Integrand size = 21, antiderivative size = 189 \[ \int f^{a+b x+c x^2} \cos \left (d+f x^2\right ) \, dx=-\frac {e^{-i d+\frac {b^2 \log ^2(f)}{4 i f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {b \log (f)-2 x (i f-c \log (f))}{2 \sqrt {i f-c \log (f)}}\right )}{4 \sqrt {i f-c \log (f)}}+\frac {e^{i d-\frac {b^2 \log ^2(f)}{4 i f+4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {b \log (f)+2 x (i f+c \log (f))}{2 \sqrt {i f+c \log (f)}}\right )}{4 \sqrt {i f+c \log (f)}} \] Output:
-1/4*exp(-I*d+b^2*ln(f)^2/(4*I*f-4*c*ln(f)))*f^a*Pi^(1/2)*erf(1/2*(b*ln(f) -2*x*(I*f-c*ln(f)))/(I*f-c*ln(f))^(1/2))/(I*f-c*ln(f))^(1/2)+1/4*exp(I*d-b ^2*ln(f)^2/(4*I*f+4*c*ln(f)))*f^a*Pi^(1/2)*erfi(1/2*(b*ln(f)+2*x*(I*f+c*ln (f)))/(I*f+c*ln(f))^(1/2))/(I*f+c*ln(f))^(1/2)
Time = 0.67 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.22 \[ \int f^{a+b x+c x^2} \cos \left (d+f x^2\right ) \, dx=-\frac {(-1)^{3/4} e^{\frac {b^2 \log ^2(f)}{4 i f-4 c \log (f)}} f^a \sqrt {\pi } \left (\text {erfi}\left (\frac {(-1)^{3/4} (2 f x+i (b+2 c x) \log (f))}{2 \sqrt {f+i c \log (f)}}\right ) (f-i c \log (f)) \sqrt {f+i c \log (f)} (-i \cos (d)-\sin (d))+e^{\frac {i b^2 f \log ^2(f)}{2 \left (f^2+c^2 \log ^2(f)\right )}} \text {erfi}\left (\frac {\sqrt [4]{-1} (2 f x-i (b+2 c x) \log (f))}{2 \sqrt {f-i c \log (f)}}\right ) \sqrt {f-i c \log (f)} (f+i c \log (f)) (\cos (d)+i \sin (d))\right )}{4 \left (f^2+c^2 \log ^2(f)\right )} \] Input:
Integrate[f^(a + b*x + c*x^2)*Cos[d + f*x^2],x]
Output:
-1/4*((-1)^(3/4)*E^((b^2*Log[f]^2)/((4*I)*f - 4*c*Log[f]))*f^a*Sqrt[Pi]*(E rfi[((-1)^(3/4)*(2*f*x + I*(b + 2*c*x)*Log[f]))/(2*Sqrt[f + I*c*Log[f]])]* (f - I*c*Log[f])*Sqrt[f + I*c*Log[f]]*((-I)*Cos[d] - Sin[d]) + E^(((I/2)*b ^2*f*Log[f]^2)/(f^2 + c^2*Log[f]^2))*Erfi[((-1)^(1/4)*(2*f*x - I*(b + 2*c* x)*Log[f]))/(2*Sqrt[f - I*c*Log[f]])]*Sqrt[f - I*c*Log[f]]*(f + I*c*Log[f] )*(Cos[d] + I*Sin[d])))/(f^2 + c^2*Log[f]^2)
Time = 0.53 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.
\(\displaystyle \int \cos \left (d+f x^2\right ) f^{a+b x+c x^2} \, dx\) |
\(\Big \downarrow \) 4976 |
\(\displaystyle \int \left (\frac {1}{2} e^{-i d-i f x^2} f^{a+b x+c x^2}+\frac {1}{2} e^{i d+i f x^2} f^{a+b x+c x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {\pi } f^a e^{i d-\frac {b^2 \log ^2(f)}{4 c \log (f)+4 i f}} \text {erfi}\left (\frac {b \log (f)+2 x (c \log (f)+i f)}{2 \sqrt {c \log (f)+i f}}\right )}{4 \sqrt {c \log (f)+i f}}-\frac {\sqrt {\pi } f^a e^{\frac {b^2 \log ^2(f)}{-4 c \log (f)+4 i f}-i d} \text {erf}\left (\frac {b \log (f)-2 x (-c \log (f)+i f)}{2 \sqrt {-c \log (f)+i f}}\right )}{4 \sqrt {-c \log (f)+i f}}\) |
Input:
Int[f^(a + b*x + c*x^2)*Cos[d + f*x^2],x]
Output:
-1/4*(E^((-I)*d + (b^2*Log[f]^2)/((4*I)*f - 4*c*Log[f]))*f^a*Sqrt[Pi]*Erf[ (b*Log[f] - 2*x*(I*f - c*Log[f]))/(2*Sqrt[I*f - c*Log[f]])])/Sqrt[I*f - c* Log[f]] + (E^(I*d - (b^2*Log[f]^2)/((4*I)*f + 4*c*Log[f]))*f^a*Sqrt[Pi]*Er fi[(b*Log[f] + 2*x*(I*f + c*Log[f]))/(2*Sqrt[I*f + c*Log[f]])])/(4*Sqrt[I* f + c*Log[f]])
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 = 0.63 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.94
method | result | size |
risch | \(-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}+4 i d \ln \left (f \right ) c +4 d f}{4 \left (-i f +c \ln \left (f \right )\right )}} \operatorname {erf}\left (-x \sqrt {i f -c \ln \left (f \right )}+\frac {\ln \left (f \right ) b}{2 \sqrt {i f -c \ln \left (f \right )}}\right )}{4 \sqrt {i f -c \ln \left (f \right )}}-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}-4 i d \ln \left (f \right ) c +4 d f}{4 \left (i f +c \ln \left (f \right )\right )}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-i f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-c \ln \left (f \right )-i f}}\right )}{4 \sqrt {-c \ln \left (f \right )-i f}}\) | \(178\) |
Input:
int(f^(c*x^2+b*x+a)*cos(f*x^2+d),x,method=_RETURNVERBOSE)
Output:
-1/4*Pi^(1/2)*f^a*exp(-1/4*(ln(f)^2*b^2+4*I*d*ln(f)*c+4*d*f)/(-I*f+c*ln(f) ))/(I*f-c*ln(f))^(1/2)*erf(-x*(I*f-c*ln(f))^(1/2)+1/2*ln(f)*b/(I*f-c*ln(f) )^(1/2))-1/4*Pi^(1/2)*f^a*exp(-1/4*(ln(f)^2*b^2-4*I*d*ln(f)*c+4*d*f)/(I*f+ c*ln(f)))/(-c*ln(f)-I*f)^(1/2)*erf(-(-c*ln(f)-I*f)^(1/2)*x+1/2*ln(f)*b/(-c *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. 311 vs. \(2 (145) = 290\).
Time = 0.09 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.65 \[ \int f^{a+b x+c x^2} \cos \left (d+f x^2\right ) \, dx=-\frac {\sqrt {\pi } {\left (c \log \left (f\right ) - i \, f\right )} \sqrt {-c \log \left (f\right ) - i \, f} \operatorname {erf}\left (\frac {{\left (2 \, f^{2} x - i \, b f \log \left (f\right ) + {\left (2 \, c^{2} x + b c\right )} \log \left (f\right )^{2}\right )} \sqrt {-c \log \left (f\right ) - i \, f}}{2 \, {\left (c^{2} \log \left (f\right )^{2} + f^{2}\right )}}\right ) e^{\left (\frac {4 \, a f^{2} \log \left (f\right ) - {\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )^{3} + 4 i \, d f^{2} + {\left (4 i \, c^{2} d + i \, b^{2} f\right )} \log \left (f\right )^{2}}{4 \, {\left (c^{2} \log \left (f\right )^{2} + f^{2}\right )}}\right )} + \sqrt {\pi } {\left (c \log \left (f\right ) + i \, f\right )} \sqrt {-c \log \left (f\right ) + i \, f} \operatorname {erf}\left (\frac {{\left (2 \, f^{2} x + i \, b f \log \left (f\right ) + {\left (2 \, c^{2} x + b c\right )} \log \left (f\right )^{2}\right )} \sqrt {-c \log \left (f\right ) + i \, f}}{2 \, {\left (c^{2} \log \left (f\right )^{2} + f^{2}\right )}}\right ) e^{\left (\frac {4 \, a f^{2} \log \left (f\right ) - {\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )^{3} - 4 i \, d f^{2} + {\left (-4 i \, c^{2} d - i \, b^{2} f\right )} \log \left (f\right )^{2}}{4 \, {\left (c^{2} \log \left (f\right )^{2} + f^{2}\right )}}\right )}}{4 \, {\left (c^{2} \log \left (f\right )^{2} + f^{2}\right )}} \] Input:
integrate(f^(c*x^2+b*x+a)*cos(f*x^2+d),x, algorithm="fricas")
Output:
-1/4*(sqrt(pi)*(c*log(f) - I*f)*sqrt(-c*log(f) - I*f)*erf(1/2*(2*f^2*x - I *b*f*log(f) + (2*c^2*x + b*c)*log(f)^2)*sqrt(-c*log(f) - I*f)/(c^2*log(f)^ 2 + f^2))*e^(1/4*(4*a*f^2*log(f) - (b^2*c - 4*a*c^2)*log(f)^3 + 4*I*d*f^2 + (4*I*c^2*d + I*b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)) + sqrt(pi)*(c*log( f) + I*f)*sqrt(-c*log(f) + I*f)*erf(1/2*(2*f^2*x + I*b*f*log(f) + (2*c^2*x + b*c)*log(f)^2)*sqrt(-c*log(f) + I*f)/(c^2*log(f)^2 + f^2))*e^(1/4*(4*a* f^2*log(f) - (b^2*c - 4*a*c^2)*log(f)^3 - 4*I*d*f^2 + (-4*I*c^2*d - I*b^2* f)*log(f)^2)/(c^2*log(f)^2 + f^2)))/(c^2*log(f)^2 + f^2)
\[ \int f^{a+b x+c x^2} \cos \left (d+f x^2\right ) \, dx=\int f^{a + b x + c x^{2}} \cos {\left (d + f x^{2} \right )}\, dx \] Input:
integrate(f**(c*x**2+b*x+a)*cos(f*x**2+d),x)
Output:
Integral(f**(a + b*x + c*x**2)*cos(d + f*x**2), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 648 vs. \(2 (145) = 290\).
Time = 0.05 (sec) , antiderivative size = 648, normalized size of antiderivative = 3.43 \[ \int f^{a+b x+c x^2} \cos \left (d+f x^2\right ) \, dx =\text {Too large to display} \] Input:
integrate(f^(c*x^2+b*x+a)*cos(f*x^2+d),x, algorithm="maxima")
Output:
1/8*(sqrt(pi)*sqrt(2*c^2*log(f)^2 + 2*f^2)*((I*f^a*cos(1/4*(4*d*f^2 + (4*c ^2*d + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)) + f^a*sin(1/4*(4*d*f^2 + (4* c^2*d + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)))*erf(1/2*(2*(c*log(f) - I*f )*x + b*log(f))/sqrt(-c*log(f) + I*f)) + (-I*f^a*cos(1/4*(4*d*f^2 + (4*c^2 *d + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)) + f^a*sin(1/4*(4*d*f^2 + (4*c^ 2*d + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)))*erf(1/2*(2*(c*log(f) + I*f)* x + b*log(f))/sqrt(-c*log(f) - I*f)))*sqrt(c*log(f) + sqrt(c^2*log(f)^2 + f^2)) - sqrt(pi)*sqrt(2*c^2*log(f)^2 + 2*f^2)*((f^a*cos(1/4*(4*d*f^2 + (4* c^2*d + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)) - I*f^a*sin(1/4*(4*d*f^2 + (4*c^2*d + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)))*erf(1/2*(2*(c*log(f) - I*f)*x + b*log(f))/sqrt(-c*log(f) + I*f)) + (f^a*cos(1/4*(4*d*f^2 + (4*c^2 *d + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)) + I*f^a*sin(1/4*(4*d*f^2 + (4* c^2*d + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)))*erf(1/2*(2*(c*log(f) + I*f )*x + b*log(f))/sqrt(-c*log(f) - I*f)))*sqrt(-c*log(f) + sqrt(c^2*log(f)^2 + f^2)))/(c^2*e^(1/4*b^2*c*log(f)^3/(c^2*log(f)^2 + f^2))*log(f)^2 + f^2* e^(1/4*b^2*c*log(f)^3/(c^2*log(f)^2 + f^2)))
\[ \int f^{a+b x+c x^2} \cos \left (d+f x^2\right ) \, dx=\int { f^{c x^{2} + b x + a} \cos \left (f x^{2} + d\right ) \,d x } \] Input:
integrate(f^(c*x^2+b*x+a)*cos(f*x^2+d),x, algorithm="giac")
Output:
integrate(f^(c*x^2 + b*x + a)*cos(f*x^2 + d), x)
Timed out. \[ \int f^{a+b x+c x^2} \cos \left (d+f x^2\right ) \, dx=\int f^{c\,x^2+b\,x+a}\,\cos \left (f\,x^2+d\right ) \,d x \] Input:
int(f^(a + b*x + c*x^2)*cos(d + f*x^2),x)
Output:
int(f^(a + b*x + c*x^2)*cos(d + f*x^2), x)
\[ \int f^{a+b x+c x^2} \cos \left (d+f x^2\right ) \, dx=f^{a} \left (\int f^{c \,x^{2}+b x} \cos \left (f \,x^{2}+d \right )d x \right ) \] Input:
int(f^(c*x^2+b*x+a)*cos(f*x^2+d),x)
Output:
f**a*int(f**(b*x + c*x**2)*cos(d + f*x**2),x)