Integrand size = 15, antiderivative size = 91 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^4} \, dx=-\frac {(i A+B) \cosh (x)}{7 (i+\sinh (x))^4}-\frac {(3 A+4 i B) \cosh (x)}{35 (i+\sinh (x))^3}+\frac {2 (3 i A-4 B) \cosh (x)}{105 (i+\sinh (x))^2}+\frac {2 (3 A+4 i B) \cosh (x)}{105 (i+\sinh (x))} \] Output:
-1/7*(I*A+B)*cosh(x)/(I+sinh(x))^4-1/35*(3*A+4*I*B)*cosh(x)/(I+sinh(x))^3+ 2/105*(3*I*A-4*B)*cosh(x)/(I+sinh(x))^2+2*(3*A+4*I*B)*cosh(x)/(105*I+105*s inh(x))
Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^4} \, dx=\frac {\cosh (x) \left (-36 i A+13 B-13 (3 A+4 i B) \sinh (x)+8 i (3 A+4 i B) \sinh ^2(x)+(6 A+8 i B) \sinh ^3(x)\right )}{105 (i+\sinh (x))^4} \] Input:
Integrate[(A + B*Sinh[x])/(I + Sinh[x])^4,x]
Output:
(Cosh[x]*((-36*I)*A + 13*B - 13*(3*A + (4*I)*B)*Sinh[x] + (8*I)*(3*A + (4* I)*B)*Sinh[x]^2 + (6*A + (8*I)*B)*Sinh[x]^3))/(105*(I + Sinh[x])^4)
Time = 0.67 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 3229, 3042, 3129, 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.
\(\displaystyle \int \frac {A+B \sinh (x)}{(\sinh (x)+i)^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A-i B \sin (i x)}{(i-i \sin (i x))^4}dx\) |
\(\Big \downarrow \) 3229 |
\(\displaystyle -\frac {1}{7} (-4 B+3 i A) \int \frac {1}{(\sinh (x)+i)^3}dx-\frac {(B+i A) \cosh (x)}{7 (\sinh (x)+i)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{7} (-4 B+3 i A) \int \frac {1}{(i-i \sin (i x))^3}dx-\frac {(B+i A) \cosh (x)}{7 (\sinh (x)+i)^4}\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle -\frac {1}{7} (-4 B+3 i A) \left (-\frac {2}{5} i \int \frac {1}{(\sinh (x)+i)^2}dx-\frac {i \cosh (x)}{5 (\sinh (x)+i)^3}\right )-\frac {(B+i A) \cosh (x)}{7 (\sinh (x)+i)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{7} (-4 B+3 i A) \left (-\frac {2}{5} i \int \frac {1}{(i-i \sin (i x))^2}dx-\frac {i \cosh (x)}{5 (\sinh (x)+i)^3}\right )-\frac {(B+i A) \cosh (x)}{7 (\sinh (x)+i)^4}\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle -\frac {1}{7} (-4 B+3 i A) \left (-\frac {2}{5} i \left (-\frac {1}{3} i \int \frac {1}{\sinh (x)+i}dx-\frac {i \cosh (x)}{3 (\sinh (x)+i)^2}\right )-\frac {i \cosh (x)}{5 (\sinh (x)+i)^3}\right )-\frac {(B+i A) \cosh (x)}{7 (\sinh (x)+i)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{7} (-4 B+3 i A) \left (-\frac {2}{5} i \left (-\frac {1}{3} i \int \frac {1}{i-i \sin (i x)}dx-\frac {i \cosh (x)}{3 (\sinh (x)+i)^2}\right )-\frac {i \cosh (x)}{5 (\sinh (x)+i)^3}\right )-\frac {(B+i A) \cosh (x)}{7 (\sinh (x)+i)^4}\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle -\frac {(B+i A) \cosh (x)}{7 (\sinh (x)+i)^4}-\frac {1}{7} (-4 B+3 i A) \left (-\frac {i \cosh (x)}{5 (\sinh (x)+i)^3}-\frac {2}{5} i \left (-\frac {\cosh (x)}{3 (\sinh (x)+i)}-\frac {i \cosh (x)}{3 (\sinh (x)+i)^2}\right )\right )\) |
Input:
Int[(A + B*Sinh[x])/(I + Sinh[x])^4,x]
Output:
-1/7*((I*A + B)*Cosh[x])/(I + Sinh[x])^4 - (((3*I)*A - 4*B)*(((-1/5*I)*Cos h[x])/(I + Sinh[x])^3 - ((2*I)/5)*(((-1/3*I)*Cosh[x])/(I + Sinh[x])^2 - Co sh[x]/(3*(I + Sinh[x])))))/7
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 = 0.65 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73
method | result | size |
risch | \(-\frac {4 \left (4 B -21 A \,{\mathrm e}^{x}-84 B \,{\mathrm e}^{2 x}-3 i A +70 i B \,{\mathrm e}^{3 x}+63 i A \,{\mathrm e}^{2 x}-28 i B \,{\mathrm e}^{x}+70 B \,{\mathrm e}^{4 x}+105 A \,{\mathrm e}^{3 x}\right )}{105 \left ({\mathrm e}^{x}+i\right )^{7}}\) | \(66\) |
default | \(-\frac {2 \left (-8 i B +8 A \right )}{7 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{7}}+\frac {2 A}{\tanh \left (\frac {x}{2}\right )+i}-\frac {24 i A +24 B}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{6}}-\frac {6 i A +2 B}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}-\frac {2 \left (32 i B -36 A \right )}{5 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{5}}-\frac {2 \left (-10 i B +18 A \right )}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}-\frac {-32 i A -24 B}{2 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{4}}\) | \(128\) |
parallelrisch | \(\frac {210 A \tanh \left (\frac {x}{2}\right )^{6}+\left (630 i A -210 B \right ) \tanh \left (\frac {x}{2}\right )^{5}+\left (-350 i B -1260 A \right ) \tanh \left (\frac {x}{2}\right )^{4}+\left (-1260 i A +560 B \right ) \tanh \left (\frac {x}{2}\right )^{3}+\left (336 i B +882 A \right ) \tanh \left (\frac {x}{2}\right )^{2}+\left (294 i A -182 B \right ) \tanh \left (\frac {x}{2}\right )-26 i B -72 A}{-2205 \tanh \left (\frac {x}{2}\right )^{5}+735 i \tanh \left (\frac {x}{2}\right )^{6}+105 \tanh \left (\frac {x}{2}\right )^{7}+3675 \tanh \left (\frac {x}{2}\right )^{3}-3675 i \tanh \left (\frac {x}{2}\right )^{4}-735 \tanh \left (\frac {x}{2}\right )+2205 i \tanh \left (\frac {x}{2}\right )^{2}-105 i}\) | \(154\) |
Input:
int((A+B*sinh(x))/(I+sinh(x))^4,x,method=_RETURNVERBOSE)
Output:
-4/105*(4*B-21*A*exp(x)-84*B*exp(x)^2-3*I*A+70*I*B*exp(x)^3+63*I*A*exp(x)^ 2-28*I*B*exp(x)+70*B*exp(x)^4+105*A*exp(x)^3)/(exp(x)+I)^7
Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.04 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^4} \, dx=-\frac {4 \, {\left (70 \, B e^{\left (4 \, x\right )} + 35 \, {\left (3 \, A + 2 i \, B\right )} e^{\left (3 \, x\right )} + 21 \, {\left (3 i \, A - 4 \, B\right )} e^{\left (2 \, x\right )} - 7 \, {\left (3 \, A + 4 i \, B\right )} e^{x} - 3 i \, A + 4 \, B\right )}}{105 \, {\left (e^{\left (7 \, x\right )} + 7 i \, e^{\left (6 \, x\right )} - 21 \, e^{\left (5 \, x\right )} - 35 i \, e^{\left (4 \, x\right )} + 35 \, e^{\left (3 \, x\right )} + 21 i \, e^{\left (2 \, x\right )} - 7 \, e^{x} - i\right )}} \] Input:
integrate((A+B*sinh(x))/(I+sinh(x))^4,x, algorithm="fricas")
Output:
-4/105*(70*B*e^(4*x) + 35*(3*A + 2*I*B)*e^(3*x) + 21*(3*I*A - 4*B)*e^(2*x) - 7*(3*A + 4*I*B)*e^x - 3*I*A + 4*B)/(e^(7*x) + 7*I*e^(6*x) - 21*e^(5*x) - 35*I*e^(4*x) + 35*e^(3*x) + 21*I*e^(2*x) - 7*e^x - I)
Time = 0.42 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.21 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^4} \, dx=\frac {12 i A - 280 B e^{4 x} - 16 B + \left (- 420 A - 280 i B\right ) e^{3 x} + \left (84 A + 112 i B\right ) e^{x} + \left (- 252 i A + 336 B\right ) e^{2 x}}{105 e^{7 x} + 735 i e^{6 x} - 2205 e^{5 x} - 3675 i e^{4 x} + 3675 e^{3 x} + 2205 i e^{2 x} - 735 e^{x} - 105 i} \] Input:
integrate((A+B*sinh(x))/(I+sinh(x))**4,x)
Output:
(12*I*A - 280*B*exp(4*x) - 16*B + (-420*A - 280*I*B)*exp(3*x) + (84*A + 11 2*I*B)*exp(x) + (-252*I*A + 336*B)*exp(2*x))/(105*exp(7*x) + 735*I*exp(6*x ) - 2205*exp(5*x) - 3675*I*exp(4*x) + 3675*exp(3*x) + 2205*I*exp(2*x) - 73 5*exp(x) - 105*I)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 469 vs. \(2 (67) = 134\).
Time = 0.04 (sec) , antiderivative size = 469, normalized size of antiderivative = 5.15 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^4} \, dx =\text {Too large to display} \] Input:
integrate((A+B*sinh(x))/(I+sinh(x))^4,x, algorithm="maxima")
Output:
4/35*A*(7*e^(-x)/(7*e^(-x) + 21*I*e^(-2*x) - 35*e^(-3*x) - 35*I*e^(-4*x) + 21*e^(-5*x) + 7*I*e^(-6*x) - e^(-7*x) - I) + 21*I*e^(-2*x)/(7*e^(-x) + 21 *I*e^(-2*x) - 35*e^(-3*x) - 35*I*e^(-4*x) + 21*e^(-5*x) + 7*I*e^(-6*x) - e ^(-7*x) - I) - 35*e^(-3*x)/(7*e^(-x) + 21*I*e^(-2*x) - 35*e^(-3*x) - 35*I* e^(-4*x) + 21*e^(-5*x) + 7*I*e^(-6*x) - e^(-7*x) - I) - I/(7*e^(-x) + 21*I *e^(-2*x) - 35*e^(-3*x) - 35*I*e^(-4*x) + 21*e^(-5*x) + 7*I*e^(-6*x) - e^( -7*x) - I)) - 8/105*B*(-14*I*e^(-x)/(7*e^(-x) + 21*I*e^(-2*x) - 35*e^(-3*x ) - 35*I*e^(-4*x) + 21*e^(-5*x) + 7*I*e^(-6*x) - e^(-7*x) - I) + 42*e^(-2* x)/(7*e^(-x) + 21*I*e^(-2*x) - 35*e^(-3*x) - 35*I*e^(-4*x) + 21*e^(-5*x) + 7*I*e^(-6*x) - e^(-7*x) - I) + 35*I*e^(-3*x)/(7*e^(-x) + 21*I*e^(-2*x) - 35*e^(-3*x) - 35*I*e^(-4*x) + 21*e^(-5*x) + 7*I*e^(-6*x) - e^(-7*x) - I) - 35*e^(-4*x)/(7*e^(-x) + 21*I*e^(-2*x) - 35*e^(-3*x) - 35*I*e^(-4*x) + 21* e^(-5*x) + 7*I*e^(-6*x) - e^(-7*x) - I) - 2/(7*e^(-x) + 21*I*e^(-2*x) - 35 *e^(-3*x) - 35*I*e^(-4*x) + 21*e^(-5*x) + 7*I*e^(-6*x) - e^(-7*x) - I))
Time = 0.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.66 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^4} \, dx=-\frac {4 \, {\left (70 \, B e^{\left (4 \, x\right )} + 105 \, A e^{\left (3 \, x\right )} + 70 i \, B e^{\left (3 \, x\right )} + 63 i \, A e^{\left (2 \, x\right )} - 84 \, B e^{\left (2 \, x\right )} - 21 \, A e^{x} - 28 i \, B e^{x} - 3 i \, A + 4 \, B\right )}}{105 \, {\left (e^{x} + i\right )}^{7}} \] Input:
integrate((A+B*sinh(x))/(I+sinh(x))^4,x, algorithm="giac")
Output:
-4/105*(70*B*e^(4*x) + 105*A*e^(3*x) + 70*I*B*e^(3*x) + 63*I*A*e^(2*x) - 8 4*B*e^(2*x) - 21*A*e^x - 28*I*B*e^x - 3*I*A + 4*B)/(e^x + I)^7
Time = 1.99 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^4} \, dx=-\frac {\frac {16\,B}{105}+4\,A\,{\mathrm {e}}^{3\,x}-{\mathrm {e}}^x\,\left (\frac {4\,A}{5}+\frac {B\,16{}\mathrm {i}}{15}\right )-\frac {16\,B\,{\mathrm {e}}^{2\,x}}{5}+\frac {8\,B\,{\mathrm {e}}^{4\,x}}{3}-\frac {A\,4{}\mathrm {i}}{35}+\frac {A\,{\mathrm {e}}^{2\,x}\,12{}\mathrm {i}}{5}+\frac {B\,{\mathrm {e}}^{3\,x}\,8{}\mathrm {i}}{3}}{{\left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}^7} \] Input:
int((A + B*sinh(x))/(sinh(x) + 1i)^4,x)
Output:
-((16*B)/105 - (A*4i)/35 + (A*exp(2*x)*12i)/5 + 4*A*exp(3*x) - exp(x)*((4* A)/5 + (B*16i)/15) - (16*B*exp(2*x))/5 + (B*exp(3*x)*8i)/3 + (8*B*exp(4*x) )/3)/(exp(x) + 1i)^7
\[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^4} \, dx=\left (\int \frac {\sinh \left (x \right )}{\sinh \left (x \right )^{4}+4 \sinh \left (x \right )^{3} i -6 \sinh \left (x \right )^{2}-4 \sinh \left (x \right ) i +1}d x \right ) b +\left (\int \frac {1}{\sinh \left (x \right )^{4}+4 \sinh \left (x \right )^{3} i -6 \sinh \left (x \right )^{2}-4 \sinh \left (x \right ) i +1}d x \right ) a \] Input:
int((A+B*sinh(x))/(I+sinh(x))^4,x)
Output:
int(sinh(x)/(sinh(x)**4 + 4*sinh(x)**3*i - 6*sinh(x)**2 - 4*sinh(x)*i + 1) ,x)*b + int(1/(sinh(x)**4 + 4*sinh(x)**3*i - 6*sinh(x)**2 - 4*sinh(x)*i + 1),x)*a