Integrand size = 22, antiderivative size = 60 \[ \int F^{c (a+b x)} \cos (d+e x) \sin (d+e x) \, dx=-\frac {F^{c (a+b x)} (2 e \cos (2 d+2 e x)-b c \log (F) \sin (2 d+2 e x))}{2 \left (4 e^2+b^2 c^2 \log ^2(F)\right )} \] Output:
-1/2*F^(c*(b*x+a))*(2*e*cos(2*e*x+2*d)-b*c*ln(F)*sin(2*e*x+2*d))/(4*e^2+b^ 2*c^2*ln(F)^2)
Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95 \[ \int F^{c (a+b x)} \cos (d+e x) \sin (d+e x) \, dx=\frac {F^{c (a+b x)} (-2 e \cos (2 (d+e x))+b c \log (F) \sin (2 (d+e x)))}{2 \left (4 e^2+b^2 c^2 \log ^2(F)\right )} \] Input:
Integrate[F^(c*(a + b*x))*Cos[d + e*x]*Sin[d + e*x],x]
Output:
(F^(c*(a + b*x))*(-2*e*Cos[2*(d + e*x)] + b*c*Log[F]*Sin[2*(d + e*x)]))/(2 *(4*e^2 + b^2*c^2*Log[F]^2))
Time = 0.27 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.45, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4972, 27, 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.
\(\displaystyle \int \sin (d+e x) \cos (d+e x) F^{c (a+b x)} \, dx\) |
\(\Big \downarrow \) 4972 |
\(\displaystyle \int \frac {1}{2} \sin (2 d+2 e x) F^{c (a+b x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int F^{c (a+b x)} \sin (2 d+2 e x)dx\) |
\(\Big \downarrow \) 4932 |
\(\displaystyle \frac {1}{2} \left (\frac {b c \log (F) \sin (2 d+2 e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}-\frac {2 e \cos (2 d+2 e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}\right )\) |
Input:
Int[F^(c*(a + b*x))*Cos[d + e*x]*Sin[d + e*x],x]
Output:
((-2*e*F^(c*(a + b*x))*Cos[2*d + 2*e*x])/(4*e^2 + b^2*c^2*Log[F]^2) + (b*c *F^(c*(a + b*x))*Log[F]*Sin[2*d + 2*e*x])/(4*e^2 + b^2*c^2*Log[F]^2))/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[Cos[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_ .) + (e_.)*(x_)]^(m_.), x_Symbol] :> Int[ExpandTrigReduce[F^(c*(a + b*x)), Sin[d + e*x]^m*Cos[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] && IGtQ[m, 0] && IGtQ[n, 0]
Time = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.98
method | result | size |
parallelrisch | \(\frac {\left (b c \sin \left (2 e x +2 d \right ) \ln \left (\sqrt {F}\right )-e \cos \left (2 e x +2 d \right )\right ) F^{c \left (b x +a \right )}}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\) | \(59\) |
risch | \(-\frac {e \,F^{c \left (b x +a \right )} \cos \left (2 e x +2 d \right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {c b \ln \left (F \right ) F^{c \left (b x +a \right )} \sin \left (2 e x +2 d \right )}{2 b^{2} c^{2} \ln \left (F \right )^{2}+8 e^{2}}\) | \(85\) |
orering | \(\frac {2 b c \ln \left (F \right ) F^{c \left (b x +a \right )} \cos \left (e x +d \right ) \sin \left (e x +d \right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {F^{c \left (b x +a \right )} b c \ln \left (F \right ) \cos \left (e x +d \right ) \sin \left (e x +d \right )-F^{c \left (b x +a \right )} e \sin \left (e x +d \right )^{2}+F^{c \left (b x +a \right )} \cos \left (e x +d \right )^{2} e}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\) | \(135\) |
norman | \(\frac {-\frac {e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {6 e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 b c \ln \left (F \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {2 b c \ln \left (F \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}\right )^{2}}\) | \(226\) |
Input:
int(F^(c*(b*x+a))*cos(e*x+d)*sin(e*x+d),x,method=_RETURNVERBOSE)
Output:
1/(4*e^2+b^2*c^2*ln(F)^2)*(b*c*sin(2*e*x+2*d)*ln(F^(1/2))-e*cos(2*e*x+2*d) )*F^(c*(b*x+a))
Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} \cos (d+e x) \sin (d+e x) \, dx=\frac {{\left (b c \cos \left (e x + d\right ) \log \left (F\right ) \sin \left (e x + d\right ) - 2 \, e \cos \left (e x + d\right )^{2} + e\right )} F^{b c x + a c}}{b^{2} c^{2} \log \left (F\right )^{2} + 4 \, e^{2}} \] Input:
integrate(F^(c*(b*x+a))*cos(e*x+d)*sin(e*x+d),x, algorithm="fricas")
Output:
(b*c*cos(e*x + d)*log(F)*sin(e*x + d) - 2*e*cos(e*x + d)^2 + e)*F^(b*c*x + a*c)/(b^2*c^2*log(F)^2 + 4*e^2)
Result contains complex when optimal does not.
Time = 1.13 (sec) , antiderivative size = 484, normalized size of antiderivative = 8.07 \[ \int F^{c (a+b x)} \cos (d+e x) \sin (d+e x) \, dx=\begin {cases} x \sin {\left (d \right )} \cos {\left (d \right )} & \text {for}\: F = 1 \wedge e = 0 \\F^{a c} x \sin {\left (d \right )} \cos {\left (d \right )} & \text {for}\: b = 0 \wedge e = 0 \\x \sin {\left (d \right )} \cos {\left (d \right )} & \text {for}\: c = 0 \wedge e = 0 \\- \frac {i F^{a c + b c x} x \sin ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )}}{4} - \frac {F^{a c + b c x} x \sin {\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )} \cos {\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )}}{2} + \frac {i F^{a c + b c x} x \cos ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )}}{4} - \frac {F^{a c + b c x} \sin {\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )} \cos {\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )}}{2 b c \log {\left (F \right )}} & \text {for}\: e = - \frac {i b c \log {\left (F \right )}}{2} \\\frac {i F^{a c + b c x} x \sin ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )}}{4} + \frac {F^{a c + b c x} x \sin {\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )} \cos {\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )}}{2} - \frac {i F^{a c + b c x} x \cos ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )}}{4} + \frac {F^{a c + b c x} \sin {\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )} \cos {\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )}}{2 b c \log {\left (F \right )}} & \text {for}\: e = \frac {i b c \log {\left (F \right )}}{2} \\\frac {F^{a c + b c x} b c \log {\left (F \right )} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{b^{2} c^{2} \log {\left (F \right )}^{2} + 4 e^{2}} + \frac {F^{a c + b c x} e \sin ^{2}{\left (d + e x \right )}}{b^{2} c^{2} \log {\left (F \right )}^{2} + 4 e^{2}} - \frac {F^{a c + b c x} e \cos ^{2}{\left (d + e x \right )}}{b^{2} c^{2} \log {\left (F \right )}^{2} + 4 e^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(F**(c*(b*x+a))*cos(e*x+d)*sin(e*x+d),x)
Output:
Piecewise((x*sin(d)*cos(d), Eq(F, 1) & Eq(e, 0)), (F**(a*c)*x*sin(d)*cos(d ), Eq(b, 0) & Eq(e, 0)), (x*sin(d)*cos(d), Eq(c, 0) & Eq(e, 0)), (-I*F**(a *c + b*c*x)*x*sin(I*b*c*x*log(F)/2 - d)**2/4 - 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 + I*F**(a*c + b*c*x)*x*cos(I *b*c*x*log(F)/2 - d)**2/4 - 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)), Eq(e, -I*b*c*log(F)/2)), (I*F**(a*c + b*c*x)*x*sin(I*b*c*x*log(F)/2 + d)**2/4 + 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 - I*F**(a*c + b*c*x)*x*cos(I*b *c*x*log(F)/2 + d)**2/4 + 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)), Eq(e, I*b*c*log(F)/2)), (F**(a*c + b* c*x)*b*c*log(F)*sin(d + e*x)*cos(d + e*x)/(b**2*c**2*log(F)**2 + 4*e**2) + F**(a*c + b*c*x)*e*sin(d + e*x)**2/(b**2*c**2*log(F)**2 + 4*e**2) - F**(a *c + b*c*x)*e*cos(d + e*x)**2/(b**2*c**2*log(F)**2 + 4*e**2), True))
Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (57) = 114\).
Time = 0.04 (sec) , antiderivative size = 223, normalized size of antiderivative = 3.72 \[ \int F^{c (a+b x)} \cos (d+e x) \sin (d+e x) \, dx=\frac {{\left (F^{a c} b c \log \left (F\right ) \sin \left (2 \, d\right ) - 2 \, F^{a c} e \cos \left (2 \, d\right )\right )} F^{b c x} \cos \left (2 \, e x\right ) - {\left (F^{a c} b c \log \left (F\right ) \sin \left (2 \, d\right ) + 2 \, F^{a c} e \cos \left (2 \, d\right )\right )} F^{b c x} \cos \left (2 \, e x + 4 \, d\right ) + {\left (F^{a c} b c \cos \left (2 \, d\right ) \log \left (F\right ) + 2 \, F^{a c} e \sin \left (2 \, d\right )\right )} F^{b c x} \sin \left (2 \, e x\right ) + {\left (F^{a c} b c \cos \left (2 \, d\right ) \log \left (F\right ) - 2 \, F^{a c} e \sin \left (2 \, d\right )\right )} F^{b c x} \sin \left (2 \, e x + 4 \, d\right )}{4 \, {\left (b^{2} c^{2} \cos \left (2 \, d\right )^{2} \log \left (F\right )^{2} + b^{2} c^{2} \log \left (F\right )^{2} \sin \left (2 \, d\right )^{2} + 4 \, {\left (\cos \left (2 \, d\right )^{2} + \sin \left (2 \, d\right )^{2}\right )} e^{2}\right )}} \] Input:
integrate(F^(c*(b*x+a))*cos(e*x+d)*sin(e*x+d),x, algorithm="maxima")
Output:
1/4*((F^(a*c)*b*c*log(F)*sin(2*d) - 2*F^(a*c)*e*cos(2*d))*F^(b*c*x)*cos(2* e*x) - (F^(a*c)*b*c*log(F)*sin(2*d) + 2*F^(a*c)*e*cos(2*d))*F^(b*c*x)*cos( 2*e*x + 4*d) + (F^(a*c)*b*c*cos(2*d)*log(F) + 2*F^(a*c)*e*sin(2*d))*F^(b*c *x)*sin(2*e*x) + (F^(a*c)*b*c*cos(2*d)*log(F) - 2*F^(a*c)*e*sin(2*d))*F^(b *c*x)*sin(2*e*x + 4*d))/(b^2*c^2*cos(2*d)^2*log(F)^2 + b^2*c^2*log(F)^2*si n(2*d)^2 + 4*(cos(2*d)^2 + sin(2*d)^2)*e^2)
Result contains complex when optimal does not.
Time = 0.16 (sec) , antiderivative size = 641, normalized size of antiderivative = 10.68 \[ \int F^{c (a+b x)} \cos (d+e x) \sin (d+e x) \, dx=\text {Too large to display} \] Input:
integrate(F^(c*(b*x+a))*cos(e*x+d)*sin(e*x+d),x, algorithm="giac")
Output:
1/2*(2*b*c*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 + 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) - (pi*b*c*sgn(F) - pi*b*c + 4*e)*cos(1/2*pi*b*c*x*s gn(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*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 - 2*e*x - 2*d)/(4*b^2*c^2*lo g(abs(F))^2 + (pi*b*c*sgn(F) - pi*b*c - 4*e)^2) - (pi*b*c*sgn(F) - pi*b*c - 4*e)*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)/(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))) - (-I*e^(1/2*I*pi*b*c*x*sg n(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*x*log(abs(F)) + a*c*log(abs(F))) - (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...
Time = 15.84 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.97 \[ \int F^{c (a+b x)} \cos (d+e x) \sin (d+e x) \, dx=-\frac {F^{a\,c+b\,c\,x}\,\left (e\,\cos \left (2\,d+2\,e\,x\right )-\frac {b\,c\,\ln \left (F\right )\,\sin \left (2\,d+2\,e\,x\right )}{2}\right )}{b^2\,c^2\,{\ln \left (F\right )}^2+4\,e^2} \] Input:
int(F^(c*(a + b*x))*cos(d + e*x)*sin(d + e*x),x)
Output:
-(F^(a*c + b*c*x)*(e*cos(2*d + 2*e*x) - (b*c*log(F)*sin(2*d + 2*e*x))/2))/ (4*e^2 + b^2*c^2*log(F)^2)
Time = 0.16 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.15 \[ \int F^{c (a+b x)} \cos (d+e x) \sin (d+e x) \, dx=\frac {f^{b c x +a c} \left (-\cos \left (e x +d \right )^{2} e +\cos \left (e x +d \right ) \mathrm {log}\left (f \right ) \sin \left (e x +d \right ) b c +\sin \left (e x +d \right )^{2} e \right )}{\mathrm {log}\left (f \right )^{2} b^{2} c^{2}+4 e^{2}} \] Input:
int(F^(c*(b*x+a))*cos(e*x+d)*sin(e*x+d),x)
Output:
(f**(a*c + b*c*x)*( - cos(d + e*x)**2*e + cos(d + e*x)*log(f)*sin(d + e*x) *b*c + sin(d + e*x)**2*e))/(log(f)**2*b**2*c**2 + 4*e**2)