Integrand size = 24, antiderivative size = 213 \[ \int f^{a+b x+c x^2} \sin \left (a+b x+e x^2\right ) \, dx=\frac {i e^{-\left ((i-\log (f)) \left (a-\frac {b^2 (i-\log (f))}{4 i e-4 c \log (f)}\right )\right )} \sqrt {\pi } \text {erf}\left (\frac {b (i-\log (f))+2 x (i e-c \log (f))}{2 \sqrt {i e-c \log (f)}}\right )}{4 \sqrt {i e-c \log (f)}}-\frac {i e^{(i+\log (f)) \left (a-\frac {b^2 (i+\log (f))}{4 i e+4 c \log (f)}\right )} \sqrt {\pi } \text {erfi}\left (\frac {b (i+\log (f))+2 x (i e+c \log (f))}{2 \sqrt {i e+c \log (f)}}\right )}{4 \sqrt {i e+c \log (f)}} \]
-1/4*I*erf(1/2*(-b*(I-ln(f))-2*x*(I*e-c*ln(f)))/(I*e-c*ln(f))^(1/2))*Pi^(1 /2)/exp((I-ln(f))*(a-b^2*(I-ln(f))/(4*I*e-4*c*ln(f))))/(I*e-c*ln(f))^(1/2) -1/4*I*exp((I+ln(f))*(a-b^2*(I+ln(f))/(4*I*e+4*c*ln(f))))*erfi(1/2*(b*(I+l n(f))+2*x*(I*e+c*ln(f)))/(I*e+c*ln(f))^(1/2))*Pi^(1/2)/(I*e+c*ln(f))^(1/2)
Time = 1.61 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.52 \[ \int f^{a+b x+c x^2} \sin \left (a+b x+e x^2\right ) \, dx=\frac {e^{-\frac {b^2 c \log ^3(f)}{2 \left (e^2+c^2 \log ^2(f)\right )}} f^{a-\frac {b^2}{2 (e-i c \log (f))}} \sqrt {\pi } \left (-e^{\frac {1}{4} b^2 \left (\frac {1}{-i e+c \log (f)}+\frac {\log ^2(f)}{i e+c \log (f)}\right )} f^{\frac {i b^2 c \log (f)}{e^2+c^2 \log ^2(f)}} \text {erfi}\left (\frac {-i (b+2 e x)+(b+2 c x) \log (f)}{2 \sqrt {-i e+c \log (f)}}\right ) (e-i c \log (f)) \sqrt {-i e+c \log (f)} (\cos (a)-i \sin (a))+e^{\frac {1}{4} b^2 \left (\frac {\log ^2(f)}{-i e+c \log (f)}+\frac {1}{i e+c \log (f)}\right )} \text {erfi}\left (\frac {-i (b+2 e x)-(b+2 c x) \log (f)}{2 \sqrt {i e+c \log (f)}}\right ) (e+i c \log (f)) \sqrt {i e+c \log (f)} (\cos (a)+i \sin (a))\right )}{4 \left (e^2+c^2 \log ^2(f)\right )} \]
(f^(a - b^2/(2*(e - I*c*Log[f])))*Sqrt[Pi]*(-(E^((b^2*(((-I)*e + c*Log[f]) ^(-1) + Log[f]^2/(I*e + c*Log[f])))/4)*f^((I*b^2*c*Log[f])/(e^2 + c^2*Log[ f]^2))*Erfi[((-I)*(b + 2*e*x) + (b + 2*c*x)*Log[f])/(2*Sqrt[(-I)*e + c*Log [f]])]*(e - I*c*Log[f])*Sqrt[(-I)*e + c*Log[f]]*(Cos[a] - I*Sin[a])) + E^( (b^2*(Log[f]^2/((-I)*e + c*Log[f]) + (I*e + c*Log[f])^(-1)))/4)*Erfi[((-I) *(b + 2*e*x) - (b + 2*c*x)*Log[f])/(2*Sqrt[I*e + c*Log[f]])]*(e + I*c*Log[ f])*Sqrt[I*e + c*Log[f]]*(Cos[a] + I*Sin[a])))/(4*E^((b^2*c*Log[f]^3)/(2*( e^2 + c^2*Log[f]^2)))*(e^2 + c^2*Log[f]^2))
Time = 0.86 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.
\(\displaystyle \int f^{a+b x+c x^2} \sin \left (a+b x+e x^2\right ) \, dx\) |
\(\Big \downarrow \) 4975 |
\(\displaystyle \int \left (\frac {1}{2} i e^{-i a-i b x-i e x^2} f^{a+b x+c x^2}-\frac {1}{2} i e^{i a+i b x+i e x^2} f^{a+b x+c x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i \sqrt {\pi } \exp \left (-\left ((-\log (f)+i) \left (a-\frac {b^2 (-\log (f)+i)}{-4 c \log (f)+4 i e}\right )\right )\right ) \text {erf}\left (\frac {b (-\log (f)+i)+2 x (-c \log (f)+i e)}{2 \sqrt {-c \log (f)+i e}}\right )}{4 \sqrt {-c \log (f)+i e}}-\frac {i \sqrt {\pi } \exp \left ((\log (f)+i) \left (a-\frac {b^2 (\log (f)+i)}{4 c \log (f)+4 i e}\right )\right ) \text {erfi}\left (\frac {b (\log (f)+i)+2 x (c \log (f)+i e)}{2 \sqrt {c \log (f)+i e}}\right )}{4 \sqrt {c \log (f)+i e}}\) |
((I/4)*Sqrt[Pi]*Erf[(b*(I - Log[f]) + 2*x*(I*e - c*Log[f]))/(2*Sqrt[I*e - c*Log[f]])])/(E^((I - Log[f])*(a - (b^2*(I - Log[f]))/((4*I)*e - 4*c*Log[f ])))*Sqrt[I*e - c*Log[f]]) - ((I/4)*E^((I + Log[f])*(a - (b^2*(I + Log[f]) )/((4*I)*e + 4*c*Log[f])))*Sqrt[Pi]*Erfi[(b*(I + Log[f]) + 2*x*(I*e + c*Lo g[f]))/(2*Sqrt[I*e + c*Log[f]])])/Sqrt[I*e + c*Log[f]]
3.2.3.3.1 Defintions of rubi rules used
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 = 0.56 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.02
method | result | size |
risch | \(\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {-\ln \left (f \right )^{2} b^{2}+4 i \ln \left (f \right ) a c -2 i \ln \left (f \right ) b^{2}-4 a e +b^{2}}{4 i e +4 c \ln \left (f \right )}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-i e}\, x +\frac {b \ln \left (f \right )+i b}{2 \sqrt {-c \ln \left (f \right )-i e}}\right )}{4 \sqrt {-c \ln \left (f \right )-i e}}-\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}+4 i \ln \left (f \right ) a c -2 i \ln \left (f \right ) b^{2}+4 a e -b^{2}}{4 \left (c \ln \left (f \right )-i e \right )}} \operatorname {erf}\left (-\sqrt {i e -c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )-i b}{2 \sqrt {i e -c \ln \left (f \right )}}\right )}{4 \sqrt {i e -c \ln \left (f \right )}}\) | \(217\) |
1/4*I*Pi^(1/2)*f^a*exp(1/4*(-ln(f)^2*b^2+4*I*ln(f)*a*c-2*I*ln(f)*b^2-4*a*e +b^2)/(I*e+c*ln(f)))/(-c*ln(f)-I*e)^(1/2)*erf(-(-c*ln(f)-I*e)^(1/2)*x+1/2* (b*ln(f)+I*b)/(-c*ln(f)-I*e)^(1/2))-1/4*I*Pi^(1/2)*f^a*exp(-1/4*(ln(f)^2*b ^2+4*I*ln(f)*a*c-2*I*ln(f)*b^2+4*a*e-b^2)/(c*ln(f)-I*e))/(I*e-c*ln(f))^(1/ 2)*erf(-(I*e-c*ln(f))^(1/2)*x+1/2*(b*ln(f)-I*b)/(I*e-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. 379 vs. \(2 (153) = 306\).
Time = 0.26 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.78 \[ \int f^{a+b x+c x^2} \sin \left (a+b x+e x^2\right ) \, dx=\frac {\sqrt {\pi } {\left (i \, c \log \left (f\right ) + e\right )} \sqrt {-c \log \left (f\right ) - i \, e} \operatorname {erf}\left (\frac {{\left (2 \, e^{2} x + {\left (2 \, c^{2} x + b c\right )} \log \left (f\right )^{2} + b e + {\left (i \, b c - i \, b e\right )} \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) - i \, e}}{2 \, {\left (c^{2} \log \left (f\right )^{2} + e^{2}\right )}}\right ) e^{\left (-\frac {{\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )^{3} + i \, b^{2} e - 4 i \, a e^{2} - {\left (-2 i \, b^{2} c + 4 i \, a c^{2} + i \, b^{2} e\right )} \log \left (f\right )^{2} - {\left (b^{2} c - 2 \, b^{2} e + 4 \, a e^{2}\right )} \log \left (f\right )}{4 \, {\left (c^{2} \log \left (f\right )^{2} + e^{2}\right )}}\right )} + \sqrt {\pi } {\left (-i \, c \log \left (f\right ) + e\right )} \sqrt {-c \log \left (f\right ) + i \, e} \operatorname {erf}\left (\frac {{\left (2 \, e^{2} x + {\left (2 \, c^{2} x + b c\right )} \log \left (f\right )^{2} + b e + {\left (-i \, b c + i \, b e\right )} \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) + i \, e}}{2 \, {\left (c^{2} \log \left (f\right )^{2} + e^{2}\right )}}\right ) e^{\left (-\frac {{\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )^{3} - i \, b^{2} e + 4 i \, a e^{2} - {\left (2 i \, b^{2} c - 4 i \, a c^{2} - i \, b^{2} e\right )} \log \left (f\right )^{2} - {\left (b^{2} c - 2 \, b^{2} e + 4 \, a e^{2}\right )} \log \left (f\right )}{4 \, {\left (c^{2} \log \left (f\right )^{2} + e^{2}\right )}}\right )}}{4 \, {\left (c^{2} \log \left (f\right )^{2} + e^{2}\right )}} \]
1/4*(sqrt(pi)*(I*c*log(f) + e)*sqrt(-c*log(f) - I*e)*erf(1/2*(2*e^2*x + (2 *c^2*x + b*c)*log(f)^2 + b*e + (I*b*c - I*b*e)*log(f))*sqrt(-c*log(f) - I* e)/(c^2*log(f)^2 + e^2))*e^(-1/4*((b^2*c - 4*a*c^2)*log(f)^3 + I*b^2*e - 4 *I*a*e^2 - (-2*I*b^2*c + 4*I*a*c^2 + I*b^2*e)*log(f)^2 - (b^2*c - 2*b^2*e + 4*a*e^2)*log(f))/(c^2*log(f)^2 + e^2)) + sqrt(pi)*(-I*c*log(f) + e)*sqrt (-c*log(f) + I*e)*erf(1/2*(2*e^2*x + (2*c^2*x + b*c)*log(f)^2 + b*e + (-I* b*c + I*b*e)*log(f))*sqrt(-c*log(f) + I*e)/(c^2*log(f)^2 + e^2))*e^(-1/4*( (b^2*c - 4*a*c^2)*log(f)^3 - I*b^2*e + 4*I*a*e^2 - (2*I*b^2*c - 4*I*a*c^2 - I*b^2*e)*log(f)^2 - (b^2*c - 2*b^2*e + 4*a*e^2)*log(f))/(c^2*log(f)^2 + e^2)))/(c^2*log(f)^2 + e^2)
\[ \int f^{a+b x+c x^2} \sin \left (a+b x+e x^2\right ) \, dx=\int f^{a + b x + c x^{2}} \sin {\left (a + b x + e x^{2} \right )}\, dx \]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.27 (sec) , antiderivative size = 1054, normalized size of antiderivative = 4.95 \[ \int f^{a+b x+c x^2} \sin \left (a+b x+e x^2\right ) \, dx=\text {Too large to display} \]
1/8*sqrt(pi)*((f^(1/4*b^2*c/(c^2*log(f)^2 + e^2))*f^a*(-I*cos(1/2*arctan2( e, -c*log(f))) + sin(1/2*arctan2(e, -c*log(f))))*cos(-1/4*(b^2*e - 4*a*e^2 + (2*b^2*c - 4*a*c^2 - b^2*e)*log(f)^2)/(c^2*log(f)^2 + e^2)) + f^(1/4*b^ 2*c/(c^2*log(f)^2 + e^2))*f^a*(cos(1/2*arctan2(e, -c*log(f))) + I*sin(1/2* arctan2(e, -c*log(f))))*sin(-1/4*(b^2*e - 4*a*e^2 + (2*b^2*c - 4*a*c^2 - b ^2*e)*log(f)^2)/(c^2*log(f)^2 + e^2)))*erf(x*conjugate(sqrt(-c*log(f) + I* e)) - 1/2*(b*log(f) + I*b)*conjugate(1/sqrt(-c*log(f) + I*e))) + (f^(1/4*b ^2*c/(c^2*log(f)^2 + e^2))*f^a*(I*cos(1/2*arctan2(e, -c*log(f))) + sin(1/2 *arctan2(e, -c*log(f))))*cos(-1/4*(b^2*e - 4*a*e^2 + (2*b^2*c - 4*a*c^2 - b^2*e)*log(f)^2)/(c^2*log(f)^2 + e^2)) + f^(1/4*b^2*c/(c^2*log(f)^2 + e^2) )*f^a*(cos(1/2*arctan2(e, -c*log(f))) - I*sin(1/2*arctan2(e, -c*log(f))))* sin(-1/4*(b^2*e - 4*a*e^2 + (2*b^2*c - 4*a*c^2 - b^2*e)*log(f)^2)/(c^2*log (f)^2 + e^2)))*erf(x*conjugate(sqrt(-c*log(f) - I*e)) - 1/2*(b*log(f) - I* b)*conjugate(1/sqrt(-c*log(f) - I*e))) + (f^(1/4*b^2*c/(c^2*log(f)^2 + e^2 ))*f^a*(I*cos(1/2*arctan2(e, -c*log(f))) + sin(1/2*arctan2(e, -c*log(f)))) *cos(-1/4*(b^2*e - 4*a*e^2 + (2*b^2*c - 4*a*c^2 - b^2*e)*log(f)^2)/(c^2*lo g(f)^2 + e^2)) + f^(1/4*b^2*c/(c^2*log(f)^2 + e^2))*f^a*(cos(1/2*arctan2(e , -c*log(f))) - I*sin(1/2*arctan2(e, -c*log(f))))*sin(-1/4*(b^2*e - 4*a*e^ 2 + (2*b^2*c - 4*a*c^2 - b^2*e)*log(f)^2)/(c^2*log(f)^2 + e^2)))*erf(1/2*( 2*(c*log(f) - I*e)*x + b*log(f) - I*b)*sqrt(-c*log(f) + I*e)/(c*log(f) ...
\[ \int f^{a+b x+c x^2} \sin \left (a+b x+e x^2\right ) \, dx=\int { f^{c x^{2} + b x + a} \sin \left (e x^{2} + b x + a\right ) \,d x } \]
Timed out. \[ \int f^{a+b x+c x^2} \sin \left (a+b x+e x^2\right ) \, dx=\int f^{c\,x^2+b\,x+a}\,\sin \left (e\,x^2+b\,x+a\right ) \,d x \]