A-3 i B) \int \frac {1}{i+\sinh (x)} \, dx\\ &=-\frac {(i A+B) \cosh (x)}{5 (i+\sinh (x))^3}-\frac {(2 A+3 i B) \cosh (x)}{15 (i+\sinh (x))^2}+\frac {(2 i A-3 B) \cosh (x)}{15 (i+\sinh (x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 50, normalized size = 0.74 \[ \frac {\cosh (x) \left ((-3 B+2 i A) \sinh ^2(x)-3 (2 A+3 i B) \sinh (x)-7 i A+3 B\right )}{15 (\sinh (x)+i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sinh[x])/(I + Sinh[x])^3,x]

[Out]

(Cosh[x]*((-7*I)*A + 3*B - 3*(2*A + (3*I)*B)*Sinh[x] + ((2*I)*A - 3*B)*Sinh[x]^2))/(15*(I + Sinh[x])^3)

________________________________________________________________________________________

fricas [A]  time = 0.57, size = 71, normalized size = 1.04 \[ -\frac {30 \, B e^{\left (3 \, x\right )} + {\left (40 \, A + 30 i \, B\right )} e^{\left (2 \, x\right )} + 10 \, {\left (2 i \, A - 3 \, B\right )} e^{x} - 4 \, A - 6 i \, B}{15 \, e^{\left (5 \, x\right )} + 75 i \, e^{\left (4 \, x\right )} - 150 \, e^{\left (3 \, x\right )} - 150 i \, e^{\left (2 \, x\right )} + 75 \, e^{x} + 15 i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(I+sinh(x))^3,x, algorithm="fricas")

[Out]

-(30*B*e^(3*x) + (40*A + 30*I*B)*e^(2*x) + 10*(2*I*A - 3*B)*e^x - 4*A - 6*I*B)/(15*e^(5*x) + 75*I*e^(4*x) - 15
0*e^(3*x) - 150*I*e^(2*x) + 75*e^x + 15*I)

________________________________________________________________________________________

giac [A]  time = 0.21, size = 46, normalized size = 0.68 \[ -\frac {30 \, B e^{\left (3 \, x\right )} + 40 \, A e^{\left (2 \, x\right )} + 30 i \, B e^{\left (2 \, x\right )} + 20 i \, A e^{x} - 30 \, B e^{x} - 4 \, A - 6 i \, B}{15 \, {\left (e^{x} + i\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(I+sinh(x))^3,x, algorithm="giac")

[Out]

-1/15*(30*B*e^(3*x) + 40*A*e^(2*x) + 30*I*B*e^(2*x) + 20*I*A*e^x - 30*B*e^x - 4*A - 6*I*B)/(e^x + I)^5

________________________________________________________________________________________

maple [A]  time = 0.04, size = 91, normalized size = 1.34 \[ \frac {2 i A}{\tanh \left (\frac {x}{2}\right )+i}-\frac {-8 i B +8 A}{2 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{4}}-\frac {2 i B -4 A}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}-\frac {2 \left (-4 i A -4 B \right )}{5 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{5}}-\frac {2 \left (8 i A +6 B \right )}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sinh(x))/(I+sinh(x))^3,x)

[Out]

2*I*A/(tanh(1/2*x)+I)-1/2*(8*A-8*I*B)/(tanh(1/2*x)+I)^4-(-4*A+2*I*B)/(tanh(1/2*x)+I)^2-2/5*(-4*I*A-4*B)/(tanh(
1/2*x)+I)^5-2/3*(8*I*A+6*B)/(tanh(1/2*x)+I)^3

________________________________________________________________________________________

maxima [B]  time = 0.34, size = 280, normalized size = 4.12 \[ A {\left (\frac {20 i \, e^{\left (-x\right )}}{75 \, e^{\left (-x\right )} + 150 i \, e^{\left (-2 \, x\right )} - 150 \, e^{\left (-3 \, x\right )} - 75 i \, e^{\left (-4 \, x\right )} + 15 \, e^{\left (-5 \, x\right )} - 15 i} - \frac {40 \, e^{\left (-2 \, x\right )}}{75 \, e^{\left (-x\right )} + 150 i \, e^{\left (-2 \, x\right )} - 150 \, e^{\left (-3 \, x\right )} - 75 i \, e^{\left (-4 \, x\right )} + 15 \, e^{\left (-5 \, x\right )} - 15 i} + \frac {4}{75 \, e^{\left (-x\right )} + 150 i \, e^{\left (-2 \, x\right )} - 150 \, e^{\left (-3 \, x\right )} - 75 i \, e^{\left (-4 \, x\right )} + 15 \, e^{\left (-5 \, x\right )} - 15 i}\right )} - \frac {1}{2} \, B {\left (\frac {20 \, e^{\left (-x\right )}}{25 \, e^{\left (-x\right )} + 50 i \, e^{\left (-2 \, x\right )} - 50 \, e^{\left (-3 \, x\right )} - 25 i \, e^{\left (-4 \, x\right )} + 5 \, e^{\left (-5 \, x\right )} - 5 i} + \frac {20 i \, e^{\left (-2 \, x\right )}}{25 \, e^{\left (-x\right )} + 50 i \, e^{\left (-2 \, x\right )} - 50 \, e^{\left (-3 \, x\right )} - 25 i \, e^{\left (-4 \, x\right )} + 5 \, e^{\left (-5 \, x\right )} - 5 i} - \frac {20 \, e^{\left (-3 \, x\right )}}{25 \, e^{\left (-x\right )} + 50 i \, e^{\left (-2 \, x\right )} - 50 \, e^{\left (-3 \, x\right )} - 25 i \, e^{\left (-4 \, x\right )} + 5 \, e^{\left (-5 \, x\right )} - 5 i} - \frac {4 i}{25 \, e^{\left (-x\right )} + 50 i \, e^{\left (-2 \, x\right )} - 50 \, e^{\left (-3 \, x\right )} - 25 i \, e^{\left (-4 \, x\right )} + 5 \, e^{\left (-5 \, x\right )} - 5 i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(I+sinh(x))^3,x, algorithm="maxima")

[Out]

A*(20*I*e^(-x)/(75*e^(-x) + 150*I*e^(-2*x) - 150*e^(-3*x) - 75*I*e^(-4*x) + 15*e^(-5*x) - 15*I) - 40*e^(-2*x)/
(75*e^(-x) + 150*I*e^(-2*x) - 150*e^(-3*x) - 75*I*e^(-4*x) + 15*e^(-5*x) - 15*I) + 4/(75*e^(-x) + 150*I*e^(-2*
x) - 150*e^(-3*x) - 75*I*e^(-4*x) + 15*e^(-5*x) - 15*I)) - 1/2*B*(20*e^(-x)/(25*e^(-x) + 50*I*e^(-2*x) - 50*e^
(-3*x) - 25*I*e^(-4*x) + 5*e^(-5*x) - 5*I) + 20*I*e^(-2*x)/(25*e^(-x) + 50*I*e^(-2*x) - 50*e^(-3*x) - 25*I*e^(
-4*x) + 5*e^(-5*x) - 5*I) - 20*e^(-3*x)/(25*e^(-x) + 50*I*e^(-2*x) - 50*e^(-3*x) - 25*I*e^(-4*x) + 5*e^(-5*x)
- 5*I) - 4*I/(25*e^(-x) + 50*I*e^(-2*x) - 50*e^(-3*x) - 25*I*e^(-4*x) + 5*e^(-5*x) - 5*I))

________________________________________________________________________________________

mupad [B]  time = 0.83, size = 52, normalized size = 0.76 \[ \frac {\frac {A\,4{}\mathrm {i}}{15}-\frac {2\,B}{5}-\frac {A\,{\mathrm {e}}^{2\,x}\,8{}\mathrm {i}}{3}+{\mathrm {e}}^x\,\left (\frac {4\,A}{3}+B\,2{}\mathrm {i}\right )+2\,B\,{\mathrm {e}}^{2\,x}-B\,{\mathrm {e}}^{3\,x}\,2{}\mathrm {i}}{{\left (-1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sinh(x))/(sinh(x) + 1i)^3,x)

[Out]

((A*4i)/15 - (2*B)/5 - (A*exp(2*x)*8i)/3 + exp(x)*((4*A)/3 + B*2i) + 2*B*exp(2*x) - B*exp(3*x)*2i)/(exp(x)*1i
- 1)^5

________________________________________________________________________________________

sympy [A]  time = 0.39, size = 83, normalized size = 1.22 \[ \frac {- 4 i A + 30 i B e^{3 x} + 6 B + \left (- 20 A - 30 i B\right ) e^{x} + \left (40 i A - 30 B\right ) e^{2 x}}{- 15 i e^{5 x} + 75 e^{4 x} + 150 i e^{3 x} - 150 e^{2 x} - 75 i e^{x} + 15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(I+sinh(x))**3,x)

[Out]

(-4*I*A + 30*I*B*exp(3*x) + 6*B + (-20*A - 30*I*B)*exp(x) + (40*I*A - 30*B)*exp(2*x))/(-15*I*exp(5*x) + 75*exp
(4*x) + 150*I*exp(3*x) - 150*exp(2*x) - 75*I*exp(x) + 15)

________________________________________________________________________________________