\(\int \frac {A+B \sinh (x)}{(i+\sinh (x))^2} \, dx\) [116]

Optimal result

Integrand size = 15, antiderivative size = 43 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^2} \, dx=-\frac {(i A+B) \cosh (x)}{3 (i+\sinh (x))^2}-\frac {(A+2 i B) \cosh (x)}{3 (i+\sinh (x))} \]

[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 = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, 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]

-1/3*((I*A + B)*Cosh[x])/(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 = 0.74 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^2} \, dx=\frac {\cosh (x) (-2 i A+B-(A+2 i B) \sinh (x))}{3 (i+\sinh (x))^2} \]

[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 = 0.86 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84

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	\(-\frac {-2 i A -2 B}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}-\frac {2 A}{\tanh \left (\frac {x}{2}\right )+i}-\frac {2 \left (2 i B -2 A \right )}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}\)	\(52\)
parallelrisch	\(\frac {\left (3 i A -3 B \right ) \cosh \left (2 x \right )+\left (-i B +A \right ) \sinh \left (2 x \right )+\left (-2 i B -10 A \right ) \sinh \left (x \right )-3 i A +3 B}{-3 i \sinh \left (2 x \right )+12 i \sinh \left (x \right )+6 \cosh \left (x \right )+3 \cosh \left (2 x \right )-9}\)	\(71\)

[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 = 0.31 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^2} \, dx=-\frac {2 \, {\left (3 \, B e^{\left (2 \, x\right )} + 3 \, {\left (A + i \, B\right )} e^{x} + i \, A - 2 \, B\right )}}{3 \, {\left (e^{\left (3 \, x\right )} + 3 i \, e^{\left (2 \, x\right )} - 3 \, e^{x} - i\right )}} \]

[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 = 53, normalized size of antiderivative = 1.23 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^2} \, dx=\frac {- 2 i A - 6 B e^{2 x} + 4 B + \left (- 6 A - 6 i B\right ) e^{x}}{3 e^{3 x} + 9 i e^{2 x} - 9 e^{x} - 3 i} \]

[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 = 3.28 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^2} \, dx=-\frac {2}{3} \, A {\left (\frac {3 \, e^{\left (-x\right )}}{3 \, e^{\left (-x\right )} + 3 i \, e^{\left (-2 \, x\right )} - e^{\left (-3 \, x\right )} - i} - \frac {i}{3 \, e^{\left (-x\right )} + 3 i \, e^{\left (-2 \, x\right )} - e^{\left (-3 \, x\right )} - i}\right )} - \frac {2}{3} \, B {\left (\frac {3 i \, e^{\left (-x\right )}}{3 \, e^{\left (-x\right )} + 3 i \, e^{\left (-2 \, x\right )} - e^{\left (-3 \, x\right )} - i} - \frac {3 \, e^{\left (-2 \, x\right )}}{3 \, e^{\left (-x\right )} + 3 i \, e^{\left (-2 \, x\right )} - e^{\left (-3 \, x\right )} - i} + \frac {2}{3 \, e^{\left (-x\right )} + 3 i \, e^{\left (-2 \, x\right )} - e^{\left (-3 \, x\right )} - i}\right )} \]

[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 = 0.74 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^2} \, dx=-\frac {2 \, {\left (3 \, B e^{\left (2 \, x\right )} + 3 \, A e^{x} + 3 i \, B e^{x} + i \, A - 2 \, B\right )}}{3 \, {\left (e^{x} + i\right )}^{3}} \]

[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 = 1.32 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^2} \, dx=-\frac {\frac {2\,A}{3}+\frac {B\,4{}\mathrm {i}}{3}-{\mathrm {e}}^x\,\left (-2\,B+A\,2{}\mathrm {i}\right )-B\,{\mathrm {e}}^{2\,x}\,2{}\mathrm {i}}{{\left (-1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )}^3} \]

[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