\(\int f^{a+b x+c x^2} \sin ^2(d+f x^2) \, dx\) [110]

Optimal result

Integrand size = 23, antiderivative size = 245 \[ \int f^{a+b x+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{-2 i d+\frac {b^2 \log ^2(f)}{8 i f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {b \log (f)-2 x (2 i f-c \log (f))}{2 \sqrt {2 i f-c \log (f)}}\right )}{8 \sqrt {2 i f-c \log (f)}}-\frac {e^{2 i d-\frac {b^2 \log ^2(f)}{8 i f+4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {b \log (f)+2 x (2 i f+c \log (f))}{2 \sqrt {2 i f+c \log (f)}}\right )}{8 \sqrt {2 i f+c \log (f)}} \] Output:

1/4*f^(a-1/4*b^2/c)*Pi^(1/2)*erfi(1/2*(2*c*x+b)*ln(f)^(1/2)/c^(1/2))/c^(1/ 
2)/ln(f)^(1/2)+1/8*exp(-2*I*d+b^2*ln(f)^2/(8*I*f-4*c*ln(f)))*f^a*Pi^(1/2)* 
erf(1/2*(b*ln(f)-2*x*(2*I*f-c*ln(f)))/(2*I*f-c*ln(f))^(1/2))/(2*I*f-c*ln(f 
))^(1/2)-1/8*exp(2*I*d-b^2*ln(f)^2/(8*I*f+4*c*ln(f)))*f^a*Pi^(1/2)*erfi(1/ 
2*(b*ln(f)+2*x*(2*I*f+c*ln(f)))/(2*I*f+c*ln(f))^(1/2))/(2*I*f+c*ln(f))^(1/ 
2)
 

Mathematica [A] (warning: unable to verify)

Time = 2.07 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.22 \[ \int f^{a+b x+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\frac {1}{8} f^a \sqrt {\pi } \left (\frac {2 f^{-\frac {b^2}{4 c}} \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{\sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt [4]{-1} e^{\frac {b^2 \log ^2(f)}{8 i f-4 c \log (f)}} \left (\text {erf}\left (\frac {\sqrt [4]{-1} (4 f x+i (b+2 c x) \log (f))}{2 \sqrt {2 f+i c \log (f)}}\right ) \sqrt {2 f+i c \log (f)} (2 i f+c \log (f)) (\cos (2 d)-i \sin (2 d))+e^{\frac {i b^2 f \log ^2(f)}{4 f^2+c^2 \log ^2(f)}} \text {erf}\left (\frac {(-1)^{3/4} (4 f x-i (b+2 c x) \log (f))}{2 \sqrt {2 f-i c \log (f)}}\right ) \sqrt {2 f-i c \log (f)} (2 f+i c \log (f)) (\cos (2 d)+i \sin (2 d))\right )}{4 f^2+c^2 \log ^2(f)}\right ) \] Input:

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

Output:

(f^a*Sqrt[Pi]*((2*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])])/(Sqrt[c]*f 
^(b^2/(4*c))*Sqrt[Log[f]]) + ((-1)^(1/4)*E^((b^2*Log[f]^2)/((8*I)*f - 4*c* 
Log[f]))*(Erf[((-1)^(1/4)*(4*f*x + I*(b + 2*c*x)*Log[f]))/(2*Sqrt[2*f + I* 
c*Log[f]])]*Sqrt[2*f + I*c*Log[f]]*((2*I)*f + c*Log[f])*(Cos[2*d] - I*Sin[ 
2*d]) + E^((I*b^2*f*Log[f]^2)/(4*f^2 + c^2*Log[f]^2))*Erf[((-1)^(3/4)*(4*f 
*x - I*(b + 2*c*x)*Log[f]))/(2*Sqrt[2*f - I*c*Log[f]])]*Sqrt[2*f - I*c*Log 
[f]]*(2*f + I*c*Log[f])*(Cos[2*d] + I*Sin[2*d])))/(4*f^2 + c^2*Log[f]^2))) 
/8
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 245, 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.

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

\(\Big \downarrow \) 4975

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {\pi } f^a e^{\frac {b^2 \log ^2(f)}{-4 c \log (f)+8 i f}-2 i d} \text {erf}\left (\frac {b \log (f)-2 x (-c \log (f)+2 i f)}{2 \sqrt {-c \log (f)+2 i f}}\right )}{8 \sqrt {-c \log (f)+2 i f}}-\frac {\sqrt {\pi } f^a e^{2 i d-\frac {b^2 \log ^2(f)}{4 c \log (f)+8 i f}} \text {erfi}\left (\frac {b \log (f)+2 x (c \log (f)+2 i f)}{2 \sqrt {c \log (f)+2 i f}}\right )}{8 \sqrt {c \log (f)+2 i f}}+\frac {\sqrt {\pi } f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}\)

Input:

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

Output:

(f^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])])/ 
(4*Sqrt[c]*Sqrt[Log[f]]) + (E^((-2*I)*d + (b^2*Log[f]^2)/((8*I)*f - 4*c*Lo 
g[f]))*f^a*Sqrt[Pi]*Erf[(b*Log[f] - 2*x*((2*I)*f - c*Log[f]))/(2*Sqrt[(2*I 
)*f - c*Log[f]])])/(8*Sqrt[(2*I)*f - c*Log[f]]) - (E^((2*I)*d - (b^2*Log[f 
]^2)/((8*I)*f + 4*c*Log[f]))*f^a*Sqrt[Pi]*Erfi[(b*Log[f] + 2*x*((2*I)*f + 
c*Log[f]))/(2*Sqrt[(2*I)*f + c*Log[f]])])/(8*Sqrt[(2*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 = 1.10 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.93

method	result	size
risch	\(\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}+8 i d \ln \left (f \right ) c +16 d f}{4 \left (c \ln \left (f \right )-2 i f \right )}} \operatorname {erf}\left (-x \sqrt {2 i f -c \ln \left (f \right )}+\frac {\ln \left (f \right ) b}{2 \sqrt {2 i f -c \ln \left (f \right )}}\right )}{8 \sqrt {2 i f -c \ln \left (f \right )}}+\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}-8 i d \ln \left (f \right ) c +16 d f}{4 \left (2 i f +c \ln \left (f \right )\right )}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-2 i f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-c \ln \left (f \right )-2 i f}}\right )}{8 \sqrt {-c \ln \left (f \right )-2 i f}}-\frac {\sqrt {\pi }\, f^{-\frac {b^{2}}{4 c}} f^{a} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 \sqrt {-c \ln \left (f \right )}}\)	\(227\)

Input:

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

Output:

1/8*Pi^(1/2)*f^a*exp(-1/4*(ln(f)^2*b^2+8*I*d*ln(f)*c+16*d*f)/(c*ln(f)-2*I* 
f))/(2*I*f-c*ln(f))^(1/2)*erf(-x*(2*I*f-c*ln(f))^(1/2)+1/2*ln(f)*b/(2*I*f- 
c*ln(f))^(1/2))+1/8*Pi^(1/2)*f^a*exp(-1/4*(ln(f)^2*b^2-8*I*d*ln(f)*c+16*d* 
f)/(2*I*f+c*ln(f)))/(-c*ln(f)-2*I*f)^(1/2)*erf(-(-c*ln(f)-2*I*f)^(1/2)*x+1 
/2*ln(f)*b/(-c*ln(f)-2*I*f)^(1/2))-1/4*Pi^(1/2)*f^(-1/4*b^2/c)*f^a/(-c*ln( 
f))^(1/2)*erf(-(-c*ln(f))^(1/2)*x+1/2*ln(f)*b/(-c*ln(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. 402 vs. \(2 (185) = 370\).

Time = 0.09 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.64 \[ \int f^{a+b x+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\frac {\sqrt {\pi } {\left (c^{2} \log \left (f\right )^{2} - 2 i \, c f \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) - 2 i \, f} \operatorname {erf}\left (\frac {{\left (8 \, f^{2} x - 2 i \, b f \log \left (f\right ) + {\left (2 \, c^{2} x + b c\right )} \log \left (f\right )^{2}\right )} \sqrt {-c \log \left (f\right ) - 2 i \, f}}{2 \, {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )}}\right ) e^{\left (\frac {16 \, a f^{2} \log \left (f\right ) - {\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )^{3} + 32 i \, d f^{2} - 2 \, {\left (-4 i \, c^{2} d - i \, b^{2} f\right )} \log \left (f\right )^{2}}{4 \, {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )}}\right )} + \sqrt {\pi } {\left (c^{2} \log \left (f\right )^{2} + 2 i \, c f \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) + 2 i \, f} \operatorname {erf}\left (\frac {{\left (8 \, f^{2} x + 2 i \, b f \log \left (f\right ) + {\left (2 \, c^{2} x + b c\right )} \log \left (f\right )^{2}\right )} \sqrt {-c \log \left (f\right ) + 2 i \, f}}{2 \, {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )}}\right ) e^{\left (\frac {16 \, a f^{2} \log \left (f\right ) - {\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )^{3} - 32 i \, d f^{2} - 2 \, {\left (4 i \, c^{2} d + i \, b^{2} f\right )} \log \left (f\right )^{2}}{4 \, {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )}}\right )} - \frac {2 \, \sqrt {\pi } {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )} \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c \log \left (f\right )}}{2 \, c}\right )}{f^{\frac {b^{2} - 4 \, a c}{4 \, c}}}}{8 \, {\left (c^{3} \log \left (f\right )^{3} + 4 \, c f^{2} \log \left (f\right )\right )}} \] Input:

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

Output:

1/8*(sqrt(pi)*(c^2*log(f)^2 - 2*I*c*f*log(f))*sqrt(-c*log(f) - 2*I*f)*erf( 
1/2*(8*f^2*x - 2*I*b*f*log(f) + (2*c^2*x + b*c)*log(f)^2)*sqrt(-c*log(f) - 
 2*I*f)/(c^2*log(f)^2 + 4*f^2))*e^(1/4*(16*a*f^2*log(f) - (b^2*c - 4*a*c^2 
)*log(f)^3 + 32*I*d*f^2 - 2*(-4*I*c^2*d - I*b^2*f)*log(f)^2)/(c^2*log(f)^2 
 + 4*f^2)) + sqrt(pi)*(c^2*log(f)^2 + 2*I*c*f*log(f))*sqrt(-c*log(f) + 2*I 
*f)*erf(1/2*(8*f^2*x + 2*I*b*f*log(f) + (2*c^2*x + b*c)*log(f)^2)*sqrt(-c* 
log(f) + 2*I*f)/(c^2*log(f)^2 + 4*f^2))*e^(1/4*(16*a*f^2*log(f) - (b^2*c - 
 4*a*c^2)*log(f)^3 - 32*I*d*f^2 - 2*(4*I*c^2*d + I*b^2*f)*log(f)^2)/(c^2*l 
og(f)^2 + 4*f^2)) - 2*sqrt(pi)*(c^2*log(f)^2 + 4*f^2)*sqrt(-c*log(f))*erf( 
1/2*(2*c*x + b)*sqrt(-c*log(f))/c)/f^(1/4*(b^2 - 4*a*c)/c))/(c^3*log(f)^3 
+ 4*c*f^2*log(f))
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 997, normalized size of antiderivative = 4.07 \[ \int f^{a+b x+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/16*(sqrt(pi)*sqrt(2*c^2*log(f)^2 + 8*f^2)*((I*f^a*f^(1/4*b^2/c)*cos(1/2 
*(16*d*f^2 + (4*c^2*d + b^2*f)*log(f)^2)/(c^2*log(f)^2 + 4*f^2)) + f^a*f^( 
1/4*b^2/c)*sin(1/2*(16*d*f^2 + (4*c^2*d + b^2*f)*log(f)^2)/(c^2*log(f)^2 + 
 4*f^2)))*erf(1/2*(2*(c*log(f) - 2*I*f)*x + b*log(f))/sqrt(-c*log(f) + 2*I 
*f)) + (-I*f^a*f^(1/4*b^2/c)*cos(1/2*(16*d*f^2 + (4*c^2*d + b^2*f)*log(f)^ 
2)/(c^2*log(f)^2 + 4*f^2)) + f^a*f^(1/4*b^2/c)*sin(1/2*(16*d*f^2 + (4*c^2* 
d + b^2*f)*log(f)^2)/(c^2*log(f)^2 + 4*f^2)))*erf(1/2*(2*(c*log(f) + 2*I*f 
)*x + b*log(f))/sqrt(-c*log(f) - 2*I*f)))*sqrt(c*log(f) + sqrt(c^2*log(f)^ 
2 + 4*f^2))*sqrt(-c*log(f)) - sqrt(pi)*sqrt(2*c^2*log(f)^2 + 8*f^2)*((f^a* 
f^(1/4*b^2/c)*cos(1/2*(16*d*f^2 + (4*c^2*d + b^2*f)*log(f)^2)/(c^2*log(f)^ 
2 + 4*f^2)) - I*f^a*f^(1/4*b^2/c)*sin(1/2*(16*d*f^2 + (4*c^2*d + b^2*f)*lo 
g(f)^2)/(c^2*log(f)^2 + 4*f^2)))*erf(1/2*(2*(c*log(f) - 2*I*f)*x + b*log(f 
))/sqrt(-c*log(f) + 2*I*f)) + (f^a*f^(1/4*b^2/c)*cos(1/2*(16*d*f^2 + (4*c^ 
2*d + b^2*f)*log(f)^2)/(c^2*log(f)^2 + 4*f^2)) + I*f^a*f^(1/4*b^2/c)*sin(1 
/2*(16*d*f^2 + (4*c^2*d + b^2*f)*log(f)^2)/(c^2*log(f)^2 + 4*f^2)))*erf(1/ 
2*(2*(c*log(f) + 2*I*f)*x + b*log(f))/sqrt(-c*log(f) - 2*I*f)))*sqrt(-c*lo 
g(f) + sqrt(c^2*log(f)^2 + 4*f^2))*sqrt(-c*log(f)) - 2*sqrt(pi)*((c^2*f^a* 
e^(1/4*b^2*c*log(f)^3/(c^2*log(f)^2 + 4*f^2))*log(f)^2 + 4*f^(a + 2)*e^(1/ 
4*b^2*c*log(f)^3/(c^2*log(f)^2 + 4*f^2)))*erf(-1/2*b*conjugate(1/sqrt(-c*l 
og(f)))*log(f) + x*conjugate(sqrt(-c*log(f)))) - (c^2*f^a*e^(1/4*b^2*c*...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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