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

Optimal. Leaf size=162 \[ \frac {1}{4} (-1)^{3/4} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {1}{4} i \left (4 d+\frac {(b \log (f)+i e)^2}{f}\right )} \text {erf}\left (\frac {\sqrt [4]{-1} (b \log (f)+i e+2 i f x)}{2 \sqrt {f}}\right )-\frac {1}{4} (-1)^{3/4} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {i (e+i b \log (f))^2}{4 f}-i d} \text {erfi}\left (\frac {\sqrt [4]{-1} (-b \log (f)+i e+2 i f x)}{2 \sqrt {f}}\right ) \]

[Out]

1/4*(-1)^(3/4)*exp(1/4*I*(4*d+(I*e+b*ln(f))^2/f))*f^(-1/2+a)*erf(1/2*(-1)^(1/4)*(I*e+2*I*f*x+b*ln(f))/f^(1/2))
*Pi^(1/2)-1/4*(-1)^(3/4)*exp(-I*d+1/4*I*(e+I*b*ln(f))^2/f)*f^(-1/2+a)*erfi(1/2*(-1)^(1/4)*(I*e+2*I*f*x-b*ln(f)
)/f^(1/2))*Pi^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.33, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {4472, 2287, 2234, 2204, 2205} \[ \frac {1}{4} (-1)^{3/4} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {1}{4} i \left (4 d+\frac {(b \log (f)+i e)^2}{f}\right )} \text {Erf}\left (\frac {\sqrt [4]{-1} (b \log (f)+i e+2 i f x)}{2 \sqrt {f}}\right )-\frac {1}{4} (-1)^{3/4} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {i (e+i b \log (f))^2}{4 f}-i d} \text {Erfi}\left (\frac {\sqrt [4]{-1} (-b \log (f)+i e+2 i f x)}{2 \sqrt {f}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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} \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}-\frac {1}{2} i e^{i d+i e x+i f x^2} f^{a+b x}\right ) \, dx\\ &=\frac {1}{2} i \int e^{-i d-i e x-i f x^2} f^{a+b x} \, dx-\frac {1}{2} i \int e^{i d+i e x+i f x^2} f^{a+b x} \, dx\\ &=\frac {1}{2} i \int \exp \left (-i d-i f x^2+a \log (f)-x (i e-b \log (f))\right ) \, dx-\frac {1}{2} i \int \exp \left (i d+i f x^2+a \log (f)+x (i e+b \log (f))\right ) \, dx\\ &=\frac {1}{2} \left (i e^{-i d+\frac {i (e+i b \log (f))^2}{4 f}} f^a\right ) \int e^{\frac {i (-i e-2 i f x+b \log (f))^2}{4 f}} \, dx-\frac {1}{2} \left (i e^{\frac {1}{4} i \left (4 d+\frac {(i e+b \log (f))^2}{f}\right )} f^a\right ) \int e^{-\frac {i (i e+2 i f x+b \log (f))^2}{4 f}} \, dx\\ &=\frac {1}{4} (-1)^{3/4} e^{\frac {1}{4} i \left (4 d+\frac {(i e+b \log (f))^2}{f}\right )} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erf}\left (\frac {\sqrt [4]{-1} (i e+2 i f x+b \log (f))}{2 \sqrt {f}}\right )-\frac {1}{4} (-1)^{3/4} e^{-i d+\frac {i (e+i b \log (f))^2}{4 f}} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt [4]{-1} (i e+2 i f x-b \log (f))}{2 \sqrt {f}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.38, size = 162, normalized size = 1.00 \[ -\frac {1}{4} \sqrt [4]{-1} \sqrt {\pi } f^{a-\frac {b e+f}{2 f}} e^{-\frac {i \left (b^2 \log ^2(f)+e^2\right )}{4 f}} \left (e^{\frac {i b^2 \log ^2(f)}{2 f}} (\cos (d)+i \sin (d)) \text {erfi}\left (\frac {\sqrt [4]{-1} (-i b \log (f)+e+2 f x)}{2 \sqrt {f}}\right )+e^{\frac {i e^2}{2 f}} (\sin (d)+i \cos (d)) \text {erfi}\left (\frac {(-1)^{3/4} (i b \log (f)+e+2 f x)}{2 \sqrt {f}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 1.36, size = 313, normalized size = 1.93 \[ \frac {i \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \relax (f)^{2} + i \, e^{2} - 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{4 \, f}\right )} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, f x + i \, b \log \relax (f) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) + i \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \relax (f)^{2} - i \, e^{2} + 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{4 \, f}\right )} \operatorname {C}\left (-\frac {\sqrt {2} {\left (2 \, f x - i \, b \log \relax (f) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) + \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \relax (f)^{2} + i \, e^{2} - 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{4 \, f}\right )} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, f x + i \, b \log \relax (f) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) - \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \relax (f)^{2} - i \, e^{2} + 4 i \, d f - 2 \, {\left (b e - 2 \, a f\right )} \log \relax (f)}{4 \, f}\right )} \operatorname {S}\left (-\frac {\sqrt {2} {\left (2 \, f x - i \, b \log \relax (f) + e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right )}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(I*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(-I*b^2*log(f)^2 + I*e^2 - 4*I*d*f - 2*(b*e - 2*a*f)*log(f))/f)*fresnel_co
s(1/2*sqrt(2)*(2*f*x + I*b*log(f) + e)*sqrt(f/pi)/f) + I*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(I*b^2*log(f)^2 - I*e^2
+ 4*I*d*f - 2*(b*e - 2*a*f)*log(f))/f)*fresnel_cos(-1/2*sqrt(2)*(2*f*x - I*b*log(f) + e)*sqrt(f/pi)/f) + sqrt(
2)*pi*sqrt(f/pi)*e^(1/4*(-I*b^2*log(f)^2 + I*e^2 - 4*I*d*f - 2*(b*e - 2*a*f)*log(f))/f)*fresnel_sin(1/2*sqrt(2
)*(2*f*x + I*b*log(f) + e)*sqrt(f/pi)/f) - sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(I*b^2*log(f)^2 - I*e^2 + 4*I*d*f - 2*
(b*e - 2*a*f)*log(f))/f)*fresnel_sin(-1/2*sqrt(2)*(2*f*x - I*b*log(f) + e)*sqrt(f/pi)/f))/f

________________________________________________________________________________________

giac [B]  time = 0.29, size = 384, normalized size = 2.37 \[ \frac {i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{8} \, \sqrt {2} {\left (4 \, x - \frac {\pi b \mathrm {sgn}\relax (f) - \pi b + 2 i \, b \log \left ({\left | f \right |}\right ) - 2 \, e}{f}\right )} {\left (-\frac {i \, f}{{\left | f \right |}} + 1\right )} \sqrt {{\left | f \right |}}\right ) e^{\left (\frac {i \, \pi ^{2} b^{2} \mathrm {sgn}\relax (f)}{8 \, f} + \frac {\pi b^{2} \log \left ({\left | f \right |}\right ) \mathrm {sgn}\relax (f)}{4 \, f} - \frac {i \, \pi ^{2} b^{2}}{8 \, f} - \frac {\pi b^{2} \log \left ({\left | f \right |}\right )}{4 \, f} + \frac {i \, b^{2} \log \left ({\left | f \right |}\right )^{2}}{4 \, f} - \frac {1}{2} i \, \pi a \mathrm {sgn}\relax (f) + \frac {i \, \pi b e \mathrm {sgn}\relax (f)}{4 \, f} + \frac {1}{2} i \, \pi a - \frac {i \, \pi b e}{4 \, f} + a \log \left ({\left | f \right |}\right ) - \frac {b e \log \left ({\left | f \right |}\right )}{2 \, f} + i \, d - \frac {i \, e^{2}}{4 \, f}\right )}}{4 \, {\left (-\frac {i \, f}{{\left | f \right |}} + 1\right )} \sqrt {{\left | f \right |}}} - \frac {i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{8} \, \sqrt {2} {\left (4 \, x + \frac {\pi b \mathrm {sgn}\relax (f) - \pi b + 2 i \, b \log \left ({\left | f \right |}\right ) + 2 \, e}{f}\right )} {\left (\frac {i \, f}{{\left | f \right |}} + 1\right )} \sqrt {{\left | f \right |}}\right ) e^{\left (-\frac {i \, \pi ^{2} b^{2} \mathrm {sgn}\relax (f)}{8 \, f} - \frac {\pi b^{2} \log \left ({\left | f \right |}\right ) \mathrm {sgn}\relax (f)}{4 \, f} + \frac {i \, \pi ^{2} b^{2}}{8 \, f} + \frac {\pi b^{2} \log \left ({\left | f \right |}\right )}{4 \, f} - \frac {i \, b^{2} \log \left ({\left | f \right |}\right )^{2}}{4 \, f} - \frac {1}{2} i \, \pi a \mathrm {sgn}\relax (f) + \frac {i \, \pi b e \mathrm {sgn}\relax (f)}{4 \, f} + \frac {1}{2} i \, \pi a - \frac {i \, \pi b e}{4 \, f} + a \log \left ({\left | f \right |}\right ) - \frac {b e \log \left ({\left | f \right |}\right )}{2 \, f} - i \, d + \frac {i \, e^{2}}{4 \, f}\right )}}{4 \, {\left (\frac {i \, f}{{\left | f \right |}} + 1\right )} \sqrt {{\left | f \right |}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I*sqrt(2)*sqrt(pi)*erf(-1/8*sqrt(2)*(4*x - (pi*b*sgn(f) - pi*b + 2*I*b*log(abs(f)) - 2*e)/f)*(-I*f/abs(f)
+ 1)*sqrt(abs(f)))*e^(1/8*I*pi^2*b^2*sgn(f)/f + 1/4*pi*b^2*log(abs(f))*sgn(f)/f - 1/8*I*pi^2*b^2/f - 1/4*pi*b^
2*log(abs(f))/f + 1/4*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/4*I*pi*b*e*sgn(f)/f + 1/2*I*pi*a - 1/4*I*p
i*b*e/f + a*log(abs(f)) - 1/2*b*e*log(abs(f))/f + I*d - 1/4*I*e^2/f)/((-I*f/abs(f) + 1)*sqrt(abs(f))) - 1/4*I*
sqrt(2)*sqrt(pi)*erf(-1/8*sqrt(2)*(4*x + (pi*b*sgn(f) - pi*b + 2*I*b*log(abs(f)) + 2*e)/f)*(I*f/abs(f) + 1)*sq
rt(abs(f)))*e^(-1/8*I*pi^2*b^2*sgn(f)/f - 1/4*pi*b^2*log(abs(f))*sgn(f)/f + 1/8*I*pi^2*b^2/f + 1/4*pi*b^2*log(
abs(f))/f - 1/4*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/4*I*pi*b*e*sgn(f)/f + 1/2*I*pi*a - 1/4*I*pi*b*e/
f + a*log(abs(f)) - 1/2*b*e*log(abs(f))/f - I*d + 1/4*I*e^2/f)/((I*f/abs(f) + 1)*sqrt(abs(f)))

________________________________________________________________________________________

maple [A]  time = 0.43, size = 152, normalized size = 0.94 \[ \frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {i \left (-e^{2}+2 i \ln \relax (f ) b e +\ln \relax (f )^{2} b^{2}+4 d f \right )}{4 f}} \erf \left (-\sqrt {-i f}\, x +\frac {i e +b \ln \relax (f )}{2 \sqrt {-i f}}\right )}{4 \sqrt {-i f}}-\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {i \left (-e^{2}-2 i \ln \relax (f ) b e +\ln \relax (f )^{2} b^{2}+4 d f \right )}{4 f}} \erf \left (-\sqrt {i f}\, x +\frac {-i e +b \ln \relax (f )}{2 \sqrt {i f}}\right )}{4 \sqrt {i f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 190, normalized size = 1.17 \[ -\frac {\sqrt {2} \sqrt {\pi } {\left ({\left (\left (i + 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \relax (f)^{2} - e^{2} + 4 \, d f}{4 \, f}\right ) - \left (i - 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \relax (f)^{2} - e^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (2 i \, f x - b \log \relax (f) + i \, e\right )} \sqrt {i \, f}}{2 \, f}\right ) + {\left (-\left (i - 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \relax (f)^{2} - e^{2} + 4 \, d f}{4 \, f}\right ) + \left (i + 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \relax (f)^{2} - e^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (2 i \, f x + b \log \relax (f) + i \, e\right )} \sqrt {-i \, f}}{2 \, f}\right )\right )}}{8 \, \sqrt {f} f^{\frac {b e}{2 \, f}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(2)*sqrt(pi)*(((I + 1)*f^a*cos(1/4*(b^2*log(f)^2 - e^2 + 4*d*f)/f) - (I - 1)*f^a*sin(1/4*(b^2*log(f)^
2 - e^2 + 4*d*f)/f))*erf(1/2*I*(2*I*f*x - b*log(f) + I*e)*sqrt(I*f)/f) + (-(I - 1)*f^a*cos(1/4*(b^2*log(f)^2 -
e^2 + 4*d*f)/f) + (I + 1)*f^a*sin(1/4*(b^2*log(f)^2 - e^2 + 4*d*f)/f))*erf(1/2*I*(2*I*f*x + b*log(f) + I*e)*s
qrt(-I*f)/f))/(sqrt(f)*f^(1/2*b*e/f))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________