\(\int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx\) [117]

Optimal result

Integrand size = 15, antiderivative size = 68 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=-\frac {(i A+B) \cosh (x)}{5 (i+\sinh (x))^3}-\frac {(2 A+3 i B) \cosh (x)}{15 (i+\sinh (x))^2}+\frac {(2 i A-3 B) \cosh (x)}{15 (i+\sinh (x))} \]

[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] (verified)

Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2829, 2729, 2727} \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=\frac {(-3 B+2 i A) \cosh (x)}{15 (\sinh (x)+i)}-\frac {(2 A+3 i B) \cosh (x)}{15 (\sinh (x)+i)^2}-\frac {(B+i A) \cosh (x)}{5 (\sinh (x)+i)^3} \]

[In]

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

[Out]

-1/5*((I*A + B)*Cosh[x])/(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 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 2729

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 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)}{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] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.74 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=\frac {\cosh (x) \left (-7 i A+3 B-3 (2 A+3 i B) \sinh (x)+(2 i A-3 B) \sinh ^2(x)\right )}{15 (i+\sinh (x))^3} \]

[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)

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {2 \left (15 B \,{\mathrm e}^{3 x}-15 B \,{\mathrm e}^{x}+15 i B \,{\mathrm e}^{2 x}-2 A +10 i A \,{\mathrm e}^{x}-3 i B +20 A \,{\mathrm e}^{2 x}\right )}{15 \left ({\mathrm e}^{x}+i\right )^{5}}\)	\(51\)
default	\(-\frac {2 i B -4 A}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}-\frac {2 \left (8 i A +6 B \right )}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}-\frac {-8 i B +8 A}{2 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{4}}-\frac {2 \left (-4 i A -4 B \right )}{5 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{5}}+\frac {2 i A}{\tanh \left (\frac {x}{2}\right )+i}\)	\(91\)
parallelrisch	\(\frac {\operatorname {sech}\left (x \right ) \left (-10 i A \sinh \left (x \right )+2 i A \sinh \left (3 x \right )-28 i A \sinh \left (2 x \right )+15 i B \cosh \left (x \right )-12 i B \cosh \left (2 x \right )-3 i B \cosh \left (3 x \right )+35 A \cosh \left (x \right )-8 A \cosh \left (2 x \right )-7 A \cosh \left (3 x \right )-15 B \sinh \left (x \right )-3 B \sinh \left (3 x \right )+12 B \sinh \left (2 x \right )-20 A \right )}{120 i \sinh \left (x \right )+30 \cosh \left (2 x \right )-90}\)	\(105\)

[In]

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

[Out]

-2/15*(15*B*exp(x)^3-15*B*exp(x)+15*I*B*exp(x)^2-2*A+10*I*A*exp(x)-3*I*B+20*A*exp(x)^2)/(exp(x)+I)^5

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.03 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=-\frac {2 \, {\left (15 \, B e^{\left (3 \, x\right )} + 5 \, {\left (4 \, A + 3 i \, B\right )} e^{\left (2 \, x\right )} + 5 \, {\left (2 i \, A - 3 \, B\right )} e^{x} - 2 \, A - 3 i \, B\right )}}{15 \, {\left (e^{\left (5 \, x\right )} + 5 i \, e^{\left (4 \, x\right )} - 10 \, e^{\left (3 \, x\right )} - 10 i \, e^{\left (2 \, x\right )} + 5 \, e^{x} + i\right )}} \]

[In]

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

[Out]

-2/15*(15*B*e^(3*x) + 5*(4*A + 3*I*B)*e^(2*x) + 5*(2*I*A - 3*B)*e^x - 2*A - 3*I*B)/(e^(5*x) + 5*I*e^(4*x) - 10
*e^(3*x) - 10*I*e^(2*x) + 5*e^x + I)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.21 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=\frac {4 A - 30 B e^{3 x} + 6 i B + \left (- 40 A - 30 i B\right ) e^{2 x} + \left (- 20 i A + 30 B\right ) e^{x}}{15 e^{5 x} + 75 i e^{4 x} - 150 e^{3 x} - 150 i e^{2 x} + 75 e^{x} + 15 i} \]

[In]

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

[Out]

(4*A - 30*B*exp(3*x) + 6*I*B + (-40*A - 30*I*B)*exp(2*x) + (-20*I*A + 30*B)*exp(x))/(15*exp(5*x) + 75*I*exp(4*
x) - 150*exp(3*x) - 150*I*exp(2*x) + 75*exp(x) + 15*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. 267 vs. \(2 (50) = 100\).

Time = 0.20 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.93 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=-\frac {2}{5} \, B {\left (\frac {5 \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} + 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} - 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - i} + \frac {5 i \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} + 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} - 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - i} - \frac {5 \, e^{\left (-3 \, x\right )}}{5 \, e^{\left (-x\right )} + 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} - 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - i} - \frac {i}{5 \, e^{\left (-x\right )} + 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} - 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - i}\right )} - \frac {4}{15} \, A {\left (-\frac {5 i \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} + 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} - 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - i} + \frac {10 \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} + 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} - 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - i} - \frac {1}{5 \, e^{\left (-x\right )} + 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} - 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - i}\right )} \]

[In]

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

[Out]

-2/5*B*(5*e^(-x)/(5*e^(-x) + 10*I*e^(-2*x) - 10*e^(-3*x) - 5*I*e^(-4*x) + e^(-5*x) - I) + 5*I*e^(-2*x)/(5*e^(-
x) + 10*I*e^(-2*x) - 10*e^(-3*x) - 5*I*e^(-4*x) + e^(-5*x) - I) - 5*e^(-3*x)/(5*e^(-x) + 10*I*e^(-2*x) - 10*e^
(-3*x) - 5*I*e^(-4*x) + e^(-5*x) - I) - I/(5*e^(-x) + 10*I*e^(-2*x) - 10*e^(-3*x) - 5*I*e^(-4*x) + e^(-5*x) -
I)) - 4/15*A*(-5*I*e^(-x)/(5*e^(-x) + 10*I*e^(-2*x) - 10*e^(-3*x) - 5*I*e^(-4*x) + e^(-5*x) - I) + 10*e^(-2*x)
/(5*e^(-x) + 10*I*e^(-2*x) - 10*e^(-3*x) - 5*I*e^(-4*x) + e^(-5*x) - I) - 1/(5*e^(-x) + 10*I*e^(-2*x) - 10*e^(
-3*x) - 5*I*e^(-4*x) + e^(-5*x) - I))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.68 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=-\frac {2 \, {\left (15 \, B e^{\left (3 \, x\right )} + 20 \, A e^{\left (2 \, x\right )} + 15 i \, B e^{\left (2 \, x\right )} + 10 i \, A e^{x} - 15 \, B e^{x} - 2 \, A - 3 i \, B\right )}}{15 \, {\left (e^{x} + i\right )}^{5}} \]

[In]

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

[Out]

-2/15*(15*B*e^(3*x) + 20*A*e^(2*x) + 15*I*B*e^(2*x) + 10*I*A*e^x - 15*B*e^x - 2*A - 3*I*B)/(e^x + I)^5

Mupad [B] (verification not implemented)

Time = 1.57 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=\frac {\frac {A\,4{}\mathrm {i}}{15}-\frac {2\,B}{5}-\frac {A\,{\mathrm {e}}^{2\,x}\,8{}\mathrm {i}}{3}+{\mathrm {e}}^x\,\left (\frac {4\,A}{3}+B\,2{}\mathrm {i}\right )+2\,B\,{\mathrm {e}}^{2\,x}-B\,{\mathrm {e}}^{3\,x}\,2{}\mathrm {i}}{{\left (-1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )}^5} \]

[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