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

Optimal. Leaf size=151 \[ -\frac {i \sqrt {\pi } f^a e^{\frac {e^2}{4 c \log (f)}-i d} \text {erfi}\left (\frac {-2 c x \log (f)+i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {i \sqrt {\pi } f^a e^{\frac {e^2}{4 c \log (f)}+i d} \text {erfi}\left (\frac {2 c x \log (f)+i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

[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)

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Rubi [A]  time = 0.21, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4472, 2287, 2234, 2204} \[ -\frac {i \sqrt {\pi } f^a e^{\frac {e^2}{4 c \log (f)}-i d} \text {Erfi}\left (\frac {-2 c x \log (f)+i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {i \sqrt {\pi } f^a e^{\frac {e^2}{4 c \log (f)}+i d} \text {Erfi}\left (\frac {2 c x \log (f)+i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + c*x^2)*Sin[d + e*x],x]

[Out]

((-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]]) - ((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[Lo
g[f]])])/(Sqrt[c]*Sqrt[Log[f]])

Rule 2204

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 2234

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 2287

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 4472

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*} \int f^{a+c x^2} \sin (d+e x) \, dx &=\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*}

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Mathematica [A]  time = 0.15, size = 119, normalized size = 0.79 \[ \frac {\sqrt {\pi } f^a e^{\frac {e^2}{4 c \log (f)}} \left (i (\cos (d)+i \sin (d)) \text {erfi}\left (\frac {-2 c x \log (f)-i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )+(\sin (d)+i \cos (d)) \text {erfi}\left (\frac {2 c x \log (f)-i e}{2 \sqrt {c} \sqrt {\log (f)}}\right )\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]

Antiderivative was successfully verified.

[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]])

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fricas [A]  time = 0.94, size = 144, normalized size = 0.95 \[ \frac {i \, \sqrt {\pi } \sqrt {-c \log \relax (f)} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \relax (f) + i \, e\right )} \sqrt {-c \log \relax (f)}}{2 \, c \log \relax (f)}\right ) e^{\left (\frac {4 \, a c \log \relax (f)^{2} + 4 i \, c d \log \relax (f) + e^{2}}{4 \, c \log \relax (f)}\right )} - i \, \sqrt {\pi } \sqrt {-c \log \relax (f)} \operatorname {erf}\left (\frac {{\left (2 \, c x \log \relax (f) - i \, e\right )} \sqrt {-c \log \relax (f)}}{2 \, c \log \relax (f)}\right ) e^{\left (\frac {4 \, a c \log \relax (f)^{2} - 4 i \, c d \log \relax (f) + e^{2}}{4 \, c \log \relax (f)}\right )}}{4 \, c \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{c x^{2} + a} \sin \left (e x + d\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.42, size = 123, normalized size = 0.81 \[ \frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {4 i d \ln \relax (f ) c +e^{2}}{4 \ln \relax (f ) c}} \erf \left (-\sqrt {-c \ln \relax (f )}\, x +\frac {i e}{2 \sqrt {-c \ln \relax (f )}}\right )}{4 \sqrt {-c \ln \relax (f )}}+\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {4 i d \ln \relax (f ) c -e^{2}}{4 \ln \relax (f ) c}} \erf \left (\sqrt {-c \ln \relax (f )}\, x +\frac {i e}{2 \sqrt {-c \ln \relax (f )}}\right )}{4 \sqrt {-c \ln \relax (f )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))

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maxima [C]  time = 0.36, size = 206, normalized size = 1.36 \[ -\frac {\sqrt {\pi } {\left (f^{a} {\left (i \, \cos \relax (d) + \sin \relax (d)\right )} \operatorname {erf}\left (x \overline {\sqrt {-c \log \relax (f)}} + \frac {1}{2} i \, e \overline {\frac {1}{\sqrt {-c \log \relax (f)}}}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \relax (f)}\right )} + f^{a} {\left (-i \, \cos \relax (d) + \sin \relax (d)\right )} \operatorname {erf}\left (x \overline {\sqrt {-c \log \relax (f)}} - \frac {1}{2} i \, e \overline {\frac {1}{\sqrt {-c \log \relax (f)}}}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \relax (f)}\right )} + f^{a} {\left (i \, \cos \relax (d) - \sin \relax (d)\right )} \operatorname {erf}\left (\frac {2 \, c x \log \relax (f) + i \, e}{2 \, \sqrt {-c \log \relax (f)}}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \relax (f)}\right )} + f^{a} {\left (-i \, \cos \relax (d) - \sin \relax (d)\right )} \operatorname {erf}\left (\frac {2 \, c x \log \relax (f) - i \, e}{2 \, \sqrt {-c \log \relax (f)}}\right ) e^{\left (\frac {e^{2}}{4 \, c \log \relax (f)}\right )}\right )} \sqrt {-c \log \relax (f)}}{8 \, c \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int f^{c\,x^2+a}\,\sin \left (d+e\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + c*x^2)*sin(d + e*x),x)

[Out]

int(f^(a + c*x^2)*sin(d + e*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + c x^{2}} \sin {\left (d + e x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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