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

Optimal. Leaf size=179 \[ \left (-\frac{1}{16}-\frac{i}{16}\right ) \sqrt{\pi } f^{a-\frac{1}{2}} e^{\frac{i (b \log (f)+2 i e)^2}{8 f}+2 i d} \text{Erf}\left (\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (b \log (f)+2 i e+4 i f x)}{\sqrt{f}}\right )-\left (\frac{1}{16}+\frac{i}{16}\right ) \sqrt{\pi } f^{a-\frac{1}{2}} e^{\frac{i (2 e+i b \log (f))^2}{8 f}-2 i d} \text{Erfi}\left (\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (-b \log (f)+2 i e+4 i f x)}{\sqrt{f}}\right )+\frac{f^{a+b x}}{2 b \log (f)} \]

[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])

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Rubi [A]  time = 0.283542, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4473, 2194, 2287, 2234, 2204, 2205} \[ \left (-\frac{1}{16}-\frac{i}{16}\right ) \sqrt{\pi } f^{a-\frac{1}{2}} e^{\frac{i (b \log (f)+2 i e)^2}{8 f}+2 i d} \text{Erf}\left (\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (b \log (f)+2 i e+4 i f x)}{\sqrt{f}}\right )-\left (\frac{1}{16}+\frac{i}{16}\right ) \sqrt{\pi } f^{a-\frac{1}{2}} e^{\frac{i (2 e+i b \log (f))^2}{8 f}-2 i d} \text{Erfi}\left (\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (-b \log (f)+2 i e+4 i f x)}{\sqrt{f}}\right )+\frac{f^{a+b x}}{2 b \log (f)} \]

Antiderivative was successfully verified.

[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 4473

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]

Rule 2194

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 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]

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]

Rubi steps

\begin{align*} \int f^{a+b x} \cos ^2\left (d+e x+f x^2\right ) \, dx &=\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]  time = 1.14282, size = 245, normalized size = 1.37 \[ \frac{f^{a-\frac{b e+f}{2 f}} e^{-\frac{i \left (b^2 \log ^2(f)+4 e^2\right )}{8 f}} \left (\sqrt [4]{-1} \sqrt{2 \pi } b \log (f) e^{\frac{i b^2 \log ^2(f)}{4 f}} (\sin (2 d)-i \cos (2 d)) \text{Erfi}\left (\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (-i b \log (f)+2 e+4 f x)}{\sqrt{f}}\right )+8 f^{b \left (\frac{e}{2 f}+x\right )+\frac{1}{2}} e^{\frac{i \left (b^2 \log ^2(f)+4 e^2\right )}{8 f}}-\sqrt [4]{-1} \sqrt{2 \pi } b e^{\frac{i e^2}{f}} \log (f) (\sin (2 d)+i \cos (2 d)) \text{Erf}\left (\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) (i b \log (f)+2 e+4 f x)}{\sqrt{f}}\right )\right )}{16 b \log (f)} \]

Antiderivative was successfully verified.

[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])

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Maple [A]  time = 0.14, size = 175, normalized size = 1. \begin{align*} -{\frac{\sqrt{2}{f}^{a}\sqrt{\pi }}{16}{{\rm e}^{{\frac{-{\frac{i}{8}} \left ( \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}-4\,i\ln \left ( f \right ) be-4\,{e}^{2}+16\,df \right ) }{f}}}}{\it Erf} \left ( -\sqrt{2}\sqrt{if}x+{\frac{ \left ( b\ln \left ( f \right ) -2\,ie \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{if}}}} \right ){\frac{1}{\sqrt{if}}}}-{\frac{{f}^{a}\sqrt{\pi }}{8}{{\rm e}^{{\frac{{\frac{i}{8}} \left ( \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}+4\,i\ln \left ( f \right ) be-4\,{e}^{2}+16\,df \right ) }{f}}}}{\it Erf} \left ( -\sqrt{-2\,if}x+{\frac{2\,ie+b\ln \left ( f \right ) }{2}{\frac{1}{\sqrt{-2\,if}}}} \right ){\frac{1}{\sqrt{-2\,if}}}}+{\frac{{f}^{bx+a}}{2\,b\ln \left ( f \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: IndexError

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Fricas [B]  time = 0.522357, size = 909, normalized size = 5.08 \begin{align*} \frac{2 \, \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 ) - 2 \, \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 ) - 2 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 ) - 2 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 ) + 8 \, f f^{b x + a}}{16 \, b f \log \left (f\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*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) - 2*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) - 2*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) - 2*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) + 8*f*f^(b*x + a))/(
b*f*log(f))

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [B]  time = 1.38031, size = 817, normalized size = 4.56 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))) - 1/2*I*(-2*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))) + 2*
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*sqrt(pi)*erf(-1/8*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*sqrt(pi)*erf(-1/8*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))