\(\int f^{a+b x} \cos ^2(d+e x+f x^2) \, dx\) [114]

Optimal result

Integrand size = 21, antiderivative size = 179 \[ \int f^{a+b x} \cos ^2\left (d+e x+f x^2\right ) \, dx=\left (-\frac {1}{16}-\frac {i}{16}\right ) e^{2 i d+\frac {i (2 i e+b \log (f))^2}{8 f}} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erf}\left (\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (2 i e+4 i f x+b \log (f))}{\sqrt {f}}\right )-\left (\frac {1}{16}+\frac {i}{16}\right ) e^{-2 i d+\frac {i (2 e+i b \log (f))^2}{8 f}} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erfi}\left (\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (2 i e+4 i f x-b \log (f))}{\sqrt {f}}\right )+\frac {f^{a+b x}}{2 b \log (f)} \]

[Out]

1/2*f^(b*x+a)/b/ln(f)-(1/16+1/16*I)*exp(2*I*d+1/8*I*(2*I*e+b*ln(f))^2/f)*f^(-1/2+a)*erf((1/4+1/4*I)*(2*I*e+4*I
*f*x+b*ln(f))/f^(1/2))*Pi^(1/2)-(1/16+1/16*I)*exp(-2*I*d+1/8*I*(2*e+I*b*ln(f))^2/f)*f^(-1/2+a)*erfi((1/4+1/4*I
)*(2*I*e+4*I*f*x-b*ln(f))/f^(1/2))*Pi^(1/2)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4561, 2225, 2325, 2266, 2235, 2236} \[ \int f^{a+b x} \cos ^2\left (d+e x+f x^2\right ) \, dx=\left (-\frac {1}{16}-\frac {i}{16}\right ) \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {i (b \log (f)+2 i e)^2}{8 f}+2 i d} \text {erf}\left (\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (b \log (f)+2 i e+4 i f x)}{\sqrt {f}}\right )-\left (\frac {1}{16}+\frac {i}{16}\right ) \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {i (2 e+i b \log (f))^2}{8 f}-2 i d} \text {erfi}\left (\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (-b \log (f)+2 i e+4 i f x)}{\sqrt {f}}\right )+\frac {f^{a+b x}}{2 b \log (f)} \]

[In]

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

[Out]

(-1/16 - I/16)*E^((2*I)*d + ((I/8)*((2*I)*e + b*Log[f])^2)/f)*f^(-1/2 + a)*Sqrt[Pi]*Erf[((1/4 + I/4)*((2*I)*e
+ (4*I)*f*x + b*Log[f]))/Sqrt[f]] - (1/16 + I/16)*E^((-2*I)*d + ((I/8)*(2*e + I*b*Log[f])^2)/f)*f^(-1/2 + a)*S
qrt[Pi]*Erfi[((1/4 + I/4)*((2*I)*e + (4*I)*f*x - b*Log[f]))/Sqrt[f]] + f^(a + b*x)/(2*b*Log[f])

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

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 2236

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 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 4561

Int[Cos[v_]^(n_.)*(F_)^(u_), x_Symbol] :> Int[ExpandTrigToExp[F^u, Cos[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} f^{a+b x}+\frac {1}{4} e^{-2 i d-2 i e x-2 i f x^2} f^{a+b x}+\frac {1}{4} e^{2 i d+2 i e x+2 i f x^2} f^{a+b x}\right ) \, dx \\ & = \frac {1}{4} \int e^{-2 i d-2 i e x-2 i f x^2} f^{a+b x} \, dx+\frac {1}{4} \int e^{2 i d+2 i e x+2 i f x^2} f^{a+b x} \, dx+\frac {1}{2} \int f^{a+b x} \, dx \\ & = \frac {f^{a+b x}}{2 b \log (f)}+\frac {1}{4} \int \exp \left (-2 i d-2 i f x^2+a \log (f)-x (2 i e-b \log (f))\right ) \, dx+\frac {1}{4} \int \exp \left (2 i d+2 i f x^2+a \log (f)+x (2 i e+b \log (f))\right ) \, dx \\ & = \frac {f^{a+b x}}{2 b \log (f)}+\frac {1}{4} \exp \left (-2 i d+a \log (f)-\frac {i (-2 i e+b \log (f))^2}{8 f}\right ) \int e^{\frac {i (-2 i e-4 i f x+b \log (f))^2}{8 f}} \, dx+\frac {1}{4} \left (e^{2 i d+\frac {i (2 i e+b \log (f))^2}{8 f}} f^a\right ) \int e^{-\frac {i (2 i e+4 i f x+b \log (f))^2}{8 f}} \, dx \\ & = \left (-\frac {1}{16}-\frac {i}{16}\right ) e^{2 i d+\frac {i (2 i e+b \log (f))^2}{8 f}} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erf}\left (\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (2 i e+4 i f x+b \log (f))}{\sqrt {f}}\right )-\left (\frac {1}{16}+\frac {i}{16}\right ) \exp \left (-\frac {1}{8} i \left (16 d+\frac {(2 i e-b \log (f))^2}{f}\right )\right ) f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erfi}\left (\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (2 i e+4 i f x-b \log (f))}{\sqrt {f}}\right )+\frac {f^{a+b x}}{2 b \log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.37 \[ \int f^{a+b x} \cos ^2\left (d+e x+f x^2\right ) \, dx=\frac {e^{-\frac {i \left (4 e^2+b^2 \log ^2(f)\right )}{8 f}} f^{a-\frac {b e+f}{2 f}} \left (8 e^{\frac {i \left (4 e^2+b^2 \log ^2(f)\right )}{8 f}} f^{\frac {1}{2}+b \left (\frac {e}{2 f}+x\right )}+\sqrt [4]{-1} b e^{\frac {i b^2 \log ^2(f)}{4 f}} \sqrt {2 \pi } \text {erfi}\left (\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (2 e+4 f x-i b \log (f))}{\sqrt {f}}\right ) \log (f) (-i \cos (2 d)+\sin (2 d))-\sqrt [4]{-1} b e^{\frac {i e^2}{f}} \sqrt {2 \pi } \text {erf}\left (\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (2 e+4 f x+i b \log (f))}{\sqrt {f}}\right ) \log (f) (i \cos (2 d)+\sin (2 d))\right )}{16 b \log (f)} \]

[In]

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

[Out]

(f^(a - (b*e + f)/(2*f))*(8*E^(((I/8)*(4*e^2 + b^2*Log[f]^2))/f)*f^(1/2 + b*(e/(2*f) + x)) + (-1)^(1/4)*b*E^((
(I/4)*b^2*Log[f]^2)/f)*Sqrt[2*Pi]*Erfi[((1/4 + I/4)*(2*e + 4*f*x - I*b*Log[f]))/Sqrt[f]]*Log[f]*((-I)*Cos[2*d]
+ Sin[2*d]) - (-1)^(1/4)*b*E^((I*e^2)/f)*Sqrt[2*Pi]*Erf[((1/4 + I/4)*(2*e + 4*f*x + I*b*Log[f]))/Sqrt[f]]*Log
[f]*(I*Cos[2*d] + Sin[2*d])))/(16*b*E^(((I/8)*(4*e^2 + b^2*Log[f]^2))/f)*Log[f])

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00

method	result	size
risch	\(-\frac {\sqrt {\pi }\, f^{a} f^{-\frac {b e}{2 f}} {\mathrm e}^{-\frac {i \left (\ln \left (f \right )^{2} b^{2}+16 d f -4 e^{2}\right )}{8 f}} \sqrt {2}\, \operatorname {erf}\left (-\sqrt {2}\, \sqrt {i f}\, x +\frac {\left (b \ln \left (f \right )-2 i e \right ) \sqrt {2}}{4 \sqrt {i f}}\right )}{16 \sqrt {i f}}-\frac {\sqrt {\pi }\, f^{a} f^{-\frac {b e}{2 f}} {\mathrm e}^{\frac {i \left (\ln \left (f \right )^{2} b^{2}+16 d f -4 e^{2}\right )}{8 f}} \operatorname {erf}\left (-\sqrt {-2 i f}\, x +\frac {2 i e +b \ln \left (f \right )}{2 \sqrt {-2 i f}}\right )}{8 \sqrt {-2 i f}}+\frac {f^{x b +a}}{2 b \ln \left (f \right )}\)	\(179\)

[In]

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

[Out]

-1/16*Pi^(1/2)*f^a*f^(-1/2/f*b*e)*exp(-1/8*I*(ln(f)^2*b^2+16*d*f-4*e^2)/f)*2^(1/2)/(I*f)^(1/2)*erf(-2^(1/2)*(I
*f)^(1/2)*x+1/4*(b*ln(f)-2*I*e)*2^(1/2)/(I*f)^(1/2))-1/8*Pi^(1/2)*f^a*f^(-1/2/f*b*e)*exp(1/8*I*(ln(f)^2*b^2+16
*d*f-4*e^2)/f)/(-2*I*f)^(1/2)*erf(-(-2*I*f)^(1/2)*x+1/2*(2*I*e+b*ln(f))/(-2*I*f)^(1/2))+1/2*f^(b*x+a)/b/ln(f)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (116) = 232\).

Time = 0.26 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.82 \[ \int f^{a+b x} \cos ^2\left (d+e x+f x^2\right ) \, dx=\frac {\pi b \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \left (f\right )^{2} + 4 i \, e^{2} - 16 i \, d f - 4 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{8 \, f}\right )} \operatorname {C}\left (\frac {{\left (4 \, f x + i \, b \log \left (f\right ) + 2 \, e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) \log \left (f\right ) - \pi b \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \left (f\right )^{2} - 4 i \, e^{2} + 16 i \, d f - 4 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{8 \, f}\right )} \operatorname {C}\left (-\frac {{\left (4 \, f x - i \, b \log \left (f\right ) + 2 \, e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) \log \left (f\right ) - i \, \pi b \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \left (f\right )^{2} + 4 i \, e^{2} - 16 i \, d f - 4 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{8 \, f}\right )} \operatorname {S}\left (\frac {{\left (4 \, f x + i \, b \log \left (f\right ) + 2 \, e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) \log \left (f\right ) - i \, \pi b \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \left (f\right )^{2} - 4 i \, e^{2} + 16 i \, d f - 4 \, {\left (b e - 2 \, a f\right )} \log \left (f\right )}{8 \, f}\right )} \operatorname {S}\left (-\frac {{\left (4 \, f x - i \, b \log \left (f\right ) + 2 \, e\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) \log \left (f\right ) + 4 \, f f^{b x + a}}{8 \, b f \log \left (f\right )} \]

[In]

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

[Out]

1/8*(pi*b*sqrt(f/pi)*e^(1/8*(-I*b^2*log(f)^2 + 4*I*e^2 - 16*I*d*f - 4*(b*e - 2*a*f)*log(f))/f)*fresnel_cos(1/2
*(4*f*x + I*b*log(f) + 2*e)*sqrt(f/pi)/f)*log(f) - pi*b*sqrt(f/pi)*e^(1/8*(I*b^2*log(f)^2 - 4*I*e^2 + 16*I*d*f
- 4*(b*e - 2*a*f)*log(f))/f)*fresnel_cos(-1/2*(4*f*x - I*b*log(f) + 2*e)*sqrt(f/pi)/f)*log(f) - I*pi*b*sqrt(f
/pi)*e^(1/8*(-I*b^2*log(f)^2 + 4*I*e^2 - 16*I*d*f - 4*(b*e - 2*a*f)*log(f))/f)*fresnel_sin(1/2*(4*f*x + I*b*lo
g(f) + 2*e)*sqrt(f/pi)/f)*log(f) - I*pi*b*sqrt(f/pi)*e^(1/8*(I*b^2*log(f)^2 - 4*I*e^2 + 16*I*d*f - 4*(b*e - 2*
a*f)*log(f))/f)*fresnel_sin(-1/2*(4*f*x - I*b*log(f) + 2*e)*sqrt(f/pi)/f)*log(f) + 4*f*f^(b*x + a))/(b*f*log(f
))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (116) = 232\).

Time = 0.32 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.34 \[ \int f^{a+b x} \cos ^2\left (d+e x+f x^2\right ) \, dx=-\frac {4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } {\left ({\left (-\left (i - 1\right ) \, b f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} - 4 \, e^{2} + 16 \, d f}{8 \, f}\right ) \log \left (f\right ) - \left (i + 1\right ) \, b f^{a} \log \left (f\right ) \sin \left (\frac {b^{2} \log \left (f\right )^{2} - 4 \, e^{2} + 16 \, d f}{8 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (4 i \, f x - b \log \left (f\right ) + 2 i \, e\right )} \sqrt {2 i \, f}}{4 \, f}\right ) + {\left (\left (i + 1\right ) \, b f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} - 4 \, e^{2} + 16 \, d f}{8 \, f}\right ) \log \left (f\right ) + \left (i - 1\right ) \, b f^{a} \log \left (f\right ) \sin \left (\frac {b^{2} \log \left (f\right )^{2} - 4 \, e^{2} + 16 \, d f}{8 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (4 i \, f x + b \log \left (f\right ) + 2 i \, e\right )} \sqrt {-2 i \, f}}{4 \, f}\right )\right )} f^{\frac {3}{2}} - 16 \, f^{a + 2} e^{\left (b x \log \left (f\right ) + \frac {b e \log \left (f\right )}{2 \, f}\right )}}{32 \, b f^{2} f^{\frac {b e}{2 \, f}} \log \left (f\right )} \]

[In]

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

[Out]

-1/32*(4^(1/4)*sqrt(2)*sqrt(pi)*((-(I - 1)*b*f^a*cos(1/8*(b^2*log(f)^2 - 4*e^2 + 16*d*f)/f)*log(f) - (I + 1)*b
*f^a*log(f)*sin(1/8*(b^2*log(f)^2 - 4*e^2 + 16*d*f)/f))*erf(1/4*I*(4*I*f*x - b*log(f) + 2*I*e)*sqrt(2*I*f)/f)
+ ((I + 1)*b*f^a*cos(1/8*(b^2*log(f)^2 - 4*e^2 + 16*d*f)/f)*log(f) + (I - 1)*b*f^a*log(f)*sin(1/8*(b^2*log(f)^
2 - 4*e^2 + 16*d*f)/f))*erf(1/4*I*(4*I*f*x + b*log(f) + 2*I*e)*sqrt(-2*I*f)/f))*f^(3/2) - 16*f^(a + 2)*e^(b*x*
log(f) + 1/2*b*e*log(f)/f))/(b*f^2*f^(1/2*b*e/f)*log(f))

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 599 vs. \(2 (116) = 232\).

Time = 0.37 (sec) , antiderivative size = 599, normalized size of antiderivative = 3.35 \[ \int f^{a+b x} \cos ^2\left (d+e x+f x^2\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

(2*b*cos(-1/2*pi*b*x*sgn(f) + 1/2*pi*b*x - 1/2*pi*a*sgn(f) + 1/2*pi*a)*log(abs(f))/(4*b^2*log(abs(f))^2 + (pi*
b*sgn(f) - pi*b)^2) - (pi*b*sgn(f) - pi*b)*sin(-1/2*pi*b*x*sgn(f) + 1/2*pi*b*x - 1/2*pi*a*sgn(f) + 1/2*pi*a)/(
4*b^2*log(abs(f))^2 + (pi*b*sgn(f) - pi*b)^2))*e^(b*x*log(abs(f)) + a*log(abs(f))) + I*(I*e^(1/2*I*pi*b*x*sgn(
f) - 1/2*I*pi*b*x + 1/2*I*pi*a*sgn(f) - 1/2*I*pi*a)/(2*I*pi*b*sgn(f) - 2*I*pi*b + 4*b*log(abs(f))) - I*e^(-1/2
*I*pi*b*x*sgn(f) + 1/2*I*pi*b*x - 1/2*I*pi*a*sgn(f) + 1/2*I*pi*a)/(-2*I*pi*b*sgn(f) + 2*I*pi*b + 4*b*log(abs(f
))))*e^(b*x*log(abs(f)) + a*log(abs(f))) + 1/8*I*sqrt(pi)*erf(-1/8*I*sqrt(f)*(8*x - (pi*b*sgn(f) - pi*b + 2*I*
b*log(abs(f)) - 4*e)/f)*(I*f/abs(f) + 1))*e^(1/16*I*pi^2*b^2*sgn(f)/f + 1/8*pi*b^2*log(abs(f))*sgn(f)/f - 1/16
*I*pi^2*b^2/f - 1/8*pi*b^2*log(abs(f))/f + 1/8*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 + 2*I*d - 1/2*I*e^2/f)/(sqrt(f)*(I*f/
abs(f) + 1)) - 1/8*I*sqrt(pi)*erf(1/8*I*sqrt(f)*(8*x + (pi*b*sgn(f) - pi*b + 2*I*b*log(abs(f)) + 4*e)/f)*(-I*f
/abs(f) + 1))*e^(-1/16*I*pi^2*b^2*sgn(f)/f - 1/8*pi*b^2*log(abs(f))*sgn(f)/f + 1/16*I*pi^2*b^2/f + 1/8*pi*b^2*
log(abs(f))/f - 1/8*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 - 2*I*d + 1/2*I*e^2/f)/(sqrt(f)*(-I*f/abs(f) + 1))

Mupad [F(-1)]

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

[In]

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

[Out]

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