\(\int F^{c (a+b x)} \sqrt {f \tan (d+e x)} \sqrt {g \tan (d+e x)} \, dx\) [135]

Optimal result

Integrand size = 34, antiderivative size = 140 \[ \int F^{c (a+b x)} \sqrt {f \tan (d+e x)} \sqrt {g \tan (d+e x)} \, dx=-\frac {i f F^{c (a+b x)} \sqrt {g \tan (d+e x)}}{b c \log (F) \sqrt {f \tan (d+e x)}}+\frac {2 i f F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \sqrt {g \tan (d+e x)}}{b c \log (F) \sqrt {f \tan (d+e x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.68 \[ \int F^{c (a+b x)} \sqrt {f \tan (d+e x)} \sqrt {g \tan (d+e x)} \, dx=\frac {i f F^{c (a+b x)} \left (-1+2 \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{2 e},1-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )\right ) \sqrt {g \tan (d+e x)}}{b c \log (F) \sqrt {f \tan (d+e x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.82, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {2031, 4942, 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.

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

\(\Big \downarrow \) 2031

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

\(\Big \downarrow \) 4942

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(I*f*(-(F^(c*(a + b*x))/(b*c*Log[F])) + (2*F^(c*(a + b*x))*Hypergeometric2 
F1[1, ((-1/2*I)*b*c*Log[F])/e, 1 - ((I/2)*b*c*Log[F])/e, -E^((2*I)*(d + e* 
x))])/(b*c*Log[F]))*Sqrt[g*Tan[d + e*x]])/Sqrt[f*Tan[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 4942

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(F^(c*(b*x+a))*(f*tan(e*x+d))^(1/2)*(g*tan(e*x+d))^(1/2),x, algor 
ithm="fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(F^(c*(b*x+a))*(f*tan(e*x+d))^(1/2)*(g*tan(e*x+d))^(1/2),x, algor 
ithm="maxima")
 

Output:

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

Giac [F]

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

integrate(F^(c*(b*x+a))*(f*tan(e*x+d))^(1/2)*(g*tan(e*x+d))^(1/2),x, algor 
ithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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