\(\int F^{c (a+b x)} (f+f \sin (d+e x))^2 \, dx\) [28]

Optimal result

Integrand size = 22, antiderivative size = 245 \[ \int F^{c (a+b x)} (f+f \sin (d+e x))^2 \, dx=\frac {f^2 F^{a c+b c x}}{b c \log (F)}-\frac {2 e f^2 F^{a c+b c x} \cos (d+e x)}{e^2+b^2 c^2 \log ^2(F)}+\frac {2 e^2 f^2 F^{a c+b c x}}{b c \log (F) \left (4 e^2+b^2 c^2 \log ^2(F)\right )}+\frac {2 b c f^2 F^{a c+b c x} \log (F) \sin (d+e x)}{e^2+b^2 c^2 \log ^2(F)}-\frac {2 e f^2 F^{a c+b c x} \cos (d+e x) \sin (d+e x)}{4 e^2+b^2 c^2 \log ^2(F)}+\frac {b c f^2 F^{a c+b c x} \log (F) \sin ^2(d+e x)}{4 e^2+b^2 c^2 \log ^2(F)} \] Output:

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

Mathematica [A] (verified)

Time = 7.41 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.73 \[ \int F^{c (a+b x)} (f+f \sin (d+e x))^2 \, dx=\frac {f^2 F^{c (a+b x)} (1+\sin (d+e x))^2 \left (\frac {3}{b c \log (F)}-\frac {4 e \cos (d+e x)}{e^2+b^2 c^2 \log ^2(F)}-\frac {b c \cos (2 (d+e x)) \log (F)}{4 e^2+b^2 c^2 \log ^2(F)}+\frac {4 b c \log (F) \sin (d+e x)}{e^2+b^2 c^2 \log ^2(F)}-\frac {2 e \sin (2 (d+e x))}{4 e^2+b^2 c^2 \log ^2(F)}\right )}{2 \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )^4} \] Input:

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

Output:

(f^2*F^(c*(a + b*x))*(1 + Sin[d + e*x])^2*(3/(b*c*Log[F]) - (4*e*Cos[d + e 
*x])/(e^2 + b^2*c^2*Log[F]^2) - (b*c*Cos[2*(d + e*x)]*Log[F])/(4*e^2 + b^2 
*c^2*Log[F]^2) + (4*b*c*Log[F]*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2) - (2 
*e*Sin[2*(d + e*x)])/(4*e^2 + b^2*c^2*Log[F]^2)))/(2*(Cos[(d + e*x)/2] + S 
in[(d + e*x)/2])^4)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {7292, 27, 7293, 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.

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

\(\Big \downarrow \) 7292

\(\displaystyle \int f^2 (\sin (d+e x)+1)^2 F^{a c+b c x}dx\)

\(\Big \downarrow \) 27

\(\displaystyle f^2 \int F^{a c+b x c} (\sin (d+e x)+1)^2dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {3 f^{2} F^{c \left (b x +a \right )}}{2 b c \ln \left (F \right )}-\frac {2 F^{c \left (b x +a \right )} e \,f^{2} \cos \left (e x +d \right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 F^{c \left (b x +a \right )} \ln \left (F \right ) b c \,f^{2} \sin \left (e x +d \right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {\ln \left (F \right ) c b \,f^{2} F^{c \left (b x +a \right )} \cos \left (2 e x +2 d \right )}{2 \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}-\frac {e \,f^{2} F^{c \left (b x +a \right )} \sin \left (2 e x +2 d \right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\)	\(194\)
parallelrisch	\(-\frac {2 \left (\frac {b^{2} c^{2} \ln \left (F \right )^{2} \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) \cos \left (2 e x +2 d \right )}{4}+\frac {c b e \ln \left (F \right ) \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) \sin \left (2 e x +2 d \right )}{2}+\left (-\sin \left (e x +d \right ) b^{2} c^{2} \ln \left (F \right )^{2}-\frac {3 b^{2} c^{2} \ln \left (F \right )^{2}}{4}+\cos \left (e x +d \right ) b c e \ln \left (F \right )-\frac {3 e^{2}}{4}\right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )\right ) f^{2} F^{c \left (b x +a \right )}}{b c \ln \left (F \right ) \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}\)	\(195\)
default	\(\frac {F^{a c} f^{2} \left (\frac {4 F^{b c x}}{b c \ln \left (F \right )}+\frac {-\frac {8 e \,{\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {8 e \,{\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{3}}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {2 \left (b^{2} c^{2} \ln \left (F \right )^{2}+2 e^{2}\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {4 \left (b^{2} c^{2} \ln \left (F \right )^{2}-2 e^{2}\right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}-\frac {2 \left (b^{2} c^{2} \ln \left (F \right )^{2}+2 e^{2}\right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{4}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}}{\left (1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}\right )^{2}}+\frac {-\frac {4 e \,{\mathrm e}^{b c x \ln \left (F \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {4 e \,{\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {8 b c \ln \left (F \right ) {\mathrm e}^{b c x \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}\right )}{2}\)	\(433\)
parts	\(\frac {f^{2} F^{c \left (b x +a \right )}}{b c \ln \left (F \right )}+\frac {-\frac {4 e \,f^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {4 e \,f^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{3}}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 e^{2} f^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {2 e^{2} f^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{4}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {4 f^{2} \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}}{\left (1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}\right )^{2}}+\frac {-\frac {2 e \,f^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 e \,f^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {4 \ln \left (F \right ) b c \,f^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}\)	\(446\)
norman	\(\frac {\frac {\left (b^{4} c^{4} \ln \left (F \right )^{4}-2 \ln \left (F \right )^{3} b^{3} c^{3} e +7 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}-8 e^{3} b c \ln \left (F \right )+6 e^{4}\right ) f^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{b c \ln \left (F \right ) \left (b^{4} c^{4} \ln \left (F \right )^{4}+5 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+4 e^{4}\right )}+\frac {f^{2} \left (b^{4} c^{4} \ln \left (F \right )^{4}+2 \ln \left (F \right )^{3} b^{3} c^{3} e +7 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+8 e^{3} b c \ln \left (F \right )+6 e^{4}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{4}}{\left (b^{4} c^{4} \ln \left (F \right )^{4}+5 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+4 e^{4}\right ) b c \ln \left (F \right )}+\frac {4 f^{2} \left (\ln \left (F \right )^{3} b^{3} c^{3}-b^{2} c^{2} \ln \left (F \right )^{2} e +4 \ln \left (F \right ) b c \,e^{2}-e^{3}\right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{b^{4} c^{4} \ln \left (F \right )^{4}+5 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+4 e^{4}}+\frac {4 \left (\ln \left (F \right )^{3} b^{3} c^{3}+b^{2} c^{2} \ln \left (F \right )^{2} e +4 \ln \left (F \right ) b c \,e^{2}+e^{3}\right ) f^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{3}}{\left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {6 \left (b^{2} c^{2} \ln \left (F \right )^{2}+2 e^{2}\right ) f^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}}{\left (1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}\right )^{2}}\)	\(520\)
orering	\(\text {Expression too large to display}\)	\(969\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.04 \[ \int F^{c (a+b x)} (f+f \sin (d+e x))^2 \, dx=-\frac {{\left (2 \, b^{3} c^{3} e f^{2} \cos \left (e x + d\right ) \log \left (F\right )^{3} + 8 \, b c e^{3} f^{2} \cos \left (e x + d\right ) \log \left (F\right ) - 6 \, e^{4} f^{2} + {\left (b^{4} c^{4} f^{2} \cos \left (e x + d\right )^{2} - 2 \, b^{4} c^{4} f^{2}\right )} \log \left (F\right )^{4} + {\left (b^{2} c^{2} e^{2} f^{2} \cos \left (e x + d\right )^{2} - 8 \, b^{2} c^{2} e^{2} f^{2}\right )} \log \left (F\right )^{2} - 2 \, {\left (b^{4} c^{4} f^{2} \log \left (F\right )^{4} - b^{3} c^{3} e f^{2} \cos \left (e x + d\right ) \log \left (F\right )^{3} + 4 \, b^{2} c^{2} e^{2} f^{2} \log \left (F\right )^{2} - b c e^{3} f^{2} \cos \left (e x + d\right ) \log \left (F\right )\right )} \sin \left (e x + d\right )\right )} F^{b c x + a c}}{b^{5} c^{5} \log \left (F\right )^{5} + 5 \, b^{3} c^{3} e^{2} \log \left (F\right )^{3} + 4 \, b c e^{4} \log \left (F\right )} \] Input:

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

Output:

-(2*b^3*c^3*e*f^2*cos(e*x + d)*log(F)^3 + 8*b*c*e^3*f^2*cos(e*x + d)*log(F 
) - 6*e^4*f^2 + (b^4*c^4*f^2*cos(e*x + d)^2 - 2*b^4*c^4*f^2)*log(F)^4 + (b 
^2*c^2*e^2*f^2*cos(e*x + d)^2 - 8*b^2*c^2*e^2*f^2)*log(F)^2 - 2*(b^4*c^4*f 
^2*log(F)^4 - b^3*c^3*e*f^2*cos(e*x + d)*log(F)^3 + 4*b^2*c^2*e^2*f^2*log( 
F)^2 - b*c*e^3*f^2*cos(e*x + d)*log(F))*sin(e*x + d))*F^(b*c*x + a*c)/(b^5 
*c^5*log(F)^5 + 5*b^3*c^3*e^2*log(F)^3 + 4*b*c*e^4*log(F))
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.82 (sec) , antiderivative size = 2467, normalized size of antiderivative = 10.07 \[ \int F^{c (a+b x)} (f+f \sin (d+e x))^2 \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((x*(f*sin(d) + f)**2, Eq(F, 1) & Eq(b, 0) & Eq(c, 0) & Eq(e, 0)) 
, (f**2*x*sin(d + e*x)**2/2 + f**2*x*cos(d + e*x)**2/2 + f**2*x - f**2*sin 
(d + e*x)*cos(d + e*x)/(2*e) - 2*f**2*cos(d + e*x)/e, Eq(F, 1)), (F**(a*c) 
*(f**2*x*sin(d + e*x)**2/2 + f**2*x*cos(d + e*x)**2/2 + f**2*x - f**2*sin( 
d + e*x)*cos(d + e*x)/(2*e) - 2*f**2*cos(d + e*x)/e), Eq(b, 0)), (f**2*x*s 
in(d + e*x)**2/2 + f**2*x*cos(d + e*x)**2/2 + f**2*x - f**2*sin(d + e*x)*c 
os(d + e*x)/(2*e) - 2*f**2*cos(d + e*x)/e, Eq(c, 0)), (-F**(a*c + b*c*x)*f 
**2*x*sin(I*b*c*x*log(F) - d) + I*F**(a*c + b*c*x)*f**2*x*cos(I*b*c*x*log( 
F) - d) + F**(a*c + b*c*x)*f**2*sin(I*b*c*x*log(F) - d)**2/(3*b*c*log(F)) 
+ 2*I*F**(a*c + b*c*x)*f**2*sin(I*b*c*x*log(F) - d)*cos(I*b*c*x*log(F) - d 
)/(3*b*c*log(F)) + F**(a*c + b*c*x)*f**2*sin(I*b*c*x*log(F) - d)/(b*c*log( 
F)) + 2*F**(a*c + b*c*x)*f**2*cos(I*b*c*x*log(F) - d)**2/(3*b*c*log(F)) - 
2*I*F**(a*c + b*c*x)*f**2*cos(I*b*c*x*log(F) - d)/(b*c*log(F)) + F**(a*c + 
 b*c*x)*f**2/(b*c*log(F)), Eq(e, -I*b*c*log(F))), (F**(a*c + b*c*x)*f**2*x 
*sin(I*b*c*x*log(F)/2 - d)**2/4 - I*F**(a*c + b*c*x)*f**2*x*sin(I*b*c*x*lo 
g(F)/2 - d)*cos(I*b*c*x*log(F)/2 - d)/2 - F**(a*c + b*c*x)*f**2*x*cos(I*b* 
c*x*log(F)/2 - d)**2/4 + 3*I*F**(a*c + b*c*x)*f**2*sin(I*b*c*x*log(F)/2 - 
d)*cos(I*b*c*x*log(F)/2 - d)/(2*b*c*log(F)) - 8*F**(a*c + b*c*x)*f**2*sin( 
I*b*c*x*log(F)/2 - d)/(3*b*c*log(F)) + F**(a*c + b*c*x)*f**2*cos(I*b*c*x*l 
og(F)/2 - d)**2/(b*c*log(F)) + 4*I*F**(a*c + b*c*x)*f**2*cos(I*b*c*x*lo...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 581 vs. \(2 (245) = 490\).

Time = 0.16 (sec) , antiderivative size = 581, normalized size of antiderivative = 2.37 \[ \int F^{c (a+b x)} (f+f \sin (d+e x))^2 \, dx =\text {Too large to display} \] Input:

integrate(F^(c*(b*x+a))*(f+f*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))*f^2/(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) - ((F^(a*c)*b*c*l 
og(F)*sin(d) + F^(a*c)*e*cos(d))*F^(b*c*x)*cos(e*x + 2*d) - (F^(a*c)*b*c*l 
og(F)*sin(d) - F^(a*c)*e*cos(d))*F^(b*c*x)*cos(e*x) - (F^(a*c)*b*c*cos(d)* 
log(F) - F^(a*c)*e*sin(d))*F^(b*c*x)*sin(e*x + 2*d) - (F^(a*c)*b*c*cos(d)* 
log(F) + F^(a*c)*e*sin(d))*F^(b*c*x)*sin(e*x))*f^2/(b^2*c^2*cos(d)^2*log(F 
)^2 + b^2*c^2*log(F)^2*sin(d)^2 + (cos(d)^2 + sin(d)^2)*e^2) + F^(b*c*x + 
a*c)*f^2/(b*c*log(F))
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.20 (sec) , antiderivative size = 1738, normalized size of antiderivative = 7.09 \[ \int F^{c (a+b x)} (f+f \sin (d+e x))^2 \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(2*b*c*f^2*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*sgn(F) - pi*b*c + 4*e)^2) + (pi*b*c*sgn(F) - pi*b*c + 4*e)*f^2*sin(1/2*p 
i*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*f^2*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*sgn(F) - pi*b*c - 4*e)^2) + (pi*b*c*sgn 
(F) - pi*b*c - 4*e)*f^2*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*sg 
n(F) - pi*b*c - 4*e)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + 3*(2*b* 
c*f^2*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)*f^2*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(abs(F)) + a*c*log(abs(F))) + 2*(2*b*c*f^2*log(abs 
(F))*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 + e*x + d)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 2*e)^2) 
 - (pi*b*c*sgn(F) - pi*b*c + 2*e)*f^2*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 + e*x + d)/(4*b^2*c^2*log(abs(F))^2...
 

Mupad [B] (verification not implemented)

Time = 20.86 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.01 \[ \int F^{c (a+b x)} (f+f \sin (d+e x))^2 \, dx=-\frac {F^{a\,c+b\,c\,x}\,f^2\,\left (\frac {b^4\,c^4\,{\ln \left (F\right )}^4\,\cos \left (2\,d+2\,e\,x\right )}{2}-\frac {3\,b^4\,c^4\,{\ln \left (F\right )}^4}{2}-2\,b^4\,c^4\,\sin \left (d+e\,x\right )\,{\ln \left (F\right )}^4-6\,e^4-\frac {15\,b^2\,c^2\,e^2\,{\ln \left (F\right )}^2}{2}+b^3\,c^3\,e\,{\ln \left (F\right )}^3\,\sin \left (2\,d+2\,e\,x\right )-8\,b^2\,c^2\,e^2\,\sin \left (d+e\,x\right )\,{\ln \left (F\right )}^2+b\,c\,e^3\,\ln \left (F\right )\,\sin \left (2\,d+2\,e\,x\right )+\frac {b^2\,c^2\,e^2\,{\ln \left (F\right )}^2\,\cos \left (2\,d+2\,e\,x\right )}{2}+2\,b^3\,c^3\,e\,\cos \left (d+e\,x\right )\,{\ln \left (F\right )}^3+8\,b\,c\,e^3\,\cos \left (d+e\,x\right )\,\ln \left (F\right )\right )}{b\,c\,\ln \left (F\right )\,\left (b^4\,c^4\,{\ln \left (F\right )}^4+5\,b^2\,c^2\,e^2\,{\ln \left (F\right )}^2+4\,e^4\right )} \] Input:

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

Output:

-(F^(a*c + b*c*x)*f^2*((b^4*c^4*log(F)^4*cos(2*d + 2*e*x))/2 - (3*b^4*c^4* 
log(F)^4)/2 - 2*b^4*c^4*sin(d + e*x)*log(F)^4 - 6*e^4 - (15*b^2*c^2*e^2*lo 
g(F)^2)/2 + b^3*c^3*e*log(F)^3*sin(2*d + 2*e*x) - 8*b^2*c^2*e^2*sin(d + e* 
x)*log(F)^2 + b*c*e^3*log(F)*sin(2*d + 2*e*x) + (b^2*c^2*e^2*log(F)^2*cos( 
2*d + 2*e*x))/2 + 2*b^3*c^3*e*cos(d + e*x)*log(F)^3 + 8*b*c*e^3*cos(d + e* 
x)*log(F)))/(b*c*log(F)*(4*e^4 + b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 
))
 

Reduce [F]

\[ \int F^{c (a+b x)} (f+f \sin (d+e x))^2 \, dx=\frac {f^{a c} f^{2} \left (-2 f^{b c x} \cos \left (e x +d \right ) \mathrm {log}\left (f \right ) b c e +2 f^{b c x} \mathrm {log}\left (f \right )^{2} \sin \left (e x +d \right ) b^{2} c^{2}+f^{b c x} \mathrm {log}\left (f \right )^{2} b^{2} c^{2}+f^{b c x} e^{2}+\left (\int f^{b c x} \sin \left (e x +d \right )^{2}d x \right ) \mathrm {log}\left (f \right )^{3} b^{3} c^{3}+\left (\int f^{b c x} \sin \left (e x +d \right )^{2}d x \right ) \mathrm {log}\left (f \right ) b c \,e^{2}\right )}{\mathrm {log}\left (f \right ) b c \left (\mathrm {log}\left (f \right )^{2} b^{2} c^{2}+e^{2}\right )} \] Input:

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

Output:

(f**(a*c)*f**2*( - 2*f**(b*c*x)*cos(d + e*x)*log(f)*b*c*e + 2*f**(b*c*x)*l 
og(f)**2*sin(d + e*x)*b**2*c**2 + f**(b*c*x)*log(f)**2*b**2*c**2 + f**(b*c 
*x)*e**2 + int(f**(b*c*x)*sin(d + e*x)**2,x)*log(f)**3*b**3*c**3 + int(f** 
(b*c*x)*sin(d + e*x)**2,x)*log(f)*b*c*e**2))/(log(f)*b*c*(log(f)**2*b**2*c 
**2 + e**2))