Integrand size = 17, antiderivative size = 101 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^4} \, dx=\frac {(i A-B) \cosh (x)}{7 (i-\sinh (x))^4}+\frac {(3 A-4 i B) \cosh (x)}{35 (i-\sinh (x))^3}-\frac {2 (3 i A+4 B) \cosh (x)}{105 (i-\sinh (x))^2}-\frac {2 (3 A-4 i B) \cosh (x)}{105 (i-\sinh (x))} \] 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*si nh(x))
Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.62 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^4} \, dx=\frac {\cosh (x) \left (36 i A+13 B+(-39 A+52 i B) \sinh (x)+(-24 i A-32 B) \sinh ^2(x)+(6 A-8 i B) \sinh ^3(x)\right )}{105 (-i+\sinh (x))^4} \] Input:
Integrate[(A + B*Sinh[x])/(I - Sinh[x])^4,x]
Output:
(Cosh[x]*((36*I)*A + 13*B + (-39*A + (52*I)*B)*Sinh[x] + ((-24*I)*A - 32*B )*Sinh[x]^2 + (6*A - (8*I)*B)*Sinh[x]^3))/(105*(-I + Sinh[x])^4)
Time = 0.64 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, 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 |
\(\displaystyle \int \frac {A-i B \sin (i x)}{(i \sin (i x)+i)^4}dx\) |
\(\Big \downarrow \) 3229 |
\(\displaystyle \frac {(-B+i A) \cosh (x)}{7 (-\sinh (x)+i)^4}-\frac {1}{7} (4 B+3 i A) \int \frac {1}{(i-\sinh (x))^3}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(-B+i A) \cosh (x)}{7 (-\sinh (x)+i)^4}-\frac {1}{7} (4 B+3 i A) \int \frac {1}{(i \sin (i x)+i)^3}dx\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {(-B+i A) \cosh (x)}{7 (-\sinh (x)+i)^4}-\frac {1}{7} (4 B+3 i A) \left (\frac {i \cosh (x)}{5 (-\sinh (x)+i)^3}-\frac {2}{5} i \int \frac {1}{(i-\sinh (x))^2}dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(-B+i A) \cosh (x)}{7 (-\sinh (x)+i)^4}-\frac {1}{7} (4 B+3 i A) \left (\frac {i \cosh (x)}{5 (-\sinh (x)+i)^3}-\frac {2}{5} i \int \frac {1}{(i \sin (i x)+i)^2}dx\right )\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {(-B+i A) \cosh (x)}{7 (-\sinh (x)+i)^4}-\frac {1}{7} (4 B+3 i A) \left (\frac {i \cosh (x)}{5 (-\sinh (x)+i)^3}-\frac {2}{5} i \left (\frac {i \cosh (x)}{3 (-\sinh (x)+i)^2}-\frac {1}{3} i \int \frac {1}{i-\sinh (x)}dx\right )\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(-B+i A) \cosh (x)}{7 (-\sinh (x)+i)^4}-\frac {1}{7} (4 B+3 i A) \left (\frac {i \cosh (x)}{5 (-\sinh (x)+i)^3}-\frac {2}{5} i \left (\frac {i \cosh (x)}{3 (-\sinh (x)+i)^2}-\frac {1}{3} i \int \frac {1}{i \sin (i x)+i}dx\right )\right )\) |
\(\Big \downarrow \) 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*(((3*I)*A + 4*B)*(((-2*I)/5)*(((I/3)*Cosh[x])/(I - Sinh[x])^2 + Cosh[ x]/(3*(I - Sinh[x]))) + ((I/5)*Cosh[x])/(I - Sinh[x])^3)) + ((I*A - B)*Cos h[x])/(7*(I - Sinh[x])^4)
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 = 0.59 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65
method | result | size |
risch | \(-\frac {4 \left (4 B -84 B \,{\mathrm e}^{2 x}-21 A \,{\mathrm e}^{x}-70 i B \,{\mathrm e}^{3 x}-63 i A \,{\mathrm e}^{2 x}+28 i B \,{\mathrm e}^{x}+70 B \,{\mathrm e}^{4 x}+105 A \,{\mathrm e}^{3 x}+3 i A \right )}{105 \left ({\mathrm e}^{x}-i\right )^{7}}\) | \(66\) |
default | \(-\frac {-6 i A +2 B}{\left (\tanh \left (\frac {x}{2}\right )-i\right )^{2}}+\frac {2 A}{\tanh \left (\frac {x}{2}\right )-i}-\frac {2 \left (8 i B +8 A \right )}{7 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{7}}-\frac {2 \left (10 i B +18 A \right )}{3 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{3}}-\frac {32 i A -24 B}{2 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{4}}-\frac {-24 i A +24 B}{3 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{6}}-\frac {2 \left (-32 i B -36 A \right )}{5 \left (\tanh \left (\frac {x}{2}\right )-i\right )^{5}}\) | \(128\) |
parallelrisch | \(\frac {-210 A \tanh \left (\frac {x}{2}\right )^{6}+\left (630 i A +210 B \right ) \tanh \left (\frac {x}{2}\right )^{5}+\left (-350 i B +1260 A \right ) \tanh \left (\frac {x}{2}\right )^{4}+\left (-1260 i A -560 B \right ) \tanh \left (\frac {x}{2}\right )^{3}+\left (336 i B -882 A \right ) \tanh \left (\frac {x}{2}\right )^{2}+\left (294 i A +182 B \right ) \tanh \left (\frac {x}{2}\right )-26 i B +72 A}{735 i \tanh \left (\frac {x}{2}\right )^{6}-105 \tanh \left (\frac {x}{2}\right )^{7}-3675 i \tanh \left (\frac {x}{2}\right )^{4}+2205 \tanh \left (\frac {x}{2}\right )^{5}+2205 i \tanh \left (\frac {x}{2}\right )^{2}-3675 \tanh \left (\frac {x}{2}\right )^{3}-105 i+735 \tanh \left (\frac {x}{2}\right )}\) | \(154\) |
Input:
int((A+B*sinh(x))/(I-sinh(x))^4,x,method=_RETURNVERBOSE)
Output:
-4/105*(4*B-84*B*exp(x)^2-21*A*exp(x)-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+3*I*A)/(exp(x)-I)^7
Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.94 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^4} \, dx=-\frac {4 \, {\left (70 \, B e^{\left (4 \, x\right )} + 35 \, {\left (3 \, A - 2 i \, B\right )} e^{\left (3 \, x\right )} + 21 \, {\left (-3 i \, A - 4 \, B\right )} e^{\left (2 \, x\right )} - 7 \, {\left (3 \, A - 4 i \, B\right )} e^{x} + 3 i \, A + 4 \, B\right )}}{105 \, {\left (e^{\left (7 \, x\right )} - 7 i \, e^{\left (6 \, x\right )} - 21 \, e^{\left (5 \, x\right )} + 35 i \, e^{\left (4 \, x\right )} + 35 \, e^{\left (3 \, x\right )} - 21 i \, e^{\left (2 \, x\right )} - 7 \, e^{x} + i\right )}} \] 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 = 0.44 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.08 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^4} \, dx=\frac {- 12 i A - 280 B e^{4 x} - 16 B + \left (- 420 A + 280 i B\right ) e^{3 x} + \left (84 A - 112 i B\right ) e^{x} + \left (252 i A + 336 B\right ) e^{2 x}}{105 e^{7 x} - 735 i e^{6 x} - 2205 e^{5 x} + 3675 i e^{4 x} + 3675 e^{3 x} - 2205 i e^{2 x} - 735 e^{x} + 105 i} \] 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 - 1 12*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 = 0.06 (sec) , antiderivative size = 469, normalized size of antiderivative = 4.64 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^4} \, dx =\text {Too large to display} \] 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) - 3 5*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 = 0.59 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^4} \, dx=-\frac {4 \, {\left (70 \, B e^{\left (4 \, x\right )} + 105 \, A e^{\left (3 \, x\right )} - 70 i \, B e^{\left (3 \, x\right )} - 63 i \, A e^{\left (2 \, x\right )} - 84 \, B e^{\left (2 \, x\right )} - 21 \, A e^{x} + 28 i \, B e^{x} + 3 i \, A + 4 \, B\right )}}{105 \, {\left (e^{x} - i\right )}^{7}} \] 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 = 1.96 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.67 \[ \int \frac {A+B \sinh (x)}{(i-\sinh (x))^4} \, dx=\frac {\frac {12\,A\,{\mathrm {e}}^{2\,x}}{5}+\frac {B\,16{}\mathrm {i}}{105}-\frac {4\,A}{35}+A\,{\mathrm {e}}^{3\,x}\,4{}\mathrm {i}-{\mathrm {e}}^x\,\left (\frac {16\,B}{15}+\frac {A\,4{}\mathrm {i}}{5}\right )-\frac {B\,{\mathrm {e}}^{2\,x}\,16{}\mathrm {i}}{5}+\frac {8\,B\,{\mathrm {e}}^{3\,x}}{3}+\frac {B\,{\mathrm {e}}^{4\,x}\,8{}\mathrm {i}}{3}}{{\left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )}^7} \] Input:
int((A + B*sinh(x))/(sinh(x) - 1i)^4,x)
Output:
((B*16i)/105 - (4*A)/35 + (12*A*exp(2*x))/5 + A*exp(3*x)*4i - exp(x)*((A*4 i)/5 + (16*B)/15) - (B*exp(2*x)*16i)/5 + (8*B*exp(3*x))/3 + (B*exp(4*x)*8i )/3)/(exp(x)*1i + 1)^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