\(\int f^{a+c x^2} \sin (d+e x) \, dx\) [85]

Optimal result

Integrand size = 16, antiderivative size = 151 \[ \int f^{a+c x^2} \sin (d+e x) \, dx=-\frac {i e^{-i d+\frac {e^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {i e^{i d+\frac {e^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[Out]

1/4*I*exp(-I*d+1/4*e^2/c/ln(f))*f^a*erfi(1/2*(-I*e+2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2))*Pi^(1/2)/c^(1/2)/ln(f)^(1
/2)-1/4*I*exp(I*d+1/4*e^2/c/ln(f))*f^a*erfi(1/2*(I*e+2*c*x*ln(f))/c^(1/2)/ln(f)^(1/2))*Pi^(1/2)/c^(1/2)/ln(f)^
(1/2)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4560, 2325, 2266, 2235} \[ \int f^{a+c x^2} \sin (d+e x) \, dx=-\frac {i \sqrt {\pi } f^a e^{\frac {e^2}{4 c \log (f)}-i d} \text {erfi}\left (\frac {-2 c x \log (f)+i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {i \sqrt {\pi } f^a e^{\frac {e^2}{4 c \log (f)}+i d} \text {erfi}\left (\frac {2 c x \log (f)+i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[In]

Int[f^(a + c*x^2)*Sin[d + e*x],x]

[Out]

((-1/4*I)*E^((-I)*d + e^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(I*e - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])])/(Sqr
t[c]*Sqrt[Log[f]]) - ((I/4)*E^(I*d + e^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(I*e + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[
Log[f]])])/(Sqrt[c]*Sqrt[Log[f]])

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2266

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2325

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

Rule 4560

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]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{2} i e^{-i d-i e x} f^{a+c x^2}-\frac {1}{2} i e^{i d+i e x} f^{a+c x^2}\right ) \, dx \\ & = \frac {1}{2} i \int e^{-i d-i e x} f^{a+c x^2} \, dx-\frac {1}{2} i \int e^{i d+i e x} f^{a+c x^2} \, dx \\ & = \frac {1}{2} i \int e^{-i d-i e x+a \log (f)+c x^2 \log (f)} \, dx-\frac {1}{2} i \int e^{i d+i e x+a \log (f)+c x^2 \log (f)} \, dx \\ & = \frac {1}{2} \left (i e^{-i d+\frac {e^2}{4 c \log (f)}} f^a\right ) \int e^{\frac {(-i e+2 c x \log (f))^2}{4 c \log (f)}} \, dx-\frac {1}{2} \left (i e^{i d+\frac {e^2}{4 c \log (f)}} f^a\right ) \int e^{\frac {(i e+2 c x \log (f))^2}{4 c \log (f)}} \, dx \\ & = -\frac {i e^{-i d+\frac {e^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {i e^{i d+\frac {e^2}{4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.79 \[ \int f^{a+c x^2} \sin (d+e x) \, dx=\frac {e^{\frac {e^2}{4 c \log (f)}} f^a \sqrt {\pi } \left (i \text {erfi}\left (\frac {-i e-2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (\cos (d)+i \sin (d))+\text {erfi}\left (\frac {-i e+2 c x \log (f)}{2 \sqrt {c} \sqrt {\log (f)}}\right ) (i \cos (d)+\sin (d))\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[In]

Integrate[f^(a + c*x^2)*Sin[d + e*x],x]

[Out]

(E^(e^2/(4*c*Log[f]))*f^a*Sqrt[Pi]*(I*Erfi[((-I)*e - 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(Cos[d] + I*Sin[d
]) + Erfi[((-I)*e + 2*c*x*Log[f])/(2*Sqrt[c]*Sqrt[Log[f]])]*(I*Cos[d] + Sin[d])))/(4*Sqrt[c]*Sqrt[Log[f]])

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {4 i d \ln \left (f \right ) c +e^{2}}{4 \ln \left (f \right ) c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {i e}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 \sqrt {-c \ln \left (f \right )}}+\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {4 i d \ln \left (f \right ) c -e^{2}}{4 \ln \left (f \right ) c}} \operatorname {erf}\left (\sqrt {-c \ln \left (f \right )}\, x +\frac {i e}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 \sqrt {-c \ln \left (f \right )}}\)	\(123\)

[In]

int(f^(c*x^2+a)*sin(e*x+d),x,method=_RETURNVERBOSE)

[Out]

1/4*I*Pi^(1/2)*f^a*exp(1/4*(4*I*d*ln(f)*c+e^2)/ln(f)/c)/(-c*ln(f))^(1/2)*erf(-(-c*ln(f))^(1/2)*x+1/2*I*e/(-c*l
n(f))^(1/2))+1/4*I*Pi^(1/2)*f^a*exp(-1/4*(4*I*d*ln(f)*c-e^2)/ln(f)/c)/(-c*ln(f))^(1/2)*erf((-c*ln(f))^(1/2)*x+
1/2*I*e/(-c*ln(f))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.95 \[ \int f^{a+c x^2} \sin (d+e x) \, dx=\frac {i \, \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \left (f\right ) + i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (\frac {4 \, a c \log \left (f\right )^{2} + 4 i \, c d \log \left (f\right ) + e^{2}}{4 \, c \log \left (f\right )}\right )} - i \, \sqrt {\pi } \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \left (f\right ) - i \, e\right )} \sqrt {-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (\frac {4 \, a c \log \left (f\right )^{2} - 4 i \, c d \log \left (f\right ) + e^{2}}{4 \, c \log \left (f\right )}\right )}}{4 \, c \log \left (f\right )} \]

[In]

integrate(f^(c*x^2+a)*sin(e*x+d),x, algorithm="fricas")

[Out]

1/4*(I*sqrt(pi)*sqrt(-c*log(f))*erf(1/2*(2*c*x*log(f) + I*e)*sqrt(-c*log(f))/(c*log(f)))*e^(1/4*(4*a*c*log(f)^
2 + 4*I*c*d*log(f) + e^2)/(c*log(f))) - I*sqrt(pi)*sqrt(-c*log(f))*erf(1/2*(2*c*x*log(f) - I*e)*sqrt(-c*log(f)
)/(c*log(f)))*e^(1/4*(4*a*c*log(f)^2 - 4*I*c*d*log(f) + e^2)/(c*log(f))))/(c*log(f))

Sympy [F]

\[ \int f^{a+c x^2} \sin (d+e x) \, dx=\int f^{a + c x^{2}} \sin {\left (d + e x \right )}\, dx \]

[In]

integrate(f**(c*x**2+a)*sin(e*x+d),x)

[Out]

Integral(f**(a + c*x**2)*sin(d + e*x), x)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.24 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.36 \[ \int f^{a+c x^2} \sin (d+e x) \, dx=-\frac {\sqrt {\pi } {\left (f^{a} {\left (i \, \cos \left (d\right ) + \sin \left (d\right )\right )} \operatorname {erf}\left (x \overline {\sqrt {-c \log \left (f\right )}} + \frac {1}{2} i \, e \overline {\frac {1}{\sqrt {-c \log \left (f\right )}}}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \left (f\right )}\right )} + f^{a} {\left (-i \, \cos \left (d\right ) + \sin \left (d\right )\right )} \operatorname {erf}\left (x \overline {\sqrt {-c \log \left (f\right )}} - \frac {1}{2} i \, e \overline {\frac {1}{\sqrt {-c \log \left (f\right )}}}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \left (f\right )}\right )} + f^{a} {\left (i \, \cos \left (d\right ) - \sin \left (d\right )\right )} \operatorname {erf}\left (\frac {2 \, c x \log \left (f\right ) + i \, e}{2 \, \sqrt {-c \log \left (f\right )}}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \left (f\right )}\right )} + f^{a} {\left (-i \, \cos \left (d\right ) - \sin \left (d\right )\right )} \operatorname {erf}\left (\frac {2 \, c x \log \left (f\right ) - i \, e}{2 \, \sqrt {-c \log \left (f\right )}}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \left (f\right )}\right )}\right )} \sqrt {-c \log \left (f\right )}}{8 \, c \log \left (f\right )} \]

[In]

integrate(f^(c*x^2+a)*sin(e*x+d),x, algorithm="maxima")

[Out]

-1/8*sqrt(pi)*(f^a*(I*cos(d) + sin(d))*erf(x*conjugate(sqrt(-c*log(f))) + 1/2*I*e*conjugate(1/sqrt(-c*log(f)))
)*e^(1/4*e^2/(c*log(f))) + f^a*(-I*cos(d) + sin(d))*erf(x*conjugate(sqrt(-c*log(f))) - 1/2*I*e*conjugate(1/sqr
t(-c*log(f))))*e^(1/4*e^2/(c*log(f))) + f^a*(I*cos(d) - sin(d))*erf(1/2*(2*c*x*log(f) + I*e)/sqrt(-c*log(f)))*
e^(1/4*e^2/(c*log(f))) + f^a*(-I*cos(d) - sin(d))*erf(1/2*(2*c*x*log(f) - I*e)/sqrt(-c*log(f)))*e^(1/4*e^2/(c*
log(f))))*sqrt(-c*log(f))/(c*log(f))

Giac [F]

\[ \int f^{a+c x^2} \sin (d+e x) \, dx=\int { f^{c x^{2} + a} \sin \left (e x + d\right ) \,d x } \]

[In]

integrate(f^(c*x^2+a)*sin(e*x+d),x, algorithm="giac")

[Out]

integrate(f^(c*x^2 + a)*sin(e*x + d), x)

Mupad [F(-1)]

Timed out. \[ \int f^{a+c x^2} \sin (d+e x) \, dx=\int f^{c\,x^2+a}\,\sin \left (d+e\,x\right ) \,d x \]

[In]

int(f^(a + c*x^2)*sin(d + e*x),x)

[Out]

int(f^(a + c*x^2)*sin(d + e*x), x)