Integrand size = 19, antiderivative size = 162 \[ \int f^{a+b x} \cos \left (d+e x+f x^2\right ) \, dx=-\frac {1}{4} \sqrt [4]{-1} e^{\frac {1}{4} i \left (4 d+\frac {(i e+b \log (f))^2}{f}\right )} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erf}\left (\frac {\sqrt [4]{-1} (i e+2 i f x+b \log (f))}{2 \sqrt {f}}\right )-\frac {1}{4} \sqrt [4]{-1} e^{-i d+\frac {i (e+i b \log (f))^2}{4 f}} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt [4]{-1} (i e+2 i f x-b \log (f))}{2 \sqrt {f}}\right ) \] Output:
-1/4*(-1)^(1/4)*exp(1/4*I*(4*d+(I*e+b*ln(f))^2/f))*f^(-1/2+a)*Pi^(1/2)*erf (1/2*(-1)^(1/4)*(I*e+2*I*f*x+b*ln(f))/f^(1/2))-1/4*(-1)^(1/4)*exp(-I*d+1/4 *I*(e+I*b*ln(f))^2/f)*f^(-1/2+a)*Pi^(1/2)*erfi(1/2*(-1)^(1/4)*(I*e+2*I*f*x -b*ln(f))/f^(1/2))
Time = 0.25 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.01 \[ \int f^{a+b x} \cos \left (d+e x+f x^2\right ) \, dx=\frac {1}{4} \sqrt [4]{-1} e^{-\frac {i \left (e^2+b^2 \log ^2(f)\right )}{4 f}} f^{a-\frac {b e+f}{2 f}} \sqrt {\pi } \left (-e^{\frac {i e^2}{2 f}} \text {erfi}\left (\frac {(-1)^{3/4} (e+2 f x+i b \log (f))}{2 \sqrt {f}}\right ) (\cos (d)-i \sin (d))+e^{\frac {i b^2 \log ^2(f)}{2 f}} \text {erfi}\left (\frac {\sqrt [4]{-1} (e+2 f x-i b \log (f))}{2 \sqrt {f}}\right ) (-i \cos (d)+\sin (d))\right ) \] Input:
Integrate[f^(a + b*x)*Cos[d + e*x + f*x^2],x]
Output:
((-1)^(1/4)*f^(a - (b*e + f)/(2*f))*Sqrt[Pi]*(-(E^(((I/2)*e^2)/f)*Erfi[((- 1)^(3/4)*(e + 2*f*x + I*b*Log[f]))/(2*Sqrt[f])]*(Cos[d] - I*Sin[d])) + E^( ((I/2)*b^2*Log[f]^2)/f)*Erfi[((-1)^(1/4)*(e + 2*f*x - I*b*Log[f]))/(2*Sqrt [f])]*((-I)*Cos[d] + Sin[d])))/(4*E^(((I/4)*(e^2 + b^2*Log[f]^2))/f))
Time = 0.45 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {4976, 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.
\(\displaystyle \int f^{a+b x} \cos \left (d+e x+f x^2\right ) \, dx\) |
\(\Big \downarrow \) 4976 |
\(\displaystyle \int \left (\frac {1}{2} f^{a+b x} e^{-i d-i e x-i f x^2}+\frac {1}{2} f^{a+b x} e^{i d+i e x+i f x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{4} \sqrt [4]{-1} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {1}{4} i \left (4 d+\frac {(b \log (f)+i e)^2}{f}\right )} \text {erf}\left (\frac {\sqrt [4]{-1} (b \log (f)+i e+2 i f x)}{2 \sqrt {f}}\right )-\frac {1}{4} \sqrt [4]{-1} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {i (e+i b \log (f))^2}{4 f}-i d} \text {erfi}\left (\frac {\sqrt [4]{-1} (-b \log (f)+i e+2 i f x)}{2 \sqrt {f}}\right )\) |
Input:
Int[f^(a + b*x)*Cos[d + e*x + f*x^2],x]
Output:
-1/4*((-1)^(1/4)*E^((I/4)*(4*d + (I*e + b*Log[f])^2/f))*f^(-1/2 + a)*Sqrt[ Pi]*Erf[((-1)^(1/4)*(I*e + (2*I)*f*x + b*Log[f]))/(2*Sqrt[f])]) - ((-1)^(1 /4)*E^((-I)*d + ((I/4)*(e + I*b*Log[f])^2)/f)*f^(-1/2 + a)*Sqrt[Pi]*Erfi[( (-1)^(1/4)*(I*e + (2*I)*f*x - b*Log[f]))/(2*Sqrt[f])])/4
Int[Cos[v_]^(n_.)*(F_)^(u_), x_Symbol] :> Int[ExpandTrigToExp[F^u, Cos[v]^n , x], x] /; FreeQ[F, x] && (LinearQ[u, x] || PolyQ[u, x, 2]) && (LinearQ[v, x] || PolyQ[v, x, 2]) && IGtQ[n, 0]
Time = 0.80 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.95
method | result | size |
risch | \(-\frac {\sqrt {\pi }\, f^{a} f^{-\frac {b e}{2 f}} {\mathrm e}^{-\frac {i \left (\ln \left (f \right )^{2} b^{2}+4 d f -e^{2}\right )}{4 f}} \operatorname {erf}\left (-\sqrt {i f}\, x +\frac {b \ln \left (f \right )-i e}{2 \sqrt {i f}}\right )}{4 \sqrt {i f}}-\frac {\sqrt {\pi }\, f^{a} f^{-\frac {b e}{2 f}} {\mathrm e}^{\frac {i \left (\ln \left (f \right )^{2} b^{2}+4 d f -e^{2}\right )}{4 f}} \operatorname {erf}\left (-\sqrt {-i f}\, x +\frac {i e +b \ln \left (f \right )}{2 \sqrt {-i f}}\right )}{4 \sqrt {-i f}}\) | \(154\) |
Input:
int(f^(b*x+a)*cos(f*x^2+e*x+d),x,method=_RETURNVERBOSE)
Output:
-1/4*Pi^(1/2)*f^a*f^(-1/2/f*b*e)*exp(-1/4*I*(ln(f)^2*b^2+4*d*f-e^2)/f)/(I* f)^(1/2)*erf(-(I*f)^(1/2)*x+1/2*(b*ln(f)-I*e)/(I*f)^(1/2))-1/4*Pi^(1/2)*f^ a*f^(-1/2/f*b*e)*exp(1/4*I*(ln(f)^2*b^2+4*d*f-e^2)/f)/(-I*f)^(1/2)*erf(-(- I*f)^(1/2)*x+1/2*(I*e+b*ln(f))/(-I*f)^(1/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (109) = 218\).
Time = 0.08 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.93 \[ \int f^{a+b x} \cos \left (d+e x+f x^2\right ) \, dx=\frac {\sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \left (f\right )^{2} + i \, e^{2} - 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{4 \, f}\right )} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, f x + i \, b \log \left (f\right ) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) - \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \left (f\right )^{2} - i \, e^{2} + 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{4 \, f}\right )} \operatorname {C}\left (-\frac {\sqrt {2} {\left (2 \, f x - i \, b \log \left (f\right ) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) - i \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \left (f\right )^{2} + i \, e^{2} - 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{4 \, f}\right )} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, f x + i \, b \log \left (f\right ) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) - i \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \left (f\right )^{2} - i \, e^{2} + 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{4 \, f}\right )} \operatorname {S}\left (-\frac {\sqrt {2} {\left (2 \, f x - i \, b \log \left (f\right ) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right )}{4 \, f} \] Input:
integrate(f^(b*x+a)*cos(f*x^2+e*x+d),x, algorithm="fricas")
Output:
1/4*(sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(-I*b^2*log(f)^2 + I*e^2 - 4*I*d*f - 2*( b*e - 2*a*f)*log(f))/f)*fresnel_cos(1/2*sqrt(2)*(2*f*x + I*b*log(f) + e)*s qrt(f/pi)/f) - sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(I*b^2*log(f)^2 - I*e^2 + 4*I* d*f - 2*(b*e - 2*a*f)*log(f))/f)*fresnel_cos(-1/2*sqrt(2)*(2*f*x - I*b*log (f) + e)*sqrt(f/pi)/f) - I*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(-I*b^2*log(f)^2 + I*e^2 - 4*I*d*f - 2*(b*e - 2*a*f)*log(f))/f)*fresnel_sin(1/2*sqrt(2)*(2*f *x + I*b*log(f) + e)*sqrt(f/pi)/f) - I*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(I*b^2 *log(f)^2 - I*e^2 + 4*I*d*f - 2*(b*e - 2*a*f)*log(f))/f)*fresnel_sin(-1/2* sqrt(2)*(2*f*x - I*b*log(f) + e)*sqrt(f/pi)/f))/f
\[ \int f^{a+b x} \cos \left (d+e x+f x^2\right ) \, dx=\int f^{a + b x} \cos {\left (d + e x + f x^{2} \right )}\, dx \] Input:
integrate(f**(b*x+a)*cos(f*x**2+e*x+d),x)
Output:
Integral(f**(a + b*x)*cos(d + e*x + f*x**2), x)
Time = 0.04 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.17 \[ \int f^{a+b x} \cos \left (d+e x+f x^2\right ) \, dx=-\frac {\sqrt {2} \sqrt {\pi } {\left ({\left (-\left (i - 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} - e^{2} + 4 \, d f}{4 \, f}\right ) - \left (i + 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} - e^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (2 i \, f x - b \log \left (f\right ) + i \, e\right )} \sqrt {i \, f}}{2 \, f}\right ) + {\left (\left (i + 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} - e^{2} + 4 \, d f}{4 \, f}\right ) + \left (i - 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} - e^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (2 i \, f x + b \log \left (f\right ) + i \, e\right )} \sqrt {-i \, f}}{2 \, f}\right )\right )}}{8 \, \sqrt {f} f^{\frac {b e}{2 \, f}}} \] Input:
integrate(f^(b*x+a)*cos(f*x^2+e*x+d),x, algorithm="maxima")
Output:
-1/8*sqrt(2)*sqrt(pi)*((-(I - 1)*f^a*cos(1/4*(b^2*log(f)^2 - e^2 + 4*d*f)/ f) - (I + 1)*f^a*sin(1/4*(b^2*log(f)^2 - e^2 + 4*d*f)/f))*erf(1/2*I*(2*I*f *x - b*log(f) + I*e)*sqrt(I*f)/f) + ((I + 1)*f^a*cos(1/4*(b^2*log(f)^2 - e ^2 + 4*d*f)/f) + (I - 1)*f^a*sin(1/4*(b^2*log(f)^2 - e^2 + 4*d*f)/f))*erf( 1/2*I*(2*I*f*x + b*log(f) + I*e)*sqrt(-I*f)/f))/(sqrt(f)*f^(1/2*b*e/f))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 378 vs. \(2 (109) = 218\).
Time = 0.19 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.33 \[ \int f^{a+b x} \cos \left (d+e x+f x^2\right ) \, dx =\text {Too large to display} \] Input:
integrate(f^(b*x+a)*cos(f*x^2+e*x+d),x, algorithm="giac")
Output:
-1/4*sqrt(2)*sqrt(pi)*erf(-1/8*sqrt(2)*(4*x - (pi*b*sgn(f) - pi*b + 2*I*b* log(abs(f)) - 2*e)/f)*(-I*f/abs(f) + 1)*sqrt(abs(f)))*e^(1/8*I*pi^2*b^2*sg n(f)/f + 1/4*pi*b^2*log(abs(f))*sgn(f)/f - 1/8*I*pi^2*b^2/f - 1/4*pi*b^2*l og(abs(f))/f + 1/4*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/4*I*pi*b* e*sgn(f)/f + 1/2*I*pi*a - 1/4*I*pi*b*e/f + a*log(abs(f)) - 1/2*b*e*log(abs (f))/f + I*d - 1/4*I*e^2/f)/((-I*f/abs(f) + 1)*sqrt(abs(f))) - 1/4*sqrt(2) *sqrt(pi)*erf(-1/8*sqrt(2)*(4*x + (pi*b*sgn(f) - pi*b + 2*I*b*log(abs(f)) + 2*e)/f)*(I*f/abs(f) + 1)*sqrt(abs(f)))*e^(-1/8*I*pi^2*b^2*sgn(f)/f - 1/4 *pi*b^2*log(abs(f))*sgn(f)/f + 1/8*I*pi^2*b^2/f + 1/4*pi*b^2*log(abs(f))/f - 1/4*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/4*I*pi*b*e*sgn(f)/f + 1/2*I*pi*a - 1/4*I*pi*b*e/f + a*log(abs(f)) - 1/2*b*e*log(abs(f))/f - I*d + 1/4*I*e^2/f)/((I*f/abs(f) + 1)*sqrt(abs(f)))
Timed out. \[ \int f^{a+b x} \cos \left (d+e x+f x^2\right ) \, dx=\int f^{a+b\,x}\,\cos \left (f\,x^2+e\,x+d\right ) \,d x \] Input:
int(f^(a + b*x)*cos(d + e*x + f*x^2),x)
Output:
int(f^(a + b*x)*cos(d + e*x + f*x^2), x)
\[ \int f^{a+b x} \cos \left (d+e x+f x^2\right ) \, dx=f^{a} \left (\int f^{b x} \cos \left (f \,x^{2}+e x +d \right )d x \right ) \] Input:
int(f^(b*x+a)*cos(f*x^2+e*x+d),x)
Output:
f**a*int(f**(b*x)*cos(d + e*x + f*x**2),x)