Integrand size = 25, antiderivative size = 196 \[ \int \frac {F^{c (a+b x)}}{(f+i f \sinh (d+e x))^2} \, dx=\frac {2 e^{\frac {1}{2} (2 d+i \pi +2 e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1+\frac {b c \log (F)}{e},2+\frac {b c \log (F)}{e},-e^{\frac {1}{2} (2 d+i \pi +2 e x)}\right ) (e-b c \log (F))}{3 e^2 f^2}+\frac {b c F^{c (a+b x)} \log (F) \text {sech}^2\left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right )}{6 e^2 f^2}+\frac {F^{c (a+b x)} \text {sech}^2\left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right ) \tanh \left (\frac {d}{2}+\frac {i \pi }{4}+\frac {e x}{2}\right )}{6 e f^2} \]
2/3*exp(d+1/2*I*Pi+e*x)*F^(c*(b*x+a))*hypergeom([2, 1+b*c*ln(F)/e],[2+b*c* ln(F)/e],-exp(d+1/2*I*Pi+e*x))*(e-b*c*ln(F))/e^2/f^2+1/6*b*c*F^(c*(b*x+a)) *ln(F)*sech(1/2*d+1/4*I*Pi+1/2*e*x)^2/e^2/f^2+1/6*F^(c*(b*x+a))*sech(1/2*d +1/4*I*Pi+1/2*e*x)^2*tanh(1/2*d+1/4*I*Pi+1/2*e*x)/e/f^2
Time = 1.30 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.28 \[ \int \frac {F^{c (a+b x)}}{(f+i f \sinh (d+e x))^2} \, dx=\frac {F^{c (a+b x)} \left (\cosh \left (\frac {1}{2} (d+e x)\right )+i \sinh \left (\frac {1}{2} (d+e x)\right )\right ) \left (e (i e+b c \log (F)) \left (\cosh \left (\frac {1}{2} (d+e x)\right )+i \sinh \left (\frac {1}{2} (d+e x)\right )\right )-(1-i) \left (-1+(1+i) \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{e},1+\frac {b c \log (F)}{e},-i e^{d+e x}\right )\right ) \left (-e^2+b^2 c^2 \log ^2(F)\right ) \left (\cosh \left (\frac {1}{2} (d+e x)\right )+i \sinh \left (\frac {1}{2} (d+e x)\right )\right )^3+2 e^2 \sinh \left (\frac {1}{2} (d+e x)\right )+2 \left (e^2-b^2 c^2 \log ^2(F)\right ) \left (\cosh \left (\frac {1}{2} (d+e x)\right )+i \sinh \left (\frac {1}{2} (d+e x)\right )\right )^2 \sinh \left (\frac {1}{2} (d+e x)\right )\right )}{3 e^3 (f+i f \sinh (d+e x))^2} \]
(F^(c*(a + b*x))*(Cosh[(d + e*x)/2] + I*Sinh[(d + e*x)/2])*(e*(I*e + b*c*L og[F])*(Cosh[(d + e*x)/2] + I*Sinh[(d + e*x)/2]) - (1 - I)*(-1 + (1 + I)*H ypergeometric2F1[1, (b*c*Log[F])/e, 1 + (b*c*Log[F])/e, (-I)*E^(d + e*x)]) *(-e^2 + b^2*c^2*Log[F]^2)*(Cosh[(d + e*x)/2] + I*Sinh[(d + e*x)/2])^3 + 2 *e^2*Sinh[(d + e*x)/2] + 2*(e^2 - b^2*c^2*Log[F]^2)*(Cosh[(d + e*x)/2] + I *Sinh[(d + e*x)/2])^2*Sinh[(d + e*x)/2]))/(3*e^3*(f + I*f*Sinh[d + e*x])^2 )
Time = 0.46 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6019, 6013, 6015}
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.
| \(\displaystyle \int \frac {F^{c (a+b x)}}{(f+i f \sinh (d+e x))^2} \, dx\) |
| \(\Big \downarrow \) 6019 |
| \(\displaystyle \frac {\int F^{c (a+b x)} \text {sech}^4\left (\frac {d}{2}+\frac {e x}{2}+\frac {i \pi }{4}\right )dx}{4 f^2}\) |
| \(\Big \downarrow \) 6013 |
| \(\displaystyle \frac {\frac {2}{3} \left (1-\frac {b^2 c^2 \log ^2(F)}{e^2}\right ) \int F^{c (a+b x)} \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}+\frac {i \pi }{4}\right )dx+\frac {2 b c \log (F) \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}+\frac {i \pi }{4}\right ) F^{c (a+b x)}}{3 e^2}+\frac {2 \tanh \left (\frac {d}{2}+\frac {e x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}+\frac {i \pi }{4}\right ) F^{c (a+b x)}}{3 e}}{4 f^2}\) |
| \(\Big \downarrow \) 6015 |
| \(\displaystyle \frac {\frac {8 e^{\frac {1}{2} (2 d+2 e x+i \pi )} F^{c (a+b x)} \left (1-\frac {b^2 c^2 \log ^2(F)}{e^2}\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {b c \log (F)}{e}+1,\frac {b c \log (F)}{e}+2,-e^{\frac {1}{2} (2 d+2 e x+i \pi )}\right )}{3 (b c \log (F)+e)}+\frac {2 b c \log (F) \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}+\frac {i \pi }{4}\right ) F^{c (a+b x)}}{3 e^2}+\frac {2 \tanh \left (\frac {d}{2}+\frac {e x}{2}+\frac {i \pi }{4}\right ) \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}+\frac {i \pi }{4}\right ) F^{c (a+b x)}}{3 e}}{4 f^2}\) |
((8*E^((2*d + I*Pi + 2*e*x)/2)*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 + (b *c*Log[F])/e, 2 + (b*c*Log[F])/e, -E^((2*d + I*Pi + 2*e*x)/2)]*(1 - (b^2*c ^2*Log[F]^2)/e^2))/(3*(e + b*c*Log[F])) + (2*b*c*F^(c*(a + b*x))*Log[F]*Se ch[d/2 + (I/4)*Pi + (e*x)/2]^2)/(3*e^2) + (2*F^(c*(a + b*x))*Sech[d/2 + (I /4)*Pi + (e*x)/2]^2*Tanh[d/2 + (I/4)*Pi + (e*x)/2])/(3*e))/(4*f^2)
3.9.96.3.1 Defintions of rubi rules used
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_), x_Symb ol] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(Sech[d + e*x]^(n - 2)/(e^2*(n - 1)* (n - 2))), x] + (Simp[F^(c*(a + b*x))*Sech[d + e*x]^(n - 1)*(Sinh[d + e*x]/ (e*(n - 1))), x] + Simp[(e^2*(n - 2)^2 - b^2*c^2*Log[F]^2)/(e^2*(n - 1)*(n - 2)) Int[F^(c*(a + b*x))*Sech[d + e*x]^(n - 2), x], x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*(n - 2)^2 - b^2*c^2*Log[F]^2, 0] && GtQ[n, 1] && NeQ[n, 2]
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_.), x_Sym bol] :> Simp[2^n*E^(n*(d + e*x))*(F^(c*(a + b*x))/(e*n + b*c*Log[F]))*Hyper geometric2F1[n, n/2 + b*c*(Log[F]/(2*e)), 1 + n/2 + b*c*(Log[F]/(2*e)), -E^ (2*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_) + (g_.)*Sinh[(d_.) + (e_.)*(x_) ])^(n_.), x_Symbol] :> Simp[2^n*f^n Int[F^(c*(a + b*x))*Cosh[d/2 - f*(Pi/ (4*g)) + e*(x/2)]^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] && Eq Q[f^2 + g^2, 0] && ILtQ[n, 0]
\[\int \frac {F^{c \left (b x +a \right )}}{\left (f +i f \sinh \left (e x +d \right )\right )^{2}}d x\]
\[ \int \frac {F^{c (a+b x)}}{(f+i f \sinh (d+e x))^2} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (i \, f \sinh \left (e x + d\right ) + f\right )}^{2}} \,d x } \]
1/3*(2*(3*e^2*e^(e*x + d) - (I*b^2*c^2*e^(2*e*x + 2*d) + 2*b^2*c^2*e^(e*x + d) - I*b^2*c^2)*log(F)^2 - I*e^2 - (I*b*c*e*e^(2*e*x + 2*d) + b*c*e*e^(e *x + d))*log(F))*F^(b*c*x + a*c) + 3*(e^3*f^2*e^(3*e*x + 3*d) - 3*I*e^3*f^ 2*e^(2*e*x + 2*d) - 3*e^3*f^2*e^(e*x + d) + I*e^3*f^2)*integral(-2/3*(-I*b ^3*c^3*log(F)^3 + I*b*c*e^2*log(F))*F^(b*c*x + a*c)/(e^3*f^2*e^(e*x + d) - I*e^3*f^2), x))/(e^3*f^2*e^(3*e*x + 3*d) - 3*I*e^3*f^2*e^(2*e*x + 2*d) - 3*e^3*f^2*e^(e*x + d) + I*e^3*f^2)
\[ \int \frac {F^{c (a+b x)}}{(f+i f \sinh (d+e x))^2} \, dx=- \frac {\int \frac {F^{a c + b c x}}{\sinh ^{2}{\left (d + e x \right )} - 2 i \sinh {\left (d + e x \right )} - 1}\, dx}{f^{2}} \]
\[ \int \frac {F^{c (a+b x)}}{(f+i f \sinh (d+e x))^2} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (i \, f \sinh \left (e x + d\right ) + f\right )}^{2}} \,d x } \]
-16*(-I*F^(a*c)*b^2*c^2*e*log(F)^2 - I*F^(a*c)*b*c*e^2*log(F))*integrate(F ^(b*c*x)/(-I*b^3*c^3*f^2*log(F)^3 + 9*I*b^2*c^2*e*f^2*log(F)^2 - 26*I*b*c* e^2*f^2*log(F) + 24*I*e^3*f^2 + (b^3*c^3*f^2*e^(5*d)*log(F)^3 - 9*b^2*c^2* e*f^2*e^(5*d)*log(F)^2 + 26*b*c*e^2*f^2*e^(5*d)*log(F) - 24*e^3*f^2*e^(5*d ))*e^(5*e*x) - 5*(I*b^3*c^3*f^2*e^(4*d)*log(F)^3 - 9*I*b^2*c^2*e*f^2*e^(4* d)*log(F)^2 + 26*I*b*c*e^2*f^2*e^(4*d)*log(F) - 24*I*e^3*f^2*e^(4*d))*e^(4 *e*x) - 10*(b^3*c^3*f^2*e^(3*d)*log(F)^3 - 9*b^2*c^2*e*f^2*e^(3*d)*log(F)^ 2 + 26*b*c*e^2*f^2*e^(3*d)*log(F) - 24*e^3*f^2*e^(3*d))*e^(3*e*x) - 10*(-I *b^3*c^3*f^2*e^(2*d)*log(F)^3 + 9*I*b^2*c^2*e*f^2*e^(2*d)*log(F)^2 - 26*I* b*c*e^2*f^2*e^(2*d)*log(F) + 24*I*e^3*f^2*e^(2*d))*e^(2*e*x) + 5*(b^3*c^3* f^2*e^d*log(F)^3 - 9*b^2*c^2*e*f^2*e^d*log(F)^2 + 26*b*c*e^2*f^2*e^d*log(F ) - 24*e^3*f^2*e^d)*e^(e*x)), x) + 4*(4*F^(a*c)*b*c*e*log(F) + 4*F^(a*c)*e ^2 - (F^(a*c)*b^2*c^2*e^(2*d)*log(F)^2 - 7*F^(a*c)*b*c*e*e^(2*d)*log(F) + 12*F^(a*c)*e^2*e^(2*d))*e^(2*e*x) + 4*(-I*F^(a*c)*b*c*e*e^d*log(F) + 4*I*F ^(a*c)*e^2*e^d)*e^(e*x))*F^(b*c*x)/(b^3*c^3*f^2*log(F)^3 - 9*b^2*c^2*e*f^2 *log(F)^2 + 26*b*c*e^2*f^2*log(F) - 24*e^3*f^2 + (b^3*c^3*f^2*e^(4*d)*log( F)^3 - 9*b^2*c^2*e*f^2*e^(4*d)*log(F)^2 + 26*b*c*e^2*f^2*e^(4*d)*log(F) - 24*e^3*f^2*e^(4*d))*e^(4*e*x) - 4*(I*b^3*c^3*f^2*e^(3*d)*log(F)^3 - 9*I*b^ 2*c^2*e*f^2*e^(3*d)*log(F)^2 + 26*I*b*c*e^2*f^2*e^(3*d)*log(F) - 24*I*e^3* f^2*e^(3*d))*e^(3*e*x) - 6*(b^3*c^3*f^2*e^(2*d)*log(F)^3 - 9*b^2*c^2*e*...
\[ \int \frac {F^{c (a+b x)}}{(f+i f \sinh (d+e x))^2} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (i \, f \sinh \left (e x + d\right ) + f\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {F^{c (a+b x)}}{(f+i f \sinh (d+e x))^2} \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (f+f\,\mathrm {sinh}\left (d+e\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]