Integrand size = 34, antiderivative size = 268 \[ \int F^{c (a+b x)} \sqrt {g \sec (d+e x)} \sqrt {f \sin (d+e x)} \, dx=\frac {\sqrt {1-e^{4 i (d+e x)}} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {i b c \log (F)}{4 e},1-\frac {i b c \log (F)}{4 e},e^{4 i (d+e x)}\right ) \sqrt {g \sec (d+e x)} \sqrt {f \sin (d+e x)}}{b c \left (1-e^{2 i (d+e x)}\right ) \log (F)}-\frac {e^{2 i (d+e x)} \sqrt {1-e^{4 i (d+e x)}} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} \left (2-\frac {i b c \log (F)}{e}\right ),\frac {1}{4} \left (6-\frac {i b c \log (F)}{e}\right ),e^{4 i (d+e x)}\right ) \sqrt {g \sec (d+e x)} \sqrt {f \sin (d+e x)}}{\left (1-e^{2 i (d+e x)}\right ) (2 i e+b c \log (F))} \] Output:
(1-exp(4*I*(e*x+d)))^(1/2)*F^(c*(b*x+a))*hypergeom([1/2, -1/4*I*b*c*ln(F)/ e],[1-1/4*I*b*c*ln(F)/e],exp(4*I*(e*x+d)))*(g*sec(e*x+d))^(1/2)*(f*sin(e*x +d))^(1/2)/b/c/(1-exp(2*I*(e*x+d)))/ln(F)-exp(2*I*(e*x+d))*(1-exp(4*I*(e*x +d)))^(1/2)*F^(c*(b*x+a))*hypergeom([1/2, 1/2-1/4*I*b*c*ln(F)/e],[3/2-1/4* I*b*c*ln(F)/e],exp(4*I*(e*x+d)))*(g*sec(e*x+d))^(1/2)*(f*sin(e*x+d))^(1/2) /(1-exp(2*I*(e*x+d)))/(2*I*e+b*c*ln(F))
\[ \int F^{c (a+b x)} \sqrt {g \sec (d+e x)} \sqrt {f \sin (d+e x)} \, dx=\int F^{c (a+b x)} \sqrt {g \sec (d+e x)} \sqrt {f \sin (d+e x)} \, dx \] Input:
Integrate[F^(c*(a + b*x))*Sqrt[g*Sec[d + e*x]]*Sqrt[f*Sin[d + e*x]],x]
Output:
Integrate[F^(c*(a + b*x))*Sqrt[g*Sec[d + e*x]]*Sqrt[f*Sin[d + e*x]], x]
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.
\(\displaystyle \int F^{c (a+b x)} \sqrt {f \sin (d+e x)} \sqrt {g \sec (d+e x)} \, dx\) |
\(\Big \downarrow \) 7271 |
\(\displaystyle \frac {\sqrt {g \sec (d+e x)} \int F^{c (a+b x)} \sqrt {\sec (d+e x)} \sqrt {f \sin (d+e x)}dx}{\sqrt {\sec (d+e x)}}\) |
\(\Big \downarrow \) 7271 |
\(\displaystyle \frac {\sqrt {f \sin (d+e x)} \sqrt {g \sec (d+e x)} \int F^{c (a+b x)} \sqrt {\sec (d+e x)} \sqrt {\sin (d+e x)}dx}{\sqrt {\sin (d+e x)} \sqrt {\sec (d+e x)}}\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {\sqrt {f \sin (d+e x)} \sqrt {g \sec (d+e x)} \int F^{a c+b x c} \sqrt {\sec (d+e x)} \sqrt {\sin (d+e x)}dx}{\sqrt {\sin (d+e x)} \sqrt {\sec (d+e x)}}\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \frac {\sqrt {f \sin (d+e x)} \sqrt {g \sec (d+e x)} \int F^{a c+b x c} \sqrt {\sec (d+e x)} \sqrt {\sin (d+e x)}dx}{\sqrt {\sin (d+e x)} \sqrt {\sec (d+e x)}}\) |
Input:
Int[F^(c*(a + b*x))*Sqrt[g*Sec[d + e*x]]*Sqrt[f*Sin[d + e*x]],x]
Output:
$Aborted
Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*v^m)^ FracPart[p]/v^(m*FracPart[p])) Int[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] && !IntegerQ[p] && !FreeQ[v, x] && !(EqQ[a, 1] && EqQ[m, 1]) && !(Eq Q[v, x] && EqQ[m, 1])
\[\int F^{c \left (b x +a \right )} \sqrt {g \sec \left (e x +d \right )}\, \sqrt {f \sin \left (e x +d \right )}d x\]
Input:
int(F^(c*(b*x+a))*(g*sec(e*x+d))^(1/2)*(f*sin(e*x+d))^(1/2),x)
Output:
int(F^(c*(b*x+a))*(g*sec(e*x+d))^(1/2)*(f*sin(e*x+d))^(1/2),x)
\[ \int F^{c (a+b x)} \sqrt {g \sec (d+e x)} \sqrt {f \sin (d+e x)} \, dx=\int { \sqrt {g \sec \left (e x + d\right )} \sqrt {f \sin \left (e x + d\right )} F^{{\left (b x + a\right )} c} \,d x } \] Input:
integrate(F^(c*(b*x+a))*(g*sec(e*x+d))^(1/2)*(f*sin(e*x+d))^(1/2),x, algor ithm="fricas")
Output:
integral(sqrt(g*sec(e*x + d))*sqrt(f*sin(e*x + d))*F^(b*c*x + a*c), x)
Timed out. \[ \int F^{c (a+b x)} \sqrt {g \sec (d+e x)} \sqrt {f \sin (d+e x)} \, dx=\text {Timed out} \] Input:
integrate(F**(c*(b*x+a))*(g*sec(e*x+d))**(1/2)*(f*sin(e*x+d))**(1/2),x)
Output:
Timed out
\[ \int F^{c (a+b x)} \sqrt {g \sec (d+e x)} \sqrt {f \sin (d+e x)} \, dx=\int { \sqrt {g \sec \left (e x + d\right )} \sqrt {f \sin \left (e x + d\right )} F^{{\left (b x + a\right )} c} \,d x } \] Input:
integrate(F^(c*(b*x+a))*(g*sec(e*x+d))^(1/2)*(f*sin(e*x+d))^(1/2),x, algor ithm="maxima")
Output:
integrate(sqrt(g*sec(e*x + d))*sqrt(f*sin(e*x + d))*F^((b*x + a)*c), x)
\[ \int F^{c (a+b x)} \sqrt {g \sec (d+e x)} \sqrt {f \sin (d+e x)} \, dx=\int { \sqrt {g \sec \left (e x + d\right )} \sqrt {f \sin \left (e x + d\right )} F^{{\left (b x + a\right )} c} \,d x } \] Input:
integrate(F^(c*(b*x+a))*(g*sec(e*x+d))^(1/2)*(f*sin(e*x+d))^(1/2),x, algor ithm="giac")
Output:
integrate(sqrt(g*sec(e*x + d))*sqrt(f*sin(e*x + d))*F^((b*x + a)*c), x)
Timed out. \[ \int F^{c (a+b x)} \sqrt {g \sec (d+e x)} \sqrt {f \sin (d+e x)} \, dx=\int F^{c\,\left (a+b\,x\right )}\,\sqrt {f\,\sin \left (d+e\,x\right )}\,\sqrt {\frac {g}{\cos \left (d+e\,x\right )}} \,d x \] Input:
int(F^(c*(a + b*x))*(f*sin(d + e*x))^(1/2)*(g/cos(d + e*x))^(1/2),x)
Output:
int(F^(c*(a + b*x))*(f*sin(d + e*x))^(1/2)*(g/cos(d + e*x))^(1/2), x)
\[ \int F^{c (a+b x)} \sqrt {g \sec (d+e x)} \sqrt {f \sin (d+e x)} \, dx=\sqrt {g}\, f^{a c +\frac {1}{2}} \left (\int f^{b c x} \sqrt {\sin \left (e x +d \right )}\, \sqrt {\sec \left (e x +d \right )}d x \right ) \] Input:
int(F^(c*(b*x+a))*(g*sec(e*x+d))^(1/2)*(f*sin(e*x+d))^(1/2),x)
Output:
sqrt(g)*f**((2*a*c + 1)/2)*int(f**(b*c*x)*sqrt(sin(d + e*x))*sqrt(sec(d + e*x)),x)