\(\int f^{a+c x^2} \sin ^3(d+e x+f x^2) \, dx\) [105]

Optimal result

Integrand size = 23, antiderivative size = 377 \[ \int f^{a+c x^2} \sin ^3\left (d+e x+f x^2\right ) \, dx=\frac {3 i e^{-i d-\frac {e^2}{4 i f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {i e+2 x (i f-c \log (f))}{2 \sqrt {i f-c \log (f)}}\right )}{16 \sqrt {i f-c \log (f)}}-\frac {i e^{-3 i d-\frac {9 e^2}{4 (3 i f-c \log (f))}} f^a \sqrt {\pi } \text {erf}\left (\frac {3 i e+2 x (3 i f-c \log (f))}{2 \sqrt {3 i f-c \log (f)}}\right )}{16 \sqrt {3 i f-c \log (f)}}-\frac {3 i e^{i d+\frac {e^2}{4 i f+4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {i e+2 x (i f+c \log (f))}{2 \sqrt {i f+c \log (f)}}\right )}{16 \sqrt {i f+c \log (f)}}+\frac {i e^{3 i d+\frac {9 e^2}{4 (3 i f+c \log (f))}} f^a \sqrt {\pi } \text {erfi}\left (\frac {3 i e+2 x (3 i f+c \log (f))}{2 \sqrt {3 i f+c \log (f)}}\right )}{16 \sqrt {3 i f+c \log (f)}} \] Output:

3/16*I*exp(-I*d-e^2/(4*I*f-4*c*ln(f)))*f^a*Pi^(1/2)*erf(1/2*(I*e+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-9*e^2 
/(12*I*f-4*c*ln(f)))*f^a*Pi^(1/2)*erf(1/2*(3*I*e+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^2/(4*I*f+4*c*ln( 
f)))*f^a*Pi^(1/2)*erfi(1/2*(I*e+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+9*e^2/(12*I*f+4*c*ln(f)))*f^a*Pi^(1/2)* 
erfi(1/2*(3*I*e+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 [A] (warning: unable to verify)

Time = 4.66 (sec) , antiderivative size = 490, normalized size of antiderivative = 1.30 \[ \int f^{a+c x^2} \sin ^3\left (d+e x+f x^2\right ) \, dx=\frac {\sqrt [4]{-1} f^a \sqrt {\pi } \left (-3 e^{\frac {e^2}{4 i f+4 c \log (f)}} \text {erfi}\left (\frac {\sqrt [4]{-1} (e+2 f x-2 i c x \log (f))}{2 \sqrt {f-i c \log (f)}}\right ) \sqrt {f-i c \log (f)} \left (9 f^3+9 i c f^2 \log (f)+c^2 f \log ^2(f)+i c^3 \log ^3(f)\right ) (\cos (d)+i \sin (d))+(f-i c \log (f)) \left (e^{\frac {9 e^2}{4 (3 i f+c \log (f))}} \text {erfi}\left (\frac {\sqrt [4]{-1} (3 e+6 f x-2 i c x \log (f))}{2 \sqrt {3 f-i c \log (f)}}\right ) \sqrt {3 f-i c \log (f)} \left (3 f^2+4 i c f \log (f)-c^2 \log ^2(f)\right ) (\cos (3 d)+i \sin (3 d))+(3 f-i c \log (f)) \left (3 e^{\frac {e^2}{-4 i f+4 c \log (f)}} \text {erfi}\left (\frac {(-1)^{3/4} (e+2 f x+2 i c x \log (f))}{2 \sqrt {f+i c \log (f)}}\right ) \sqrt {f+i c \log (f)} (-3 i f+c \log (f)) (\cos (d)-i \sin (d))+e^{\frac {9 e^2}{4 (-3 i f+c \log (f))}} \text {erfi}\left (\frac {(-1)^{3/4} (3 e+6 f x+2 i c x \log (f))}{2 \sqrt {3 f+i c \log (f)}}\right ) (f+i c \log (f)) \sqrt {3 f+i c \log (f)} (i \cos (3 d)+\sin (3 d))\right )\right )\right )}{16 \left (9 f^4+10 c^2 f^2 \log ^2(f)+c^4 \log ^4(f)\right )} \] Input:

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

Output:

((-1)^(1/4)*f^a*Sqrt[Pi]*(-3*E^(e^2/((4*I)*f + 4*c*Log[f]))*Erfi[((-1)^(1/ 
4)*(e + 2*f*x - (2*I)*c*x*Log[f]))/(2*Sqrt[f - I*c*Log[f]])]*Sqrt[f - I*c* 
Log[f]]*(9*f^3 + (9*I)*c*f^2*Log[f] + c^2*f*Log[f]^2 + I*c^3*Log[f]^3)*(Co 
s[d] + I*Sin[d]) + (f - I*c*Log[f])*(E^((9*e^2)/(4*((3*I)*f + c*Log[f])))* 
Erfi[((-1)^(1/4)*(3*e + 6*f*x - (2*I)*c*x*Log[f]))/(2*Sqrt[3*f - I*c*Log[f 
]])]*Sqrt[3*f - I*c*Log[f]]*(3*f^2 + (4*I)*c*f*Log[f] - c^2*Log[f]^2)*(Cos 
[3*d] + I*Sin[3*d]) + (3*f - I*c*Log[f])*(3*E^(e^2/((-4*I)*f + 4*c*Log[f]) 
)*Erfi[((-1)^(3/4)*(e + 2*f*x + (2*I)*c*x*Log[f]))/(2*Sqrt[f + I*c*Log[f]] 
)]*Sqrt[f + I*c*Log[f]]*((-3*I)*f + c*Log[f])*(Cos[d] - I*Sin[d]) + E^((9* 
e^2)/(4*((-3*I)*f + c*Log[f])))*Erfi[((-1)^(3/4)*(3*e + 6*f*x + (2*I)*c*x* 
Log[f]))/(2*Sqrt[3*f + I*c*Log[f]])]*(f + I*c*Log[f])*Sqrt[3*f + I*c*Log[f 
]]*(I*Cos[3*d] + Sin[3*d])))))/(16*(9*f^4 + 10*c^2*f^2*Log[f]^2 + c^4*Log[ 
f]^4))
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, 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.

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

\(\Big \downarrow \) 4975

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {i \sqrt {\pi } f^a \exp \left (-\frac {9 e^2}{4 (-c \log (f)+3 i f)}-3 i d\right ) \text {erf}\left (\frac {2 x (-c \log (f)+3 i f)+3 i e}{2 \sqrt {-c \log (f)+3 i f}}\right )}{16 \sqrt {-c \log (f)+3 i f}}+\frac {3 i \sqrt {\pi } f^a e^{-\frac {e^2}{-4 c \log (f)+4 i f}-i d} \text {erf}\left (\frac {2 x (-c \log (f)+i f)+i e}{2 \sqrt {-c \log (f)+i f}}\right )}{16 \sqrt {-c \log (f)+i f}}-\frac {3 i \sqrt {\pi } f^a e^{\frac {e^2}{4 c \log (f)+4 i f}+i d} \text {erfi}\left (\frac {2 x (c \log (f)+i f)+i e}{2 \sqrt {c \log (f)+i f}}\right )}{16 \sqrt {c \log (f)+i f}}+\frac {i \sqrt {\pi } f^a e^{\frac {9 e^2}{4 (c \log (f)+3 i f)}+3 i d} \text {erfi}\left (\frac {2 x (c \log (f)+3 i f)+3 i e}{2 \sqrt {c \log (f)+3 i f}}\right )}{16 \sqrt {c \log (f)+3 i f}}\)

Input:

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

Output:

(((3*I)/16)*E^((-I)*d - e^2/((4*I)*f - 4*c*Log[f]))*f^a*Sqrt[Pi]*Erf[(I*e 
+ 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 - (9*e^2)/(4*((3*I)*f - c*Log[f])))*f^a*Sqrt[Pi]*Erf[( 
(3*I)*e + 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^2/((4*I)*f + 4*c*Log[f]))*f^a*S 
qrt[Pi]*Erfi[(I*e + 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 + (9*e^2)/(4*((3*I)*f + c*Log[f])))*f 
^a*Sqrt[Pi]*Erfi[((3*I)*e + 2*x*((3*I)*f + c*Log[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 = 4.14 (sec) , antiderivative size = 338, normalized size of antiderivative = 0.90

method	result	size
risch	\(-\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {3 i d \ln \left (f \right ) c -9 d f +\frac {9 e^{2}}{4}}{3 i f +c \ln \left (f \right )}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-3 i f}\, x +\frac {3 i e}{2 \sqrt {-c \ln \left (f \right )-3 i f}}\right )}{16 \sqrt {-c \ln \left (f \right )-3 i f}}-\frac {i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {3 \left (4 i d \ln \left (f \right ) c +12 d f -3 e^{2}\right )}{4 \left (c \ln \left (f \right )-3 i f \right )}} \operatorname {erf}\left (x \sqrt {3 i f -c \ln \left (f \right )}+\frac {3 i e}{2 \sqrt {3 i f -c \ln \left (f \right )}}\right )}{16 \sqrt {3 i f -c \ln \left (f \right )}}+\frac {3 i \sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {4 i d \ln \left (f \right ) c +4 d f -e^{2}}{4 \left (-i f +c \ln \left (f \right )\right )}} \operatorname {erf}\left (x \sqrt {i f -c \ln \left (f \right )}+\frac {i e}{2 \sqrt {i f -c \ln \left (f \right )}}\right )}{16 \sqrt {i f -c \ln \left (f \right )}}+\frac {3 i \sqrt {\pi }\, f^{a} {\mathrm e}^{\frac {4 i d \ln \left (f \right ) c -4 d f +e^{2}}{4 i f +4 c \ln \left (f \right )}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-i f}\, x +\frac {i e}{2 \sqrt {-c \ln \left (f \right )-i f}}\right )}{16 \sqrt {-c \ln \left (f \right )-i f}}\)	\(338\)

Input:

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

Output:

-1/16*I*Pi^(1/2)*f^a*exp(3/4*(4*I*d*ln(f)*c-12*d*f+3*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+3/2*I*e/(-c*ln(f)-3* 
I*f)^(1/2))-1/16*I*Pi^(1/2)*f^a*exp(-3/4*(4*I*d*ln(f)*c+12*d*f-3*e^2)/(c*l 
n(f)-3*I*f))/(3*I*f-c*ln(f))^(1/2)*erf(x*(3*I*f-c*ln(f))^(1/2)+3/2*I*e/(3* 
I*f-c*ln(f))^(1/2))+3/16*I*Pi^(1/2)*f^a*exp(-1/4*(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*I*e/(I* 
f-c*ln(f))^(1/2))+3/16*I*Pi^(1/2)*f^a*exp(1/4*(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/(-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. 713 vs. \(2 (269) = 538\).

Time = 0.12 (sec) , antiderivative size = 713, normalized size of antiderivative = 1.89 \[ \int f^{a+c x^2} \sin ^3\left (d+e x+f x^2\right ) \, dx =\text {Too large to display} \] Input:

integrate(f^(c*x^2+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*(2*c^2*x*log(f)^2 + 18*f^2*x + 3*I*c*e* 
log(f) + 9*e*f)*sqrt(-c*log(f) - 3*I*f)/(c^2*log(f)^2 + 9*f^2))*e^(1/4*(4* 
a*c^2*log(f)^3 + 12*I*c^2*d*log(f)^2 - 27*I*e^2*f + 108*I*d*f^2 + 9*(c*e^2 
 + 4*a*f^2)*log(f))/(c^2*log(f)^2 + 9*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* 
(2*c^2*x*log(f)^2 + 18*f^2*x - 3*I*c*e*log(f) + 9*e*f)*sqrt(-c*log(f) + 3* 
I*f)/(c^2*log(f)^2 + 9*f^2))*e^(1/4*(4*a*c^2*log(f)^3 - 12*I*c^2*d*log(f)^ 
2 + 27*I*e^2*f - 108*I*d*f^2 + 9*(c*e^2 + 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*c^2*x*log(f)^2 + 2*f^2*x + I*c* 
e*log(f) + e*f)*sqrt(-c*log(f) - I*f)/(c^2*log(f)^2 + f^2))*e^(1/4*(4*a*c^ 
2*log(f)^3 + 4*I*c^2*d*log(f)^2 - I*e^2*f + 4*I*d*f^2 + (c*e^2 + 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*c^2*x*log(f 
)^2 + 2*f^2*x - I*c*e*log(f) + e*f)*sqrt(-c*log(f) + I*f)/(c^2*log(f)^2 + 
f^2))*e^(1/4*(4*a*c^2*log(f)^3 - 4*I*c^2*d*log(f)^2 + I*e^2*f - 4*I*d*f^2 
+ (c*e^2 + 4*a*f^2)*log(f))/(c^2*log(f)^2 + f^2)))/(c^4*log(f)^4 + 10*c^2* 
f^2*log(f)^2 + 9*f^4)
 

Sympy [F]

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

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

Output:

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

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. 2175 vs. \(2 (269) = 538\).

Time = 0.11 (sec) , antiderivative size = 2175, normalized size of antiderivative = 5.77 \[ \int f^{a+c x^2} \sin ^3\left (d+e x+f x^2\right ) \, dx=\text {Too large to display} \] Input:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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