\(\int F^{c (a+b x)} \sqrt {f \sin (d+e x)} \sqrt {g \sin (d+e x)} \, dx\) [129]

Optimal result

Integrand size = 34, antiderivative size = 74 \[ \int F^{c (a+b x)} \sqrt {f \sin (d+e x)} \sqrt {g \sin (d+e x)} \, dx=-\frac {f F^{c (a+b x)} \sqrt {g \sin (d+e x)} (e \cos (d+e x)-b c \log (F) \sin (d+e x))}{\left (e^2+b^2 c^2 \log ^2(F)\right ) \sqrt {f \sin (d+e x)}} \] Output:

-f*F^(c*(b*x+a))*(g*sin(e*x+d))^(1/2)*(e*cos(e*x+d)-b*c*ln(F)*sin(e*x+d))/ 
(e^2+b^2*c^2*ln(F)^2)/(f*sin(e*x+d))^(1/2)
 

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.89 \[ \int F^{c (a+b x)} \sqrt {f \sin (d+e x)} \sqrt {g \sin (d+e x)} \, dx=\frac {F^{c (a+b x)} (-e \cot (d+e x)+b c \log (F)) \sqrt {f \sin (d+e x)} \sqrt {g \sin (d+e x)}}{e^2+b^2 c^2 \log ^2(F)} \] Input:

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

Output:

(F^(c*(a + b*x))*(-(e*Cot[d + e*x]) + b*c*Log[F])*Sqrt[f*Sin[d + e*x]]*Sqr 
t[g*Sin[d + e*x]])/(e^2 + b^2*c^2*Log[F]^2)
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.34, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2031, 4932}

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.

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

\(\Big \downarrow \) 2031

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

\(\Big \downarrow \) 4932

\(\displaystyle \frac {f \sqrt {g \sin (d+e x)} \left (\frac {b c \log (F) \sin (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+e^2}-\frac {e \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+e^2}\right )}{\sqrt {f \sin (d+e x)}}\)

Input:

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

Output:

(f*Sqrt[g*Sin[d + e*x]]*(-((e*F^(c*(a + b*x))*Cos[d + e*x])/(e^2 + b^2*c^2 
*Log[F]^2)) + (b*c*F^(c*(a + b*x))*Log[F]*Sin[d + e*x])/(e^2 + b^2*c^2*Log 
[F]^2)))/Sqrt[f*Sin[d + e*x]]
 

Defintions of rubi rules used

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 4932

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> 
 Simp[b*c*Log[F]*F^(c*(a + b*x))*(Sin[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x 
] - Simp[e*F^(c*(a + b*x))*(Cos[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x] /; F 
reeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.41 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.65

method	result	size
risch	\(\frac {\sqrt {f \sin \left (e x +d \right )}\, \sqrt {g \sin \left (e x +d \right )}\, {\mathrm e}^{2 i \left (e x +d \right )} F^{c \left (b x +a \right )}}{\left ({\mathrm e}^{2 i \left (e x +d \right )}-1\right ) \left (i e +b c \ln \left (F \right )\right )}-\frac {\sqrt {f \sin \left (e x +d \right )}\, \sqrt {g \sin \left (e x +d \right )}\, F^{c \left (b x +a \right )}}{\left ({\mathrm e}^{2 i \left (e x +d \right )}-1\right ) \left (b c \ln \left (F \right )-i e \right )}\)	\(122\)

Input:

int(F^(c*(b*x+a))*(f*sin(e*x+d))^(1/2)*(g*sin(e*x+d))^(1/2),x,method=_RETU 
RNVERBOSE)
 

Output:

(f*sin(e*x+d))^(1/2)/(exp(2*I*(e*x+d))-1)*(g*sin(e*x+d))^(1/2)*exp(2*I*(e* 
x+d))/(I*e+b*c*ln(F))*F^(c*(b*x+a))-(f*sin(e*x+d))^(1/2)/(exp(2*I*(e*x+d)) 
-1)*(g*sin(e*x+d))^(1/2)/(b*c*ln(F)-I*e)*F^(c*(b*x+a))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.04 \[ \int F^{c (a+b x)} \sqrt {f \sin (d+e x)} \sqrt {g \sin (d+e x)} \, dx=\frac {{\left (b c \log \left (F\right ) \sin \left (e x + d\right ) - e \cos \left (e x + d\right )\right )} \sqrt {f \sin \left (e x + d\right )} \sqrt {g \sin \left (e x + d\right )} F^{b c x + a c}}{{\left (b^{2} c^{2} \log \left (F\right )^{2} + e^{2}\right )} \sin \left (e x + d\right )} \] Input:

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

Output:

(b*c*log(F)*sin(e*x + d) - e*cos(e*x + d))*sqrt(f*sin(e*x + d))*sqrt(g*sin 
(e*x + d))*F^(b*c*x + a*c)/((b^2*c^2*log(F)^2 + e^2)*sin(e*x + d))
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 13.58 (sec) , antiderivative size = 512, normalized size of antiderivative = 6.92 \[ \int F^{c (a+b x)} \sqrt {f \sin (d+e x)} \sqrt {g \sin (d+e x)} \, dx=\begin {cases} \frac {F^{a c - \frac {i e x}{\log {\left (F \right )}}} x \sqrt {f \sin {\left (d + e x \right )}} \sqrt {g \sin {\left (d + e x \right )}}}{2} - \frac {i F^{a c - \frac {i e x}{\log {\left (F \right )}}} x \sqrt {f \sin {\left (d + e x \right )}} \sqrt {g \sin {\left (d + e x \right )}} \cos {\left (d + e x \right )}}{2 \sin {\left (d + e x \right )}} - \frac {i F^{a c - \frac {i e x}{\log {\left (F \right )}}} \sqrt {f \sin {\left (d + e x \right )}} \sqrt {g \sin {\left (d + e x \right )}}}{2 e} - \frac {F^{a c - \frac {i e x}{\log {\left (F \right )}}} \sqrt {f \sin {\left (d + e x \right )}} \sqrt {g \sin {\left (d + e x \right )}} \cos {\left (d + e x \right )}}{e \sin {\left (d + e x \right )}} & \text {for}\: b = - \frac {i e}{c \log {\left (F \right )}} \\\frac {F^{a c + \frac {i e x}{\log {\left (F \right )}}} x \sqrt {f \sin {\left (d + e x \right )}} \sqrt {g \sin {\left (d + e x \right )}}}{2} + \frac {i F^{a c + \frac {i e x}{\log {\left (F \right )}}} x \sqrt {f \sin {\left (d + e x \right )}} \sqrt {g \sin {\left (d + e x \right )}} \cos {\left (d + e x \right )}}{2 \sin {\left (d + e x \right )}} - \frac {F^{a c + \frac {i e x}{\log {\left (F \right )}}} \sqrt {f \sin {\left (d + e x \right )}} \sqrt {g \sin {\left (d + e x \right )}} \cos {\left (d + e x \right )}}{2 e \sin {\left (d + e x \right )}} & \text {for}\: b = \frac {i e}{c \log {\left (F \right )}} \\0 & \text {for}\: d = - e x \vee d = - e x + \pi \\\frac {F^{a c + b c x} b c \sqrt {f \sin {\left (d + e x \right )}} \sqrt {g \sin {\left (d + e x \right )}} \log {\left (F \right )} \sin {\left (d + e x \right )}}{b^{2} c^{2} \log {\left (F \right )}^{2} \sin {\left (d + e x \right )} + e^{2} \sin {\left (d + e x \right )}} - \frac {F^{a c + b c x} e \sqrt {f \sin {\left (d + e x \right )}} \sqrt {g \sin {\left (d + e x \right )}} \cos {\left (d + e x \right )}}{b^{2} c^{2} \log {\left (F \right )}^{2} \sin {\left (d + e x \right )} + e^{2} \sin {\left (d + e x \right )}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((F**(a*c - I*e*x/log(F))*x*sqrt(f*sin(d + e*x))*sqrt(g*sin(d + e 
*x))/2 - I*F**(a*c - I*e*x/log(F))*x*sqrt(f*sin(d + e*x))*sqrt(g*sin(d + e 
*x))*cos(d + e*x)/(2*sin(d + e*x)) - I*F**(a*c - I*e*x/log(F))*sqrt(f*sin( 
d + e*x))*sqrt(g*sin(d + e*x))/(2*e) - F**(a*c - I*e*x/log(F))*sqrt(f*sin( 
d + e*x))*sqrt(g*sin(d + e*x))*cos(d + e*x)/(e*sin(d + e*x)), Eq(b, -I*e/( 
c*log(F)))), (F**(a*c + I*e*x/log(F))*x*sqrt(f*sin(d + e*x))*sqrt(g*sin(d 
+ e*x))/2 + I*F**(a*c + I*e*x/log(F))*x*sqrt(f*sin(d + e*x))*sqrt(g*sin(d 
+ e*x))*cos(d + e*x)/(2*sin(d + e*x)) - F**(a*c + I*e*x/log(F))*sqrt(f*sin 
(d + e*x))*sqrt(g*sin(d + e*x))*cos(d + e*x)/(2*e*sin(d + e*x)), Eq(b, I*e 
/(c*log(F)))), (0, Eq(d, -e*x) | Eq(d, -e*x + pi)), (F**(a*c + b*c*x)*b*c* 
sqrt(f*sin(d + e*x))*sqrt(g*sin(d + e*x))*log(F)*sin(d + e*x)/(b**2*c**2*l 
og(F)**2*sin(d + e*x) + e**2*sin(d + e*x)) - F**(a*c + b*c*x)*e*sqrt(f*sin 
(d + e*x))*sqrt(g*sin(d + e*x))*cos(d + e*x)/(b**2*c**2*log(F)**2*sin(d + 
e*x) + e**2*sin(d + e*x)), True))
 

Maxima [F(-2)]

Exception generated. \[ \int F^{c (a+b x)} \sqrt {f \sin (d+e x)} \sqrt {g \sin (d+e x)} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume((2*%i*e)/(log(F)*b*c>0)', see `a 
ssume?` fo
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 155744 vs. \(2 (69) = 138\).

Time = 5.66 (sec) , antiderivative size = 155744, normalized size of antiderivative = 2104.65 \[ \int F^{c (a+b x)} \sqrt {f \sin (d+e x)} \sqrt {g \sin (d+e x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(2*pi^3*b^3*c^3*f*g*abs(F)^(b*c*x)*abs(F)^(a*c)*sqrt(abs(f))*sqrt(abs( 
g))*sgn(F)*tan(1/4*pi*b*c*x*sgn(F) - 1/4*pi*b*c*x + 1/2*e*x + 1/2*d)^2*tan 
(1/4*pi*a*c*sgn(F) - 1/4*pi*a*c)^2*tan(1/2*e*x + 1/2*d)^3 + 2*pi^3*b^3*c^3 
*g*abs(F)^(b*c*x)*abs(F)^(a*c)*abs(f)^(3/2)*sqrt(abs(g))*sgn(F)*tan(1/4*pi 
*b*c*x*sgn(F) - 1/4*pi*b*c*x + 1/2*e*x + 1/2*d)^2*tan(1/4*pi*a*c*sgn(F) - 
1/4*pi*a*c)^2*tan(1/2*e*x + 1/2*d)^3 + 2*pi^3*b^3*c^3*f*abs(F)^(b*c*x)*abs 
(F)^(a*c)*sqrt(abs(f))*abs(g)^(3/2)*sgn(F)*tan(1/4*pi*b*c*x*sgn(F) - 1/4*p 
i*b*c*x + 1/2*e*x + 1/2*d)^2*tan(1/4*pi*a*c*sgn(F) - 1/4*pi*a*c)^2*tan(1/2 
*e*x + 1/2*d)^3 - 2*pi^3*b^3*c^3*abs(F)^(b*c*x)*abs(F)^(a*c)*abs(f)^(3/2)* 
abs(g)^(3/2)*sgn(F)*tan(1/4*pi*b*c*x*sgn(F) - 1/4*pi*b*c*x + 1/2*e*x + 1/2 
*d)^2*tan(1/4*pi*a*c*sgn(F) - 1/4*pi*a*c)^2*tan(1/2*e*x + 1/2*d)^3 + 2*pi^ 
2*b^3*c^3*f*g*abs(F)^(b*c*x)*abs(F)^(a*c)*sqrt(abs(f))*sqrt(abs(g))*log(ab 
s(F))*sgn(F)*tan(1/4*pi*b*c*x*sgn(F) - 1/4*pi*b*c*x + 1/2*e*x + 1/2*d)^2*t 
an(1/4*pi*a*c*sgn(F) - 1/4*pi*a*c)^2*tan(1/2*e*x + 1/2*d)^3 - 2*pi^2*b^3*c 
^3*g*abs(F)^(b*c*x)*abs(F)^(a*c)*abs(f)^(3/2)*sqrt(abs(g))*log(abs(F))*sgn 
(F)*tan(1/4*pi*b*c*x*sgn(F) - 1/4*pi*b*c*x + 1/2*e*x + 1/2*d)^2*tan(1/4*pi 
*a*c*sgn(F) - 1/4*pi*a*c)^2*tan(1/2*e*x + 1/2*d)^3 - 2*pi^2*b^3*c^3*f*abs( 
F)^(b*c*x)*abs(F)^(a*c)*sqrt(abs(f))*abs(g)^(3/2)*log(abs(F))*sgn(F)*tan(1 
/4*pi*b*c*x*sgn(F) - 1/4*pi*b*c*x + 1/2*e*x + 1/2*d)^2*tan(1/4*pi*a*c*sgn( 
F) - 1/4*pi*a*c)^2*tan(1/2*e*x + 1/2*d)^3 - 2*pi^2*b^3*c^3*abs(F)^(b*c*...
 

Mupad [B] (verification not implemented)

Time = 15.78 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.23 \[ \int F^{c (a+b x)} \sqrt {f \sin (d+e x)} \sqrt {g \sin (d+e x)} \, dx=\frac {F^{c\,\left (a+b\,x\right )}\,\sqrt {f\,\sin \left (d+e\,x\right )}\,\sqrt {g\,\sin \left (d+e\,x\right )}\,\left (b\,c\,\ln \left (F\right )-e\,\sin \left (2\,d+2\,e\,x\right )+b\,c\,\ln \left (F\right )\,\left (2\,{\sin \left (d+e\,x\right )}^2-1\right )\right )}{2\,{\sin \left (d+e\,x\right )}^2\,\left (b^2\,c^2\,{\ln \left (F\right )}^2+e^2\right )} \] Input:

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

Output:

(F^(c*(a + b*x))*(f*sin(d + e*x))^(1/2)*(g*sin(d + e*x))^(1/2)*(b*c*log(F) 
 - e*sin(2*d + 2*e*x) + b*c*log(F)*(2*sin(d + e*x)^2 - 1)))/(2*sin(d + e*x 
)^2*(e^2 + b^2*c^2*log(F)^2))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.70 \[ \int F^{c (a+b x)} \sqrt {f \sin (d+e x)} \sqrt {g \sin (d+e x)} \, dx=\frac {\sqrt {g}\, f^{b c x +a c +\frac {1}{2}} \left (-\cos \left (e x +d \right ) e +\mathrm {log}\left (f \right ) \sin \left (e x +d \right ) b c \right )}{\mathrm {log}\left (f \right )^{2} b^{2} c^{2}+e^{2}} \] Input:

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

Output:

(sqrt(g)*f**((2*a*c + 2*b*c*x + 1)/2)*( - cos(d + e*x)*e + log(f)*sin(d + 
e*x)*b*c))/(log(f)**2*b**2*c**2 + e**2)