\(\int \frac {A+B \sinh (x)}{(i-\sinh (x))^2} \, dx\) [120]

Optimal result

Integrand size = 17, antiderivative size = 49 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^2} \, dx=\frac {(i A-B) \cosh (x)}{3 (i-\sinh (x))^2}+\frac {(A-2 i B) \cosh (x)}{3 (i-\sinh (x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2829, 2727} \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^2} \, dx=\frac {(A-2 i B) \cosh (x)}{3 (-\sinh (x)+i)}+\frac {(-B+i A) \cosh (x)}{3 (-\sinh (x)+i)^2} \]

[In]

Int[(A + B*Sinh[x])/(I - Sinh[x])^2,x]

[Out]

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

Rule 2727

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 2829

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.65 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^2} \, dx=\frac {\cosh (x) (2 i A+B-(A-2 i B) \sinh (x))}{3 (-i+\sinh (x))^2} \]

[In]

Integrate[(A + B*Sinh[x])/(I - Sinh[x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.73

method	result	size
risch	\(-\frac {2 \left (3 A \,{\mathrm e}^{x}-3 i B \,{\mathrm e}^{x}+3 B \,{\mathrm e}^{2 x}-i A -2 B \right )}{3 \left ({\mathrm e}^{x}-i\right )^{3}}\)	\(36\)
default	\(-\frac {2 i A -2 B}{\left (-i+\tanh \left (\frac {x}{2}\right )\right )^{2}}-\frac {2 A}{-i+\tanh \left (\frac {x}{2}\right )}-\frac {2 \left (-2 i B -2 A \right )}{3 \left (-i+\tanh \left (\frac {x}{2}\right )\right )^{3}}\)	\(52\)
parallelrisch	\(\frac {\left (3 i A +3 B \right ) \cosh \left (2 x \right )+\left (-i B -A \right ) \sinh \left (2 x \right )+\left (-2 i B +10 A \right ) \sinh \left (x \right )-3 i A -3 B}{12 i \sinh \left (x \right )-3 i \sinh \left (2 x \right )-6 \cosh \left (x \right )-3 \cosh \left (2 x \right )+9}\)	\(73\)

[In]

int((A+B*sinh(x))/(I-sinh(x))^2,x,method=_RETURNVERBOSE)

[Out]

-2/3*(3*A*exp(x)-3*I*B*exp(x)+3*B*exp(x)^2-I*A-2*B)/(exp(x)-I)^3

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.88 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^2} \, dx=-\frac {2 \, {\left (3 \, B e^{\left (2 \, x\right )} + 3 \, {\left (A - i \, B\right )} e^{x} - i \, A - 2 \, B\right )}}{3 \, {\left (e^{\left (3 \, x\right )} - 3 i \, e^{\left (2 \, x\right )} - 3 \, e^{x} + i\right )}} \]

[In]

integrate((A+B*sinh(x))/(I-sinh(x))^2,x, algorithm="fricas")

[Out]

-2/3*(3*B*e^(2*x) + 3*(A - I*B)*e^x - I*A - 2*B)/(e^(3*x) - 3*I*e^(2*x) - 3*e^x + I)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.04 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^2} \, dx=\frac {2 i A - 6 B e^{2 x} + 4 B + \left (- 6 A + 6 i B\right ) e^{x}}{3 e^{3 x} - 9 i e^{2 x} - 9 e^{x} + 3 i} \]

[In]

integrate((A+B*sinh(x))/(I-sinh(x))**2,x)

[Out]

(2*I*A - 6*B*exp(2*x) + 4*B + (-6*A + 6*I*B)*exp(x))/(3*exp(3*x) - 9*I*exp(2*x) - 9*exp(x) + 3*I)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (31) = 62\).

Time = 0.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.88 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^2} \, dx=-\frac {2}{3} \, A {\left (\frac {3 \, e^{\left (-x\right )}}{3 \, e^{\left (-x\right )} - 3 i \, e^{\left (-2 \, x\right )} - e^{\left (-3 \, x\right )} + i} + \frac {i}{3 \, e^{\left (-x\right )} - 3 i \, e^{\left (-2 \, x\right )} - e^{\left (-3 \, x\right )} + i}\right )} - \frac {2}{3} \, B {\left (-\frac {3 i \, e^{\left (-x\right )}}{3 \, e^{\left (-x\right )} - 3 i \, e^{\left (-2 \, x\right )} - e^{\left (-3 \, x\right )} + i} - \frac {3 \, e^{\left (-2 \, x\right )}}{3 \, e^{\left (-x\right )} - 3 i \, e^{\left (-2 \, x\right )} - e^{\left (-3 \, x\right )} + i} + \frac {2}{3 \, e^{\left (-x\right )} - 3 i \, e^{\left (-2 \, x\right )} - e^{\left (-3 \, x\right )} + i}\right )} \]

[In]

integrate((A+B*sinh(x))/(I-sinh(x))^2,x, algorithm="maxima")

[Out]

-2/3*A*(3*e^(-x)/(3*e^(-x) - 3*I*e^(-2*x) - e^(-3*x) + I) + I/(3*e^(-x) - 3*I*e^(-2*x) - e^(-3*x) + I)) - 2/3*
B*(-3*I*e^(-x)/(3*e^(-x) - 3*I*e^(-2*x) - e^(-3*x) + I) - 3*e^(-2*x)/(3*e^(-x) - 3*I*e^(-2*x) - e^(-3*x) + I)
+ 2/(3*e^(-x) - 3*I*e^(-2*x) - e^(-3*x) + I))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.65 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^2} \, dx=-\frac {2 \, {\left (3 \, B e^{\left (2 \, x\right )} + 3 \, A e^{x} - 3 i \, B e^{x} - i \, A - 2 \, B\right )}}{3 \, {\left (e^{x} - i\right )}^{3}} \]

[In]

integrate((A+B*sinh(x))/(I-sinh(x))^2,x, algorithm="giac")

[Out]

-2/3*(3*B*e^(2*x) + 3*A*e^x - 3*I*B*e^x - I*A - 2*B)/(e^x - I)^3

Mupad [B] (verification not implemented)

Time = 1.31 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.76 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^2} \, dx=\frac {\frac {2\,A}{3}-\frac {B\,4{}\mathrm {i}}{3}+{\mathrm {e}}^x\,\left (2\,B+A\,2{}\mathrm {i}\right )+B\,{\mathrm {e}}^{2\,x}\,2{}\mathrm {i}}{{\left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )}^3} \]

[In]

int((A + B*sinh(x))/(sinh(x) - 1i)^2,x)

[Out]

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