\(\int F^{c (a+b x)} \sqrt {f \tanh (d+e x)} \sqrt {g \tanh (d+e x)} \, dx\) [143]

Optimal result

Integrand size = 34, antiderivative size = 130 \[ \int F^{c (a+b x)} \sqrt {f \tanh (d+e x)} \sqrt {g \tanh (d+e x)} \, dx=\frac {f F^{c (a+b x)} \sqrt {g \tanh (d+e x)}}{b c \log (F) \sqrt {f \tanh (d+e x)}}-\frac {2 f F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},1+\frac {b c \log (F)}{2 e},-e^{2 d+2 e x}\right ) \sqrt {g \tanh (d+e x)}}{b c \log (F) \sqrt {f \tanh (d+e x)}} \] Output:

f*F^(c*(b*x+a))*(g*tanh(e*x+d))^(1/2)/b/c/ln(F)/(f*tanh(e*x+d))^(1/2)-2*f* 
F^(c*(b*x+a))*hypergeom([1, 1/2*b*c*ln(F)/e],[1+1/2*b*c*ln(F)/e],-exp(2*e* 
x+2*d))*(g*tanh(e*x+d))^(1/2)/b/c/ln(F)/(f*tanh(e*x+d))^(1/2)
 

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.67 \[ \int F^{c (a+b x)} \sqrt {f \tanh (d+e x)} \sqrt {g \tanh (d+e x)} \, dx=-\frac {f F^{c (a+b x)} \left (-1+2 \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},1+\frac {b c \log (F)}{2 e},-e^{2 (d+e x)}\right )\right ) \sqrt {g \tanh (d+e x)}}{b c \log (F) \sqrt {f \tanh (d+e x)}} \] Input:

Integrate[F^(c*(a + b*x))*Sqrt[f*Tanh[d + e*x]]*Sqrt[g*Tanh[d + e*x]],x]
 

Output:

-((f*F^(c*(a + b*x))*(-1 + 2*Hypergeometric2F1[1, (b*c*Log[F])/(2*e), 1 + 
(b*c*Log[F])/(2*e), -E^(2*(d + e*x))])*Sqrt[g*Tanh[d + e*x]])/(b*c*Log[F]* 
Sqrt[f*Tanh[d + e*x]]))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.81, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {2031, 6007, 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.

\(\displaystyle \int F^{c (a+b x)} \sqrt {f \tanh (d+e x)} \sqrt {g \tanh (d+e x)} \, dx\)

\(\Big \downarrow \) 2031

\(\displaystyle \frac {f \sqrt {g \tanh (d+e x)} \int F^{c (a+b x)} \tanh (d+e x)dx}{\sqrt {f \tanh (d+e x)}}\)

\(\Big \downarrow \) 6007

\(\displaystyle \frac {f \sqrt {g \tanh (d+e x)} \int \left (F^{c (a+b x)}-\frac {2 F^{c (a+b x)}}{1+e^{2 (d+e x)}}\right )dx}{\sqrt {f \tanh (d+e x)}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {f \sqrt {g \tanh (d+e x)} \left (\frac {F^{c (a+b x)}}{b c \log (F)}-\frac {2 F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},\frac {b c \log (F)}{2 e}+1,-e^{2 (d+e x)}\right )}{b c \log (F)}\right )}{\sqrt {f \tanh (d+e x)}}\)

Input:

Int[F^(c*(a + b*x))*Sqrt[f*Tanh[d + e*x]]*Sqrt[g*Tanh[d + e*x]],x]
 

Output:

(f*(F^(c*(a + b*x))/(b*c*Log[F]) - (2*F^(c*(a + b*x))*Hypergeometric2F1[1, 
 (b*c*Log[F])/(2*e), 1 + (b*c*Log[F])/(2*e), -E^(2*(d + e*x))])/(b*c*Log[F 
]))*Sqrt[g*Tanh[d + e*x]])/Sqrt[f*Tanh[d + e*x]]
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2031

Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m + 1/ 
2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v])   Int[v^(m + n)*Fx, x], x] /; FreeQ[{a 
, b, m}, x] &&  !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]
 

rule 6007

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Tanh[(d_.) + (e_.)*(x_)]^(n_.), x_Sym 
bol] :> Int[ExpandIntegrand[F^(c*(a + b*x))*((-1 + E^(2*(d + e*x)))^n/(1 + 
E^(2*(d + e*x)))^n), x], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]
 

Maple [F]

\[\int F^{c \left (b x +a \right )} \sqrt {f \tanh \left (e x +d \right )}\, \sqrt {g \tanh \left (e x +d \right )}d x\]

Input:

int(F^(c*(b*x+a))*(f*tanh(e*x+d))^(1/2)*(g*tanh(e*x+d))^(1/2),x)
 

Output:

int(F^(c*(b*x+a))*(f*tanh(e*x+d))^(1/2)*(g*tanh(e*x+d))^(1/2),x)
 

Fricas [F]

\[ \int F^{c (a+b x)} \sqrt {f \tanh (d+e x)} \sqrt {g \tanh (d+e x)} \, dx=\int { \sqrt {f \tanh \left (e x + d\right )} \sqrt {g \tanh \left (e x + d\right )} F^{{\left (b x + a\right )} c} \,d x } \] Input:

integrate(F^(c*(b*x+a))*(f*tanh(e*x+d))^(1/2)*(g*tanh(e*x+d))^(1/2),x, alg 
orithm="fricas")
 

Output:

integral(sqrt(f*tanh(e*x + d))*sqrt(g*tanh(e*x + d))*F^(b*c*x + a*c), x)
 

Sympy [F]

\[ \int F^{c (a+b x)} \sqrt {f \tanh (d+e x)} \sqrt {g \tanh (d+e x)} \, dx=\int F^{c \left (a + b x\right )} \sqrt {f \tanh {\left (d + e x \right )}} \sqrt {g \tanh {\left (d + e x \right )}}\, dx \] Input:

integrate(F**(c*(b*x+a))*(f*tanh(e*x+d))**(1/2)*(g*tanh(e*x+d))**(1/2),x)
 

Output:

Integral(F**(c*(a + b*x))*sqrt(f*tanh(d + e*x))*sqrt(g*tanh(d + e*x)), x)
 

Maxima [F]

\[ \int F^{c (a+b x)} \sqrt {f \tanh (d+e x)} \sqrt {g \tanh (d+e x)} \, dx=\int { \sqrt {f \tanh \left (e x + d\right )} \sqrt {g \tanh \left (e x + d\right )} F^{{\left (b x + a\right )} c} \,d x } \] Input:

integrate(F^(c*(b*x+a))*(f*tanh(e*x+d))^(1/2)*(g*tanh(e*x+d))^(1/2),x, alg 
orithm="maxima")
 

Output:

integrate(sqrt(f*tanh(e*x + d))*sqrt(g*tanh(e*x + d))*F^((b*x + a)*c), x)
 

Giac [F]

\[ \int F^{c (a+b x)} \sqrt {f \tanh (d+e x)} \sqrt {g \tanh (d+e x)} \, dx=\int { \sqrt {f \tanh \left (e x + d\right )} \sqrt {g \tanh \left (e x + d\right )} F^{{\left (b x + a\right )} c} \,d x } \] Input:

integrate(F^(c*(b*x+a))*(f*tanh(e*x+d))^(1/2)*(g*tanh(e*x+d))^(1/2),x, alg 
orithm="giac")
 

Output:

integrate(sqrt(f*tanh(e*x + d))*sqrt(g*tanh(e*x + d))*F^((b*x + a)*c), x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \sqrt {f \tanh (d+e x)} \sqrt {g \tanh (d+e x)} \, dx=\int F^{c\,\left (a+b\,x\right )}\,\sqrt {f\,\mathrm {tanh}\left (d+e\,x\right )}\,\sqrt {g\,\mathrm {tanh}\left (d+e\,x\right )} \,d x \] Input:

int(F^(c*(a + b*x))*(f*tanh(d + e*x))^(1/2)*(g*tanh(d + e*x))^(1/2),x)
 

Output:

int(F^(c*(a + b*x))*(f*tanh(d + e*x))^(1/2)*(g*tanh(d + e*x))^(1/2), x)
 

Reduce [F]

\[ \int F^{c (a+b x)} \sqrt {f \tanh (d+e x)} \sqrt {g \tanh (d+e x)} \, dx=\sqrt {g}\, f^{a c +\frac {1}{2}} \left (\int f^{b c x} \tanh \left (e x +d \right )d x \right ) \] Input:

int(F^(c*(b*x+a))*(f*tanh(e*x+d))^(1/2)*(g*tanh(e*x+d))^(1/2),x)
 

Output:

sqrt(g)*f**((2*a*c + 1)/2)*int(f**(b*c*x)*tanh(d + e*x),x)