Integrand size = 16, antiderivative size = 217 \[ \int f^{a+b x+c x^2} x^3 \, dx=-\frac {f^{a+b x+c x^2}}{2 c^2 \log ^2(f)}+\frac {3 b f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{8 c^{5/2} \log ^{\frac {3}{2}}(f)}+\frac {b^2 f^{a+b x+c x^2}}{8 c^3 \log (f)}-\frac {b f^{a+b x+c x^2} x}{4 c^2 \log (f)}+\frac {f^{a+b x+c x^2} x^2}{2 c \log (f)}-\frac {b^3 f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{16 c^{7/2} \sqrt {\log (f)}} \] Output:
-1/2*f^(c*x^2+b*x+a)/c^2/ln(f)^2+3/8*b*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^(5/2)/ln(f)^(3/2)+1/8*b^2*f^(c*x^2+b*x+a)/ c^3/ln(f)-1/4*b*f^(c*x^2+b*x+a)*x/c^2/ln(f)+1/2*f^(c*x^2+b*x+a)*x^2/c/ln(f )-1/16*b^3*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^(7/2)/ln(f)^(1/2)
Time = 0.37 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.56 \[ \int f^{a+b x+c x^2} x^3 \, dx=\frac {f^{a-\frac {b^2}{4 c}} \left (b \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right ) \sqrt {\log (f)} \left (6 c-b^2 \log (f)\right )+2 \sqrt {c} f^{\frac {(b+2 c x)^2}{4 c}} \left (-4 c+\left (b^2-2 b c x+4 c^2 x^2\right ) \log (f)\right )\right )}{16 c^{7/2} \log ^2(f)} \] Input:
Integrate[f^(a + b*x + c*x^2)*x^3,x]
Output:
(f^(a - b^2/(4*c))*(b*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c]) ]*Sqrt[Log[f]]*(6*c - b^2*Log[f]) + 2*Sqrt[c]*f^((b + 2*c*x)^2/(4*c))*(-4* c + (b^2 - 2*b*c*x + 4*c^2*x^2)*Log[f])))/(16*c^(7/2)*Log[f]^2)
Time = 1.45 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.36, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {2671, 2670, 2664, 2633, 2671, 2664, 2633, 2670, 2664, 2633}
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.
\(\displaystyle \int x^3 f^{a+b x+c x^2} \, dx\) |
\(\Big \downarrow \) 2671 |
\(\displaystyle -\frac {b \int f^{c x^2+b x+a} x^2dx}{2 c}-\frac {\int f^{c x^2+b x+a} xdx}{c \log (f)}+\frac {x^2 f^{a+b x+c x^2}}{2 c \log (f)}\) |
\(\Big \downarrow \) 2670 |
\(\displaystyle -\frac {b \int f^{c x^2+b x+a} x^2dx}{2 c}-\frac {\frac {f^{a+b x+c x^2}}{2 c \log (f)}-\frac {b \int f^{c x^2+b x+a}dx}{2 c}}{c \log (f)}+\frac {x^2 f^{a+b x+c x^2}}{2 c \log (f)}\) |
\(\Big \downarrow \) 2664 |
\(\displaystyle -\frac {\frac {f^{a+b x+c x^2}}{2 c \log (f)}-\frac {b f^{a-\frac {b^2}{4 c}} \int f^{\frac {(b+2 c x)^2}{4 c}}dx}{2 c}}{c \log (f)}-\frac {b \int f^{c x^2+b x+a} x^2dx}{2 c}+\frac {x^2 f^{a+b x+c x^2}}{2 c \log (f)}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle -\frac {b \int f^{c x^2+b x+a} x^2dx}{2 c}-\frac {\frac {f^{a+b x+c x^2}}{2 c \log (f)}-\frac {\sqrt {\pi } b f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2} \sqrt {\log (f)}}}{c \log (f)}+\frac {x^2 f^{a+b x+c x^2}}{2 c \log (f)}\) |
\(\Big \downarrow \) 2671 |
\(\displaystyle -\frac {b \left (-\frac {b \int f^{c x^2+b x+a} xdx}{2 c}-\frac {\int f^{c x^2+b x+a}dx}{2 c \log (f)}+\frac {x f^{a+b x+c x^2}}{2 c \log (f)}\right )}{2 c}-\frac {\frac {f^{a+b x+c x^2}}{2 c \log (f)}-\frac {\sqrt {\pi } b f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2} \sqrt {\log (f)}}}{c \log (f)}+\frac {x^2 f^{a+b x+c x^2}}{2 c \log (f)}\) |
\(\Big \downarrow \) 2664 |
\(\displaystyle -\frac {b \left (-\frac {f^{a-\frac {b^2}{4 c}} \int f^{\frac {(b+2 c x)^2}{4 c}}dx}{2 c \log (f)}-\frac {b \int f^{c x^2+b x+a} xdx}{2 c}+\frac {x f^{a+b x+c x^2}}{2 c \log (f)}\right )}{2 c}-\frac {\frac {f^{a+b x+c x^2}}{2 c \log (f)}-\frac {\sqrt {\pi } b f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2} \sqrt {\log (f)}}}{c \log (f)}+\frac {x^2 f^{a+b x+c x^2}}{2 c \log (f)}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle -\frac {b \left (-\frac {b \int f^{c x^2+b x+a} xdx}{2 c}-\frac {\sqrt {\pi } f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2} \log ^{\frac {3}{2}}(f)}+\frac {x f^{a+b x+c x^2}}{2 c \log (f)}\right )}{2 c}-\frac {\frac {f^{a+b x+c x^2}}{2 c \log (f)}-\frac {\sqrt {\pi } b f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2} \sqrt {\log (f)}}}{c \log (f)}+\frac {x^2 f^{a+b x+c x^2}}{2 c \log (f)}\) |
\(\Big \downarrow \) 2670 |
\(\displaystyle -\frac {b \left (-\frac {b \left (\frac {f^{a+b x+c x^2}}{2 c \log (f)}-\frac {b \int f^{c x^2+b x+a}dx}{2 c}\right )}{2 c}-\frac {\sqrt {\pi } f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2} \log ^{\frac {3}{2}}(f)}+\frac {x f^{a+b x+c x^2}}{2 c \log (f)}\right )}{2 c}-\frac {\frac {f^{a+b x+c x^2}}{2 c \log (f)}-\frac {\sqrt {\pi } b f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2} \sqrt {\log (f)}}}{c \log (f)}+\frac {x^2 f^{a+b x+c x^2}}{2 c \log (f)}\) |
\(\Big \downarrow \) 2664 |
\(\displaystyle -\frac {b \left (-\frac {b \left (\frac {f^{a+b x+c x^2}}{2 c \log (f)}-\frac {b f^{a-\frac {b^2}{4 c}} \int f^{\frac {(b+2 c x)^2}{4 c}}dx}{2 c}\right )}{2 c}-\frac {\sqrt {\pi } f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2} \log ^{\frac {3}{2}}(f)}+\frac {x f^{a+b x+c x^2}}{2 c \log (f)}\right )}{2 c}-\frac {\frac {f^{a+b x+c x^2}}{2 c \log (f)}-\frac {\sqrt {\pi } b f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2} \sqrt {\log (f)}}}{c \log (f)}+\frac {x^2 f^{a+b x+c x^2}}{2 c \log (f)}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle -\frac {b \left (-\frac {b \left (\frac {f^{a+b x+c x^2}}{2 c \log (f)}-\frac {\sqrt {\pi } b f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2} \sqrt {\log (f)}}\right )}{2 c}-\frac {\sqrt {\pi } f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2} \log ^{\frac {3}{2}}(f)}+\frac {x f^{a+b x+c x^2}}{2 c \log (f)}\right )}{2 c}-\frac {\frac {f^{a+b x+c x^2}}{2 c \log (f)}-\frac {\sqrt {\pi } b f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 c^{3/2} \sqrt {\log (f)}}}{c \log (f)}+\frac {x^2 f^{a+b x+c x^2}}{2 c \log (f)}\) |
Input:
Int[f^(a + b*x + c*x^2)*x^3,x]
Output:
-1/2*(b*(-1/2*(b*(f^(a + b*x + c*x^2)/(2*c*Log[f]) - (b*f^(a - b^2/(4*c))* Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*Sqrt[c])])/(4*c^(3/2)*Sqrt[Log [f]])))/c - (f^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2 *Sqrt[c])])/(4*c^(3/2)*Log[f]^(3/2)) + (f^(a + b*x + c*x^2)*x)/(2*c*Log[f] )))/c + (f^(a + b*x + c*x^2)*x^2)/(2*c*Log[f]) - (f^(a + b*x + c*x^2)/(2*c *Log[f]) - (b*f^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/( 2*Sqrt[c])])/(4*c^(3/2)*Sqrt[Log[f]]))/(c*Log[f])
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt [Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{ F, a, b, c, d}, x] && PosQ[b]
Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[F^(a - b^2/ (4*c)) Int[F^((b + 2*c*x)^2/(4*c)), x], x] /; FreeQ[{F, a, b, c}, x]
Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol ] :> Simp[e*(F^(a + b*x + c*x^2)/(2*c*Log[F])), x] - Simp[(b*e - 2*c*d)/(2* c) Int[F^(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e}, x] && NeQ [b*e - 2*c*d, 0]
Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_))^(m_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*(F^(a + b*x + c*x^2)/(2*c*Log[F])), x] + (-Simp[(b*e - 2*c*d)/(2*c) Int[(d + e*x)^(m - 1)*F^(a + b*x + c*x^2), x] , x] - Simp[(m - 1)*(e^2/(2*c*Log[F])) Int[(d + e*x)^(m - 2)*F^(a + b*x + c*x^2), x], x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[b*e - 2*c*d, 0] && GtQ[m, 1]
Time = 0.08 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {x^{2} f^{c \,x^{2}} f^{b x} f^{a}}{2 c \ln \left (f \right )}-\frac {b x \,f^{c \,x^{2}} f^{b x} f^{a}}{4 c^{2} \ln \left (f \right )}+\frac {b^{2} f^{c \,x^{2}} f^{b x} f^{a}}{8 c^{3} \ln \left (f \right )}+\frac {b^{3} \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{16 c^{3} \sqrt {-c \ln \left (f \right )}}-\frac {3 b \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{8 c^{2} \ln \left (f \right ) \sqrt {-c \ln \left (f \right )}}-\frac {f^{c \,x^{2}} f^{b x} f^{a}}{2 c^{2} \ln \left (f \right )^{2}}\) | \(218\) |
Input:
int(f^(c*x^2+b*x+a)*x^3,x,method=_RETURNVERBOSE)
Output:
1/2/c/ln(f)*x^2*f^(c*x^2)*f^(b*x)*f^a-1/4*b/c^2/ln(f)*x*f^(c*x^2)*f^(b*x)* f^a+1/8*b^2/c^3/ln(f)*f^(c*x^2)*f^(b*x)*f^a+1/16*b^3/c^3*Pi^(1/2)*f^a*f^(- 1/4*b^2/c)/(-c*ln(f))^(1/2)*erf(-(-c*ln(f))^(1/2)*x+1/2*b*ln(f)/(-c*ln(f)) ^(1/2))-3/8*b/c^2/ln(f)*Pi^(1/2)*f^a*f^(-1/4*b^2/c)/(-c*ln(f))^(1/2)*erf(- (-c*ln(f))^(1/2)*x+1/2*b*ln(f)/(-c*ln(f))^(1/2))-1/2/c^2/ln(f)^2*f^(c*x^2) *f^(b*x)*f^a
Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.53 \[ \int f^{a+b x+c x^2} x^3 \, dx=-\frac {2 \, {\left (4 \, c^{2} - {\left (4 \, c^{3} x^{2} - 2 \, b c^{2} x + b^{2} c\right )} \log \left (f\right )\right )} f^{c x^{2} + b x + a} - \frac {\sqrt {\pi } {\left (b^{3} \log \left (f\right ) - 6 \, b c\right )} \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c \log \left (f\right )}}{2 \, c}\right )}{f^{\frac {b^{2} - 4 \, a c}{4 \, c}}}}{16 \, c^{4} \log \left (f\right )^{2}} \] Input:
integrate(f^(c*x^2+b*x+a)*x^3,x, algorithm="fricas")
Output:
-1/16*(2*(4*c^2 - (4*c^3*x^2 - 2*b*c^2*x + b^2*c)*log(f))*f^(c*x^2 + b*x + a) - sqrt(pi)*(b^3*log(f) - 6*b*c)*sqrt(-c*log(f))*erf(1/2*(2*c*x + b)*sq rt(-c*log(f))/c)/f^(1/4*(b^2 - 4*a*c)/c))/(c^4*log(f)^2)
\[ \int f^{a+b x+c x^2} x^3 \, dx=\int f^{a + b x + c x^{2}} x^{3}\, dx \] Input:
integrate(f**(c*x**2+b*x+a)*x**3,x)
Output:
Integral(f**(a + b*x + c*x**2)*x**3, x)
Time = 0.14 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.93 \[ \int f^{a+b x+c x^2} x^3 \, dx=-\frac {{\left (\frac {\sqrt {\pi } {\left (2 \, c x + b\right )} b^{3} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}}\right ) - 1\right )} \log \left (f\right )^{4}}{\sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}} \left (c \log \left (f\right )\right )^{\frac {7}{2}}} - \frac {12 \, {\left (2 \, c x + b\right )}^{3} b \Gamma \left (\frac {3}{2}, -\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{4 \, c}\right ) \log \left (f\right )^{4}}{\left (-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}\right )^{\frac {3}{2}} \left (c \log \left (f\right )\right )^{\frac {7}{2}}} - \frac {6 \, b^{2} c f^{\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}} \log \left (f\right )^{3}}{\left (c \log \left (f\right )\right )^{\frac {7}{2}}} + \frac {8 \, c^{2} \Gamma \left (2, -\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{4 \, c}\right ) \log \left (f\right )^{2}}{\left (c \log \left (f\right )\right )^{\frac {7}{2}}}\right )} f^{a - \frac {b^{2}}{4 \, c}}}{16 \, \sqrt {c \log \left (f\right )}} \] Input:
integrate(f^(c*x^2+b*x+a)*x^3,x, algorithm="maxima")
Output:
-1/16*(sqrt(pi)*(2*c*x + b)*b^3*(erf(1/2*sqrt(-(2*c*x + b)^2*log(f)/c)) - 1)*log(f)^4/(sqrt(-(2*c*x + b)^2*log(f)/c)*(c*log(f))^(7/2)) - 12*(2*c*x + b)^3*b*gamma(3/2, -1/4*(2*c*x + b)^2*log(f)/c)*log(f)^4/((-(2*c*x + b)^2* log(f)/c)^(3/2)*(c*log(f))^(7/2)) - 6*b^2*c*f^(1/4*(2*c*x + b)^2/c)*log(f) ^3/(c*log(f))^(7/2) + 8*c^2*gamma(2, -1/4*(2*c*x + b)^2*log(f)/c)*log(f)^2 /(c*log(f))^(7/2))*f^(a - 1/4*b^2/c)/sqrt(c*log(f))
Time = 0.14 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.63 \[ \int f^{a+b x+c x^2} x^3 \, dx=\frac {\frac {\sqrt {\pi } {\left (b^{3} \log \left (f\right ) - 6 \, b c\right )} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right ) - 4 \, a c \log \left (f\right )}{4 \, c}\right )}}{\sqrt {-c \log \left (f\right )} \log \left (f\right )} + \frac {2 \, {\left (c^{2} {\left (2 \, x + \frac {b}{c}\right )}^{2} \log \left (f\right ) - 3 \, b c {\left (2 \, x + \frac {b}{c}\right )} \log \left (f\right ) + 3 \, b^{2} \log \left (f\right ) - 4 \, c\right )} e^{\left (c x^{2} \log \left (f\right ) + b x \log \left (f\right ) + a \log \left (f\right )\right )}}{\log \left (f\right )^{2}}}{16 \, c^{3}} \] Input:
integrate(f^(c*x^2+b*x+a)*x^3,x, algorithm="giac")
Output:
1/16*(sqrt(pi)*(b^3*log(f) - 6*b*c)*erf(-1/2*sqrt(-c*log(f))*(2*x + b/c))* e^(-1/4*(b^2*log(f) - 4*a*c*log(f))/c)/(sqrt(-c*log(f))*log(f)) + 2*(c^2*( 2*x + b/c)^2*log(f) - 3*b*c*(2*x + b/c)*log(f) + 3*b^2*log(f) - 4*c)*e^(c* x^2*log(f) + b*x*log(f) + a*log(f))/log(f)^2)/c^3
Time = 0.28 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.71 \[ \int f^{a+b x+c x^2} x^3 \, dx=\frac {f^a\,f^{c\,x^2}\,f^{b\,x}\,x^2}{2\,c\,\ln \left (f\right )}-f^a\,f^{c\,x^2}\,f^{b\,x}\,\left (\frac {1}{2\,c^2\,{\ln \left (f\right )}^2}-\frac {b^2}{8\,c^3\,\ln \left (f\right )}\right )+\frac {f^{a-\frac {b^2}{4\,c}}\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {\frac {b\,\ln \left (f\right )}{2}+c\,x\,\ln \left (f\right )}{\sqrt {c\,\ln \left (f\right )}}\right )\,\left (\frac {3\,b\,c}{8}-\frac {b^3\,\ln \left (f\right )}{16}\right )}{c^3\,\ln \left (f\right )\,\sqrt {c\,\ln \left (f\right )}}-\frac {b\,f^a\,f^{c\,x^2}\,f^{b\,x}\,x}{4\,c^2\,\ln \left (f\right )} \] Input:
int(f^(a + b*x + c*x^2)*x^3,x)
Output:
(f^a*f^(c*x^2)*f^(b*x)*x^2)/(2*c*log(f)) - f^a*f^(c*x^2)*f^(b*x)*(1/(2*c^2 *log(f)^2) - b^2/(8*c^3*log(f))) + (f^(a - b^2/(4*c))*pi^(1/2)*erfi(((b*lo g(f))/2 + c*x*log(f))/(c*log(f))^(1/2))*((3*b*c)/8 - (b^3*log(f))/16))/(c^ 3*log(f)*(c*log(f))^(1/2)) - (b*f^a*f^(c*x^2)*f^(b*x)*x)/(4*c^2*log(f))
Time = 0.22 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.13 \[ \int f^{a+b x+c x^2} x^3 \, dx=\frac {f^{a} \left (\sqrt {\pi }\, \mathrm {erf}\left (\frac {\mathrm {log}\left (f \right ) b i +2 \,\mathrm {log}\left (f \right ) c i x}{2 \sqrt {c}\, \sqrt {\mathrm {log}\left (f \right )}}\right ) \mathrm {log}\left (f \right )^{2} b^{3} i -6 \sqrt {\pi }\, \mathrm {erf}\left (\frac {\mathrm {log}\left (f \right ) b i +2 \,\mathrm {log}\left (f \right ) c i x}{2 \sqrt {c}\, \sqrt {\mathrm {log}\left (f \right )}}\right ) \mathrm {log}\left (f \right ) b c i +2 f^{c \,x^{2}+b x} e^{\frac {\mathrm {log}\left (f \right ) b^{2}}{4 c}} \sqrt {c}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right ) b^{2}-4 f^{c \,x^{2}+b x} e^{\frac {\mathrm {log}\left (f \right ) b^{2}}{4 c}} \sqrt {c}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right ) b c x +8 f^{c \,x^{2}+b x} e^{\frac {\mathrm {log}\left (f \right ) b^{2}}{4 c}} \sqrt {c}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right ) c^{2} x^{2}-8 f^{c \,x^{2}+b x} e^{\frac {\mathrm {log}\left (f \right ) b^{2}}{4 c}} \sqrt {c}\, \sqrt {\mathrm {log}\left (f \right )}\, c \right )}{16 e^{\frac {\mathrm {log}\left (f \right ) b^{2}}{4 c}} \sqrt {c}\, \sqrt {\mathrm {log}\left (f \right )}\, \mathrm {log}\left (f \right )^{2} c^{3}} \] Input:
int(f^(c*x^2+b*x+a)*x^3,x)
Output:
(f**a*(sqrt(pi)*erf((log(f)*b*i + 2*log(f)*c*i*x)/(2*sqrt(c)*sqrt(log(f))) )*log(f)**2*b**3*i - 6*sqrt(pi)*erf((log(f)*b*i + 2*log(f)*c*i*x)/(2*sqrt( c)*sqrt(log(f))))*log(f)*b*c*i + 2*f**(b*x + c*x**2)*e**((log(f)*b**2)/(4* c))*sqrt(c)*sqrt(log(f))*log(f)*b**2 - 4*f**(b*x + c*x**2)*e**((log(f)*b** 2)/(4*c))*sqrt(c)*sqrt(log(f))*log(f)*b*c*x + 8*f**(b*x + c*x**2)*e**((log (f)*b**2)/(4*c))*sqrt(c)*sqrt(log(f))*log(f)*c**2*x**2 - 8*f**(b*x + c*x** 2)*e**((log(f)*b**2)/(4*c))*sqrt(c)*sqrt(log(f))*c))/(16*e**((log(f)*b**2) /(4*c))*sqrt(c)*sqrt(log(f))*log(f)**2*c**3)