Integrand size = 21, antiderivative size = 340 \[ \int f^{a+b x} \sin ^3\left (d+e x+f x^2\right ) \, dx=\frac {3}{16} (-1)^{3/4} e^{\frac {1}{4} i \left (4 d+\frac {(i e+b \log (f))^2}{f}\right )} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erf}\left (\frac {\sqrt [4]{-1} (i e+2 i f x+b \log (f))}{2 \sqrt {f}}\right )+\left (\frac {1}{16}-\frac {i}{16}\right ) e^{3 i d+\frac {i (3 i e+b \log (f))^2}{12 f}} f^{-\frac {1}{2}+a} \sqrt {\frac {\pi }{6}} \text {erf}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (3 i e+6 i f x+b \log (f))}{\sqrt {6} \sqrt {f}}\right )-\frac {3}{16} (-1)^{3/4} e^{-i d+\frac {i (e+i b \log (f))^2}{4 f}} f^{-\frac {1}{2}+a} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt [4]{-1} (i e+2 i f x-b \log (f))}{2 \sqrt {f}}\right )-\left (\frac {1}{16}-\frac {i}{16}\right ) e^{-3 i d+\frac {i (3 e+i b \log (f))^2}{12 f}} f^{-\frac {1}{2}+a} \sqrt {\frac {\pi }{6}} \text {erfi}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (3 i e+6 i f x-b \log (f))}{\sqrt {6} \sqrt {f}}\right ) \] Output:
3/16*(-1)^(3/4)*exp(1/4*I*(4*d+(I*e+b*ln(f))^2/f))*f^(-1/2+a)*Pi^(1/2)*erf (1/2*(-1)^(1/4)*(I*e+2*I*f*x+b*ln(f))/f^(1/2))+(1/96-1/96*I)*exp(3*I*d+1/1 2*I*(3*I*e+b*ln(f))^2/f)*f^(-1/2+a)*6^(1/2)*Pi^(1/2)*erf((1/12+1/12*I)*(3* I*e+6*I*f*x+b*ln(f))*6^(1/2)/f^(1/2))-3/16*(-1)^(3/4)*exp(-I*d+1/4*I*(e+I* b*ln(f))^2/f)*f^(-1/2+a)*Pi^(1/2)*erfi(1/2*(-1)^(1/4)*(I*e+2*I*f*x-b*ln(f) )/f^(1/2))+(-1/96+1/96*I)*exp(-3*I*d+1/12*I*(3*e+I*b*ln(f))^2/f)*f^(-1/2+a )*6^(1/2)*Pi^(1/2)*erfi((1/12+1/12*I)*(3*I*e+6*I*f*x-b*ln(f))*6^(1/2)/f^(1 /2))
Time = 1.06 (sec) , antiderivative size = 323, normalized size of antiderivative = 0.95 \[ \int f^{a+b x} \sin ^3\left (d+e x+f x^2\right ) \, dx=\frac {1}{48} (-1)^{3/4} e^{-\frac {i \left (3 e^2+b^2 \log ^2(f)\right )}{4 f}} f^{a-\frac {b e+f}{2 f}} \sqrt {\pi } \left (9 i e^{\frac {i \left (e^2+b^2 \log ^2(f)\right )}{2 f}} \text {erfi}\left (\frac {\sqrt [4]{-1} (e+2 f x-i b \log (f))}{2 \sqrt {f}}\right ) (\cos (d)+i \sin (d))+e^{\frac {i e^2}{f}} \left (-9 \text {erfi}\left (\frac {(-1)^{3/4} (e+2 f x+i b \log (f))}{2 \sqrt {f}}\right ) (\cos (d)-i \sin (d))+\sqrt {3} e^{\frac {i \left (3 e^2+b^2 \log ^2(f)\right )}{6 f}} \text {erfi}\left (\frac {(-1)^{3/4} (3 e+6 f x+i b \log (f))}{2 \sqrt {3} \sqrt {f}}\right ) (\cos (3 d)-i \sin (3 d))\right )+\sqrt {3} e^{\frac {i b^2 \log ^2(f)}{3 f}} \text {erfi}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (3 e+6 f x-i b \log (f))}{\sqrt {6} \sqrt {f}}\right ) (-i \cos (3 d)+\sin (3 d))\right ) \] Input:
Integrate[f^(a + b*x)*Sin[d + e*x + f*x^2]^3,x]
Output:
((-1)^(3/4)*f^(a - (b*e + f)/(2*f))*Sqrt[Pi]*((9*I)*E^(((I/2)*(e^2 + b^2*L og[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*e^2)/f)*(-9*Erfi[((-1)^(3/4)*(e + 2*f*x + I*b*Log[f ]))/(2*Sqrt[f])]*(Cos[d] - I*Sin[d]) + Sqrt[3]*E^(((I/6)*(3*e^2 + b^2*Log[ f]^2))/f)*Erfi[((-1)^(3/4)*(3*e + 6*f*x + I*b*Log[f]))/(2*Sqrt[3]*Sqrt[f]) ]*(Cos[3*d] - I*Sin[3*d])) + Sqrt[3]*E^(((I/3)*b^2*Log[f]^2)/f)*Erfi[((1/2 + I/2)*(3*e + 6*f*x - I*b*Log[f]))/(Sqrt[6]*Sqrt[f])]*((-I)*Cos[3*d] + Si n[3*d])))/(48*E^(((I/4)*(3*e^2 + b^2*Log[f]^2))/f))
Time = 0.82 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.
| \(\displaystyle \int f^{a+b x} \sin ^3\left (d+e x+f x^2\right ) \, dx\) |
| \(\Big \downarrow \) 4975 |
| \(\displaystyle \int \left (\frac {3}{8} i f^{a+b x} \exp \left (-3 i \left (d+e x+f x^2\right )+2 i d+2 i e x+2 i f x^2\right )-\frac {3}{8} i f^{a+b x} \exp \left (-3 i \left (d+e x+f x^2\right )+4 i d+4 i e x+4 i f x^2\right )+\frac {1}{8} i f^{a+b x} \exp \left (-3 i \left (d+e x+f x^2\right )+6 i d+6 i e x+6 i f x^2\right )-\frac {1}{8} i f^{a+b x} e^{-3 i \left (d+e x+f x^2\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3}{16} (-1)^{3/4} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {1}{4} i \left (4 d+\frac {(b \log (f)+i e)^2}{f}\right )} \text {erf}\left (\frac {\sqrt [4]{-1} (b \log (f)+i e+2 i f x)}{2 \sqrt {f}}\right )+\left (\frac {1}{16}-\frac {i}{16}\right ) \sqrt {\frac {\pi }{6}} f^{a-\frac {1}{2}} e^{\frac {i (b \log (f)+3 i e)^2}{12 f}+3 i d} \text {erf}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (b \log (f)+3 i e+6 i f x)}{\sqrt {6} \sqrt {f}}\right )-\frac {3}{16} (-1)^{3/4} \sqrt {\pi } f^{a-\frac {1}{2}} e^{\frac {i (e+i b \log (f))^2}{4 f}-i d} \text {erfi}\left (\frac {\sqrt [4]{-1} (-b \log (f)+i e+2 i f x)}{2 \sqrt {f}}\right )-\left (\frac {1}{16}-\frac {i}{16}\right ) \sqrt {\frac {\pi }{6}} f^{a-\frac {1}{2}} e^{\frac {i (3 e+i b \log (f))^2}{12 f}-3 i d} \text {erfi}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (-b \log (f)+3 i e+6 i f x)}{\sqrt {6} \sqrt {f}}\right )\) |
Input:
Int[f^(a + b*x)*Sin[d + e*x + f*x^2]^3,x]
Output:
(3*(-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])])/16 + (1/16 - I/16)*E^((3*I)*d + ((I/12)*((3*I)*e + b*Log[f])^2)/f)*f^(-1/2 + a)*Sqrt[Pi /6]*Erf[((1/2 + I/2)*((3*I)*e + (6*I)*f*x + b*Log[f]))/(Sqrt[6]*Sqrt[f])] - (3*(-1)^(3/4)*E^((-I)*d + ((I/4)*(e + I*b*Log[f])^2)/f)*f^(-1/2 + a)*Sqr t[Pi]*Erfi[((-1)^(1/4)*(I*e + (2*I)*f*x - b*Log[f]))/(2*Sqrt[f])])/16 - (1 /16 - I/16)*E^((-3*I)*d + ((I/12)*(3*e + I*b*Log[f])^2)/f)*f^(-1/2 + a)*Sq rt[Pi/6]*Erfi[((1/2 + I/2)*((3*I)*e + (6*I)*f*x - b*Log[f]))/(Sqrt[6]*Sqrt [f])]
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]
Time = 4.34 (sec) , antiderivative size = 319, normalized size of antiderivative = 0.94
| method | result | size |
| risch | \(-\frac {i \sqrt {\pi }\, f^{a} f^{-\frac {b e}{2 f}} {\mathrm e}^{\frac {i \left (\ln \left (f \right )^{2} b^{2}+36 d f -9 e^{2}\right )}{12 f}} \operatorname {erf}\left (-\sqrt {-3 i f}\, x +\frac {3 i e +b \ln \left (f \right )}{2 \sqrt {-3 i f}}\right )}{16 \sqrt {-3 i f}}+\frac {i \sqrt {\pi }\, f^{a} f^{-\frac {b e}{2 f}} {\mathrm e}^{-\frac {i \left (\ln \left (f \right )^{2} b^{2}+36 d f -9 e^{2}\right )}{12 f}} \sqrt {3}\, \operatorname {erf}\left (-\sqrt {3}\, \sqrt {i f}\, x +\frac {\left (b \ln \left (f \right )-3 i e \right ) \sqrt {3}}{6 \sqrt {i f}}\right )}{48 \sqrt {i f}}-\frac {3 i \sqrt {\pi }\, f^{a} f^{-\frac {b e}{2 f}} {\mathrm e}^{-\frac {i \left (\ln \left (f \right )^{2} b^{2}+4 d f -e^{2}\right )}{4 f}} \operatorname {erf}\left (-\sqrt {i f}\, x +\frac {b \ln \left (f \right )-i e}{2 \sqrt {i f}}\right )}{16 \sqrt {i f}}+\frac {3 i \sqrt {\pi }\, f^{a} f^{-\frac {b e}{2 f}} {\mathrm e}^{\frac {i \left (\ln \left (f \right )^{2} b^{2}+4 d f -e^{2}\right )}{4 f}} \operatorname {erf}\left (-\sqrt {-i f}\, x +\frac {i e +b \ln \left (f \right )}{2 \sqrt {-i f}}\right )}{16 \sqrt {-i f}}\) | \(319\) |
Input:
int(f^(b*x+a)*sin(f*x^2+e*x+d)^3,x,method=_RETURNVERBOSE)
Output:
-1/16*I*Pi^(1/2)*f^a*f^(-1/2/f*b*e)*exp(1/12*I*(ln(f)^2*b^2+36*d*f-9*e^2)/ f)/(-3*I*f)^(1/2)*erf(-(-3*I*f)^(1/2)*x+1/2*(3*I*e+b*ln(f))/(-3*I*f)^(1/2) )+1/48*I*Pi^(1/2)*f^a*f^(-1/2/f*b*e)*exp(-1/12*I*(ln(f)^2*b^2+36*d*f-9*e^2 )/f)*3^(1/2)/(I*f)^(1/2)*erf(-3^(1/2)*(I*f)^(1/2)*x+1/6*(b*ln(f)-3*I*e)*3^ (1/2)/(I*f)^(1/2))-3/16*I*Pi^(1/2)*f^a*f^(-1/2/f*b*e)*exp(-1/4*I*(ln(f)^2* b^2+4*d*f-e^2)/f)/(I*f)^(1/2)*erf(-(I*f)^(1/2)*x+1/2*(b*ln(f)-I*e)/(I*f)^( 1/2))+3/16*I*Pi^(1/2)*f^a*f^(-1/2/f*b*e)*exp(1/4*I*(ln(f)^2*b^2+4*d*f-e^2) /f)/(-I*f)^(1/2)*erf(-(-I*f)^(1/2)*x+1/2*(I*e+b*ln(f))/(-I*f)^(1/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 629 vs. \(2 (220) = 440\).
Time = 0.09 (sec) , antiderivative size = 629, normalized size of antiderivative = 1.85 \[ \int f^{a+b x} \sin ^3\left (d+e x+f x^2\right ) \, dx =\text {Too large to display} \] Input:
integrate(f^(b*x+a)*sin(f*x^2+e*x+d)^3,x, algorithm="fricas")
Output:
1/48*(-I*sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(-I*b^2*log(f)^2 + 9*I*e^2 - 36*I*d *f - 6*(b*e - 2*a*f)*log(f))/f)*fresnel_cos(1/6*sqrt(6)*(6*f*x + I*b*log(f ) + 3*e)*sqrt(f/pi)/f) - I*sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(I*b^2*log(f)^2 - 9*I*e^2 + 36*I*d*f - 6*(b*e - 2*a*f)*log(f))/f)*fresnel_cos(-1/6*sqrt(6)* (6*f*x - I*b*log(f) + 3*e)*sqrt(f/pi)/f) + 9*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) + 9*I*sqrt(2)*pi*sq rt(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) - sqr t(6)*pi*sqrt(f/pi)*e^(1/12*(-I*b^2*log(f)^2 + 9*I*e^2 - 36*I*d*f - 6*(b*e - 2*a*f)*log(f))/f)*fresnel_sin(1/6*sqrt(6)*(6*f*x + I*b*log(f) + 3*e)*sqr t(f/pi)/f) + sqrt(6)*pi*sqrt(f/pi)*e^(1/12*(I*b^2*log(f)^2 - 9*I*e^2 + 36* I*d*f - 6*(b*e - 2*a*f)*log(f))/f)*fresnel_sin(-1/6*sqrt(6)*(6*f*x - I*b*l og(f) + 3*e)*sqrt(f/pi)/f) + 9*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) - 9*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
\[ \int f^{a+b x} \sin ^3\left (d+e x+f x^2\right ) \, dx=\int f^{a + b x} \sin ^{3}{\left (d + e x + f x^{2} \right )}\, dx \] Input:
integrate(f**(b*x+a)*sin(f*x**2+e*x+d)**3,x)
Output:
Integral(f**(a + b*x)*sin(d + e*x + f*x**2)**3, x)
Time = 0.15 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.11 \[ \int f^{a+b x} \sin ^3\left (d+e x+f x^2\right ) \, dx=\frac {9^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } {\left ({\left (\left (i + 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} - 9 \, e^{2} + 36 \, d f}{12 \, f}\right ) - \left (i - 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} - 9 \, e^{2} + 36 \, d f}{12 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (6 i \, f x - b \log \left (f\right ) + 3 i \, e\right )} \sqrt {3 i \, f}}{6 \, f}\right ) + {\left (-\left (i - 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} - 9 \, e^{2} + 36 \, d f}{12 \, f}\right ) + \left (i + 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} - 9 \, e^{2} + 36 \, d f}{12 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (6 i \, f x + b \log \left (f\right ) + 3 i \, e\right )} \sqrt {-3 i \, f}}{6 \, f}\right )\right )} f^{\frac {3}{2}} - 9 \, \sqrt {2} \sqrt {\pi } {\left ({\left (\left (i + 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} - e^{2} + 4 \, d f}{4 \, f}\right ) - \left (i - 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} - e^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (2 i \, f x - b \log \left (f\right ) + i \, e\right )} \sqrt {i \, f}}{2 \, f}\right ) + {\left (-\left (i - 1\right ) \, f^{a} \cos \left (\frac {b^{2} \log \left (f\right )^{2} - e^{2} + 4 \, d f}{4 \, f}\right ) + \left (i + 1\right ) \, f^{a} \sin \left (\frac {b^{2} \log \left (f\right )^{2} - e^{2} + 4 \, d f}{4 \, f}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (2 i \, f x + b \log \left (f\right ) + i \, e\right )} \sqrt {-i \, f}}{2 \, f}\right )\right )} f^{\frac {3}{2}}}{96 \, f^{2} f^{\frac {b e}{2 \, f}}} \] Input:
integrate(f^(b*x+a)*sin(f*x^2+e*x+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 - 9*e^ 2 + 36*d*f)/f) - (I - 1)*f^a*sin(1/12*(b^2*log(f)^2 - 9*e^2 + 36*d*f)/f))* erf(1/6*I*(6*I*f*x - b*log(f) + 3*I*e)*sqrt(3*I*f)/f) + (-(I - 1)*f^a*cos( 1/12*(b^2*log(f)^2 - 9*e^2 + 36*d*f)/f) + (I + 1)*f^a*sin(1/12*(b^2*log(f) ^2 - 9*e^2 + 36*d*f)/f))*erf(1/6*I*(6*I*f*x + b*log(f) + 3*I*e)*sqrt(-3*I* f)/f))*f^(3/2) - 9*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)*sqrt(-I*f)/f))*f^(3/2))/(f^ 2*f^(1/2*b*e/f))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 751 vs. \(2 (220) = 440\).
Time = 0.30 (sec) , antiderivative size = 751, normalized size of antiderivative = 2.21 \[ \int f^{a+b x} \sin ^3\left (d+e x+f x^2\right ) \, dx=\text {Too large to display} \] Input:
integrate(f^(b*x+a)*sin(f*x^2+e*x+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)) - 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*pi*b*e/f + a*log(abs(f)) - 1/2*b*e*log(a bs(f))/f + I*d - 1/4*I*e^2/f)/((-I*f/abs(f) + 1)*sqrt(abs(f))) - 1/48*I*sq rt(6)*sqrt(pi)*erf(-1/24*sqrt(6)*sqrt(f)*(12*x - (pi*b*sgn(f) - pi*b + 2*I *b*log(abs(f)) - 6*e)/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) + 1/4*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 + 3*I*d - 3/4*I*e^2/f)/(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)) + 6*e)/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) + 1/4*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 - 3*I*d + 3/4*I*e^2/f)/(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)) + 2*e)/f)*(I*f/a bs(f) + 1)*sqrt(abs(f)))*e^(-1/8*I*pi^2*b^2*sgn(f)/f - 1/4*pi*b^2*log(a...
Timed out. \[ \int f^{a+b x} \sin ^3\left (d+e x+f x^2\right ) \, dx=\int f^{a+b\,x}\,{\sin \left (f\,x^2+e\,x+d\right )}^3 \,d x \] Input:
int(f^(a + b*x)*sin(d + e*x + f*x^2)^3,x)
Output:
int(f^(a + b*x)*sin(d + e*x + f*x^2)^3, x)
\[ \int f^{a+b x} \sin ^3\left (d+e x+f x^2\right ) \, dx=f^{a} \left (\int f^{b x} \sin \left (f \,x^{2}+e x +d \right )^{3}d x \right ) \] Input:
int(f^(b*x+a)*sin(f*x^2+e*x+d)^3,x)
Output:
f**a*int(f**(b*x)*sin(d + e*x + f*x**2)**3,x)