Integrand size = 24, antiderivative size = 163 \[ \int F^{c (a+b x)} \sinh ^2(d+e x) \tanh (d+e x) \, dx=-\frac {F^{c (a+b x)}}{b c \log (F)}-\frac {7 e^{-2 d-2 e x} F^{c (a+b x)}}{4 (2 e-b c \log (F))}+\frac {2 e^{-2 d-2 e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (-2+\frac {b c \log (F)}{e}\right ),\frac {b c \log (F)}{2 e},-e^{2 d+2 e x}\right )}{2 e-b c \log (F)}+\frac {e^{2 d+2 e x} F^{c (a+b x)}}{4 (2 e+b c \log (F))} \] Output:
-F^(c*(b*x+a))/b/c/ln(F)-7*exp(-2*e*x-2*d)*F^(c*(b*x+a))/(8*e-4*b*c*ln(F)) +2*exp(-2*e*x-2*d)*F^(c*(b*x+a))*hypergeom([1, -1+1/2*b*c*ln(F)/e],[1/2*b* c*ln(F)/e],-exp(2*e*x+2*d))/(2*e-b*c*ln(F))+exp(2*e*x+2*d)*F^(c*(b*x+a))/( 8*e+4*b*c*ln(F))
Time = 0.53 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.87 \[ \int F^{c (a+b x)} \sinh ^2(d+e x) \tanh (d+e x) \, dx=\frac {F^{c (a+b x)} \left (8 e^2-2 b c e \cosh (2 (d+e x)) \log (F)-2 b^2 c^2 \log ^2(F)+4 \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},1+\frac {b c \log (F)}{2 e},-e^{2 (d+e x)}\right ) \left (-4 e^2+b^2 c^2 \log ^2(F)\right )+b^2 c^2 \log ^2(F) \sinh (2 (d+e x))\right )}{-8 b c e^2 \log (F)+2 b^3 c^3 \log ^3(F)} \] Input:
Integrate[F^(c*(a + b*x))*Sinh[d + e*x]^2*Tanh[d + e*x],x]
Output:
(F^(c*(a + b*x))*(8*e^2 - 2*b*c*e*Cosh[2*(d + e*x)]*Log[F] - 2*b^2*c^2*Log [F]^2 + 4*Hypergeometric2F1[1, (b*c*Log[F])/(2*e), 1 + (b*c*Log[F])/(2*e), -E^(2*(d + e*x))]*(-4*e^2 + b^2*c^2*Log[F]^2) + b^2*c^2*Log[F]^2*Sinh[2*( d + e*x)]))/(-8*b*c*e^2*Log[F] + 2*b^3*c^3*Log[F]^3)
Time = 0.55 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6037, 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.
\(\displaystyle \int \sinh ^2(d+e x) \tanh (d+e x) F^{c (a+b x)} \, dx\) |
\(\Big \downarrow \) 6037 |
\(\displaystyle \int \left (\frac {7}{4} e^{-2 d-2 e x} F^{a c+b c x}-e^{2 (d+e x)-2 d-2 e x} F^{a c+b c x}+\frac {1}{4} e^{4 (d+e x)-2 d-2 e x} F^{a c+b c x}-\frac {2 e^{-2 d-2 e x} F^{a c+b c x}}{e^{2 (d+e x)}+1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 e^{-2 d-2 e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (\frac {b c \log (F)}{e}-2\right ),\frac {b c \log (F)}{2 e},-e^{2 (d+e x)}\right )}{2 e-b c \log (F)}-\frac {7 F^{a c} e^{-x (2 e-b c \log (F))-2 d}}{4 (2 e-b c \log (F))}+\frac {F^{a c} e^{x (b c \log (F)+2 e)+2 d}}{4 (b c \log (F)+2 e)}-\frac {F^{a c+b c x}}{b c \log (F)}\) |
Input:
Int[F^(c*(a + b*x))*Sinh[d + e*x]^2*Tanh[d + e*x],x]
Output:
-(F^(a*c + b*c*x)/(b*c*Log[F])) - (7*E^(-2*d - x*(2*e - b*c*Log[F]))*F^(a* c))/(4*(2*e - b*c*Log[F])) + (2*E^(-2*d - 2*e*x)*F^(a*c + b*c*x)*Hypergeom etric2F1[1, (-2 + (b*c*Log[F])/e)/2, (b*c*Log[F])/(2*e), -E^(2*(d + e*x))] )/(2*e - b*c*Log[F]) + (E^(2*d + x*(2*e + b*c*Log[F]))*F^(a*c))/(4*(2*e + b*c*Log[F]))
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(G_)[(d_.) + (e_.)*(x_)]^(m_.)*(H_)[( d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^(c*(a + b*x)), G[d + e*x]^m*H[d + e*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] && IGtQ[ m, 0] && IGtQ[n, 0] && HyperbolicQ[G] && HyperbolicQ[H]
\[\int F^{c \left (b x +a \right )} \sinh \left (e x +d \right )^{2} \tanh \left (e x +d \right )d x\]
Input:
int(F^(c*(b*x+a))*sinh(e*x+d)^2*tanh(e*x+d),x)
Output:
int(F^(c*(b*x+a))*sinh(e*x+d)^2*tanh(e*x+d),x)
\[ \int F^{c (a+b x)} \sinh ^2(d+e x) \tanh (d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \sinh \left (e x + d\right )^{2} \tanh \left (e x + d\right ) \,d x } \] Input:
integrate(F^(c*(b*x+a))*sinh(e*x+d)^2*tanh(e*x+d),x, algorithm="fricas")
Output:
integral(F^(b*c*x + a*c)*sinh(e*x + d)^2*tanh(e*x + d), x)
\[ \int F^{c (a+b x)} \sinh ^2(d+e x) \tanh (d+e x) \, dx=\int F^{c \left (a + b x\right )} \sinh ^{2}{\left (d + e x \right )} \tanh {\left (d + e x \right )}\, dx \] Input:
integrate(F**(c*(b*x+a))*sinh(e*x+d)**2*tanh(e*x+d),x)
Output:
Integral(F**(c*(a + b*x))*sinh(d + e*x)**2*tanh(d + e*x), x)
\[ \int F^{c (a+b x)} \sinh ^2(d+e x) \tanh (d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \sinh \left (e x + d\right )^{2} \tanh \left (e x + d\right ) \,d x } \] Input:
integrate(F^(c*(b*x+a))*sinh(e*x+d)^2*tanh(e*x+d),x, algorithm="maxima")
Output:
4*F^(a*c)*e*integrate(F^(b*c*x)/((b*c*e^(6*d)*log(F) - 4*e*e^(6*d))*e^(6*e *x) + 2*(b*c*e^(4*d)*log(F) - 4*e*e^(4*d))*e^(4*e*x) + (b*c*e^(2*d)*log(F) - 4*e*e^(2*d))*e^(2*e*x)), x) - 1/4*(F^(a*c)*b^3*c^3*log(F)^3 + 14*F^(a*c )*b^2*c^2*e*log(F)^2 + 24*F^(a*c)*b*c*e^2*log(F) - (F^(a*c)*b^3*c^3*e^(6*d )*log(F)^3 - 6*F^(a*c)*b^2*c^2*e*e^(6*d)*log(F)^2 + 8*F^(a*c)*b*c*e^2*e^(6 *d)*log(F))*e^(6*e*x) + (3*F^(a*c)*b^3*c^3*e^(4*d)*log(F)^3 - 10*F^(a*c)*b ^2*c^2*e*e^(4*d)*log(F)^2 - 24*F^(a*c)*b*c*e^2*e^(4*d)*log(F) + 64*F^(a*c) *e^3*e^(4*d))*e^(4*e*x) - (3*F^(a*c)*b^3*c^3*e^(2*d)*log(F)^3 + 2*F^(a*c)* b^2*c^2*e*e^(2*d)*log(F)^2 - 40*F^(a*c)*b*c*e^2*e^(2*d)*log(F) - 64*F^(a*c )*e^3*e^(2*d))*e^(2*e*x))*F^(b*c*x)/((b^4*c^4*e^(4*d)*log(F)^4 - 4*b^3*c^3 *e*e^(4*d)*log(F)^3 - 4*b^2*c^2*e^2*e^(4*d)*log(F)^2 + 16*b*c*e^3*e^(4*d)* log(F))*e^(4*e*x) + (b^4*c^4*e^(2*d)*log(F)^4 - 4*b^3*c^3*e*e^(2*d)*log(F) ^3 - 4*b^2*c^2*e^2*e^(2*d)*log(F)^2 + 16*b*c*e^3*e^(2*d)*log(F))*e^(2*e*x) )
\[ \int F^{c (a+b x)} \sinh ^2(d+e x) \tanh (d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \sinh \left (e x + d\right )^{2} \tanh \left (e x + d\right ) \,d x } \] Input:
integrate(F^(c*(b*x+a))*sinh(e*x+d)^2*tanh(e*x+d),x, algorithm="giac")
Output:
integrate(F^((b*x + a)*c)*sinh(e*x + d)^2*tanh(e*x + d), x)
Timed out. \[ \int F^{c (a+b x)} \sinh ^2(d+e x) \tanh (d+e x) \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\mathrm {sinh}\left (d+e\,x\right )}^2\,\mathrm {tanh}\left (d+e\,x\right ) \,d x \] Input:
int(F^(c*(a + b*x))*sinh(d + e*x)^2*tanh(d + e*x),x)
Output:
int(F^(c*(a + b*x))*sinh(d + e*x)^2*tanh(d + e*x), x)
\[ \int F^{c (a+b x)} \sinh ^2(d+e x) \tanh (d+e x) \, dx=f^{a c} \left (\int f^{b c x} \sinh \left (e x +d \right )^{2} \tanh \left (e x +d \right )d x \right ) \] Input:
int(F^(c*(b*x+a))*sinh(e*x+d)^2*tanh(e*x+d),x)
Output:
f**(a*c)*int(f**(b*c*x)*sinh(d + e*x)**2*tanh(d + e*x),x)