\(\int f^{a+b x} \sin ^3(d+f x^2) \, dx\) [93]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 18, antiderivative size = 298 \[ \int f^{a+b x} \sin ^3\left (d+f x^2\right ) \, dx=\frac {3}{16} (-1)^{3/4} e^{\frac {1}{4} i \left (4 d+\frac {b^2 \log ^2(f)}{f}\right )} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erf}\left (\frac {\sqrt [4]{-1} (2 i f x+b \log (f))}{2 \sqrt {f}}\right )+\left (\frac {1}{16}-\frac {i}{16}\right ) e^{3 i d+\frac {i b^2 \log ^2(f)}{12 f}} f^{-\frac {1}{2}+a} \sqrt {\frac {\pi }{6}} \text {erf}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (6 i f x+b \log (f))}{\sqrt {6} \sqrt {f}}\right )-\frac {3}{16} (-1)^{3/4} e^{-\frac {1}{4} i \left (4 d+\frac {b^2 \log ^2(f)}{f}\right )} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt [4]{-1} (2 i f x-b \log (f))}{2 \sqrt {f}}\right )-\left (\frac {1}{16}-\frac {i}{16}\right ) e^{-\frac {1}{12} i \left (36 d+\frac {b^2 \log ^2(f)}{f}\right )} f^{-\frac {1}{2}+a} \sqrt {\frac {\pi }{6}} \text {erfi}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (6 i f x-b \log (f))}{\sqrt {6} \sqrt {f}}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 268, normalized size of antiderivative = 0.90 \[ \int f^{a+b x} \sin ^3\left (d+f x^2\right ) \, dx=\frac {1}{48} (-1)^{3/4} e^{-\frac {i b^2 \log ^2(f)}{4 f}} f^{-\frac {1}{2}+a} \sqrt {\pi } \left (-9 \text {erfi}\left (\frac {(-1)^{3/4} (2 f x+i b \log (f))}{2 \sqrt {f}}\right ) (\cos (d)-i \sin (d))+9 i e^{\frac {i b^2 \log ^2(f)}{2 f}} \text {erfi}\left (\frac {\sqrt [4]{-1} (2 f x-i b \log (f))}{2 \sqrt {f}}\right ) (\cos (d)+i \sin (d))+\sqrt {3} e^{\frac {i b^2 \log ^2(f)}{6 f}} \left (\text {erfi}\left (\frac {(-1)^{3/4} (6 f x+i b \log (f))}{2 \sqrt {3} \sqrt {f}}\right ) (\cos (3 d)-i \sin (3 d))+e^{\frac {i b^2 \log ^2(f)}{6 f}} \text {erfi}\left (\frac {(6+6 i) f x+(1-i) b \log (f)}{2 \sqrt {6} \sqrt {f}}\right ) (-i \cos (3 d)+\sin (3 d))\right )\right ) \] Input:

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

Output:

((-1)^(3/4)*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*I)*E^(((I/2)*b^2*Log[f]^2)/f)*Erf 
i[((-1)^(1/4)*(2*f*x - I*b*Log[f]))/(2*Sqrt[f])]*(Cos[d] + I*Sin[d]) + Sqr 
t[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)*Er 
fi[((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))
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4975, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int f^{a+b x} \sin ^3\left (d+f x^2\right ) \, dx\)

\(\Big \downarrow \) 4975

\(\displaystyle \int \left (\frac {3}{8} i e^{-i d-i f x^2} f^{a+b x}-\frac {3}{8} i e^{i d+i f x^2} f^{a+b x}-\frac {1}{8} i e^{-3 i d-3 i f x^2} f^{a+b x}+\frac {1}{8} i e^{3 i d+3 i f x^2} f^{a+b x}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[f^(a + b*x)*Sin[d + f*x^2]^3,x]
 

Output:

(3*(-1)^(3/4)*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)^(3/4)*f^(-1/2 + a)*S 
qrt[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))
 

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4975
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]
 
Maple [A] (verified)

Time = 2.36 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.80

method result size
risch \(-\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {i \left (\ln \left (f \right )^{2} b^{2}+36 d f \right )}{12 f}} \operatorname {erf}\left (-\sqrt {-3 i f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-3 i f}}\right )}{16 \sqrt {-3 i f}}+\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {i \left (\ln \left (f \right )^{2} b^{2}+36 d f \right )}{12 f}} \sqrt {3}\, \operatorname {erf}\left (-\sqrt {3}\, \sqrt {i f}\, x +\frac {\ln \left (f \right ) b \sqrt {3}}{6 \sqrt {i f}}\right )}{48 \sqrt {i f}}-\frac {3 i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {i \left (\ln \left (f \right )^{2} b^{2}+4 d f \right )}{4 f}} \operatorname {erf}\left (-\sqrt {i f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {i f}}\right )}{16 \sqrt {i f}}+\frac {3 i \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {i \left (\ln \left (f \right )^{2} b^{2}+4 d f \right )}{4 f}} \operatorname {erf}\left (-\sqrt {-i f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-i f}}\right )}{16 \sqrt {-i f}}\) \(239\)

Input:

int(f^(b*x+a)*sin(f*x^2+d)^3,x,method=_RETURNVERBOSE)
 

Output:

-1/16*I*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))+1/48*I*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*I*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))+3/1 
6*I*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))
 

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. 525 vs. \(2 (196) = 392\).

Time = 0.08 (sec) , antiderivative size = 525, normalized size of antiderivative = 1.76 \[ \int f^{a+b x} \sin ^3\left (d+f x^2\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

1/48*(-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_cos(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_cos(-1/6*sqrt(6)*(6*f*x - I*b*log(f))*sqrt(f/pi)/f) + 9*I*sqr 
t(2)*pi*sqrt(f/pi)*e^(1/4*(-I*b^2*log(f)^2 + 4*a*f*log(f) - 4*I*d*f)/f)*fr 
esnel_cos(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_cos 
(-1/2*sqrt(2)*(2*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_sin(1/6*sqr 
t(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_sin(-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*lo 
g(f)^2 + 4*a*f*log(f) - 4*I*d*f)/f)*fresnel_sin(1/2*sqrt(2)*(2*f*x + I*b*l 
og(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_sin(-1/2*sqrt(2)*(2*f*x - I*b*log(f))*sqr 
t(f/pi)/f))/f
 

Sympy [F]

\[ \int f^{a+b x} \sin ^3\left (d+f x^2\right ) \, dx=\int f^{a + b x} \sin ^{3}{\left (d + f x^{2} \right )}\, dx \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.01 \[ \int f^{a+b x} \sin ^3\left (d+f x^2\right ) \, dx=\frac {9^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } {\left ({\left (-\left (i + 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} + 36 \, d f}{12 \, f}\right ) + \left (i - 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} + 36 \, d f}{12 \, f}\right )\right )} \operatorname {erf}\left (\frac {6 i \, f x - b \log \left (f\right )}{2 \, \sqrt {3 i \, f}}\right ) + {\left (-\left (i - 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} + 36 \, d f}{12 \, f}\right ) + \left (i + 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} + 36 \, d f}{12 \, f}\right )\right )} \operatorname {erf}\left (\frac {6 i \, f x + b \log \left (f\right )}{2 \, \sqrt {-3 i \, f}}\right )\right )} f^{\frac {3}{2}} - 9 \, \sqrt {2} \sqrt {\pi } {\left ({\left (-\left (i + 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} + 4 \, d f}{4 \, f}\right ) + \left (i - 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {2 i \, f x - b \log \left (f\right )}{2 \, \sqrt {i \, f}}\right ) + {\left (-\left (i - 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} + 4 \, d f}{4 \, f}\right ) + \left (i + 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {2 i \, f x + b \log \left (f\right )}{2 \, \sqrt {-i \, f}}\right )\right )} f^{\frac {3}{2}}}{96 \, f^{2}} \] Input:

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

Output:

1/96*(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) - 9*sqrt(2)*sqrt(pi)*((-(I + 1)*f^a*cos( 
1/4*(b^2*log(f)^2 + 4*d*f)/f) + (I - 1)*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)) + (-(I - 1)*f^a*cos(1/4*(b^2* 
log(f)^2 + 4*d*f)/f) + (I + 1)*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
 

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. 595 vs. \(2 (196) = 392\).

Time = 0.23 (sec) , antiderivative size = 595, normalized size of antiderivative = 2.00 \[ \int f^{a+b x} \sin ^3\left (d+f x^2\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

3/16*I*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(abs(f))*sgn(f)/f - 1/8*I*pi^2*b^2/f - 1/4*pi*b^2*log(a 
bs(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*I*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(a 
bs(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*I*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 + 1/2 
4*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*I*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(abs(f))*sgn(f)/f + 1/8*I*pi^2*b^2/f + 1/4*p 
i*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)))
 

Mupad [F(-1)]

Timed out. \[ \int f^{a+b x} \sin ^3\left (d+f x^2\right ) \, dx=\int f^{a+b\,x}\,{\sin \left (f\,x^2+d\right )}^3 \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int f^{a+b x} \sin ^3\left (d+f x^2\right ) \, dx=f^{a} \left (\int f^{b x} \sin \left (f \,x^{2}+d \right )^{3}d x \right ) \] Input:

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

Output:

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