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

Optimal. Leaf size=76 \[ -\frac{(3 B+2 i A) \cosh (x)}{15 (-\sinh (x)+i)}+\frac{(2 A-3 i B) \cosh (x)}{15 (-\sinh (x)+i)^2}+\frac{(-B+i A) \cosh (x)}{5 (-\sinh (x)+i)^3} \]

[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)*Co
sh[x])/(15*(I - Sinh[x]))

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Rubi [A]  time = 0.0585895, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, 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)*Co
sh[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 = 0.215178, size = 92, normalized size = 1.21 \[ -\frac{5 (2 A-3 i B) \cosh \left (\frac{3 x}{2}\right )-20 i A \sinh \left (\frac{x}{2}\right )+2 i A \sinh \left (\frac{5 x}{2}\right )-15 B \sinh \left (\frac{x}{2}\right )+3 B \sinh \left (\frac{5 x}{2}\right )+15 i B \cosh \left (\frac{x}{2}\right )}{30 \left (\cosh \left (\frac{x}{2}\right )+i \sinh \left (\frac{x}{2}\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((15*I)*B*Cosh[x/2] + 5*(2*A - (3*I)*B)*Cosh[(3*x)/2] - (20*I)*A*Sinh[x/2] - 15*B*Sinh[x/2] + (2*I)*A*Sinh[(5
*x)/2] + 3*B*Sinh[(5*x)/2])/(30*(Cosh[x/2] + I*Sinh[x/2])^5)

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Maple [A]  time = 0.035, size = 91, normalized size = 1.2 \begin{align*} -{(4\,A+2\,iB) \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-2}}+{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{-8\,A-8\,iB}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-4}}-{\frac{16\,iA-12\,B}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) -i \right ) ^{-3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.26862, size = 379, normalized size = 4.99 \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 = 1.73298, size = 208, normalized size = 2.74 \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) - 15
0*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 = 1.3382, size = 62, normalized size = 0.82 \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