3.112 \(\int f^{a+b x} \cos ^3(d+f x^2) \, dx\)

Optimal. Leaf size=298 \[ -\frac {3}{16} \sqrt [4]{-1} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {1}{4} i \left (\frac {b^2 \log ^2(f)}{f}+4 d\right )} \text {erf}\left (\frac {\sqrt [4]{-1} (b \log (f)+2 i f x)}{2 \sqrt {f}}\right )-\left (\frac {1}{16}+\frac {i}{16}\right ) \sqrt {\frac {\pi }{6}} f^{a-\frac {1}{2}} e^{\frac {i b^2 \log ^2(f)}{12 f}+3 i d} \text {erf}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (b \log (f)+6 i f x)}{\sqrt {6} \sqrt {f}}\right )-\frac {3}{16} \sqrt [4]{-1} \sqrt {\pi } f^{a-\frac {1}{2}} e^{-\frac {1}{4} i \left (\frac {b^2 \log ^2(f)}{f}+4 d\right )} \text {erfi}\left (\frac {\sqrt [4]{-1} (-b \log (f)+2 i f x)}{2 \sqrt {f}}\right )-\left (\frac {1}{16}+\frac {i}{16}\right ) \sqrt {\frac {\pi }{6}} f^{a-\frac {1}{2}} e^{-\frac {1}{12} i \left (\frac {b^2 \log ^2(f)}{f}+36 d\right )} \text {erfi}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (-b \log (f)+6 i f x)}{\sqrt {6} \sqrt {f}}\right ) \]

[Out]

(-1/96-1/96*I)*exp(3*I*d+1/12*I*b^2*ln(f)^2/f)*f^(-1/2+a)*erf((1/12+1/12*I)*(6*I*f*x+b*ln(f))*6^(1/2)/f^(1/2))
*6^(1/2)*Pi^(1/2)-(1/96+1/96*I)*f^(-1/2+a)*erfi((1/12+1/12*I)*(6*I*f*x-b*ln(f))*6^(1/2)/f^(1/2))*6^(1/2)*Pi^(1
/2)/exp(1/12*I*(36*d+b^2*ln(f)^2/f))-3/16*(-1)^(1/4)*exp(1/4*I*(4*d+b^2*ln(f)^2/f))*f^(-1/2+a)*erf(1/2*(-1)^(1
/4)*(2*I*f*x+b*ln(f))/f^(1/2))*Pi^(1/2)-3/16*(-1)^(1/4)*f^(-1/2+a)*erfi(1/2*(-1)^(1/4)*(2*I*f*x-b*ln(f))/f^(1/
2))*Pi^(1/2)/exp(1/4*I*(4*d+b^2*ln(f)^2/f))

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Rubi [A]  time = 0.33, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {4473, 2287, 2234, 2204, 2205} \[ -\frac {3}{16} \sqrt [4]{-1} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {1}{4} i \left (\frac {b^2 \log ^2(f)}{f}+4 d\right )} \text {Erf}\left (\frac {\sqrt [4]{-1} (b \log (f)+2 i f x)}{2 \sqrt {f}}\right )-\left (\frac {1}{16}+\frac {i}{16}\right ) \sqrt {\frac {\pi }{6}} f^{a-\frac {1}{2}} e^{\frac {i b^2 \log ^2(f)}{12 f}+3 i d} \text {Erf}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (b \log (f)+6 i f x)}{\sqrt {6} \sqrt {f}}\right )-\frac {3}{16} \sqrt [4]{-1} \sqrt {\pi } f^{a-\frac {1}{2}} e^{-\frac {1}{4} i \left (\frac {b^2 \log ^2(f)}{f}+4 d\right )} \text {Erfi}\left (\frac {\sqrt [4]{-1} (-b \log (f)+2 i f x)}{2 \sqrt {f}}\right )-\left (\frac {1}{16}+\frac {i}{16}\right ) \sqrt {\frac {\pi }{6}} f^{a-\frac {1}{2}} e^{-\frac {1}{12} i \left (\frac {b^2 \log ^2(f)}{f}+36 d\right )} \text {Erfi}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (-b \log (f)+6 i f x)}{\sqrt {6} \sqrt {f}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x)*Cos[d + f*x^2]^3,x]

[Out]

(-3*(-1)^(1/4)*E^((I/4)*(4*d + (b^2*Log[f]^2)/f))*f^(-1/2 + a)*Sqrt[Pi]*Erf[((-1)^(1/4)*((2*I)*f*x + b*Log[f])
)/(2*Sqrt[f])])/16 - (1/16 + I/16)*E^((3*I)*d + ((I/12)*b^2*Log[f]^2)/f)*f^(-1/2 + a)*Sqrt[Pi/6]*Erf[((1/2 + I
/2)*((6*I)*f*x + b*Log[f]))/(Sqrt[6]*Sqrt[f])] - (3*(-1)^(1/4)*f^(-1/2 + a)*Sqrt[Pi]*Erfi[((-1)^(1/4)*((2*I)*f
*x - b*Log[f]))/(2*Sqrt[f])])/(16*E^((I/4)*(4*d + (b^2*Log[f]^2)/f))) - ((1/16 + I/16)*f^(-1/2 + a)*Sqrt[Pi/6]
*Erfi[((1/2 + I/2)*((6*I)*f*x - b*Log[f]))/(Sqrt[6]*Sqrt[f])])/E^((I/12)*(36*d + (b^2*Log[f]^2)/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 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]

Rubi steps

\begin {align*} \int f^{a+b x} \cos ^3\left (d+f x^2\right ) \, dx &=\int \left (\frac {3}{8} e^{-i d-i f x^2} f^{a+b x}+\frac {3}{8} e^{i d+i f x^2} f^{a+b x}+\frac {1}{8} e^{-3 i d-3 i f x^2} f^{a+b x}+\frac {1}{8} e^{3 i d+3 i f x^2} f^{a+b x}\right ) \, dx\\ &=\frac {1}{8} \int e^{-3 i d-3 i f x^2} f^{a+b x} \, dx+\frac {1}{8} \int e^{3 i d+3 i f x^2} f^{a+b x} \, dx+\frac {3}{8} \int e^{-i d-i f x^2} f^{a+b x} \, dx+\frac {3}{8} \int e^{i d+i f x^2} f^{a+b x} \, dx\\ &=\frac {1}{8} \int e^{-3 i d-3 i f x^2+a \log (f)+b x \log (f)} \, dx+\frac {1}{8} \int e^{3 i d+3 i f x^2+a \log (f)+b x \log (f)} \, dx+\frac {3}{8} \int e^{-i d-i f x^2+a \log (f)+b x \log (f)} \, dx+\frac {3}{8} \int e^{i d+i f x^2+a \log (f)+b x \log (f)} \, dx\\ &=\frac {1}{8} \left (e^{3 i d+\frac {i b^2 \log ^2(f)}{12 f}} f^a\right ) \int e^{-\frac {i (6 i f x+b \log (f))^2}{12 f}} \, dx+\frac {1}{8} \left (3 e^{-\frac {1}{4} i \left (4 d+\frac {b^2 \log ^2(f)}{f}\right )} f^a\right ) \int e^{\frac {i (-2 i f x+b \log (f))^2}{4 f}} \, dx+\frac {1}{8} \left (3 e^{\frac {1}{4} i \left (4 d+\frac {b^2 \log ^2(f)}{f}\right )} f^a\right ) \int e^{-\frac {i (2 i f x+b \log (f))^2}{4 f}} \, dx+\frac {1}{8} \left (e^{-\frac {1}{12} i \left (36 d+\frac {b^2 \log ^2(f)}{f}\right )} f^a\right ) \int e^{\frac {i (-6 i f x+b \log (f))^2}{12 f}} \, dx\\ &=-\frac {3}{16} \sqrt [4]{-1} e^{\frac {1}{4} i \left (4 d+\frac {b^2 \log ^2(f)}{f}\right )} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erf}\left (\frac {\sqrt [4]{-1} (2 i f x+b \log (f))}{2 \sqrt {f}}\right )-\left (\frac {1}{16}+\frac {i}{16}\right ) e^{3 i d+\frac {i b^2 \log ^2(f)}{12 f}} f^{-\frac {1}{2}+a} \sqrt {\frac {\pi }{6}} \text {erf}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (6 i f x+b \log (f))}{\sqrt {6} \sqrt {f}}\right )-\frac {3}{16} \sqrt [4]{-1} e^{-\frac {1}{4} i \left (4 d+\frac {b^2 \log ^2(f)}{f}\right )} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt [4]{-1} (2 i f x-b \log (f))}{2 \sqrt {f}}\right )-\left (\frac {1}{16}+\frac {i}{16}\right ) e^{-\frac {1}{12} i \left (36 d+\frac {b^2 \log ^2(f)}{f}\right )} f^{-\frac {1}{2}+a} \sqrt {\frac {\pi }{6}} \text {erfi}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (6 i f x-b \log (f))}{\sqrt {6} \sqrt {f}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.88, size = 267, normalized size = 0.90 \[ \frac {1}{48} \sqrt [4]{-1} \sqrt {\pi } f^{a-\frac {1}{2}} e^{-\frac {i b^2 \log ^2(f)}{4 f}} \left (9 e^{\frac {i b^2 \log ^2(f)}{2 f}} (\sin (d)-i \cos (d)) \text {erfi}\left (\frac {\sqrt [4]{-1} (2 f x-i b \log (f))}{2 \sqrt {f}}\right )+\sqrt {3} e^{\frac {i b^2 \log ^2(f)}{6 f}} \left (e^{\frac {i b^2 \log ^2(f)}{6 f}} (\sin (3 d)-i \cos (3 d)) \text {erfi}\left (\frac {(1-i) b \log (f)+(6+6 i) f x}{2 \sqrt {6} \sqrt {f}}\right )-(\cos (3 d)-i \sin (3 d)) \text {erfi}\left (\frac {(-1)^{3/4} (6 f x+i b \log (f))}{2 \sqrt {3} \sqrt {f}}\right )\right )-9 (\cos (d)-i \sin (d)) \text {erfi}\left (\frac {(-1)^{3/4} (2 f x+i b \log (f))}{2 \sqrt {f}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x)*Cos[d + f*x^2]^3,x]

[Out]

((-1)^(1/4)*f^(-1/2 + a)*Sqrt[Pi]*(-9*Erfi[((-1)^(3/4)*(2*f*x + I*b*Log[f]))/(2*Sqrt[f])]*(Cos[d] - I*Sin[d])
+ 9*E^(((I/2)*b^2*Log[f]^2)/f)*Erfi[((-1)^(1/4)*(2*f*x - I*b*Log[f]))/(2*Sqrt[f])]*((-I)*Cos[d] + Sin[d]) + Sq
rt[3]*E^(((I/6)*b^2*Log[f]^2)/f)*(-(Erfi[((-1)^(3/4)*(6*f*x + I*b*Log[f]))/(2*Sqrt[3]*Sqrt[f])]*(Cos[3*d] - I*
Sin[3*d])) + E^(((I/6)*b^2*Log[f]^2)/f)*Erfi[((6 + 6*I)*f*x + (1 - I)*b*Log[f])/(2*Sqrt[6]*Sqrt[f])]*((-I)*Cos
[3*d] + Sin[3*d]))))/(48*E^(((I/4)*b^2*Log[f]^2)/f))

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fricas [B]  time = 1.02, size = 525, normalized size = 1.76 \[ \frac {\sqrt {6} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \relax (f)^{2} + 12 \, a f \log \relax (f) - 36 i \, d f}{12 \, f}\right )} \operatorname {C}\left (\frac {\sqrt {6} {\left (6 \, f x + i \, b \log \relax (f)\right )} \sqrt {\frac {f}{\pi }}}{6 \, f}\right ) - \sqrt {6} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \relax (f)^{2} + 12 \, a f \log \relax (f) + 36 i \, d f}{12 \, f}\right )} \operatorname {C}\left (-\frac {\sqrt {6} {\left (6 \, f x - i \, b \log \relax (f)\right )} \sqrt {\frac {f}{\pi }}}{6 \, f}\right ) + 9 \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \relax (f)^{2} + 4 \, a f \log \relax (f) - 4 i \, d f}{4 \, f}\right )} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, f x + i \, b \log \relax (f)\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) - 9 \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \relax (f)^{2} + 4 \, a f \log \relax (f) + 4 i \, d f}{4 \, f}\right )} \operatorname {C}\left (-\frac {\sqrt {2} {\left (2 \, f x - i \, b \log \relax (f)\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) - i \, \sqrt {6} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \relax (f)^{2} + 12 \, a f \log \relax (f) - 36 i \, d f}{12 \, f}\right )} \operatorname {S}\left (\frac {\sqrt {6} {\left (6 \, f x + i \, b \log \relax (f)\right )} \sqrt {\frac {f}{\pi }}}{6 \, f}\right ) - i \, \sqrt {6} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \relax (f)^{2} + 12 \, a f \log \relax (f) + 36 i \, d f}{12 \, f}\right )} \operatorname {S}\left (-\frac {\sqrt {6} {\left (6 \, f x - i \, b \log \relax (f)\right )} \sqrt {\frac {f}{\pi }}}{6 \, f}\right ) - 9 i \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {-i \, b^{2} \log \relax (f)^{2} + 4 \, a f \log \relax (f) - 4 i \, d f}{4 \, f}\right )} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, f x + i \, b \log \relax (f)\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right ) - 9 i \, \sqrt {2} \pi \sqrt {\frac {f}{\pi }} e^{\left (\frac {i \, b^{2} \log \relax (f)^{2} + 4 \, a f \log \relax (f) + 4 i \, d f}{4 \, f}\right )} \operatorname {S}\left (-\frac {\sqrt {2} {\left (2 \, f x - i \, b \log \relax (f)\right )} \sqrt {\frac {f}{\pi }}}{2 \, f}\right )}{48 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x+a)*cos(f*x^2+d)^3,x, algorithm="fricas")

[Out]

1/48*(sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(-I*b^2*log(f)^2 + 12*a*f*log(f) - 36*I*d*f)/f)*fresnel_cos(1/6*sqrt(6)*(6
*f*x + I*b*log(f))*sqrt(f/pi)/f) - sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(I*b^2*log(f)^2 + 12*a*f*log(f) + 36*I*d*f)/f
)*fresnel_cos(-1/6*sqrt(6)*(6*f*x - I*b*log(f))*sqrt(f/pi)/f) + 9*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(-I*b^2*log(f)^
2 + 4*a*f*log(f) - 4*I*d*f)/f)*fresnel_cos(1/2*sqrt(2)*(2*f*x + I*b*log(f))*sqrt(f/pi)/f) - 9*sqrt(2)*pi*sqrt(
f/pi)*e^(1/4*(I*b^2*log(f)^2 + 4*a*f*log(f) + 4*I*d*f)/f)*fresnel_cos(-1/2*sqrt(2)*(2*f*x - I*b*log(f))*sqrt(f
/pi)/f) - I*sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(-I*b^2*log(f)^2 + 12*a*f*log(f) - 36*I*d*f)/f)*fresnel_sin(1/6*sqrt
(6)*(6*f*x + I*b*log(f))*sqrt(f/pi)/f) - I*sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(I*b^2*log(f)^2 + 12*a*f*log(f) + 36*
I*d*f)/f)*fresnel_sin(-1/6*sqrt(6)*(6*f*x - I*b*log(f))*sqrt(f/pi)/f) - 9*I*sqrt(2)*pi*sqrt(f/pi)*e^(1/4*(-I*b
^2*log(f)^2 + 4*a*f*log(f) - 4*I*d*f)/f)*fresnel_sin(1/2*sqrt(2)*(2*f*x + I*b*log(f))*sqrt(f/pi)/f) - 9*I*sqrt
(2)*pi*sqrt(f/pi)*e^(1/4*(I*b^2*log(f)^2 + 4*a*f*log(f) + 4*I*d*f)/f)*fresnel_sin(-1/2*sqrt(2)*(2*f*x - I*b*lo
g(f))*sqrt(f/pi)/f))/f

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giac [B]  time = 0.40, size = 595, normalized size = 2.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x+a)*cos(f*x^2+d)^3,x, algorithm="giac")

[Out]

-3/16*sqrt(2)*sqrt(pi)*erf(-1/8*sqrt(2)*(4*x - (pi*b*sgn(f) - pi*b + 2*I*b*log(abs(f)))/f)*(-I*f/abs(f) + 1)*s
qrt(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) + 1/2*I*pi*a + a*log(abs(f)) + I*d)/((-I*f/abs(f) +
1)*sqrt(abs(f))) - 1/48*sqrt(6)*sqrt(pi)*erf(-1/24*sqrt(6)*sqrt(f)*(12*x - (pi*b*sgn(f) - pi*b + 2*I*b*log(abs
(f)))/f)*(-I*f/abs(f) + 1))*e^(1/24*I*pi^2*b^2*sgn(f)/f + 1/12*pi*b^2*log(abs(f))*sgn(f)/f - 1/24*I*pi^2*b^2/f
- 1/12*pi*b^2*log(abs(f))/f + 1/12*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/2*I*pi*a + a*log(abs(f)) + 3
*I*d)/(sqrt(f)*(-I*f/abs(f) + 1)) - 1/48*sqrt(6)*sqrt(pi)*erf(-1/24*sqrt(6)*sqrt(f)*(12*x + (pi*b*sgn(f) - pi*
b + 2*I*b*log(abs(f)))/f)*(I*f/abs(f) + 1))*e^(-1/24*I*pi^2*b^2*sgn(f)/f - 1/12*pi*b^2*log(abs(f))*sgn(f)/f +
1/24*I*pi^2*b^2/f + 1/12*pi*b^2*log(abs(f))/f - 1/12*I*b^2*log(abs(f))^2/f - 1/2*I*pi*a*sgn(f) + 1/2*I*pi*a +
a*log(abs(f)) - 3*I*d)/(sqrt(f)*(I*f/abs(f) + 1)) - 3/16*sqrt(2)*sqrt(pi)*erf(-1/8*sqrt(2)*(4*x + (pi*b*sgn(f)
- pi*b + 2*I*b*log(abs(f)))/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(ab
s(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) +
1/2*I*pi*a + a*log(abs(f)) - I*d)/((I*f/abs(f) + 1)*sqrt(abs(f)))

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maple [A]  time = 0.63, size = 235, normalized size = 0.79 \[ -\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {i \left (\ln \relax (f )^{2} b^{2}+36 d f \right )}{12 f}} \sqrt {3}\, \erf \left (-\sqrt {3}\, \sqrt {i f}\, x +\frac {\ln \relax (f ) b \sqrt {3}}{6 \sqrt {i f}}\right )}{48 \sqrt {i f}}-\frac {3 \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {i \left (\ln \relax (f )^{2} b^{2}+4 d f \right )}{4 f}} \erf \left (-\sqrt {i f}\, x +\frac {\ln \relax (f ) b}{2 \sqrt {i f}}\right )}{16 \sqrt {i f}}-\frac {3 \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {i \left (\ln \relax (f )^{2} b^{2}+4 d f \right )}{4 f}} \erf \left (-\sqrt {-i f}\, x +\frac {\ln \relax (f ) b}{2 \sqrt {-i f}}\right )}{16 \sqrt {-i f}}-\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {i \left (\ln \relax (f )^{2} b^{2}+36 d f \right )}{12 f}} \erf \left (-\sqrt {-3 i f}\, x +\frac {\ln \relax (f ) b}{2 \sqrt {-3 i f}}\right )}{16 \sqrt {-3 i f}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x+a)*cos(f*x^2+d)^3,x)

[Out]

-1/48*Pi^(1/2)*f^a*exp(-1/12*I*(ln(f)^2*b^2+36*d*f)/f)*3^(1/2)/(I*f)^(1/2)*erf(-3^(1/2)*(I*f)^(1/2)*x+1/6*ln(f
)*b*3^(1/2)/(I*f)^(1/2))-3/16*Pi^(1/2)*f^a*exp(-1/4*I*(ln(f)^2*b^2+4*d*f)/f)/(I*f)^(1/2)*erf(-(I*f)^(1/2)*x+1/
2*ln(f)*b/(I*f)^(1/2))-3/16*Pi^(1/2)*f^a*exp(1/4*I*(ln(f)^2*b^2+4*d*f)/f)/(-I*f)^(1/2)*erf(-(-I*f)^(1/2)*x+1/2
*ln(f)*b/(-I*f)^(1/2))-1/16*Pi^(1/2)*f^a*exp(1/12*I*(ln(f)^2*b^2+36*d*f)/f)/(-3*I*f)^(1/2)*erf(-(-3*I*f)^(1/2)
*x+1/2*ln(f)*b/(-3*I*f)^(1/2))

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maxima [A]  time = 0.45, size = 302, normalized size = 1.01 \[ -\frac {3 \cdot 9^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } {\left ({\left (\left (i - 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \relax (f)^{2} + 36 \, d f}{12 \, f}\right ) + \left (i + 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \relax (f)^{2} + 36 \, d f}{12 \, f}\right )\right )} \operatorname {erf}\left (\frac {6 i \, f x - b \log \relax (f)}{2 \, \sqrt {3 i \, f}}\right ) + {\left (\left (i + 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \relax (f)^{2} + 36 \, d f}{12 \, f}\right ) + \left (i - 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \relax (f)^{2} + 36 \, d f}{12 \, f}\right )\right )} \operatorname {erf}\left (\frac {6 i \, f x + b \log \relax (f)}{2 \, \sqrt {-3 i \, f}}\right )\right )} f^{\frac {3}{2}} + \sqrt {2} \sqrt {\pi } {\left ({\left (\left (27 i - 27\right ) \, f^{a} \cos \left (\frac {b^{2} \log \relax (f)^{2} + 4 \, d f}{4 \, f}\right ) + \left (27 i + 27\right ) \, f^{a} \sin \left (\frac {b^{2} \log \relax (f)^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {2 i \, f x - b \log \relax (f)}{2 \, \sqrt {i \, f}}\right ) + {\left (\left (27 i + 27\right ) \, f^{a} \cos \left (\frac {b^{2} \log \relax (f)^{2} + 4 \, d f}{4 \, f}\right ) + \left (27 i - 27\right ) \, f^{a} \sin \left (\frac {b^{2} \log \relax (f)^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {2 i \, f x + b \log \relax (f)}{2 \, \sqrt {-i \, f}}\right )\right )} f^{\frac {3}{2}}}{288 \, f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x+a)*cos(f*x^2+d)^3,x, algorithm="maxima")

[Out]

-1/288*(3*9^(1/4)*sqrt(2)*sqrt(pi)*(((I - 1)*f^a*cos(1/12*(b^2*log(f)^2 + 36*d*f)/f) + (I + 1)*f^a*sin(1/12*(b
^2*log(f)^2 + 36*d*f)/f))*erf(1/2*(6*I*f*x - b*log(f))/sqrt(3*I*f)) + ((I + 1)*f^a*cos(1/12*(b^2*log(f)^2 + 36
*d*f)/f) + (I - 1)*f^a*sin(1/12*(b^2*log(f)^2 + 36*d*f)/f))*erf(1/2*(6*I*f*x + b*log(f))/sqrt(-3*I*f)))*f^(3/2
) + sqrt(2)*sqrt(pi)*(((27*I - 27)*f^a*cos(1/4*(b^2*log(f)^2 + 4*d*f)/f) + (27*I + 27)*f^a*sin(1/4*(b^2*log(f)
^2 + 4*d*f)/f))*erf(1/2*(2*I*f*x - b*log(f))/sqrt(I*f)) + ((27*I + 27)*f^a*cos(1/4*(b^2*log(f)^2 + 4*d*f)/f) +
(27*I - 27)*f^a*sin(1/4*(b^2*log(f)^2 + 4*d*f)/f))*erf(1/2*(2*I*f*x + b*log(f))/sqrt(-I*f)))*f^(3/2))/f^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int f^{a+b\,x}\,{\cos \left (f\,x^2+d\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x)*cos(d + f*x^2)^3,x)

[Out]

int(f^(a + b*x)*cos(d + f*x^2)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + b x} \cos ^{3}{\left (d + f x^{2} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x+a)*cos(f*x**2+d)**3,x)

[Out]

Integral(f**(a + b*x)*cos(d + f*x**2)**3, x)

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