\(\int f^{a+b x+c x^2} x^3 \, dx\) [426]

Optimal result

Integrand size = 16, antiderivative size = 217 \[ \int f^{a+b x+c x^2} x^3 \, dx=-\frac {f^{a+b x+c x^2}}{2 c^2 \log ^2(f)}+\frac {3 b f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{8 c^{5/2} \log ^{\frac {3}{2}}(f)}+\frac {b^2 f^{a+b x+c x^2}}{8 c^3 \log (f)}-\frac {b f^{a+b x+c x^2} x}{4 c^2 \log (f)}+\frac {f^{a+b x+c x^2} x^2}{2 c \log (f)}-\frac {b^3 f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{16 c^{7/2} \sqrt {\log (f)}} \]

[Out]

-1/2*f^(c*x^2+b*x+a)/c^2/ln(f)^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)+3/8*b*f^(a-1/4*b^2/c)*erfi(1/2*(2*c*x+b)*ln(f)^(1/2)/c^(1/2))*Pi^(1/2)/c^(5/2)/ln(f)^(3
/2)-1/16*b^3*f^(a-1/4*b^2/c)*erfi(1/2*(2*c*x+b)*ln(f)^(1/2)/c^(1/2))*Pi^(1/2)/c^(7/2)/ln(f)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2273, 2272, 2266, 2235} \[ \int f^{a+b x+c x^2} x^3 \, dx=\frac {3 \sqrt {\pi } b f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{8 c^{5/2} \log ^{\frac {3}{2}}(f)}+\frac {b^2 f^{a+b x+c x^2}}{8 c^3 \log (f)}-\frac {\sqrt {\pi } b^3 f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{16 c^{7/2} \sqrt {\log (f)}}-\frac {f^{a+b x+c x^2}}{2 c^2 \log ^2(f)}-\frac {b x f^{a+b x+c x^2}}{4 c^2 \log (f)}+\frac {x^2 f^{a+b x+c x^2}}{2 c \log (f)} \]

[In]

Int[f^(a + b*x + c*x^2)*x^3,x]

[Out]

-1/2*f^(a + b*x + c*x^2)/(c^2*Log[f]^2) + (3*b*f^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[Log[f]])/(2*S
qrt[c])])/(8*c^(5/2)*Log[f]^(3/2)) + (b^2*f^(a + b*x + c*x^2))/(8*c^3*Log[f]) - (b*f^(a + b*x + c*x^2)*x)/(4*c
^2*Log[f]) + (f^(a + b*x + c*x^2)*x^2)/(2*c*Log[f]) - (b^3*f^(a - b^2/(4*c))*Sqrt[Pi]*Erfi[((b + 2*c*x)*Sqrt[L
og[f]])/(2*Sqrt[c])])/(16*c^(7/2)*Sqrt[Log[f]])

Rule 2235

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]

Rule 2266

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2272

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] - Dist[(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]

Rule 2273

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
(F^(a + b*x + c*x^2)/(2*c*Log[F])), x] + (-Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*F^(a + b*x + c*x^2)
, x], x] - Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {f^{a+b x+c x^2} x^2}{2 c \log (f)}-\frac {b \int f^{a+b x+c x^2} x^2 \, dx}{2 c}-\frac {\int f^{a+b x+c x^2} x \, dx}{c \log (f)} \\ & = -\frac {f^{a+b x+c x^2}}{2 c^2 \log ^2(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^2 \int f^{a+b x+c x^2} x \, dx}{4 c^2}+\frac {b \int f^{a+b x+c x^2} \, dx}{4 c^2 \log (f)}+\frac {b \int f^{a+b x+c x^2} \, dx}{2 c^2 \log (f)} \\ & = -\frac {f^{a+b x+c x^2}}{2 c^2 \log ^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 \int f^{a+b x+c x^2} \, dx}{8 c^3}+\frac {\left (b f^{a-\frac {b^2}{4 c}}\right ) \int f^{\frac {(b+2 c x)^2}{4 c}} \, dx}{4 c^2 \log (f)}+\frac {\left (b f^{a-\frac {b^2}{4 c}}\right ) \int f^{\frac {(b+2 c x)^2}{4 c}} \, dx}{2 c^2 \log (f)} \\ & = -\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 {\left (b^3 f^{a-\frac {b^2}{4 c}}\right ) \int f^{\frac {(b+2 c x)^2}{4 c}} \, dx}{8 c^3} \\ & = -\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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.56 \[ \int f^{a+b x+c x^2} x^3 \, dx=\frac {f^{a-\frac {b^2}{4 c}} \left (b \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right ) \sqrt {\log (f)} \left (6 c-b^2 \log (f)\right )+2 \sqrt {c} f^{\frac {(b+2 c x)^2}{4 c}} \left (-4 c+\left (b^2-2 b c x+4 c^2 x^2\right ) \log (f)\right )\right )}{16 c^{7/2} \log ^2(f)} \]

[In]

Integrate[f^(a + b*x + c*x^2)*x^3,x]

[Out]

(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)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {x^{2} f^{c \,x^{2}} f^{b x} f^{a}}{2 c \ln \left (f \right )}-\frac {b x \,f^{c \,x^{2}} f^{b x} f^{a}}{4 c^{2} \ln \left (f \right )}+\frac {b^{2} f^{c \,x^{2}} f^{b x} f^{a}}{8 c^{3} \ln \left (f \right )}+\frac {b^{3} \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{16 c^{3} \sqrt {-c \ln \left (f \right )}}-\frac {3 b \sqrt {\pi }\, f^{a} f^{-\frac {b^{2}}{4 c}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {b \ln \left (f \right )}{2 \sqrt {-c \ln \left (f \right )}}\right )}{8 c^{2} \ln \left (f \right ) \sqrt {-c \ln \left (f \right )}}-\frac {f^{c \,x^{2}} f^{b x} f^{a}}{2 c^{2} \ln \left (f \right )^{2}}\)	\(218\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.53 \[ \int f^{a+b x+c x^2} x^3 \, dx=-\frac {2 \, {\left (4 \, c^{2} - {\left (4 \, c^{3} x^{2} - 2 \, b c^{2} x + b^{2} c\right )} \log \left (f\right )\right )} f^{c x^{2} + b x + a} - \frac {\sqrt {\pi } {\left (b^{3} \log \left (f\right ) - 6 \, b c\right )} \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c \log \left (f\right )}}{2 \, c}\right )}{f^{\frac {b^{2} - 4 \, a c}{4 \, c}}}}{16 \, c^{4} \log \left (f\right )^{2}} \]

[In]

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

[Out]

-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)*sqrt(-c*log(f))/c)/f^(1/4*(b^2 - 4*a*c)/c))/(c^4*log(f)^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.93 \[ \int f^{a+b x+c x^2} x^3 \, dx=-\frac {{\left (\frac {\sqrt {\pi } {\left (2 \, c x + b\right )} b^{3} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}}\right ) - 1\right )} \log \left (f\right )^{4}}{\sqrt {-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}} \left (c \log \left (f\right )\right )^{\frac {7}{2}}} - \frac {12 \, {\left (2 \, c x + b\right )}^{3} b \Gamma \left (\frac {3}{2}, -\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{4 \, c}\right ) \log \left (f\right )^{4}}{\left (-\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{c}\right )^{\frac {3}{2}} \left (c \log \left (f\right )\right )^{\frac {7}{2}}} - \frac {6 \, b^{2} c f^{\frac {{\left (2 \, c x + b\right )}^{2}}{4 \, c}} \log \left (f\right )^{3}}{\left (c \log \left (f\right )\right )^{\frac {7}{2}}} + \frac {8 \, c^{2} \Gamma \left (2, -\frac {{\left (2 \, c x + b\right )}^{2} \log \left (f\right )}{4 \, c}\right ) \log \left (f\right )^{2}}{\left (c \log \left (f\right )\right )^{\frac {7}{2}}}\right )} f^{a - \frac {b^{2}}{4 \, c}}}{16 \, \sqrt {c \log \left (f\right )}} \]

[In]

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

[Out]

-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))

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.63 \[ \int f^{a+b x+c x^2} x^3 \, dx=\frac {\frac {\sqrt {\pi } {\left (b^{3} \log \left (f\right ) - 6 \, b c\right )} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-c \log \left (f\right )} {\left (2 \, x + \frac {b}{c}\right )}\right ) e^{\left (-\frac {b^{2} \log \left (f\right ) - 4 \, a c \log \left (f\right )}{4 \, c}\right )}}{\sqrt {-c \log \left (f\right )} \log \left (f\right )} + \frac {2 \, {\left (c^{2} {\left (2 \, x + \frac {b}{c}\right )}^{2} \log \left (f\right ) - 3 \, b c {\left (2 \, x + \frac {b}{c}\right )} \log \left (f\right ) + 3 \, b^{2} \log \left (f\right ) - 4 \, c\right )} e^{\left (c x^{2} \log \left (f\right ) + b x \log \left (f\right ) + a \log \left (f\right )\right )}}{\log \left (f\right )^{2}}}{16 \, c^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.71 \[ \int f^{a+b x+c x^2} x^3 \, dx=\frac {f^a\,f^{c\,x^2}\,f^{b\,x}\,x^2}{2\,c\,\ln \left (f\right )}-f^a\,f^{c\,x^2}\,f^{b\,x}\,\left (\frac {1}{2\,c^2\,{\ln \left (f\right )}^2}-\frac {b^2}{8\,c^3\,\ln \left (f\right )}\right )+\frac {f^{a-\frac {b^2}{4\,c}}\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {\frac {b\,\ln \left (f\right )}{2}+c\,x\,\ln \left (f\right )}{\sqrt {c\,\ln \left (f\right )}}\right )\,\left (\frac {3\,b\,c}{8}-\frac {b^3\,\ln \left (f\right )}{16}\right )}{c^3\,\ln \left (f\right )\,\sqrt {c\,\ln \left (f\right )}}-\frac {b\,f^a\,f^{c\,x^2}\,f^{b\,x}\,x}{4\,c^2\,\ln \left (f\right )} \]

[In]

int(f^(a + b*x + c*x^2)*x^3,x)

[Out]

(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*log(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))