\(\int f^{a+b x+c x^2} \sin ^3(d+e x+f x^2) \, dx\) [114]

Optimal result

Integrand size = 26, antiderivative size = 430 \[ \int f^{a+b x+c x^2} \sin ^3\left (d+e x+f x^2\right ) \, dx=\frac {3 i e^{-i d-\frac {(e+i b \log (f))^2}{4 i f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {i e-b \log (f)+2 x (i f-c \log (f))}{2 \sqrt {i f-c \log (f)}}\right )}{16 \sqrt {i f-c \log (f)}}-\frac {i e^{-3 i d-\frac {(3 e+i b \log (f))^2}{4 (3 i f-c \log (f))}} f^a \sqrt {\pi } \text {erf}\left (\frac {3 i e-b \log (f)+2 x (3 i f-c \log (f))}{2 \sqrt {3 i f-c \log (f)}}\right )}{16 \sqrt {3 i f-c \log (f)}}-\frac {3 i e^{i d+\frac {(e-i b \log (f))^2}{4 i f+4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e+b \log (f)+2 x (i f+c \log (f))}{2 \sqrt {i f+c \log (f)}}\right )}{16 \sqrt {i f+c \log (f)}}+\frac {i e^{3 i d-\frac {(3 i e+b \log (f))^2}{4 (3 i f+c \log (f))}} f^a \sqrt {\pi } \text {erfi}\left (\frac {3 i e+b \log (f)+2 x (3 i f+c \log (f))}{2 \sqrt {3 i f+c \log (f)}}\right )}{16 \sqrt {3 i f+c \log (f)}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(3835\) vs. \(2(430)=860\).

Time = 6.83 (sec) , antiderivative size = 3835, normalized size of antiderivative = 8.92 \[ \int f^{a+b x+c x^2} \sin ^3\left (d+e x+f x^2\right ) \, dx=\text {Result too large to show} \] Input:

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

Output:

(f^a*Sqrt[Pi]*(-27*(-1)^(3/4)*E^(((I/4)*(-e^2 + (2*I)*b*e*Log[f] + b^2*Log 
[f]^2))/(f - I*c*Log[f]))*f^3*Cos[d]*Erfi[((-1)^(1/4)*(e + 2*f*x - I*b*Log 
[f] - (2*I)*c*x*Log[f]))/(2*Sqrt[f - I*c*Log[f]])]*Sqrt[f - I*c*Log[f]] + 
27*(-1)^(1/4)*c*E^(((I/4)*(-e^2 + (2*I)*b*e*Log[f] + b^2*Log[f]^2))/(f - I 
*c*Log[f]))*f^2*Cos[d]*Erfi[((-1)^(1/4)*(e + 2*f*x - I*b*Log[f] - (2*I)*c* 
x*Log[f]))/(2*Sqrt[f - I*c*Log[f]])]*Log[f]*Sqrt[f - I*c*Log[f]] - 3*(-1)^ 
(3/4)*c^2*E^(((I/4)*(-e^2 + (2*I)*b*e*Log[f] + b^2*Log[f]^2))/(f - I*c*Log 
[f]))*f*Cos[d]*Erfi[((-1)^(1/4)*(e + 2*f*x - I*b*Log[f] - (2*I)*c*x*Log[f] 
))/(2*Sqrt[f - I*c*Log[f]])]*Log[f]^2*Sqrt[f - I*c*Log[f]] + 3*(-1)^(1/4)* 
c^3*E^(((I/4)*(-e^2 + (2*I)*b*e*Log[f] + b^2*Log[f]^2))/(f - I*c*Log[f]))* 
Cos[d]*Erfi[((-1)^(1/4)*(e + 2*f*x - I*b*Log[f] - (2*I)*c*x*Log[f]))/(2*Sq 
rt[f - I*c*Log[f]])]*Log[f]^3*Sqrt[f - I*c*Log[f]] + 3*(-1)^(3/4)*E^(((I/4 
)*(-9*e^2 + (6*I)*b*e*Log[f] + b^2*Log[f]^2))/(3*f - I*c*Log[f]))*f^3*Cos[ 
3*d]*Erfi[((-1)^(1/4)*(3*e + 6*f*x - I*b*Log[f] - (2*I)*c*x*Log[f]))/(2*Sq 
rt[3*f - I*c*Log[f]])]*Sqrt[3*f - I*c*Log[f]] - (-1)^(1/4)*c*E^(((I/4)*(-9 
*e^2 + (6*I)*b*e*Log[f] + b^2*Log[f]^2))/(3*f - I*c*Log[f]))*f^2*Cos[3*d]* 
Erfi[((-1)^(1/4)*(3*e + 6*f*x - I*b*Log[f] - (2*I)*c*x*Log[f]))/(2*Sqrt[3* 
f - I*c*Log[f]])]*Log[f]*Sqrt[3*f - I*c*Log[f]] + 3*(-1)^(3/4)*c^2*E^(((I/ 
4)*(-9*e^2 + (6*I)*b*e*Log[f] + b^2*Log[f]^2))/(3*f - I*c*Log[f]))*f*Cos[3 
*d]*Erfi[((-1)^(1/4)*(3*e + 6*f*x - I*b*Log[f] - (2*I)*c*x*Log[f]))/(2*...
 

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, 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.

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

\(\Big \downarrow \) 4975

\(\displaystyle \int \left (\frac {3}{8} i f^{a+b x+c x^2} \exp \left (-3 i \left (d+e x+f x^2\right )+2 i d+2 i e x+2 i f x^2\right )-\frac {3}{8} i f^{a+b x+c x^2} \exp \left (-3 i \left (d+e x+f x^2\right )+4 i d+4 i e x+4 i f x^2\right )+\frac {1}{8} i f^{a+b x+c x^2} \exp \left (-3 i \left (d+e x+f x^2\right )+6 i d+6 i e x+6 i f x^2\right )-\frac {1}{8} i e^{-3 i \left (d+e x+f x^2\right )} f^{a+b x+c x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 i \sqrt {\pi } f^a \exp \left (-\frac {(e+i b \log (f))^2}{-4 c \log (f)+4 i f}-i d\right ) \text {erf}\left (\frac {-b \log (f)+2 x (-c \log (f)+i f)+i e}{2 \sqrt {-c \log (f)+i f}}\right )}{16 \sqrt {-c \log (f)+i f}}-\frac {i \sqrt {\pi } f^a \exp \left (-\frac {(3 e+i b \log (f))^2}{4 (-c \log (f)+3 i f)}-3 i d\right ) \text {erf}\left (\frac {-b \log (f)+2 x (-c \log (f)+3 i f)+3 i e}{2 \sqrt {-c \log (f)+3 i f}}\right )}{16 \sqrt {-c \log (f)+3 i f}}-\frac {3 i \sqrt {\pi } f^a \exp \left (\frac {(e-i b \log (f))^2}{4 c \log (f)+4 i f}+i d\right ) \text {erfi}\left (\frac {b \log (f)+2 x (c \log (f)+i f)+i e}{2 \sqrt {c \log (f)+i f}}\right )}{16 \sqrt {c \log (f)+i f}}+\frac {i \sqrt {\pi } f^a \exp \left (3 i d-\frac {(b \log (f)+3 i e)^2}{4 (c \log (f)+3 i f)}\right ) \text {erfi}\left (\frac {b \log (f)+2 x (c \log (f)+3 i f)+3 i e}{2 \sqrt {c \log (f)+3 i f}}\right )}{16 \sqrt {c \log (f)+3 i f}}\)

Input:

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

Output:

(((3*I)/16)*E^((-I)*d - (e + I*b*Log[f])^2/((4*I)*f - 4*c*Log[f]))*f^a*Sqr 
t[Pi]*Erf[(I*e - b*Log[f] + 2*x*(I*f - c*Log[f]))/(2*Sqrt[I*f - c*Log[f]]) 
])/Sqrt[I*f - c*Log[f]] - ((I/16)*E^((-3*I)*d - (3*e + I*b*Log[f])^2/(4*(( 
3*I)*f - c*Log[f])))*f^a*Sqrt[Pi]*Erf[((3*I)*e - b*Log[f] + 2*x*((3*I)*f - 
 c*Log[f]))/(2*Sqrt[(3*I)*f - c*Log[f]])])/Sqrt[(3*I)*f - c*Log[f]] - (((3 
*I)/16)*E^(I*d + (e - I*b*Log[f])^2/((4*I)*f + 4*c*Log[f]))*f^a*Sqrt[Pi]*E 
rfi[(I*e + b*Log[f] + 2*x*(I*f + c*Log[f]))/(2*Sqrt[I*f + c*Log[f]])])/Sqr 
t[I*f + c*Log[f]] + ((I/16)*E^((3*I)*d - ((3*I)*e + b*Log[f])^2/(4*((3*I)* 
f + c*Log[f])))*f^a*Sqrt[Pi]*Erfi[((3*I)*e + b*Log[f] + 2*x*((3*I)*f + c*L 
og[f]))/(2*Sqrt[(3*I)*f + c*Log[f]])])/Sqrt[(3*I)*f + c*Log[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 = 4.51 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.00

method	result	size
risch	\(-\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}+6 i \ln \left (f \right ) b e -12 i d \ln \left (f \right ) c +36 d f -9 e^{2}}{4 \left (3 i f +c \ln \left (f \right )\right )}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-3 i f}\, x +\frac {3 i e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )-3 i f}}\right )}{16 \sqrt {-c \ln \left (f \right )-3 i f}}+\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}-6 i \ln \left (f \right ) b e +12 i d \ln \left (f \right ) c +36 d f -9 e^{2}}{4 \left (c \ln \left (f \right )-3 i f \right )}} \operatorname {erf}\left (-x \sqrt {3 i f -c \ln \left (f \right )}+\frac {b \ln \left (f \right )-3 i e}{2 \sqrt {3 i f -c \ln \left (f \right )}}\right )}{16 \sqrt {3 i f -c \ln \left (f \right )}}-\frac {3 i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}-2 i \ln \left (f \right ) b e +4 i d \ln \left (f \right ) c +4 d f -e^{2}}{4 \left (-i f +c \ln \left (f \right )\right )}} \operatorname {erf}\left (-x \sqrt {i f -c \ln \left (f \right )}+\frac {b \ln \left (f \right )-i e}{2 \sqrt {i f -c \ln \left (f \right )}}\right )}{16 \sqrt {i f -c \ln \left (f \right )}}+\frac {3 i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}+2 i \ln \left (f \right ) b e -4 i d \ln \left (f \right ) c +4 d f -e^{2}}{4 \left (i f +c \ln \left (f \right )\right )}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-i f}\, x +\frac {i e +b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )-i f}}\right )}{16 \sqrt {-c \ln \left (f \right )-i f}}\)	\(430\)

Input:

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

Output:

-1/16*I*Pi^(1/2)*f^a*exp(-1/4*(ln(f)^2*b^2+6*I*ln(f)*b*e-12*I*d*ln(f)*c+36 
*d*f-9*e^2)/(3*I*f+c*ln(f)))/(-c*ln(f)-3*I*f)^(1/2)*erf(-(-c*ln(f)-3*I*f)^ 
(1/2)*x+1/2*(3*I*e+b*ln(f))/(-c*ln(f)-3*I*f)^(1/2))+1/16*I*Pi^(1/2)*f^a*ex 
p(-1/4*(ln(f)^2*b^2-6*I*ln(f)*b*e+12*I*d*ln(f)*c+36*d*f-9*e^2)/(c*ln(f)-3* 
I*f))/(3*I*f-c*ln(f))^(1/2)*erf(-x*(3*I*f-c*ln(f))^(1/2)+1/2*(b*ln(f)-3*I* 
e)/(3*I*f-c*ln(f))^(1/2))-3/16*I*Pi^(1/2)*f^a*exp(-1/4*(ln(f)^2*b^2-2*I*ln 
(f)*b*e+4*I*d*ln(f)*c+4*d*f-e^2)/(-I*f+c*ln(f)))/(I*f-c*ln(f))^(1/2)*erf(- 
x*(I*f-c*ln(f))^(1/2)+1/2*(b*ln(f)-I*e)/(I*f-c*ln(f))^(1/2))+3/16*I*Pi^(1/ 
2)*f^a*exp(-1/4*(ln(f)^2*b^2+2*I*ln(f)*b*e-4*I*d*ln(f)*c+4*d*f-e^2)/(I*f+c 
*ln(f)))/(-c*ln(f)-I*f)^(1/2)*erf(-(-c*ln(f)-I*f)^(1/2)*x+1/2*(I*e+b*ln(f) 
)/(-c*ln(f)-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. 865 vs. \(2 (312) = 624\).

Time = 0.11 (sec) , antiderivative size = 865, normalized size of antiderivative = 2.01 \[ \int f^{a+b x+c x^2} \sin ^3\left (d+e x+f x^2\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

1/16*(sqrt(pi)*(-I*c^3*log(f)^3 - 3*c^2*f*log(f)^2 - I*c*f^2*log(f) - 3*f^ 
3)*sqrt(-c*log(f) - 3*I*f)*erf(1/2*(18*f^2*x + (2*c^2*x + b*c)*log(f)^2 + 
9*e*f - 3*(-I*c*e + I*b*f)*log(f))*sqrt(-c*log(f) - 3*I*f)/(c^2*log(f)^2 + 
 9*f^2))*e^(-1/4*((b^2*c - 4*a*c^2)*log(f)^3 + 27*I*e^2*f - 108*I*d*f^2 + 
3*(-4*I*c^2*d + 2*I*b*c*e - I*b^2*f)*log(f)^2 - 9*(c*e^2 - 2*b*e*f + 4*a*f 
^2)*log(f))/(c^2*log(f)^2 + 9*f^2)) - 3*sqrt(pi)*(-I*c^3*log(f)^3 - c^2*f* 
log(f)^2 - 9*I*c*f^2*log(f) - 9*f^3)*sqrt(-c*log(f) - I*f)*erf(1/2*(2*f^2* 
x + (2*c^2*x + b*c)*log(f)^2 + e*f + (I*c*e - I*b*f)*log(f))*sqrt(-c*log(f 
) - I*f)/(c^2*log(f)^2 + f^2))*e^(-1/4*((b^2*c - 4*a*c^2)*log(f)^3 + I*e^2 
*f - 4*I*d*f^2 - (4*I*c^2*d - 2*I*b*c*e + I*b^2*f)*log(f)^2 - (c*e^2 - 2*b 
*e*f + 4*a*f^2)*log(f))/(c^2*log(f)^2 + f^2)) - 3*sqrt(pi)*(I*c^3*log(f)^3 
 - c^2*f*log(f)^2 + 9*I*c*f^2*log(f) - 9*f^3)*sqrt(-c*log(f) + I*f)*erf(1/ 
2*(2*f^2*x + (2*c^2*x + b*c)*log(f)^2 + e*f + (-I*c*e + I*b*f)*log(f))*sqr 
t(-c*log(f) + I*f)/(c^2*log(f)^2 + f^2))*e^(-1/4*((b^2*c - 4*a*c^2)*log(f) 
^3 - I*e^2*f + 4*I*d*f^2 - (-4*I*c^2*d + 2*I*b*c*e - I*b^2*f)*log(f)^2 - ( 
c*e^2 - 2*b*e*f + 4*a*f^2)*log(f))/(c^2*log(f)^2 + f^2)) + sqrt(pi)*(I*c^3 
*log(f)^3 - 3*c^2*f*log(f)^2 + I*c*f^2*log(f) - 3*f^3)*sqrt(-c*log(f) + 3* 
I*f)*erf(1/2*(18*f^2*x + (2*c^2*x + b*c)*log(f)^2 + 9*e*f - 3*(I*c*e - I*b 
*f)*log(f))*sqrt(-c*log(f) + 3*I*f)/(c^2*log(f)^2 + 9*f^2))*e^(-1/4*((b^2* 
c - 4*a*c^2)*log(f)^3 - 27*I*e^2*f + 108*I*d*f^2 + 3*(4*I*c^2*d - 2*I*b...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

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. 4343 vs. \(2 (312) = 624\).

Time = 0.13 (sec) , antiderivative size = 4343, normalized size of antiderivative = 10.10 \[ \int f^{a+b x+c x^2} \sin ^3\left (d+e x+f x^2\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/32*(sqrt(pi)*sqrt(2*c^2*log(f)^2 + 18*f^2)*(((c^2*f^a*e^(1/4*b^2*c*log( 
f)^3/(c^2*log(f)^2 + f^2) + 9/4*c*e^2*log(f)/(c^2*log(f)^2 + 9*f^2) + 1/2* 
b*e*f*log(f)/(c^2*log(f)^2 + f^2))*log(f)^2 + f^(a + 2)*e^(1/4*b^2*c*log(f 
)^3/(c^2*log(f)^2 + f^2) + 9/4*c*e^2*log(f)/(c^2*log(f)^2 + 9*f^2) + 1/2*b 
*e*f*log(f)/(c^2*log(f)^2 + f^2)))*cos(-3/4*(9*e^2*f - 36*d*f^2 - (4*c^2*d 
 - 2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(f)^2 + 9*f^2)) + (-I*c^2*f^a*e^(1/4 
*b^2*c*log(f)^3/(c^2*log(f)^2 + f^2) + 9/4*c*e^2*log(f)/(c^2*log(f)^2 + 9* 
f^2) + 1/2*b*e*f*log(f)/(c^2*log(f)^2 + f^2))*log(f)^2 - I*f^(a + 2)*e^(1/ 
4*b^2*c*log(f)^3/(c^2*log(f)^2 + f^2) + 9/4*c*e^2*log(f)/(c^2*log(f)^2 + 9 
*f^2) + 1/2*b*e*f*log(f)/(c^2*log(f)^2 + f^2)))*sin(-3/4*(9*e^2*f - 36*d*f 
^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(f)^2)/(c^2*log(f)^2 + 9*f^2)))*erf(1/ 
2*(2*(c*log(f) - 3*I*f)*x + b*log(f) - 3*I*e)*sqrt(-c*log(f) + 3*I*f)/(c*l 
og(f) - 3*I*f)) + ((c^2*f^a*e^(1/4*b^2*c*log(f)^3/(c^2*log(f)^2 + f^2) + 9 
/4*c*e^2*log(f)/(c^2*log(f)^2 + 9*f^2) + 1/2*b*e*f*log(f)/(c^2*log(f)^2 + 
f^2))*log(f)^2 + f^(a + 2)*e^(1/4*b^2*c*log(f)^3/(c^2*log(f)^2 + f^2) + 9/ 
4*c*e^2*log(f)/(c^2*log(f)^2 + 9*f^2) + 1/2*b*e*f*log(f)/(c^2*log(f)^2 + f 
^2)))*cos(-3/4*(9*e^2*f - 36*d*f^2 - (4*c^2*d - 2*b*c*e + b^2*f)*log(f)^2) 
/(c^2*log(f)^2 + 9*f^2)) + (I*c^2*f^a*e^(1/4*b^2*c*log(f)^3/(c^2*log(f)^2 
+ f^2) + 9/4*c*e^2*log(f)/(c^2*log(f)^2 + 9*f^2) + 1/2*b*e*f*log(f)/(c^2*l 
og(f)^2 + f^2))*log(f)^2 + I*f^(a + 2)*e^(1/4*b^2*c*log(f)^3/(c^2*log(f...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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