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

Optimal. Leaf size=212 \[ \frac {i \sqrt {\pi } f^a \exp \left (-\frac {(e+i b \log (f))^2}{-4 c \log (f)+4 i f}-i d\right ) \text {erf}\left (\frac {-b \log (f)+2 x (-c \log (f)+i f)+i e}{2 \sqrt {-c \log (f)+i f}}\right )}{4 \sqrt {-c \log (f)+i f}}-\frac {i \sqrt {\pi } f^a \exp \left (\frac {(e-i b \log (f))^2}{4 c \log (f)+4 i f}+i d\right ) \text {erfi}\left (\frac {b \log (f)+2 x (c \log (f)+i f)+i e}{2 \sqrt {c \log (f)+i f}}\right )}{4 \sqrt {c \log (f)+i f}} \]

[Out]

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

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Rubi [A]  time = 0.57, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {4472, 2287, 2234, 2205, 2204} \[ \frac {i \sqrt {\pi } f^a \exp \left (-\frac {(e+i b \log (f))^2}{-4 c \log (f)+4 i f}-i d\right ) \text {Erf}\left (\frac {-b \log (f)+2 x (-c \log (f)+i f)+i e}{2 \sqrt {-c \log (f)+i f}}\right )}{4 \sqrt {-c \log (f)+i f}}-\frac {i \sqrt {\pi } f^a \exp \left (\frac {(e-i b \log (f))^2}{4 c \log (f)+4 i f}+i d\right ) \text {Erfi}\left (\frac {b \log (f)+2 x (c \log (f)+i f)+i e}{2 \sqrt {c \log (f)+i f}}\right )}{4 \sqrt {c \log (f)+i f}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/4)*E^((-I)*d - (e + I*b*Log[f])^2/((4*I)*f - 4*c*Log[f]))*f^a*Sqrt[Pi]*Erf[(I*e - b*Log[f] + 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 - I*b*Log[f])^2/((4*I)*f + 4*c*L
og[f]))*f^a*Sqrt[Pi]*Erfi[(I*e + b*Log[f] + 2*x*(I*f + c*Log[f]))/(2*Sqrt[I*f + c*Log[f]])])/Sqrt[I*f + c*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 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[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+b x+c x^2} \sin \left (d+e x+f x^2\right ) \, dx &=\int \left (\frac {1}{2} i e^{-i d-i e x-i f x^2} f^{a+b x+c x^2}-\frac {1}{2} i e^{i d+i e x+i f x^2} f^{a+b x+c x^2}\right ) \, dx\\ &=\frac {1}{2} i \int e^{-i d-i e x-i f x^2} f^{a+b x+c x^2} \, dx-\frac {1}{2} i \int e^{i d+i e x+i f x^2} f^{a+b x+c x^2} \, dx\\ &=\frac {1}{2} i \int \exp \left (-i d+a \log (f)-x (i e-b \log (f))-x^2 (i f-c \log (f))\right ) \, dx-\frac {1}{2} i \int \exp \left (i d+a \log (f)+x (i e+b \log (f))+x^2 (i f+c \log (f))\right ) \, dx\\ &=\frac {1}{2} \left (i \exp \left (-i d-\frac {(e+i b \log (f))^2}{4 i f-4 c \log (f)}\right ) f^a\right ) \int \exp \left (\frac {(-i e+b \log (f)+2 x (-i f+c \log (f)))^2}{4 (-i f+c \log (f))}\right ) \, dx-\frac {1}{2} \left (i \exp \left (i d+\frac {(e-i b \log (f))^2}{4 i f+4 c \log (f)}\right ) f^a\right ) \int \exp \left (\frac {(i e+b \log (f)+2 x (i f+c \log (f)))^2}{4 (i f+c \log (f))}\right ) \, dx\\ &=\frac {i \exp \left (-i d-\frac {(e+i b \log (f))^2}{4 i f-4 c \log (f)}\right ) f^a \sqrt {\pi } \text {erf}\left (\frac {i e-b \log (f)+2 x (i f-c \log (f))}{2 \sqrt {i f-c \log (f)}}\right )}{4 \sqrt {i f-c \log (f)}}-\frac {i \exp \left (i d+\frac {(e-i b \log (f))^2}{4 i f+4 c \log (f)}\right ) f^a \sqrt {\pi } \text {erfi}\left (\frac {i e+b \log (f)+2 x (i f+c \log (f))}{2 \sqrt {i f+c \log (f)}}\right )}{4 \sqrt {i f+c \log (f)}}\\ \end {align*}

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Mathematica [A]  time = 2.20, size = 347, normalized size = 1.64 \[ -\frac {\sqrt [4]{-1} \sqrt {\pi } f^{\frac {f (a f-b e)+a c^2 \log ^2(f)}{c^2 \log ^2(f)+f^2}} \exp \left (-\frac {1}{4} i \left (\frac {b^2 \log ^2(f)}{f+i c \log (f)}+\frac {e^2}{f-i c \log (f)}\right )\right ) \left (\sqrt {f-i c \log (f)} (f+i c \log (f)) (\cos (d)+i \sin (d)) e^{\frac {i b^2 f \log ^2(f)}{2 \left (c^2 \log ^2(f)+f^2\right )}} f^{\frac {b e}{2 f+2 i c \log (f)}} \text {erfi}\left (\frac {\sqrt [4]{-1} (-i \log (f) (b+2 c x)+e+2 f x)}{2 \sqrt {f-i c \log (f)}}\right )+(f-i c \log (f)) \sqrt {f+i c \log (f)} (\sin (d)+i \cos (d)) e^{\frac {i e^2 f}{2 \left (c^2 \log ^2(f)+f^2\right )}} f^{\frac {b e}{2 f-2 i c \log (f)}} \text {erfi}\left (\frac {(-1)^{3/4} (i \log (f) (b+2 c x)+e+2 f x)}{2 \sqrt {f+i c \log (f)}}\right )\right )}{4 \left (c^2 \log ^2(f)+f^2\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/4*((-1)^(1/4)*f^((f*(-(b*e) + a*f) + a*c^2*Log[f]^2)/(f^2 + c^2*Log[f]^2))*Sqrt[Pi]*(E^(((I/2)*b^2*f*Log[f]
^2)/(f^2 + c^2*Log[f]^2))*f^((b*e)/(2*f + (2*I)*c*Log[f]))*Erfi[((-1)^(1/4)*(e + 2*f*x - I*(b + 2*c*x)*Log[f])
)/(2*Sqrt[f - I*c*Log[f]])]*Sqrt[f - I*c*Log[f]]*(f + I*c*Log[f])*(Cos[d] + I*Sin[d]) + E^(((I/2)*e^2*f)/(f^2
+ c^2*Log[f]^2))*f^((b*e)/(2*f - (2*I)*c*Log[f]))*Erfi[((-1)^(3/4)*(e + 2*f*x + I*(b + 2*c*x)*Log[f]))/(2*Sqrt
[f + I*c*Log[f]])]*(f - I*c*Log[f])*Sqrt[f + I*c*Log[f]]*(I*Cos[d] + Sin[d])))/(E^((I/4)*(e^2/(f - I*c*Log[f])
+ (b^2*Log[f]^2)/(f + I*c*Log[f])))*(f^2 + c^2*Log[f]^2))

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fricas [B]  time = 1.04, size = 375, normalized size = 1.77 \[ \frac {\sqrt {\pi } {\left (i \, c \log \relax (f) + f\right )} \sqrt {-c \log \relax (f) - i \, f} \operatorname {erf}\left (\frac {{\left (2 \, f^{2} x + {\left (2 \, c^{2} x + b c\right )} \log \relax (f)^{2} + e f + {\left (i \, c e - i \, b f\right )} \log \relax (f)\right )} \sqrt {-c \log \relax (f) - i \, f}}{2 \, {\left (c^{2} \log \relax (f)^{2} + f^{2}\right )}}\right ) e^{\left (-\frac {{\left (b^{2} c - 4 \, a c^{2}\right )} \log \relax (f)^{3} + i \, e^{2} f - 4 i \, d f^{2} - {\left (4 i \, c^{2} d - 2 i \, b c e + i \, b^{2} f\right )} \log \relax (f)^{2} - {\left (c e^{2} - 2 \, b e f + 4 \, a f^{2}\right )} \log \relax (f)}{4 \, {\left (c^{2} \log \relax (f)^{2} + f^{2}\right )}}\right )} + \sqrt {\pi } {\left (-i \, c \log \relax (f) + f\right )} \sqrt {-c \log \relax (f) + i \, f} \operatorname {erf}\left (\frac {{\left (2 \, f^{2} x + {\left (2 \, c^{2} x + b c\right )} \log \relax (f)^{2} + e f + {\left (-i \, c e + i \, b f\right )} \log \relax (f)\right )} \sqrt {-c \log \relax (f) + i \, f}}{2 \, {\left (c^{2} \log \relax (f)^{2} + f^{2}\right )}}\right ) e^{\left (-\frac {{\left (b^{2} c - 4 \, a c^{2}\right )} \log \relax (f)^{3} - i \, e^{2} f + 4 i \, d f^{2} - {\left (-4 i \, c^{2} d + 2 i \, b c e - i \, b^{2} f\right )} \log \relax (f)^{2} - {\left (c e^{2} - 2 \, b e f + 4 \, a f^{2}\right )} \log \relax (f)}{4 \, {\left (c^{2} \log \relax (f)^{2} + f^{2}\right )}}\right )}}{4 \, {\left (c^{2} \log \relax (f)^{2} + f^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(pi)*(I*c*log(f) + f)*sqrt(-c*log(f) - I*f)*erf(1/2*(2*f^2*x + (2*c^2*x + b*c)*log(f)^2 + e*f + (I*c*
e - I*b*f)*log(f))*sqrt(-c*log(f) - I*f)/(c^2*log(f)^2 + f^2))*e^(-1/4*((b^2*c - 4*a*c^2)*log(f)^3 + I*e^2*f -
4*I*d*f^2 - (4*I*c^2*d - 2*I*b*c*e + I*b^2*f)*log(f)^2 - (c*e^2 - 2*b*e*f + 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*f^2*x + (2*c^2*x + b*c)*log(f)^2 + e*f + (
-I*c*e + I*b*f)*log(f))*sqrt(-c*log(f) + I*f)/(c^2*log(f)^2 + f^2))*e^(-1/4*((b^2*c - 4*a*c^2)*log(f)^3 - I*e^
2*f + 4*I*d*f^2 - (-4*I*c^2*d + 2*I*b*c*e - I*b^2*f)*log(f)^2 - (c*e^2 - 2*b*e*f + 4*a*f^2)*log(f))/(c^2*log(f
)^2 + f^2)))/(c^2*log(f)^2 + f^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.62, size = 216, normalized size = 1.02 \[ \frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {4 d f -e^{2}+2 i \ln \relax (f ) b e -4 i d \ln \relax (f ) c +\ln \relax (f )^{2} b^{2}}{4 \left (i f +c \ln \relax (f )\right )}} \erf \left (-\sqrt {-i f -c \ln \relax (f )}\, x +\frac {i e +b \ln \relax (f )}{2 \sqrt {-i f -c \ln \relax (f )}}\right )}{4 \sqrt {-i f -c \ln \relax (f )}}-\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {4 d f -e^{2}-2 i \ln \relax (f ) b e +4 i d \ln \relax (f ) c +\ln \relax (f )^{2} b^{2}}{4 \left (-i f +c \ln \relax (f )\right )}} \erf \left (-x \sqrt {i f -c \ln \relax (f )}+\frac {-i e +b \ln \relax (f )}{2 \sqrt {i f -c \ln \relax (f )}}\right )}{4 \sqrt {i f -c \ln \relax (f )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.39, size = 1007, normalized size = 4.75 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*(e^2*f - 4*d*f^2
- (4*c^2*d - 2*b*c*e + b^2*f)*log(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*(e^2*f - 4*d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)))*erf(1/2*(2*(c*log(f) - I*f
)*x + b*log(f) - I*e)*sqrt(-c*log(f) + I*f)/(c*log(f) - I*f)) + (f^(1/4*c*e^2/(c^2*log(f)^2 + f^2))*f^a*cos(-1
/4*(e^2*f - 4*d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(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*(e^2*f - 4*d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)))*erf(1/
2*(2*(c*log(f) + I*f)*x + b*log(f) + I*e)*sqrt(-c*log(f) - I*f)/(c*log(f) + I*f)))*sqrt(c*log(f) + sqrt(c^2*lo
g(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*(e^
2*f - 4*d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(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*(e^2*f - 4*d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(f)^2 + f^2)))*erf(1/2*(2*(c*
log(f) - I*f)*x + b*log(f) - I*e)*sqrt(-c*log(f) + I*f)/(c*log(f) - I*f)) + (-I*f^(1/4*c*e^2/(c^2*log(f)^2 + f
^2))*f^a*cos(-1/4*(e^2*f - 4*d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(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*(e^2*f - 4*d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(f)^2 +
f^2)))*erf(1/2*(2*(c*log(f) + I*f)*x + b*log(f) + I*e)*sqrt(-c*log(f) - I*f)/(c*log(f) + I*f)))*sqrt(-c*log(f)
+ sqrt(c^2*log(f)^2 + f^2)))/(c^2*e^(1/4*b^2*c*log(f)^3/(c^2*log(f)^2 + f^2) + 1/2*b*e*f*log(f)/(c^2*log(f)^2
+ f^2))*log(f)^2 + f^2*e^(1/4*b^2*c*log(f)^3/(c^2*log(f)^2 + f^2) + 1/2*b*e*f*log(f)/(c^2*log(f)^2 + f^2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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