Integrand size = 21, antiderivative size = 187 \[ \int f^{a+c x^2} \sin \left (d+e x+f x^2\right ) \, dx=\frac {i e^{-i d-\frac {e^2}{4 i f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {i e+2 x (i f-c \log (f))}{2 \sqrt {i f-c \log (f)}}\right )}{4 \sqrt {i f-c \log (f)}}-\frac {i e^{i d+\frac {e^2}{4 i f+4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e+2 x (i f+c \log (f))}{2 \sqrt {i f+c \log (f)}}\right )}{4 \sqrt {i f+c \log (f)}} \] Output:
1/4*I*exp(-I*d-e^2/(4*I*f-4*c*ln(f)))*f^a*Pi^(1/2)*erf(1/2*(I*e+2*x*(I*f-c *ln(f)))/(I*f-c*ln(f))^(1/2))/(I*f-c*ln(f))^(1/2)-1/4*I*exp(I*d+e^2/(4*I*f +4*c*ln(f)))*f^a*Pi^(1/2)*erfi(1/2*(I*e+2*x*(I*f+c*ln(f)))/(I*f+c*ln(f))^( 1/2))/(I*f+c*ln(f))^(1/2)
Time = 0.65 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.16 \[ \int f^{a+c x^2} \sin \left (d+e x+f x^2\right ) \, dx=-\frac {(-1)^{3/4} e^{\frac {e^2}{4 i f+4 c \log (f)}} f^a \sqrt {\pi } \left (e^{\frac {i e^2 f}{2 \left (f^2+c^2 \log ^2(f)\right )}} \text {erfi}\left (\frac {(-1)^{3/4} (e+2 f x+2 i c x \log (f))}{2 \sqrt {f+i c \log (f)}}\right ) (f-i c \log (f)) \sqrt {f+i c \log (f)} (\cos (d)-i \sin (d))+\text {erfi}\left (\frac {\sqrt [4]{-1} (e+2 f x-2 i c x \log (f))}{2 \sqrt {f-i c \log (f)}}\right ) \sqrt {f-i c \log (f)} (-i f+c \log (f)) (\cos (d)+i \sin (d))\right )}{4 \left (f^2+c^2 \log ^2(f)\right )} \] Input:
Integrate[f^(a + c*x^2)*Sin[d + e*x + f*x^2],x]
Output:
-1/4*((-1)^(3/4)*E^(e^2/((4*I)*f + 4*c*Log[f]))*f^a*Sqrt[Pi]*(E^(((I/2)*e^ 2*f)/(f^2 + c^2*Log[f]^2))*Erfi[((-1)^(3/4)*(e + 2*f*x + (2*I)*c*x*Log[f]) )/(2*Sqrt[f + I*c*Log[f]])]*(f - I*c*Log[f])*Sqrt[f + I*c*Log[f]]*(Cos[d] - I*Sin[d]) + Erfi[((-1)^(1/4)*(e + 2*f*x - (2*I)*c*x*Log[f]))/(2*Sqrt[f - I*c*Log[f]])]*Sqrt[f - I*c*Log[f]]*((-I)*f + c*Log[f])*(Cos[d] + I*Sin[d] )))/(f^2 + c^2*Log[f]^2)
Time = 0.54 (sec) , antiderivative size = 187, 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 = {4975, 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.
\(\displaystyle \int f^{a+c x^2} \sin \left (d+e x+f x^2\right ) \, dx\) |
\(\Big \downarrow \) 4975 |
\(\displaystyle \int \left (\frac {1}{2} i f^{a+c x^2} e^{-i d-i e x-i f x^2}-\frac {1}{2} i f^{a+c x^2} e^{i d+i e x+i f x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i \sqrt {\pi } f^a e^{-\frac {e^2}{-4 c \log (f)+4 i f}-i d} \text {erf}\left (\frac {2 x (-c \log (f)+i f)+i e}{2 \sqrt {-c \log (f)+i f}}\right )}{4 \sqrt {-c \log (f)+i f}}-\frac {i \sqrt {\pi } f^a e^{\frac {e^2}{4 c \log (f)+4 i f}+i d} \text {erfi}\left (\frac {2 x (c \log (f)+i f)+i e}{2 \sqrt {c \log (f)+i f}}\right )}{4 \sqrt {c \log (f)+i f}}\) |
Input:
Int[f^(a + c*x^2)*Sin[d + e*x + f*x^2],x]
Output:
((I/4)*E^((-I)*d - e^2/((4*I)*f - 4*c*Log[f]))*f^a*Sqrt[Pi]*Erf[(I*e + 2*x *(I*f - c*Log[f]))/(2*Sqrt[I*f - c*Log[f]])])/Sqrt[I*f - c*Log[f]] - ((I/4 )*E^(I*d + e^2/((4*I)*f + 4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(I*e + 2*x*(I*f + c*Log[f]))/(2*Sqrt[I*f + c*Log[f]])])/Sqrt[I*f + c*Log[f]]
Int[(F_)^(u_)*Sin[v_]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^u, Sin[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 = 0.64 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.90
method | result | size |
risch | \(\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {4 i d \ln \left (f \right ) c -4 d f +e^{2}}{4 i f +4 c \ln \left (f \right )}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-i f}\, x +\frac {i e}{2 \sqrt {-c \ln \left (f \right )-i f}}\right )}{4 \sqrt {-c \ln \left (f \right )-i f}}+\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {4 i d \ln \left (f \right ) c +4 d f -e^{2}}{4 \left (-i f +c \ln \left (f \right )\right )}} \operatorname {erf}\left (x \sqrt {i f -c \ln \left (f \right )}+\frac {i e}{2 \sqrt {i f -c \ln \left (f \right )}}\right )}{4 \sqrt {i f -c \ln \left (f \right )}}\) | \(169\) |
Input:
int(f^(c*x^2+a)*sin(f*x^2+e*x+d),x,method=_RETURNVERBOSE)
Output:
1/4*I*Pi^(1/2)*f^a*exp(1/4*(4*I*d*ln(f)*c-4*d*f+e^2)/(I*f+c*ln(f)))/(-c*ln (f)-I*f)^(1/2)*erf(-(-c*ln(f)-I*f)^(1/2)*x+1/2*I*e/(-c*ln(f)-I*f)^(1/2))+1 /4*I*Pi^(1/2)*f^a*exp(-1/4*(4*I*d*ln(f)*c+4*d*f-e^2)/(-I*f+c*ln(f)))/(I*f- c*ln(f))^(1/2)*erf(x*(I*f-c*ln(f))^(1/2)+1/2*I*e/(I*f-c*ln(f))^(1/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (135) = 270\).
Time = 0.09 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.60 \[ \int f^{a+c x^2} \sin \left (d+e x+f x^2\right ) \, dx=\frac {\sqrt {\pi } {\left (i \, c \log \left (f\right ) + f\right )} \sqrt {-c \log \left (f\right ) - i \, f} \operatorname {erf}\left (\frac {{\left (2 \, c^{2} x \log \left (f\right )^{2} + 2 \, f^{2} x + i \, c e \log \left (f\right ) + e f\right )} \sqrt {-c \log \left (f\right ) - i \, f}}{2 \, {\left (c^{2} \log \left (f\right )^{2} + f^{2}\right )}}\right ) e^{\left (\frac {4 \, a c^{2} \log \left (f\right )^{3} + 4 i \, c^{2} d \log \left (f\right )^{2} - i \, e^{2} f + 4 i \, d f^{2} + {\left (c e^{2} + 4 \, a f^{2}\right )} \log \left (f\right )}{4 \, {\left (c^{2} \log \left (f\right )^{2} + f^{2}\right )}}\right )} + \sqrt {\pi } {\left (-i \, c \log \left (f\right ) + f\right )} \sqrt {-c \log \left (f\right ) + i \, f} \operatorname {erf}\left (\frac {{\left (2 \, c^{2} x \log \left (f\right )^{2} + 2 \, f^{2} x - i \, c e \log \left (f\right ) + e f\right )} \sqrt {-c \log \left (f\right ) + i \, f}}{2 \, {\left (c^{2} \log \left (f\right )^{2} + f^{2}\right )}}\right ) e^{\left (\frac {4 \, a c^{2} \log \left (f\right )^{3} - 4 i \, c^{2} d \log \left (f\right )^{2} + i \, e^{2} f - 4 i \, d f^{2} + {\left (c e^{2} + 4 \, a f^{2}\right )} \log \left (f\right )}{4 \, {\left (c^{2} \log \left (f\right )^{2} + f^{2}\right )}}\right )}}{4 \, {\left (c^{2} \log \left (f\right )^{2} + f^{2}\right )}} \] Input:
integrate(f^(c*x^2+a)*sin(f*x^2+e*x+d),x, algorithm="fricas")
Output:
1/4*(sqrt(pi)*(I*c*log(f) + f)*sqrt(-c*log(f) - I*f)*erf(1/2*(2*c^2*x*log( f)^2 + 2*f^2*x + I*c*e*log(f) + e*f)*sqrt(-c*log(f) - I*f)/(c^2*log(f)^2 + f^2))*e^(1/4*(4*a*c^2*log(f)^3 + 4*I*c^2*d*log(f)^2 - I*e^2*f + 4*I*d*f^2 + (c*e^2 + 4*a*f^2)*log(f))/(c^2*log(f)^2 + f^2)) + sqrt(pi)*(-I*c*log(f) + f)*sqrt(-c*log(f) + I*f)*erf(1/2*(2*c^2*x*log(f)^2 + 2*f^2*x - I*c*e*lo g(f) + e*f)*sqrt(-c*log(f) + I*f)/(c^2*log(f)^2 + f^2))*e^(1/4*(4*a*c^2*lo g(f)^3 - 4*I*c^2*d*log(f)^2 + I*e^2*f - 4*I*d*f^2 + (c*e^2 + 4*a*f^2)*log( f))/(c^2*log(f)^2 + f^2)))/(c^2*log(f)^2 + f^2)
\[ \int f^{a+c x^2} \sin \left (d+e x+f x^2\right ) \, dx=\int f^{a + c x^{2}} \sin {\left (d + e x + f x^{2} \right )}\, dx \] Input:
integrate(f**(c*x**2+a)*sin(f*x**2+e*x+d),x)
Output:
Integral(f**(a + c*x**2)*sin(d + e*x + f*x**2), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 760 vs. \(2 (135) = 270\).
Time = 0.07 (sec) , antiderivative size = 760, normalized size of antiderivative = 4.06 \[ \int f^{a+c x^2} \sin \left (d+e x+f x^2\right ) \, dx =\text {Too large to display} \] Input:
integrate(f^(c*x^2+a)*sin(f*x^2+e*x+d),x, algorithm="maxima")
Output:
-1/8*(sqrt(pi)*sqrt(2*c^2*log(f)^2 + 2*f^2)*((f^(1/4*c*e^2/(c^2*log(f)^2 + f^2))*f^a*cos(1/4*(4*c^2*d*log(f)^2 - e^2*f + 4*d*f^2)/(c^2*log(f)^2 + f^ 2)) - I*f^(1/4*c*e^2/(c^2*log(f)^2 + f^2))*f^a*sin(1/4*(4*c^2*d*log(f)^2 - e^2*f + 4*d*f^2)/(c^2*log(f)^2 + f^2)))*erf(1/2*(2*(c*log(f) - I*f)*x - I *e)/sqrt(-c*log(f) + I*f)) + (f^(1/4*c*e^2/(c^2*log(f)^2 + f^2))*f^a*cos(1 /4*(4*c^2*d*log(f)^2 - e^2*f + 4*d*f^2)/(c^2*log(f)^2 + f^2)) + I*f^(1/4*c *e^2/(c^2*log(f)^2 + f^2))*f^a*sin(1/4*(4*c^2*d*log(f)^2 - e^2*f + 4*d*f^2 )/(c^2*log(f)^2 + f^2)))*erf(1/2*(2*(c*log(f) + I*f)*x + I*e)/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)*((I*f^(1/4*c*e^2/(c^2*log(f)^2 + f^2))*f^a*cos(1/4*(4* c^2*d*log(f)^2 - e^2*f + 4*d*f^2)/(c^2*log(f)^2 + f^2)) + f^(1/4*c*e^2/(c^ 2*log(f)^2 + f^2))*f^a*sin(1/4*(4*c^2*d*log(f)^2 - e^2*f + 4*d*f^2)/(c^2*l og(f)^2 + f^2)))*erf(1/2*(2*(c*log(f) - I*f)*x - I*e)/sqrt(-c*log(f) + I*f )) + (-I*f^(1/4*c*e^2/(c^2*log(f)^2 + f^2))*f^a*cos(1/4*(4*c^2*d*log(f)^2 - e^2*f + 4*d*f^2)/(c^2*log(f)^2 + f^2)) + f^(1/4*c*e^2/(c^2*log(f)^2 + f^ 2))*f^a*sin(1/4*(4*c^2*d*log(f)^2 - e^2*f + 4*d*f^2)/(c^2*log(f)^2 + f^2)) )*erf(1/2*(2*(c*log(f) + I*f)*x + I*e)/sqrt(-c*log(f) - I*f)))*sqrt(-c*log (f) + sqrt(c^2*log(f)^2 + f^2)))/(c^2*log(f)^2 + f^2)
\[ \int f^{a+c x^2} \sin \left (d+e x+f x^2\right ) \, dx=\int { f^{c x^{2} + a} \sin \left (f x^{2} + e x + d\right ) \,d x } \] Input:
integrate(f^(c*x^2+a)*sin(f*x^2+e*x+d),x, algorithm="giac")
Output:
integrate(f^(c*x^2 + a)*sin(f*x^2 + e*x + d), x)
Timed out. \[ \int f^{a+c x^2} \sin \left (d+e x+f x^2\right ) \, dx=\int f^{c\,x^2+a}\,\sin \left (f\,x^2+e\,x+d\right ) \,d x \] Input:
int(f^(a + c*x^2)*sin(d + e*x + f*x^2),x)
Output:
int(f^(a + c*x^2)*sin(d + e*x + f*x^2), x)
\[ \int f^{a+c x^2} \sin \left (d+e x+f x^2\right ) \, dx=f^{a} \left (\int f^{c \,x^{2}} \sin \left (f \,x^{2}+e x +d \right )d x \right ) \] Input:
int(f^(c*x^2+a)*sin(f*x^2+e*x+d),x)
Output:
f**a*int(f**(c*x**2)*sin(d + e*x + f*x**2),x)