Integrand size = 24, antiderivative size = 97 \[ \int \frac {\text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {3 \arctan (\sinh (c+d x))}{8 a d}-\frac {i}{8 d (a-i a \sinh (c+d x))}+\frac {i}{4 d (a+i a \sinh (c+d x))}+\frac {i a^3}{8 d \left (a^2+i a^2 \sinh (c+d x)\right )^2} \] Output:
3/8*arctan(sinh(d*x+c))/a/d-1/8*I/d/(a-I*a*sinh(d*x+c))+1/4*I/d/(a+I*a*sin h(d*x+c))+1/8*I*a^3/d/(a^2+I*a^2*sinh(d*x+c))^2
Time = 0.07 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.04 \[ \int \frac {\text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {\text {sech}^2(c+d x) \left (2-3 i \arctan (\sinh (c+d x))+3 (-i+\arctan (\sinh (c+d x))) \sinh (c+d x)+(3-3 i \arctan (\sinh (c+d x))) \sinh ^2(c+d x)+3 \arctan (\sinh (c+d x)) \sinh ^3(c+d x)\right )}{8 a d (-i+\sinh (c+d x))} \] Input:
Integrate[Sech[c + d*x]^3/(a + I*a*Sinh[c + d*x]),x]
Output:
(Sech[c + d*x]^2*(2 - (3*I)*ArcTan[Sinh[c + d*x]] + 3*(-I + ArcTan[Sinh[c + d*x]])*Sinh[c + d*x] + (3 - (3*I)*ArcTan[Sinh[c + d*x]])*Sinh[c + d*x]^2 + 3*ArcTan[Sinh[c + d*x]]*Sinh[c + d*x]^3))/(8*a*d*(-I + Sinh[c + d*x]))
Time = 0.30 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3146, 54, 2009}
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.
\(\displaystyle \int \frac {\text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (i c+i d x)^3 (a+a \sin (i c+i d x))}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle -\frac {i a^3 \int \frac {1}{(a-i a \sinh (c+d x))^2 (i \sinh (c+d x) a+a)^3}d(i a \sinh (c+d x))}{d}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle -\frac {i a^3 \int \left (\frac {1}{8 a^3 (a-i a \sinh (c+d x))^2}+\frac {1}{4 a^3 (i \sinh (c+d x) a+a)^2}+\frac {1}{4 a^2 (i \sinh (c+d x) a+a)^3}+\frac {3}{8 a^3 \left (\sinh ^2(c+d x) a^2+a^2\right )}\right )d(i a \sinh (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i a^3 \left (\frac {3 i \arctan (\sinh (c+d x))}{8 a^4}+\frac {1}{8 a^3 (a-i a \sinh (c+d x))}-\frac {1}{4 a^3 (a+i a \sinh (c+d x))}-\frac {1}{8 a^2 (a+i a \sinh (c+d x))^2}\right )}{d}\) |
Input:
Int[Sech[c + d*x]^3/(a + I*a*Sinh[c + d*x]),x]
Output:
((-I)*a^3*((((3*I)/8)*ArcTan[Sinh[c + d*x]])/a^4 + 1/(8*a^3*(a - I*a*Sinh[ c + d*x])) - 1/(8*a^2*(a + I*a*Sinh[c + d*x])^2) - 1/(4*a^3*(a + I*a*Sinh[ c + d*x]))))/d
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 39.33 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.29
method | result | size |
risch | \(\frac {6 i {\mathrm e}^{2 d x +2 c}-6 i {\mathrm e}^{4 d x +4 c}+3 \,{\mathrm e}^{5 d x +5 c}+2 \,{\mathrm e}^{3 d x +3 c}+3 \,{\mathrm e}^{d x +c}}{4 \left ({\mathrm e}^{d x +c}+i\right )^{2} \left ({\mathrm e}^{d x +c}-i\right )^{4} d a}-\frac {3 i \ln \left ({\mathrm e}^{d x +c}-i\right )}{8 d a}+\frac {3 i \ln \left ({\mathrm e}^{d x +c}+i\right )}{8 d a}\) | \(125\) |
derivativedivides | \(\frac {\frac {i}{2 \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {3 i \ln \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {3 i}{2 \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {1}{\left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {1}{-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {i}{4 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{2}}+\frac {3 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )}{8}-\frac {1}{4 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )}}{d a}\) | \(141\) |
default | \(\frac {\frac {i}{2 \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}-\frac {3 i \ln \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {3 i}{2 \left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {1}{\left (-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {1}{-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {i}{4 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{2}}+\frac {3 i \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )}{8}-\frac {1}{4 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )}}{d a}\) | \(141\) |
Input:
int(sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/4*(6*I*exp(2*d*x+2*c)-6*I*exp(4*d*x+4*c)+3*exp(5*d*x+5*c)+2*exp(3*d*x+3* c)+3*exp(d*x+c))/(exp(d*x+c)+I)^2/(exp(d*x+c)-I)^4/d/a-3/8*I/d/a*ln(exp(d* x+c)-I)+3/8*I/d/a*ln(exp(d*x+c)+I)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (77) = 154\).
Time = 0.11 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.96 \[ \int \frac {\text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {3 \, {\left (-i \, e^{\left (6 \, d x + 6 \, c\right )} - 2 \, e^{\left (5 \, d x + 5 \, c\right )} - i \, e^{\left (4 \, d x + 4 \, c\right )} - 4 \, e^{\left (3 \, d x + 3 \, c\right )} + i \, e^{\left (2 \, d x + 2 \, c\right )} - 2 \, e^{\left (d x + c\right )} + i\right )} \log \left (e^{\left (d x + c\right )} + i\right ) + 3 \, {\left (i \, e^{\left (6 \, d x + 6 \, c\right )} + 2 \, e^{\left (5 \, d x + 5 \, c\right )} + i \, e^{\left (4 \, d x + 4 \, c\right )} + 4 \, e^{\left (3 \, d x + 3 \, c\right )} - i \, e^{\left (2 \, d x + 2 \, c\right )} + 2 \, e^{\left (d x + c\right )} - i\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - 6 \, e^{\left (5 \, d x + 5 \, c\right )} + 12 i \, e^{\left (4 \, d x + 4 \, c\right )} - 4 \, e^{\left (3 \, d x + 3 \, c\right )} - 12 i \, e^{\left (2 \, d x + 2 \, c\right )} - 6 \, e^{\left (d x + c\right )}}{8 \, {\left (a d e^{\left (6 \, d x + 6 \, c\right )} - 2 i \, a d e^{\left (5 \, d x + 5 \, c\right )} + a d e^{\left (4 \, d x + 4 \, c\right )} - 4 i \, a d e^{\left (3 \, d x + 3 \, c\right )} - a d e^{\left (2 \, d x + 2 \, c\right )} - 2 i \, a d e^{\left (d x + c\right )} - a d\right )}} \] Input:
integrate(sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")
Output:
-1/8*(3*(-I*e^(6*d*x + 6*c) - 2*e^(5*d*x + 5*c) - I*e^(4*d*x + 4*c) - 4*e^ (3*d*x + 3*c) + I*e^(2*d*x + 2*c) - 2*e^(d*x + c) + I)*log(e^(d*x + c) + I ) + 3*(I*e^(6*d*x + 6*c) + 2*e^(5*d*x + 5*c) + I*e^(4*d*x + 4*c) + 4*e^(3* d*x + 3*c) - I*e^(2*d*x + 2*c) + 2*e^(d*x + c) - I)*log(e^(d*x + c) - I) - 6*e^(5*d*x + 5*c) + 12*I*e^(4*d*x + 4*c) - 4*e^(3*d*x + 3*c) - 12*I*e^(2* d*x + 2*c) - 6*e^(d*x + c))/(a*d*e^(6*d*x + 6*c) - 2*I*a*d*e^(5*d*x + 5*c) + a*d*e^(4*d*x + 4*c) - 4*I*a*d*e^(3*d*x + 3*c) - a*d*e^(2*d*x + 2*c) - 2 *I*a*d*e^(d*x + c) - a*d)
\[ \int \frac {\text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=- \frac {i \int \frac {\operatorname {sech}^{3}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx}{a} \] Input:
integrate(sech(d*x+c)**3/(a+I*a*sinh(d*x+c)),x)
Output:
-I*Integral(sech(c + d*x)**3/(sinh(c + d*x) - I), x)/a
Exception generated. \[ \int \frac {\text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (77) = 154\).
Time = 0.12 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.78 \[ \int \frac {\text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {-\frac {6 i \, \log \left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} + 2 i\right )}{a} + \frac {6 i \, \log \left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} - 2 i\right )}{a} - \frac {2 \, {\left (3 \, e^{\left (d x + c\right )} - 3 \, e^{\left (-d x - c\right )} + 10 i\right )}}{a {\left (i \, e^{\left (d x + c\right )} - i \, e^{\left (-d x - c\right )} - 2\right )}} + \frac {-9 i \, {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} - 52 \, e^{\left (d x + c\right )} + 52 \, e^{\left (-d x - c\right )} + 84 i}{a {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} - 2 i\right )}^{2}}}{32 \, d} \] Input:
integrate(sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="giac")
Output:
-1/32*(-6*I*log(e^(d*x + c) - e^(-d*x - c) + 2*I)/a + 6*I*log(e^(d*x + c) - e^(-d*x - c) - 2*I)/a - 2*(3*e^(d*x + c) - 3*e^(-d*x - c) + 10*I)/(a*(I* e^(d*x + c) - I*e^(-d*x - c) - 2)) + (-9*I*(e^(d*x + c) - e^(-d*x - c))^2 - 52*e^(d*x + c) + 52*e^(-d*x - c) + 84*I)/(a*(e^(d*x + c) - e^(-d*x - c) - 2*I)^2))/d
Time = 1.92 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.41 \[ \int \frac {\text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {a^2\,d^2}}{a\,d}\right )}{4\,\sqrt {a^2\,d^2}}+\frac {1}{2\,a\,d\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )}+\frac {1}{4\,a\,d\,\left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )}-\frac {1{}\mathrm {i}}{4\,a\,d\,{\left ({\mathrm {e}}^{c+d\,x}+1{}\mathrm {i}\right )}^2}-\frac {1{}\mathrm {i}}{a\,d\,{\left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}^3}+\frac {1{}\mathrm {i}}{2\,a\,d\,{\left (1+{\mathrm {e}}^{c+d\,x}\,1{}\mathrm {i}\right )}^4} \] Input:
int(1/(cosh(c + d*x)^3*(a + a*sinh(c + d*x)*1i)),x)
Output:
(3*atan((exp(d*x)*exp(c)*(a^2*d^2)^(1/2))/(a*d)))/(4*(a^2*d^2)^(1/2)) + 1/ (2*a*d*(exp(c + d*x) - 1i)) + 1/(4*a*d*(exp(c + d*x) + 1i)) - 1i/(4*a*d*(e xp(c + d*x) + 1i)^2) - 1i/(a*d*(exp(c + d*x)*1i + 1)^3) + 1i/(2*a*d*(exp(c + d*x)*1i + 1)^4)
\[ \int \frac {\text {sech}^3(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {\int \frac {\mathrm {sech}\left (d x +c \right )^{3}}{\sinh \left (d x +c \right ) i +1}d x}{a} \] Input:
int(sech(d*x+c)^3/(a+I*a*sinh(d*x+c)),x)
Output:
int(sech(c + d*x)**3/(sinh(c + d*x)*i + 1),x)/a