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]

-((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]))

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Rubi [A]  time = 0.0535386, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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)]

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 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]

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 = 0.0374888, 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)

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Maple [A]  time = 0.029, size = 91, normalized size = 1.3 \begin{align*} -{\frac{8\,A-8\,iB}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-4}}+{2\,iA \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}}-{\frac{-8\,iA-8\,B}{5} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-5}}-{\frac{16\,iA+12\,B}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}-{(-4\,A+2\,iB) \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.27767, size = 378, normalized size = 5.56 \begin{align*} 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 )} \end{align*}

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))

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Fricas [A]  time = 1.62266, size = 209, normalized size = 3.07 \begin{align*} -\frac{30 \, B e^{\left (3 \, x\right )} + 10 \,{\left (4 \, A + 3 i \, B\right )} e^{\left (2 \, x\right )} -{\left (-20 i \, A + 30 \, 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} \end{align*}

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) + 10*(4*A + 3*I*B)*e^(2*x) - (-20*I*A + 30*B)*e^x - 4*A - 6*I*B)/(15*e^(5*x) + 75*I*e^(4*x) - 1
50*e^(3*x) - 150*I*e^(2*x) + 75*e^x + 15*I)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.20797, size = 62, normalized size = 0.91 \begin{align*} -\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}} \end{align*}

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