Integrand size = 34, antiderivative size = 44 \[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \tan (d+e x)} \, dx=\frac {F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \tan (d+e x)}}{b c \log (F)} \] Output:
F^(c*(b*x+a))*(g*cot(e*x+d))^(1/2)*(f*tan(e*x+d))^(1/2)/b/c/ln(F)
Time = 1.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \tan (d+e x)} \, dx=\frac {F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \tan (d+e x)}}{b c \log (F)} \] Input:
Integrate[F^(c*(a + b*x))*Sqrt[g*Cot[d + e*x]]*Sqrt[f*Tan[d + e*x]],x]
Output:
(F^(c*(a + b*x))*Sqrt[g*Cot[d + e*x]]*Sqrt[f*Tan[d + e*x]])/(b*c*Log[F])
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.
\(\displaystyle \int F^{c (a+b x)} \sqrt {f \tan (d+e x)} \sqrt {g \cot (d+e x)} \, dx\) |
\(\Big \downarrow \) 7271 |
\(\displaystyle \frac {\sqrt {g \cot (d+e x)} \int F^{c (a+b x)} \sqrt {\cot (d+e x)} \sqrt {f \tan (d+e x)}dx}{\sqrt {\cot (d+e x)}}\) |
\(\Big \downarrow \) 7271 |
\(\displaystyle \sqrt {\tan (d+e x)} \sqrt {\cot (d+e x)} \sqrt {f \tan (d+e x)} \sqrt {g \cot (d+e x)} \int F^{c (a+b x)} \sqrt {\cot (d+e x)} \sqrt {\tan (d+e x)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \sqrt {\tan (d+e x)} \sqrt {\cot (d+e x)} \sqrt {f \tan (d+e x)} \sqrt {g \cot (d+e x)} \int F^{a c+b x c} \sqrt {\cot (d+e x)} \sqrt {\tan (d+e x)}dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \sqrt {\tan (d+e x)} \sqrt {\cot (d+e x)} \sqrt {f \tan (d+e x)} \sqrt {g \cot (d+e x)} \int F^{a c+b x c} \sqrt {\cot (d+e x)} \sqrt {\tan (d+e x)}dx\) |
Input:
Int[F^(c*(a + b*x))*Sqrt[g*Cot[d + e*x]]*Sqrt[f*Tan[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])
Time = 0.72 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93
method | result | size |
orering | \(\frac {F^{c \left (b x +a \right )} \sqrt {g \cot \left (e x +d \right )}\, \sqrt {f \tan \left (e x +d \right )}}{b c \ln \left (F \right )}\) | \(41\) |
risch | \(\frac {\sqrt {\frac {i g \left (1+{\mathrm e}^{2 i \left (e x +d \right )}\right )}{{\mathrm e}^{2 i \left (e x +d \right )}-1}}\, \sqrt {-\frac {i f \left ({\mathrm e}^{2 i \left (e x +d \right )}-1\right )}{1+{\mathrm e}^{2 i \left (e x +d \right )}}}\, F^{c \left (b x +a \right )}}{b c \ln \left (F \right )}\) | \(81\) |
Input:
int(F^(c*(b*x+a))*(g*cot(e*x+d))^(1/2)*(f*tan(e*x+d))^(1/2),x,method=_RETU RNVERBOSE)
Output:
F^(c*(b*x+a))*(g*cot(e*x+d))^(1/2)*(f*tan(e*x+d))^(1/2)/b/c/ln(F)
Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98 \[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \tan (d+e x)} \, dx=\frac {\sqrt {f \tan \left (e x + d\right )} F^{b c x + a c} \sqrt {\frac {g}{\tan \left (e x + d\right )}}}{b c \log \left (F\right )} \] Input:
integrate(F^(c*(b*x+a))*(g*cot(e*x+d))^(1/2)*(f*tan(e*x+d))^(1/2),x, algor ithm="fricas")
Output:
sqrt(f*tan(e*x + d))*F^(b*c*x + a*c)*sqrt(g/tan(e*x + d))/(b*c*log(F))
Timed out. \[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \tan (d+e x)} \, dx=\text {Timed out} \] Input:
integrate(F**(c*(b*x+a))*(g*cot(e*x+d))**(1/2)*(f*tan(e*x+d))**(1/2),x)
Output:
Timed out
\[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \tan (d+e x)} \, dx=\int { \sqrt {g \cot \left (e x + d\right )} \sqrt {f \tan \left (e x + d\right )} F^{{\left (b x + a\right )} c} \,d x } \] Input:
integrate(F^(c*(b*x+a))*(g*cot(e*x+d))^(1/2)*(f*tan(e*x+d))^(1/2),x, algor ithm="maxima")
Output:
integrate(sqrt(g*cot(e*x + d))*sqrt(f*tan(e*x + d))*F^((b*x + a)*c), x)
\[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \tan (d+e x)} \, dx=\int { \sqrt {g \cot \left (e x + d\right )} \sqrt {f \tan \left (e x + d\right )} F^{{\left (b x + a\right )} c} \,d x } \] Input:
integrate(F^(c*(b*x+a))*(g*cot(e*x+d))^(1/2)*(f*tan(e*x+d))^(1/2),x, algor ithm="giac")
Output:
integrate(sqrt(g*cot(e*x + d))*sqrt(f*tan(e*x + d))*F^((b*x + a)*c), x)
Time = 18.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 2.14 \[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \tan (d+e x)} \, dx=\frac {F^{a\,c+b\,c\,x}\,\sqrt {-\frac {f\,\left ({\mathrm {e}}^{d\,2{}\mathrm {i}+e\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{d\,2{}\mathrm {i}+e\,x\,2{}\mathrm {i}}+1}}\,\sqrt {\frac {g\,\left ({\mathrm {e}}^{d\,2{}\mathrm {i}+e\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}{{\mathrm {e}}^{d\,2{}\mathrm {i}+e\,x\,2{}\mathrm {i}}-1}}}{b\,c\,\ln \left (F\right )} \] Input:
int(F^(c*(a + b*x))*(g*cot(d + e*x))^(1/2)*(f*tan(d + e*x))^(1/2),x)
Output:
(F^(a*c + b*c*x)*(-(f*(exp(d*2i + e*x*2i)*1i - 1i))/(exp(d*2i + e*x*2i) + 1))^(1/2)*((g*(exp(d*2i + e*x*2i)*1i + 1i))/(exp(d*2i + e*x*2i) - 1))^(1/2 ))/(b*c*log(F))
\[ \int F^{c (a+b x)} \sqrt {g \cot (d+e x)} \sqrt {f \tan (d+e x)} \, dx=\sqrt {g}\, f^{a c +\frac {1}{2}} \left (\int f^{b c x} \sqrt {\tan \left (e x +d \right )}\, \sqrt {\cot \left (e x +d \right )}d x \right ) \] Input:
int(F^(c*(b*x+a))*(g*cot(e*x+d))^(1/2)*(f*tan(e*x+d))^(1/2),x)
Output:
sqrt(g)*f**((2*a*c + 1)/2)*int(f**(b*c*x)*sqrt(tan(d + e*x))*sqrt(cot(d + e*x)),x)