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

Optimal. Leaf size=176 \[ -\frac{i \sqrt{\pi } f^a e^{\frac{(e+i b \log (f))^2}{4 c \log (f)}-i d} \text{Erfi}\left (\frac{-b \log (f)-2 c x \log (f)+i e}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{4 \sqrt{c} \sqrt{\log (f)}}-\frac{i \sqrt{\pi } f^a e^{\frac{(e-i b \log (f))^2}{4 c \log (f)}+i d} \text{Erfi}\left (\frac{b \log (f)+2 c x \log (f)+i e}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{4 \sqrt{c} \sqrt{\log (f)}} \]

[Out]

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

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Rubi [A]  time = 0.334657, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {4472, 2287, 2234, 2204} \[ -\frac{i \sqrt{\pi } f^a e^{\frac{(e+i b \log (f))^2}{4 c \log (f)}-i d} \text{Erfi}\left (\frac{-b \log (f)-2 c x \log (f)+i e}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{4 \sqrt{c} \sqrt{\log (f)}}-\frac{i \sqrt{\pi } f^a e^{\frac{(e-i b \log (f))^2}{4 c \log (f)}+i d} \text{Erfi}\left (\frac{b \log (f)+2 c x \log (f)+i e}{2 \sqrt{c} \sqrt{\log (f)}}\right )}{4 \sqrt{c} \sqrt{\log (f)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.332312, size = 155, normalized size = 0.88 \[ \frac{\sqrt{\pi } f^{a-\frac{b^2}{4 c}} e^{\frac{e (e-2 i b \log (f))}{4 c \log (f)}} \left (i (\cos (d)+i \sin (d)) \text{Erfi}\left (\frac{-\log (f) (b+2 c x)-i e}{2 \sqrt{c} \sqrt{\log (f)}}\right )+e^{\frac{i b e}{c}} (\sin (d)+i \cos (d)) \text{Erfi}\left (\frac{\log (f) (b+2 c x)-i e}{2 \sqrt{c} \sqrt{\log (f)}}\right )\right )}{4 \sqrt{c} \sqrt{\log (f)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^((e*(e - (2*I)*b*Log[f]))/(4*c*Log[f]))*f^(a - b^2/(4*c))*Sqrt[Pi]*(I*Erfi[((-I)*e - (b + 2*c*x)*Log[f])/(2
*Sqrt[c]*Sqrt[Log[f]])]*(Cos[d] + I*Sin[d]) + E^((I*b*e)/c)*Erfi[((-I)*e + (b + 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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Maple [A]  time = 0.234, size = 170, normalized size = 1. \begin{align*}{{\frac{i}{4}}{f}^{a}\sqrt{\pi }{{\rm e}^{-{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}+2\,i\ln \left ( f \right ) be-4\,id\ln \left ( f \right ) c-{e}^{2}}{4\,c\ln \left ( f \right ) }}}}{\it Erf} \left ( -\sqrt{-c\ln \left ( f \right ) }x+{\frac{ie+b\ln \left ( f \right ) }{2}{\frac{1}{\sqrt{-c\ln \left ( f \right ) }}}} \right ){\frac{1}{\sqrt{-c\ln \left ( f \right ) }}}}-{{\frac{i}{4}}{f}^{a}\sqrt{\pi }{{\rm e}^{-{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}-2\,i\ln \left ( f \right ) be+4\,id\ln \left ( f \right ) c-{e}^{2}}{4\,c\ln \left ( f \right ) }}}}{\it Erf} \left ( -\sqrt{-c\ln \left ( f \right ) }x+{\frac{b\ln \left ( f \right ) -ie}{2}{\frac{1}{\sqrt{-c\ln \left ( f \right ) }}}} \right ){\frac{1}{\sqrt{-c\ln \left ( f \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: IndexError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: IndexError

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Fricas [A]  time = 0.497979, size = 487, normalized size = 2.77 \begin{align*} \frac{i \, \sqrt{\pi } \sqrt{-c \log \left (f\right )} \operatorname{erf}\left (\frac{{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) + i \, e\right )} \sqrt{-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac{{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - e^{2} -{\left (4 i \, c d - 2 i \, b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )} - i \, \sqrt{\pi } \sqrt{-c \log \left (f\right )} \operatorname{erf}\left (\frac{{\left ({\left (2 \, c x + b\right )} \log \left (f\right ) - i \, e\right )} \sqrt{-c \log \left (f\right )}}{2 \, c \log \left (f\right )}\right ) e^{\left (-\frac{{\left (b^{2} - 4 \, a c\right )} \log \left (f\right )^{2} - e^{2} -{\left (-4 i \, c d + 2 i \, b e\right )} \log \left (f\right )}{4 \, c \log \left (f\right )}\right )}}{4 \, c \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int f^{a + b x + c x^{2}} \sin{\left (d + e x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int f^{c x^{2} + b x + a} \sin \left (e x + d\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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