\(\int f^{c (a+b x)^2} x^3 \, dx\) [134]

Optimal result

Integrand size = 15, antiderivative size = 203 \[ \int f^{c (a+b x)^2} x^3 \, dx=-\frac {f^{c (a+b x)^2}}{2 b^4 c^2 \log ^2(f)}+\frac {3 a \sqrt {\pi } \text {erfi}\left (\sqrt {c} (a+b x) \sqrt {\log (f)}\right )}{4 b^4 c^{3/2} \log ^{\frac {3}{2}}(f)}+\frac {3 a^2 f^{c (a+b x)^2}}{2 b^4 c \log (f)}-\frac {3 a f^{c (a+b x)^2} (a+b x)}{2 b^4 c \log (f)}+\frac {f^{c (a+b x)^2} (a+b x)^2}{2 b^4 c \log (f)}-\frac {a^3 \sqrt {\pi } \text {erfi}\left (\sqrt {c} (a+b x) \sqrt {\log (f)}\right )}{2 b^4 \sqrt {c} \sqrt {\log (f)}} \] Output:

-1/2*f^(c*(b*x+a)^2)/b^4/c^2/ln(f)^2+3/4*a*Pi^(1/2)*erfi(c^(1/2)*(b*x+a)*l 
n(f)^(1/2))/b^4/c^(3/2)/ln(f)^(3/2)+3/2*a^2*f^(c*(b*x+a)^2)/b^4/c/ln(f)-3/ 
2*a*f^(c*(b*x+a)^2)*(b*x+a)/b^4/c/ln(f)+1/2*f^(c*(b*x+a)^2)*(b*x+a)^2/b^4/ 
c/ln(f)-1/2*a^3*Pi^(1/2)*erfi(c^(1/2)*(b*x+a)*ln(f)^(1/2))/b^4/c^(1/2)/ln( 
f)^(1/2)
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.47 \[ \int f^{c (a+b x)^2} x^3 \, dx=\frac {a \sqrt {c} \sqrt {\pi } \text {erfi}\left (\sqrt {c} (a+b x) \sqrt {\log (f)}\right ) \sqrt {\log (f)} \left (3-2 a^2 c \log (f)\right )+2 f^{c (a+b x)^2} \left (-1+c \left (a^2-a b x+b^2 x^2\right ) \log (f)\right )}{4 b^4 c^2 \log ^2(f)} \] Input:

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

Output:

(a*Sqrt[c]*Sqrt[Pi]*Erfi[Sqrt[c]*(a + b*x)*Sqrt[Log[f]]]*Sqrt[Log[f]]*(3 - 
 2*a^2*c*Log[f]) + 2*f^(c*(a + b*x)^2)*(-1 + c*(a^2 - a*b*x + b^2*x^2)*Log 
[f]))/(4*b^4*c^2*Log[f]^2)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2656, 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.

\(\displaystyle \int x^3 f^{c (a+b x)^2} \, dx\)

\(\Big \downarrow \) 2656

\(\displaystyle \int \left (-\frac {a^3 f^{c (a+b x)^2}}{b^3}+\frac {3 a^2 (a+b x) f^{c (a+b x)^2}}{b^3}+\frac {(a+b x)^3 f^{c (a+b x)^2}}{b^3}-\frac {3 a (a+b x)^2 f^{c (a+b x)^2}}{b^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\sqrt {\pi } a^3 \text {erfi}\left (\sqrt {c} \sqrt {\log (f)} (a+b x)\right )}{2 b^4 \sqrt {c} \sqrt {\log (f)}}+\frac {3 a^2 f^{c (a+b x)^2}}{2 b^4 c \log (f)}+\frac {3 \sqrt {\pi } a \text {erfi}\left (\sqrt {c} \sqrt {\log (f)} (a+b x)\right )}{4 b^4 c^{3/2} \log ^{\frac {3}{2}}(f)}-\frac {f^{c (a+b x)^2}}{2 b^4 c^2 \log ^2(f)}+\frac {(a+b x)^2 f^{c (a+b x)^2}}{2 b^4 c \log (f)}-\frac {3 a (a+b x) f^{c (a+b x)^2}}{2 b^4 c \log (f)}\)

Input:

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

Output:

-1/2*f^(c*(a + b*x)^2)/(b^4*c^2*Log[f]^2) + (3*a*Sqrt[Pi]*Erfi[Sqrt[c]*(a 
+ b*x)*Sqrt[Log[f]]])/(4*b^4*c^(3/2)*Log[f]^(3/2)) + (3*a^2*f^(c*(a + b*x) 
^2))/(2*b^4*c*Log[f]) - (3*a*f^(c*(a + b*x)^2)*(a + b*x))/(2*b^4*c*Log[f]) 
 + (f^(c*(a + b*x)^2)*(a + b*x)^2)/(2*b^4*c*Log[f]) - (a^3*Sqrt[Pi]*Erfi[S 
qrt[c]*(a + b*x)*Sqrt[Log[f]]])/(2*b^4*Sqrt[c]*Sqrt[Log[f]])
 

Defintions of rubi rules used

rule 2009

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

rule 2656

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(Px_), x_Symbol] :> Int[ 
ExpandLinearProduct[F^(a + b*(c + d*x)^n), Px, c, d, x], x] /; FreeQ[{F, a, 
 b, c, d, n}, x] && PolynomialQ[Px, x]
 

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.23

method	result	size
risch	\(\frac {x^{2} f^{b^{2} c \,x^{2}} f^{2 a b c x} f^{a^{2} c}}{2 c \,b^{2} \ln \left (f \right )}-\frac {a x \,f^{b^{2} c \,x^{2}} f^{2 a b c x} f^{a^{2} c}}{2 b^{3} c \ln \left (f \right )}+\frac {a^{2} f^{b^{2} c \,x^{2}} f^{2 a b c x} f^{a^{2} c}}{2 b^{4} c \ln \left (f \right )}+\frac {a^{3} \sqrt {\pi }\, \operatorname {erf}\left (-b \sqrt {-c \ln \left (f \right )}\, x +\frac {a c \ln \left (f \right )}{\sqrt {-c \ln \left (f \right )}}\right )}{2 b^{4} \sqrt {-c \ln \left (f \right )}}-\frac {3 a \sqrt {\pi }\, \operatorname {erf}\left (-b \sqrt {-c \ln \left (f \right )}\, x +\frac {a c \ln \left (f \right )}{\sqrt {-c \ln \left (f \right )}}\right )}{4 b^{4} c \ln \left (f \right ) \sqrt {-c \ln \left (f \right )}}-\frac {f^{b^{2} c \,x^{2}} f^{2 a b c x} f^{a^{2} c}}{2 c^{2} b^{4} \ln \left (f \right )^{2}}\)	\(249\)

Input:

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

Output:

1/2/c/b^2/ln(f)*x^2*f^(b^2*c*x^2)*f^(2*a*b*c*x)*f^(a^2*c)-1/2*a/b^3/c/ln(f 
)*x*f^(b^2*c*x^2)*f^(2*a*b*c*x)*f^(a^2*c)+1/2*a^2/b^4/c/ln(f)*f^(b^2*c*x^2 
)*f^(2*a*b*c*x)*f^(a^2*c)+1/2*a^3/b^4*Pi^(1/2)/(-c*ln(f))^(1/2)*erf(-b*(-c 
*ln(f))^(1/2)*x+a*c*ln(f)/(-c*ln(f))^(1/2))-3/4*a/b^4/c/ln(f)*Pi^(1/2)/(-c 
*ln(f))^(1/2)*erf(-b*(-c*ln(f))^(1/2)*x+a*c*ln(f)/(-c*ln(f))^(1/2))-1/2/c^ 
2/b^4/ln(f)^2*f^(b^2*c*x^2)*f^(2*a*b*c*x)*f^(a^2*c)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.56 \[ \int f^{c (a+b x)^2} x^3 \, dx=\frac {\sqrt {\pi } {\left (2 \, a^{3} c \log \left (f\right ) - 3 \, a\right )} \sqrt {-b^{2} c \log \left (f\right )} \operatorname {erf}\left (\frac {\sqrt {-b^{2} c \log \left (f\right )} {\left (b x + a\right )}}{b}\right ) + 2 \, {\left ({\left (b^{3} c x^{2} - a b^{2} c x + a^{2} b c\right )} \log \left (f\right ) - b\right )} f^{b^{2} c x^{2} + 2 \, a b c x + a^{2} c}}{4 \, b^{5} c^{2} \log \left (f\right )^{2}} \] Input:

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

Output:

1/4*(sqrt(pi)*(2*a^3*c*log(f) - 3*a)*sqrt(-b^2*c*log(f))*erf(sqrt(-b^2*c*l 
og(f))*(b*x + a)/b) + 2*((b^3*c*x^2 - a*b^2*c*x + a^2*b*c)*log(f) - b)*f^( 
b^2*c*x^2 + 2*a*b*c*x + a^2*c))/(b^5*c^2*log(f)^2)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.30 \[ \int f^{c (a+b x)^2} x^3 \, dx=-\frac {\frac {\sqrt {\pi } {\left (b^{2} c x + a b c\right )} a^{3} c^{3} {\left (\operatorname {erf}\left (\sqrt {-\frac {{\left (b^{2} c x + a b c\right )}^{2} \log \left (f\right )}{b^{2} c}}\right ) - 1\right )} \log \left (f\right )^{4}}{\left (c \log \left (f\right )\right )^{\frac {7}{2}} b^{4} \sqrt {-\frac {{\left (b^{2} c x + a b c\right )}^{2} \log \left (f\right )}{b^{2} c}}} - \frac {3 \, a^{2} c^{3} f^{\frac {{\left (b^{2} c x + a b c\right )}^{2}}{b^{2} c}} \log \left (f\right )^{3}}{\left (c \log \left (f\right )\right )^{\frac {7}{2}} b^{3}} - \frac {3 \, {\left (b^{2} c x + a b c\right )}^{3} a c \Gamma \left (\frac {3}{2}, -\frac {{\left (b^{2} c x + a b c\right )}^{2} \log \left (f\right )}{b^{2} c}\right ) \log \left (f\right )^{4}}{\left (c \log \left (f\right )\right )^{\frac {7}{2}} b^{6} \left (-\frac {{\left (b^{2} c x + a b c\right )}^{2} \log \left (f\right )}{b^{2} c}\right )^{\frac {3}{2}}} + \frac {c^{2} \Gamma \left (2, -\frac {{\left (b^{2} c x + a b c\right )}^{2} \log \left (f\right )}{b^{2} c}\right ) \log \left (f\right )^{2}}{\left (c \log \left (f\right )\right )^{\frac {7}{2}} b^{3}}}{2 \, \sqrt {c \log \left (f\right )} b} \] Input:

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

Output:

-1/2*(sqrt(pi)*(b^2*c*x + a*b*c)*a^3*c^3*(erf(sqrt(-(b^2*c*x + a*b*c)^2*lo 
g(f)/(b^2*c))) - 1)*log(f)^4/((c*log(f))^(7/2)*b^4*sqrt(-(b^2*c*x + a*b*c) 
^2*log(f)/(b^2*c))) - 3*a^2*c^3*f^((b^2*c*x + a*b*c)^2/(b^2*c))*log(f)^3/( 
(c*log(f))^(7/2)*b^3) - 3*(b^2*c*x + a*b*c)^3*a*c*gamma(3/2, -(b^2*c*x + a 
*b*c)^2*log(f)/(b^2*c))*log(f)^4/((c*log(f))^(7/2)*b^6*(-(b^2*c*x + a*b*c) 
^2*log(f)/(b^2*c))^(3/2)) + c^2*gamma(2, -(b^2*c*x + a*b*c)^2*log(f)/(b^2* 
c))*log(f)^2/((c*log(f))^(7/2)*b^3))/(sqrt(c*log(f))*b)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.67 \[ \int f^{c (a+b x)^2} x^3 \, dx=\frac {\frac {\sqrt {\pi } {\left (2 \, a^{3} c \log \left (f\right ) - 3 \, a\right )} \operatorname {erf}\left (-\sqrt {-c \log \left (f\right )} b {\left (x + \frac {a}{b}\right )}\right )}{\sqrt {-c \log \left (f\right )} b c \log \left (f\right )} + \frac {2 \, {\left (b^{2} c {\left (x + \frac {a}{b}\right )}^{2} \log \left (f\right ) - 3 \, a b c {\left (x + \frac {a}{b}\right )} \log \left (f\right ) + 3 \, a^{2} c \log \left (f\right ) - 1\right )} e^{\left (b^{2} c x^{2} \log \left (f\right ) + 2 \, a b c x \log \left (f\right ) + a^{2} c \log \left (f\right )\right )}}{b c^{2} \log \left (f\right )^{2}}}{4 \, b^{3}} \] Input:

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

Output:

1/4*(sqrt(pi)*(2*a^3*c*log(f) - 3*a)*erf(-sqrt(-c*log(f))*b*(x + a/b))/(sq 
rt(-c*log(f))*b*c*log(f)) + 2*(b^2*c*(x + a/b)^2*log(f) - 3*a*b*c*(x + a/b 
)*log(f) + 3*a^2*c*log(f) - 1)*e^(b^2*c*x^2*log(f) + 2*a*b*c*x*log(f) + a^ 
2*c*log(f))/(b*c^2*log(f)^2))/b^3
 

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.84 \[ \int f^{c (a+b x)^2} x^3 \, dx=\frac {f^{b^2\,c\,x^2}\,f^{a^2\,c}\,f^{2\,a\,b\,c\,x}\,x^2}{2\,b^2\,c\,\ln \left (f\right )}-\frac {\sqrt {\pi }\,\mathrm {erfi}\left (\sqrt {c\,\ln \left (f\right )}\,\left (a+b\,x\right )\right )\,\left (\frac {a^3}{b^4}-\frac {3\,a}{2\,b^4\,c\,\ln \left (f\right )}\right )}{2\,\sqrt {c\,\ln \left (f\right )}}+\frac {f^{b^2\,c\,x^2}\,f^{a^2\,c}\,f^{2\,a\,b\,c\,x}\,\left (\frac {a^2\,c\,\ln \left (f\right )}{2}-\frac {1}{2}\right )}{b^4\,c^2\,{\ln \left (f\right )}^2}-\frac {a\,f^{b^2\,c\,x^2}\,f^{a^2\,c}\,f^{2\,a\,b\,c\,x}\,x}{2\,b^3\,c\,\ln \left (f\right )} \] Input:

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

Output:

(f^(b^2*c*x^2)*f^(a^2*c)*f^(2*a*b*c*x)*x^2)/(2*b^2*c*log(f)) - (pi^(1/2)*e 
rfi((c*log(f))^(1/2)*(a + b*x))*(a^3/b^4 - (3*a)/(2*b^4*c*log(f))))/(2*(c* 
log(f))^(1/2)) + (f^(b^2*c*x^2)*f^(a^2*c)*f^(2*a*b*c*x)*((a^2*c*log(f))/2 
- 1/2))/(b^4*c^2*log(f)^2) - (a*f^(b^2*c*x^2)*f^(a^2*c)*f^(2*a*b*c*x)*x)/( 
2*b^3*c*log(f))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.15 \[ \int f^{c (a+b x)^2} x^3 \, dx=\frac {2 \sqrt {\pi }\, \mathrm {erf}\left (\frac {\mathrm {log}\left (f \right ) a c i +\mathrm {log}\left (f \right ) b c i x}{\sqrt {c}\, \sqrt {\mathrm {log}\left (f \right )}}\right ) \mathrm {log}\left (f \right )^{2} a^{3} c^{2} i -3 \sqrt {\pi }\, \mathrm {erf}\left (\frac {\mathrm {log}\left (f \right ) a c i +\mathrm {log}\left (f \right ) b c i x}{\sqrt {c}\, \sqrt {\mathrm {log}\left (f \right )}}\right ) \mathrm {log}\left (f \right ) a c i +2 f^{b^{2} c \,x^{2}+2 a b c x +a^{2} c} \sqrt {c}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right ) a^{2} c -2 f^{b^{2} c \,x^{2}+2 a b c x +a^{2} c} \sqrt {c}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right ) a b c x +2 f^{b^{2} c \,x^{2}+2 a b c x +a^{2} c} \sqrt {c}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right ) b^{2} c \,x^{2}-2 f^{b^{2} c \,x^{2}+2 a b c x +a^{2} c} \sqrt {c}\, \sqrt {\mathrm {log}\left (f \right )}}{4 \sqrt {c}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right )^{2} b^{4} c^{2}} \] Input:

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

Output:

(2*sqrt(pi)*erf((log(f)*a*c*i + log(f)*b*c*i*x)/(sqrt(c)*sqrt(log(f))))*lo 
g(f)**2*a**3*c**2*i - 3*sqrt(pi)*erf((log(f)*a*c*i + log(f)*b*c*i*x)/(sqrt 
(c)*sqrt(log(f))))*log(f)*a*c*i + 2*f**(a**2*c + 2*a*b*c*x + b**2*c*x**2)* 
sqrt(c)*sqrt(log(f))*log(f)*a**2*c - 2*f**(a**2*c + 2*a*b*c*x + b**2*c*x** 
2)*sqrt(c)*sqrt(log(f))*log(f)*a*b*c*x + 2*f**(a**2*c + 2*a*b*c*x + b**2*c 
*x**2)*sqrt(c)*sqrt(log(f))*log(f)*b**2*c*x**2 - 2*f**(a**2*c + 2*a*b*c*x 
+ b**2*c*x**2)*sqrt(c)*sqrt(log(f)))/(4*sqrt(c)*sqrt(log(f))*log(f)**2*b** 
4*c**2)