Integrand size = 17, antiderivative size = 76 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^3} \, dx=\frac {(i A-B) \cosh (x)}{5 (i-\sinh (x))^3}+\frac {(2 A-3 i B) \cosh (x)}{15 (i-\sinh (x))^2}-\frac {(2 i A+3 B) \cosh (x)}{15 (i-\sinh (x))} \] Output:
1/5*(I*A-B)*cosh(x)/(I-sinh(x))^3+1/15*(2*A-3*I*B)*cosh(x)/(I-sinh(x))^2-( 2*I*A+3*B)*cosh(x)/(15*I-15*sinh(x))
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.64 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^3} \, dx=\frac {\cosh (x) \left (-7 i A-3 B+(6 A-9 i B) \sinh (x)+(2 i A+3 B) \sinh ^2(x)\right )}{15 (-i+\sinh (x))^3} \] Input:
Integrate[(A + B*Sinh[x])/(I - Sinh[x])^3,x]
Output:
(Cosh[x]*((-7*I)*A - 3*B + (6*A - (9*I)*B)*Sinh[x] + ((2*I)*A + 3*B)*Sinh[ x]^2))/(15*(-I + Sinh[x])^3)
Time = 0.62 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3042, 3229, 3042, 3129, 3042, 3127}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {A+B \sinh (x)}{(-\sinh (x)+i)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A-i B \sin (i x)}{(i \sin (i x)+i)^3}dx\) |
\(\Big \downarrow \) 3229 |
\(\displaystyle \frac {(-B+i A) \cosh (x)}{5 (-\sinh (x)+i)^3}-\frac {1}{5} (3 B+2 i A) \int \frac {1}{(i-\sinh (x))^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(-B+i A) \cosh (x)}{5 (-\sinh (x)+i)^3}-\frac {1}{5} (3 B+2 i A) \int \frac {1}{(i \sin (i x)+i)^2}dx\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {(-B+i A) \cosh (x)}{5 (-\sinh (x)+i)^3}-\frac {1}{5} (3 B+2 i A) \left (\frac {i \cosh (x)}{3 (-\sinh (x)+i)^2}-\frac {1}{3} i \int \frac {1}{i-\sinh (x)}dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(-B+i A) \cosh (x)}{5 (-\sinh (x)+i)^3}-\frac {1}{5} (3 B+2 i A) \left (\frac {i \cosh (x)}{3 (-\sinh (x)+i)^2}-\frac {1}{3} i \int \frac {1}{i \sin (i x)+i}dx\right )\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {(-B+i A) \cosh (x)}{5 (-\sinh (x)+i)^3}-\frac {1}{5} (3 B+2 i A) \left (\frac {\cosh (x)}{3 (-\sinh (x)+i)}+\frac {i \cosh (x)}{3 (-\sinh (x)+i)^2}\right )\) |
Input:
Int[(A + B*Sinh[x])/(I - Sinh[x])^3,x]
Output:
-1/5*(((2*I)*A + 3*B)*(((I/3)*Cosh[x])/(I - Sinh[x])^2 + Cosh[x]/(3*(I - S inh[x])))) + ((I*A - B)*Cosh[x])/(5*(I - Sinh[x])^3)
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]
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] + Simp[(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]
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] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)) I nt[(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)]
Time = 0.44 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.67
method | result | size |
risch | \(\frac {2 B \,{\mathrm e}^{3 x}+\frac {8 A \,{\mathrm e}^{2 x}}{3}-2 B \,{\mathrm e}^{x}-2 i B \,{\mathrm e}^{2 x}-\frac {4 A}{15}-\frac {4 i A \,{\mathrm e}^{x}}{3}+\frac {2 i B}{5}}{\left ({\mathrm e}^{x}-i\right )^{5}}\) | \(51\) |
default | \(-\frac {2 \left (-4 i A +4 B \right )}{5 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{5}}-\frac {2 i B +4 A}{\left (\tanh \left (\frac {x}{2}\right )-i\right )^{2}}-\frac {2 \left (8 i A -6 B \right )}{3 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{3}}-\frac {-8 i B -8 A}{2 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{4}}+\frac {2 i A}{\tanh \left (\frac {x}{2}\right )-i}\) | \(91\) |
parallelrisch | \(\frac {\left (3 i B -6 A \right ) \tanh \left (\frac {x}{2}\right )^{5}+15 B \tanh \left (\frac {x}{2}\right )^{4}+20 i A \tanh \left (\frac {x}{2}\right )^{2}+\left (-15 i B +10 A \right ) \tanh \left (\frac {x}{2}\right )-8 i A -3 B}{75 i \tanh \left (\frac {x}{2}\right )^{4}-15 \tanh \left (\frac {x}{2}\right )^{5}-150 i \tanh \left (\frac {x}{2}\right )^{2}+150 \tanh \left (\frac {x}{2}\right )^{3}+15 i-75 \tanh \left (\frac {x}{2}\right )}\) | \(102\) |
Input:
int((A+B*sinh(x))/(I-sinh(x))^3,x,method=_RETURNVERBOSE)
Output:
2/15*(15*B*exp(x)^3+20*A*exp(x)^2-15*B*exp(x)-15*I*B*exp(x)^2-2*A-10*I*A*e xp(x)+3*I*B)/(exp(x)-I)^5
Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^3} \, dx=\frac {2 \, {\left (15 \, B e^{\left (3 \, x\right )} + 5 \, {\left (4 \, A - 3 i \, B\right )} e^{\left (2 \, x\right )} - 5 \, {\left (2 i \, A + 3 \, B\right )} e^{x} - 2 \, A + 3 i \, B\right )}}{15 \, {\left (e^{\left (5 \, x\right )} - 5 i \, e^{\left (4 \, x\right )} - 10 \, e^{\left (3 \, x\right )} + 10 i \, e^{\left (2 \, x\right )} + 5 \, e^{x} - i\right )}} \] Input:
integrate((A+B*sinh(x))/(I-sinh(x))^3,x, algorithm="fricas")
Output:
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)
Time = 0.25 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.08 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^3} \, dx=\frac {- 4 A + 30 B e^{3 x} + 6 i B + \left (40 A - 30 i B\right ) e^{2 x} + \left (- 20 i A - 30 B\right ) e^{x}}{15 e^{5 x} - 75 i e^{4 x} - 150 e^{3 x} + 150 i e^{2 x} + 75 e^{x} - 15 i} \] Input:
integrate((A+B*sinh(x))/(I-sinh(x))**3,x)
Output:
(-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) + 7 5*exp(x) - 15*I)
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 = 0.05 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.51 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^3} \, dx=\frac {2}{5} \, B {\left (\frac {5 \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} - 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} + 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + i} - \frac {5 i \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} - 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} + 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + i} - \frac {5 \, e^{\left (-3 \, x\right )}}{5 \, e^{\left (-x\right )} - 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} + 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + i} + \frac {i}{5 \, e^{\left (-x\right )} - 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} + 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + i}\right )} + \frac {4}{15} \, A {\left (\frac {5 i \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} - 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} + 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + i} + \frac {10 \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} - 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} + 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + i} - \frac {1}{5 \, e^{\left (-x\right )} - 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} + 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + i}\right )} \] Input:
integrate((A+B*sinh(x))/(I-sinh(x))^3,x, algorithm="maxima")
Output:
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))
Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.61 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^3} \, dx=\frac {2 \, {\left (15 \, B e^{\left (3 \, x\right )} + 20 \, A e^{\left (2 \, x\right )} - 15 i \, B e^{\left (2 \, x\right )} - 10 i \, A e^{x} - 15 \, B e^{x} - 2 \, A + 3 i \, B\right )}}{15 \, {\left (e^{x} - i\right )}^{5}} \] Input:
integrate((A+B*sinh(x))/(I-sinh(x))^3,x, algorithm="giac")
Output:
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
Time = 1.73 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.68 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^3} \, dx=\frac {2\,B\,{\mathrm {e}}^{2\,x}-\frac {2\,B}{5}+\frac {A\,{\mathrm {e}}^{2\,x}\,8{}\mathrm {i}}{3}+{\mathrm {e}}^x\,\left (\frac {4\,A}{3}-B\,2{}\mathrm {i}\right )-\frac {A\,4{}\mathrm {i}}{15}+B\,{\mathrm {e}}^{3\,x}\,2{}\mathrm {i}}{{\left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )}^5} \] Input:
int(-(A + B*sinh(x))/(sinh(x) - 1i)^3,x)
Output:
((A*exp(2*x)*8i)/3 - (2*B)/5 - (A*4i)/15 + exp(x)*((4*A)/3 - B*2i) + 2*B*e xp(2*x) + B*exp(3*x)*2i)/(exp(x)*1i + 1)^5
\[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^3} \, dx=-\left (\int \frac {\sinh \left (x \right )}{\sinh \left (x \right )^{3}-3 \sinh \left (x \right )^{2} i -3 \sinh \left (x \right )+i}d x \right ) b -\left (\int \frac {1}{\sinh \left (x \right )^{3}-3 \sinh \left (x \right )^{2} i -3 \sinh \left (x \right )+i}d x \right ) a \] Input:
int((A+B*sinh(x))/(I-sinh(x))^3,x)
Output:
- (int(sinh(x)/(sinh(x)**3 - 3*sinh(x)**2*i - 3*sinh(x) + i),x)*b + int(1 /(sinh(x)**3 - 3*sinh(x)**2*i - 3*sinh(x) + i),x)*a)