Integrand size = 16, antiderivative size = 72 \[ \int F^{c (a+b x)} \cos (d+e x) \, dx=\frac {b c F^{c (a+b x)} \cos (d+e x) \log (F)}{e^2+b^2 c^2 \log ^2(F)}+\frac {e F^{c (a+b x)} \sin (d+e x)}{e^2+b^2 c^2 \log ^2(F)} \] Output:
b*c*F^(c*(b*x+a))*cos(e*x+d)*ln(F)/(e^2+b^2*c^2*ln(F)^2)+e*F^(c*(b*x+a))*s in(e*x+d)/(e^2+b^2*c^2*ln(F)^2)
Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.65 \[ \int F^{c (a+b x)} \cos (d+e x) \, dx=\frac {F^{c (a+b x)} (b c \cos (d+e x) \log (F)+e \sin (d+e x))}{e^2+b^2 c^2 \log ^2(F)} \] Input:
Integrate[F^(c*(a + b*x))*Cos[d + e*x],x]
Output:
(F^(c*(a + b*x))*(b*c*Cos[d + e*x]*Log[F] + e*Sin[d + e*x]))/(e^2 + b^2*c^ 2*Log[F]^2)
Time = 0.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4933}
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.
\(\displaystyle \int \cos (d+e x) F^{c (a+b x)} \, dx\) |
\(\Big \downarrow \) 4933 |
\(\displaystyle \frac {e \sin (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+e^2}+\frac {b c \log (F) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+e^2}\) |
Input:
Int[F^(c*(a + b*x))*Cos[d + e*x],x]
Output:
(b*c*F^(c*(a + b*x))*Cos[d + e*x]*Log[F])/(e^2 + b^2*c^2*Log[F]^2) + (e*F^ (c*(a + b*x))*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2)
Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(Cos[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x ] + Simp[e*F^(c*(a + b*x))*(Sin[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]
Time = 0.49 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.67
method | result | size |
parallelrisch | \(\frac {F^{c \left (b x +a \right )} \left (\ln \left (F \right ) \cos \left (e x +d \right ) b c +e \sin \left (e x +d \right )\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\) | \(48\) |
risch | \(\frac {b c \,F^{c \left (b x +a \right )} \cos \left (e x +d \right ) \ln \left (F \right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {e \,F^{c \left (b x +a \right )} \sin \left (e x +d \right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\) | \(73\) |
orering | \(\frac {2 b c \,F^{c \left (b x +a \right )} \cos \left (e x +d \right ) \ln \left (F \right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {F^{c \left (b x +a \right )} b c \ln \left (F \right ) \cos \left (e x +d \right )-F^{c \left (b x +a \right )} e \sin \left (e x +d \right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\) | \(98\) |
norman | \(\frac {\frac {b c \ln \left (F \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 e \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}-\frac {b c \ln \left (F \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )} \tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}{e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}}{1+\tan \left (\frac {d}{2}+\frac {e x}{2}\right )^{2}}\) | \(133\) |
Input:
int(F^(c*(b*x+a))*cos(e*x+d),x,method=_RETURNVERBOSE)
Output:
F^(c*(b*x+a))*(ln(F)*cos(e*x+d)*b*c+e*sin(e*x+d))/(e^2+b^2*c^2*ln(F)^2)
Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.67 \[ \int F^{c (a+b x)} \cos (d+e x) \, dx=\frac {{\left (b c \cos \left (e x + d\right ) \log \left (F\right ) + e \sin \left (e x + d\right )\right )} F^{b c x + a c}}{b^{2} c^{2} \log \left (F\right )^{2} + e^{2}} \] Input:
integrate(F^(c*(b*x+a))*cos(e*x+d),x, algorithm="fricas")
Output:
(b*c*cos(e*x + d)*log(F) + e*sin(e*x + d))*F^(b*c*x + a*c)/(b^2*c^2*log(F) ^2 + e^2)
Result contains complex when optimal does not.
Time = 0.64 (sec) , antiderivative size = 316, normalized size of antiderivative = 4.39 \[ \int F^{c (a+b x)} \cos (d+e x) \, dx=\begin {cases} x \cos {\left (d \right )} & \text {for}\: F = 1 \wedge e = 0 \\F^{a c} x \cos {\left (d \right )} & \text {for}\: b = 0 \wedge e = 0 \\x \cos {\left (d \right )} & \text {for}\: c = 0 \wedge e = 0 \\\frac {i F^{a c + b c x} x \sin {\left (i b c x \log {\left (F \right )} - d \right )}}{2} + \frac {F^{a c + b c x} x \cos {\left (i b c x \log {\left (F \right )} - d \right )}}{2} - \frac {i F^{a c + b c x} \sin {\left (i b c x \log {\left (F \right )} - d \right )}}{b c \log {\left (F \right )}} - \frac {F^{a c + b c x} \cos {\left (i b c x \log {\left (F \right )} - d \right )}}{2 b c \log {\left (F \right )}} & \text {for}\: e = - i b c \log {\left (F \right )} \\\frac {i F^{a c + b c x} x \sin {\left (i b c x \log {\left (F \right )} + d \right )}}{2} + \frac {F^{a c + b c x} x \cos {\left (i b c x \log {\left (F \right )} + d \right )}}{2} - \frac {i F^{a c + b c x} \sin {\left (i b c x \log {\left (F \right )} + d \right )}}{2 b c \log {\left (F \right )}} & \text {for}\: e = i b c \log {\left (F \right )} \\\frac {F^{a c + b c x} b c \log {\left (F \right )} \cos {\left (d + e x \right )}}{b^{2} c^{2} \log {\left (F \right )}^{2} + e^{2}} + \frac {F^{a c + b c x} e \sin {\left (d + e x \right )}}{b^{2} c^{2} \log {\left (F \right )}^{2} + e^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(F**(c*(b*x+a))*cos(e*x+d),x)
Output:
Piecewise((x*cos(d), Eq(F, 1) & Eq(e, 0)), (F**(a*c)*x*cos(d), Eq(b, 0) & Eq(e, 0)), (x*cos(d), Eq(c, 0) & Eq(e, 0)), (I*F**(a*c + b*c*x)*x*sin(I*b* c*x*log(F) - d)/2 + F**(a*c + b*c*x)*x*cos(I*b*c*x*log(F) - d)/2 - I*F**(a *c + b*c*x)*sin(I*b*c*x*log(F) - d)/(b*c*log(F)) - F**(a*c + b*c*x)*cos(I* b*c*x*log(F) - d)/(2*b*c*log(F)), Eq(e, -I*b*c*log(F))), (I*F**(a*c + b*c* x)*x*sin(I*b*c*x*log(F) + d)/2 + F**(a*c + b*c*x)*x*cos(I*b*c*x*log(F) + d )/2 - I*F**(a*c + b*c*x)*sin(I*b*c*x*log(F) + d)/(2*b*c*log(F)), Eq(e, I*b *c*log(F))), (F**(a*c + b*c*x)*b*c*log(F)*cos(d + e*x)/(b**2*c**2*log(F)** 2 + e**2) + F**(a*c + b*c*x)*e*sin(d + e*x)/(b**2*c**2*log(F)**2 + e**2), True))
Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (72) = 144\).
Time = 0.05 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.67 \[ \int F^{c (a+b x)} \cos (d+e x) \, dx=\frac {{\left (F^{a c} b c \cos \left (d\right ) \log \left (F\right ) - F^{a c} e \sin \left (d\right )\right )} F^{b c x} \cos \left (e x + 2 \, d\right ) + {\left (F^{a c} b c \cos \left (d\right ) \log \left (F\right ) + F^{a c} e \sin \left (d\right )\right )} F^{b c x} \cos \left (e x\right ) + {\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) + F^{a c} e \cos \left (d\right )\right )} F^{b c x} \sin \left (e x + 2 \, d\right ) - {\left (F^{a c} b c \log \left (F\right ) \sin \left (d\right ) - F^{a c} e \cos \left (d\right )\right )} F^{b c x} \sin \left (e x\right )}{2 \, {\left (b^{2} c^{2} \cos \left (d\right )^{2} \log \left (F\right )^{2} + b^{2} c^{2} \log \left (F\right )^{2} \sin \left (d\right )^{2} + {\left (\cos \left (d\right )^{2} + \sin \left (d\right )^{2}\right )} e^{2}\right )}} \] Input:
integrate(F^(c*(b*x+a))*cos(e*x+d),x, algorithm="maxima")
Output:
1/2*((F^(a*c)*b*c*cos(d)*log(F) - F^(a*c)*e*sin(d))*F^(b*c*x)*cos(e*x + 2* d) + (F^(a*c)*b*c*cos(d)*log(F) + F^(a*c)*e*sin(d))*F^(b*c*x)*cos(e*x) + ( F^(a*c)*b*c*log(F)*sin(d) + F^(a*c)*e*cos(d))*F^(b*c*x)*sin(e*x + 2*d) - ( F^(a*c)*b*c*log(F)*sin(d) - F^(a*c)*e*cos(d))*F^(b*c*x)*sin(e*x))/(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 )
Result contains complex when optimal does not.
Time = 0.14 (sec) , antiderivative size = 631, normalized size of antiderivative = 8.76 \[ \int F^{c (a+b x)} \cos (d+e x) \, dx=\text {Too large to display} \] Input:
integrate(F^(c*(b*x+a))*cos(e*x+d),x, algorithm="giac")
Output:
(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 + e*x + d)*log(abs(F))/(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)*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(ab s(F))^2 + (pi*b*c*sgn(F) - pi*b*c + 2*e)^2))*e^(b*c*x*log(abs(F)) + a*c*lo g(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*sg n(F) - 1/2*pi*a*c - e*x - d)*log(abs(F))/(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)*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))*e^(b*c*x*log(abs (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 + I*e*x + I*d)/(2*I*pi*b*c*sgn(F) - 2* I*pi*b*c + 4*b*c*log(abs(F)) + 4*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 - I*e*x - I*d)/(-2*I*pi*b* c*sgn(F) + 2*I*pi*b*c + 4*b*c*log(abs(F)) - 4*I*e))*e^(b*c*x*log(abs(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 - I*e*x - I*d)/(2*I*pi*b*c*sgn(F) - 2*I*pi*b *c + 4*b*c*log(abs(F)) - 4*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 + I*e*x + I*d)/(-2*I*pi*b*c*sgn( F) + 2*I*pi*b*c + 4*b*c*log(abs(F)) + 4*I*e))*e^(b*c*x*log(abs(F)) + a*...
Time = 18.94 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.67 \[ \int F^{c (a+b x)} \cos (d+e x) \, dx=\frac {F^{a\,c+b\,c\,x}\,\left (e\,\sin \left (d+e\,x\right )+b\,c\,\cos \left (d+e\,x\right )\,\ln \left (F\right )\right )}{b^2\,c^2\,{\ln \left (F\right )}^2+e^2} \] Input:
int(F^(c*(a + b*x))*cos(d + e*x),x)
Output:
(F^(a*c + b*c*x)*(e*sin(d + e*x) + b*c*cos(d + e*x)*log(F)))/(e^2 + b^2*c^ 2*log(F)^2)
Time = 0.15 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.67 \[ \int F^{c (a+b x)} \cos (d+e x) \, dx=\frac {f^{b c x +a c} \left (\cos \left (e x +d \right ) \mathrm {log}\left (f \right ) b c +\sin \left (e x +d \right ) e \right )}{\mathrm {log}\left (f \right )^{2} b^{2} c^{2}+e^{2}} \] Input:
int(F^(c*(b*x+a))*cos(e*x+d),x)
Output:
(f**(a*c + b*c*x)*(cos(d + e*x)*log(f)*b*c + sin(d + e*x)*e))/(log(f)**2*b **2*c**2 + e**2)