\(\int F^{c (a+b x)} \sin ^2(d+e x) \, dx\) [3]

Optimal result

Integrand size = 18, antiderivative size = 128 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \, dx=\frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (4 e^2+b^2 c^2 \log ^2(F)\right )}-\frac {2 e F^{c (a+b x)} \cos (d+e x) \sin (d+e x)}{4 e^2+b^2 c^2 \log ^2(F)}+\frac {b c F^{c (a+b x)} \log (F) \sin ^2(d+e x)}{4 e^2+b^2 c^2 \log ^2(F)} \] Output:

2*e^2*F^(c*(b*x+a))/b/c/ln(F)/(4*e^2+b^2*c^2*ln(F)^2)-2*e*F^(c*(b*x+a))*co 
s(e*x+d)*sin(e*x+d)/(4*e^2+b^2*c^2*ln(F)^2)+b*c*F^(c*(b*x+a))*ln(F)*sin(e* 
x+d)^2/(4*e^2+b^2*c^2*ln(F)^2)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.67 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \, dx=\frac {F^{c (a+b x)} \left (4 e^2+b^2 c^2 \log ^2(F)-b^2 c^2 \cos (2 (d+e x)) \log ^2(F)-2 b c e \log (F) \sin (2 (d+e x))\right )}{8 b c e^2 \log (F)+2 b^3 c^3 \log ^3(F)} \] Input:

Integrate[F^(c*(a + b*x))*Sin[d + e*x]^2,x]
 

Output:

(F^(c*(a + b*x))*(4*e^2 + b^2*c^2*Log[F]^2 - b^2*c^2*Cos[2*(d + e*x)]*Log[ 
F]^2 - 2*b*c*e*Log[F]*Sin[2*(d + e*x)]))/(8*b*c*e^2*Log[F] + 2*b^3*c^3*Log 
[F]^3)
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4934, 2624}

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.

\(\displaystyle \int \sin ^2(d+e x) F^{c (a+b x)} \, dx\)

\(\Big \downarrow \) 4934

\(\displaystyle \frac {2 e^2 \int F^{c (a+b x)}dx}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {b c \log (F) \sin ^2(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}-\frac {2 e \sin (d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}\)

\(\Big \downarrow \) 2624

\(\displaystyle \frac {b c \log (F) \sin ^2(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}-\frac {2 e \sin (d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (b^2 c^2 \log ^2(F)+4 e^2\right )}\)

Input:

Int[F^(c*(a + b*x))*Sin[d + e*x]^2,x]
 

Output:

(2*e^2*F^(c*(a + b*x)))/(b*c*Log[F]*(4*e^2 + b^2*c^2*Log[F]^2)) - (2*e*F^( 
c*(a + b*x))*Cos[d + e*x]*Sin[d + e*x])/(4*e^2 + b^2*c^2*Log[F]^2) + (b*c* 
F^(c*(a + b*x))*Log[F]*Sin[d + e*x]^2)/(4*e^2 + b^2*c^2*Log[F]^2)
 

Defintions of rubi rules used

rule 2624

Int[((F_)^(v_))^(n_.), x_Symbol] :> Simp[(F^v)^n/(n*Log[F]*D[v, x]), x] /; 
FreeQ[{F, n}, x] && LinearQ[v, x]
 

rule 4934

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(n_), x_Symbo 
l] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(Sin[d + e*x]^n/(e^2*n^2 + b^2*c^2*Lo 
g[F]^2)), x] + (-Simp[e*n*F^(c*(a + b*x))*Cos[d + e*x]*(Sin[d + e*x]^(n - 1 
)/(e^2*n^2 + b^2*c^2*Log[F]^2)), x] + Simp[(n*(n - 1)*e^2)/(e^2*n^2 + b^2*c 
^2*Log[F]^2)   Int[F^(c*(a + b*x))*Sin[d + e*x]^(n - 2), x], x]) /; FreeQ[{ 
F, a, b, c, d, e}, x] && NeQ[e^2*n^2 + b^2*c^2*Log[F]^2, 0] && GtQ[n, 1]
 

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.74

method	result	size
parallelrisch	\(-\frac {F^{c \left (b x +a \right )} \left (\frac {b^{2} c^{2} \ln \left (F \right )^{2} \cos \left (2 e x +2 d \right )}{2}-\frac {b^{2} c^{2} \ln \left (F \right )^{2}}{2}+e \sin \left (2 e x +2 d \right ) b c \ln \left (F \right )-2 e^{2}\right )}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}\)	\(95\)
risch	\(\frac {F^{c \left (b x +a \right )}}{2 b c \ln \left (F \right )}-\frac {F^{c \left (b x +a \right )} b c \ln \left (F \right ) \cos \left (2 e x +2 d \right )}{2 \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}-\frac {e \,F^{c \left (b x +a \right )} \sin \left (2 e x +2 d \right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\)	\(106\)
orering	\(\frac {\left (3 b^{2} c^{2} \ln \left (F \right )^{2}+4 e^{2}\right ) F^{c \left (b x +a \right )} \sin \left (e x +d \right )^{2}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}-\frac {3 \left (F^{c \left (b x +a \right )} b c \ln \left (F \right ) \sin \left (e x +d \right )^{2}+2 F^{c \left (b x +a \right )} \sin \left (e x +d \right ) e \cos \left (e x +d \right )\right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {F^{c \left (b x +a \right )} b^{2} c^{2} \ln \left (F \right )^{2} \sin \left (e x +d \right )^{2}+4 F^{c \left (b x +a \right )} b c \ln \left (F \right ) \sin \left (e x +d \right ) e \cos \left (e x +d \right )+2 F^{c \left (b x +a \right )} e^{2} \cos \left (e x +d \right )^{2}-2 F^{c \left (b x +a \right )} \sin \left (e x +d \right )^{2} e^{2}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}\)	\(266\)
norman	\(\frac {-\frac {4 e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {4 e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{3}}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 e^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {2 e^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{4}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {4 \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}}{\left (1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}\right )^{2}}\)	\(268\)

Input:

int(F^(c*(b*x+a))*sin(e*x+d)^2,x,method=_RETURNVERBOSE)
 

Output:

-F^(c*(b*x+a))*(1/2*b^2*c^2*ln(F)^2*cos(2*e*x+2*d)-1/2*b^2*c^2*ln(F)^2+e*s 
in(2*e*x+2*d)*b*c*ln(F)-2*e^2)/b/c/ln(F)/(4*e^2+b^2*c^2*ln(F)^2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.70 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \, dx=-\frac {{\left (2 \, b c e \cos \left (e x + d\right ) \log \left (F\right ) \sin \left (e x + d\right ) + {\left (b^{2} c^{2} \cos \left (e x + d\right )^{2} - b^{2} c^{2}\right )} \log \left (F\right )^{2} - 2 \, e^{2}\right )} F^{b c x + a c}}{b^{3} c^{3} \log \left (F\right )^{3} + 4 \, b c e^{2} \log \left (F\right )} \] Input:

integrate(F^(c*(b*x+a))*sin(e*x+d)^2,x, algorithm="fricas")
 

Output:

-(2*b*c*e*cos(e*x + d)*log(F)*sin(e*x + d) + (b^2*c^2*cos(e*x + d)^2 - b^2 
*c^2)*log(F)^2 - 2*e^2)*F^(b*c*x + a*c)/(b^3*c^3*log(F)^3 + 4*b*c*e^2*log( 
F))
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.20 (sec) , antiderivative size = 743, normalized size of antiderivative = 5.80 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \, dx =\text {Too large to display} \] Input:

integrate(F**(c*(b*x+a))*sin(e*x+d)**2,x)
 

Output:

Piecewise((x*sin(d)**2, Eq(F, 1) & Eq(b, 0) & Eq(c, 0) & Eq(e, 0)), (x*sin 
(d + e*x)**2/2 + x*cos(d + e*x)**2/2 - sin(d + e*x)*cos(d + e*x)/(2*e), Eq 
(F, 1)), (F**(a*c)*(x*sin(d + e*x)**2/2 + x*cos(d + e*x)**2/2 - sin(d + e* 
x)*cos(d + e*x)/(2*e)), Eq(b, 0)), (x*sin(d + e*x)**2/2 + x*cos(d + e*x)** 
2/2 - sin(d + e*x)*cos(d + e*x)/(2*e), Eq(c, 0)), (F**(a*c + b*c*x)*x*sin( 
I*b*c*x*log(F)/2 - d)**2/4 - I*F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F)/2 - d 
)*cos(I*b*c*x*log(F)/2 - d)/2 - F**(a*c + b*c*x)*x*cos(I*b*c*x*log(F)/2 - 
d)**2/4 + 3*I*F**(a*c + b*c*x)*sin(I*b*c*x*log(F)/2 - d)*cos(I*b*c*x*log(F 
)/2 - d)/(2*b*c*log(F)) + F**(a*c + b*c*x)*cos(I*b*c*x*log(F)/2 - d)**2/(b 
*c*log(F)), Eq(e, -I*b*c*log(F)/2)), (F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F 
)/2 + d)**2/4 - I*F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F)/2 + d)*cos(I*b*c*x 
*log(F)/2 + d)/2 - F**(a*c + b*c*x)*x*cos(I*b*c*x*log(F)/2 + d)**2/4 + 3*I 
*F**(a*c + b*c*x)*sin(I*b*c*x*log(F)/2 + d)*cos(I*b*c*x*log(F)/2 + d)/(2*b 
*c*log(F)) + F**(a*c + b*c*x)*cos(I*b*c*x*log(F)/2 + d)**2/(b*c*log(F)), E 
q(e, I*b*c*log(F)/2)), (F**(a*c + b*c*x)*b**2*c**2*log(F)**2*sin(d + e*x)* 
*2/(b**3*c**3*log(F)**3 + 4*b*c*e**2*log(F)) - 2*F**(a*c + b*c*x)*b*c*e*lo 
g(F)*sin(d + e*x)*cos(d + e*x)/(b**3*c**3*log(F)**3 + 4*b*c*e**2*log(F)) + 
 2*F**(a*c + b*c*x)*e**2*sin(d + e*x)**2/(b**3*c**3*log(F)**3 + 4*b*c*e**2 
*log(F)) + 2*F**(a*c + b*c*x)*e**2*cos(d + e*x)**2/(b**3*c**3*log(F)**3 + 
4*b*c*e**2*log(F)), True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 356 vs. \(2 (128) = 256\).

Time = 0.10 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.78 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \, dx=-\frac {{\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right ) \log \left (F\right )^{2} + 2 \, F^{a c} b c e \log \left (F\right ) \sin \left (2 \, d\right )\right )} F^{b c x} \cos \left (2 \, e x\right ) + {\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right ) \log \left (F\right )^{2} - 2 \, F^{a c} b c e \log \left (F\right ) \sin \left (2 \, d\right )\right )} F^{b c x} \cos \left (2 \, e x + 4 \, d\right ) - {\left (F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (2 \, d\right ) - 2 \, F^{a c} b c e \cos \left (2 \, d\right ) \log \left (F\right )\right )} F^{b c x} \sin \left (2 \, e x\right ) + {\left (F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (2 \, d\right ) + 2 \, F^{a c} b c e \cos \left (2 \, d\right ) \log \left (F\right )\right )} F^{b c x} \sin \left (2 \, e x + 4 \, d\right ) - 2 \, {\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right )^{2} \log \left (F\right )^{2} + F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (2 \, d\right )^{2} + 4 \, {\left (F^{a c} \cos \left (2 \, d\right )^{2} + F^{a c} \sin \left (2 \, d\right )^{2}\right )} e^{2}\right )} F^{b c x}}{4 \, {\left (b^{3} c^{3} \cos \left (2 \, d\right )^{2} \log \left (F\right )^{3} + b^{3} c^{3} \log \left (F\right )^{3} \sin \left (2 \, d\right )^{2} + 4 \, {\left (b c \cos \left (2 \, d\right )^{2} \log \left (F\right ) + b c \log \left (F\right ) \sin \left (2 \, d\right )^{2}\right )} e^{2}\right )}} \] Input:

integrate(F^(c*(b*x+a))*sin(e*x+d)^2,x, algorithm="maxima")
 

Output:

-1/4*((F^(a*c)*b^2*c^2*cos(2*d)*log(F)^2 + 2*F^(a*c)*b*c*e*log(F)*sin(2*d) 
)*F^(b*c*x)*cos(2*e*x) + (F^(a*c)*b^2*c^2*cos(2*d)*log(F)^2 - 2*F^(a*c)*b* 
c*e*log(F)*sin(2*d))*F^(b*c*x)*cos(2*e*x + 4*d) - (F^(a*c)*b^2*c^2*log(F)^ 
2*sin(2*d) - 2*F^(a*c)*b*c*e*cos(2*d)*log(F))*F^(b*c*x)*sin(2*e*x) + (F^(a 
*c)*b^2*c^2*log(F)^2*sin(2*d) + 2*F^(a*c)*b*c*e*cos(2*d)*log(F))*F^(b*c*x) 
*sin(2*e*x + 4*d) - 2*(F^(a*c)*b^2*c^2*cos(2*d)^2*log(F)^2 + F^(a*c)*b^2*c 
^2*log(F)^2*sin(2*d)^2 + 4*(F^(a*c)*cos(2*d)^2 + F^(a*c)*sin(2*d)^2)*e^2)* 
F^(b*c*x))/(b^3*c^3*cos(2*d)^2*log(F)^3 + b^3*c^3*log(F)^3*sin(2*d)^2 + 4* 
(b*c*cos(2*d)^2*log(F) + b*c*log(F)*sin(2*d)^2)*e^2)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 915, normalized size of antiderivative = 7.15 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \, dx=\text {Too large to display} \] Input:

integrate(F^(c*(b*x+a))*sin(e*x+d)^2,x, algorithm="giac")
 

Output:

-1/2*(2*b*c*cos(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1 
/2*pi*a*c + 2*e*x + 2*d)*log(abs(F))/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sg 
n(F) - pi*b*c + 4*e)^2) + (pi*b*c*sgn(F) - pi*b*c + 4*e)*sin(1/2*pi*b*c*x* 
sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*pi*a*c + 2*e*x + 2*d)/(4*b 
^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 4*e)^2))*e^(b*c*x*log(abs 
(F)) + a*c*log(abs(F))) - 1/2*(2*b*c*cos(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c* 
x + 1/2*pi*a*c*sgn(F) - 1/2*pi*a*c - 2*e*x - 2*d)*log(abs(F))/(4*b^2*c^2*l 
og(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 4*e)^2) + (pi*b*c*sgn(F) - pi*b*c 
 - 4*e)*sin(1/2*pi*b*c*x*sgn(F) - 1/2*pi*b*c*x + 1/2*pi*a*c*sgn(F) - 1/2*p 
i*a*c - 2*e*x - 2*d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 
4*e)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + (2*b*c*cos(-1/2*pi*b*c* 
x*sgn(F) + 1/2*pi*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a*c)*log(abs(F))/(4*b 
^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c)^2) - (pi*b*c*sgn(F) - pi*b 
*c)*sin(-1/2*pi*b*c*x*sgn(F) + 1/2*pi*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a 
*c)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c)^2))*e^(b*c*x*log(a 
bs(F)) + a*c*log(abs(F))) + I*(-I*e^(1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi*b*c* 
x + 1/2*I*pi*a*c*sgn(F) - 1/2*I*pi*a*c + 2*I*e*x + 2*I*d)/(4*I*pi*b*c*sgn( 
F) - 4*I*pi*b*c + 8*b*c*log(abs(F)) + 16*I*e) + I*e^(-1/2*I*pi*b*c*x*sgn(F 
) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*sgn(F) + 1/2*I*pi*a*c - 2*I*e*x - 2*I*d) 
/(-4*I*pi*b*c*sgn(F) + 4*I*pi*b*c + 8*b*c*log(abs(F)) - 16*I*e))*e^(b*c...
 

Mupad [B] (verification not implemented)

Time = 20.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.74 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \, dx=\frac {F^{a\,c+b\,c\,x}\,\left (2\,e^2+\frac {b^2\,c^2\,{\ln \left (F\right )}^2}{2}-\frac {b^2\,c^2\,{\ln \left (F\right )}^2\,\cos \left (2\,d+2\,e\,x\right )}{2}-b\,c\,e\,\ln \left (F\right )\,\sin \left (2\,d+2\,e\,x\right )\right )}{b\,c\,\ln \left (F\right )\,\left (b^2\,c^2\,{\ln \left (F\right )}^2+4\,e^2\right )} \] Input:

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

Output:

(F^(a*c + b*c*x)*(2*e^2 + (b^2*c^2*log(F)^2)/2 - (b^2*c^2*log(F)^2*cos(2*d 
 + 2*e*x))/2 - b*c*e*log(F)*sin(2*d + 2*e*x)))/(b*c*log(F)*(4*e^2 + b^2*c^ 
2*log(F)^2))
 

Reduce [F]

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

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

Output:

f**(a*c)*int(f**(b*c*x)*sin(d + e*x)**2,x)