\(\int f^{c (a+b x)^2} x^3 \, dx\) [195]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2258, 2235, 2240, 2243} \[ \int f^{c (a+b x)^2} x^3 \, dx=-\frac {\sqrt {\pi } a^3 \text {erfi}\left (\sqrt {c} \sqrt {\log (f)} (a+b x)\right )}{2 b^4 \sqrt {c} \sqrt {\log (f)}}+\frac {3 a^2 f^{c (a+b x)^2}}{2 b^4 c \log (f)}+\frac {3 \sqrt {\pi } a \text {erfi}\left (\sqrt {c} \sqrt {\log (f)} (a+b x)\right )}{4 b^4 c^{3/2} \log ^{\frac {3}{2}}(f)}-\frac {f^{c (a+b x)^2}}{2 b^4 c^2 \log ^2(f)}+\frac {(a+b x)^2 f^{c (a+b x)^2}}{2 b^4 c \log (f)}-\frac {3 a (a+b x) f^{c (a+b x)^2}}{2 b^4 c \log (f)} \]

[In]

Int[f^(c*(a + b*x)^2)*x^3,x]

[Out]

-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[Sqrt[c]*(a + b*x)*Sqrt[Log[f]]])/(2
*b^4*Sqrt[c]*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 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2243

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
- n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*Log[F])), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2258

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

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.47 \[ \int f^{c (a+b x)^2} x^3 \, dx=\frac {a \sqrt {c} \sqrt {\pi } \text {erfi}\left (\sqrt {c} (a+b x) \sqrt {\log (f)}\right ) \sqrt {\log (f)} \left (3-2 a^2 c \log (f)\right )+2 f^{c (a+b x)^2} \left (-1+c \left (a^2-a b x+b^2 x^2\right ) \log (f)\right )}{4 b^4 c^2 \log ^2(f)} \]

[In]

Integrate[f^(c*(a + b*x)^2)*x^3,x]

[Out]

(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)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.23

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

[In]

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

[Out]

1/2/b^2/c/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/b^4/c^2/ln(f)^2*f^(b^2*c*x^2)*f^(2*a*b*c*x)*f^(a^2*c)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.56 \[ \int f^{c (a+b x)^2} x^3 \, dx=\frac {\sqrt {\pi } {\left (2 \, a^{3} c \log \left (f\right ) - 3 \, a\right )} \sqrt {-b^{2} c \log \left (f\right )} \operatorname {erf}\left (\frac {\sqrt {-b^{2} c \log \left (f\right )} {\left (b x + a\right )}}{b}\right ) + 2 \, {\left ({\left (b^{3} c x^{2} - a b^{2} c x + a^{2} b c\right )} \log \left (f\right ) - b\right )} f^{b^{2} c x^{2} + 2 \, a b c x + a^{2} c}}{4 \, b^{5} c^{2} \log \left (f\right )^{2}} \]

[In]

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

[Out]

1/4*(sqrt(pi)*(2*a^3*c*log(f) - 3*a)*sqrt(-b^2*c*log(f))*erf(sqrt(-b^2*c*log(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]

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

[In]

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

[Out]

Integral(f**(c*(a + b*x)**2)*x**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.30 \[ \int f^{c (a+b x)^2} x^3 \, dx=-\frac {\frac {\sqrt {\pi } {\left (b^{2} c x + a b c\right )} a^{3} c^{3} {\left (\operatorname {erf}\left (\sqrt {-\frac {{\left (b^{2} c x + a b c\right )}^{2} \log \left (f\right )}{b^{2} c}}\right ) - 1\right )} \log \left (f\right )^{4}}{\left (c \log \left (f\right )\right )^{\frac {7}{2}} b^{4} \sqrt {-\frac {{\left (b^{2} c x + a b c\right )}^{2} \log \left (f\right )}{b^{2} c}}} - \frac {3 \, a^{2} c^{3} f^{\frac {{\left (b^{2} c x + a b c\right )}^{2}}{b^{2} c}} \log \left (f\right )^{3}}{\left (c \log \left (f\right )\right )^{\frac {7}{2}} b^{3}} - \frac {3 \, {\left (b^{2} c x + a b c\right )}^{3} a c \Gamma \left (\frac {3}{2}, -\frac {{\left (b^{2} c x + a b c\right )}^{2} \log \left (f\right )}{b^{2} c}\right ) \log \left (f\right )^{4}}{\left (c \log \left (f\right )\right )^{\frac {7}{2}} b^{6} \left (-\frac {{\left (b^{2} c x + a b c\right )}^{2} \log \left (f\right )}{b^{2} c}\right )^{\frac {3}{2}}} + \frac {c^{2} \Gamma \left (2, -\frac {{\left (b^{2} c x + a b c\right )}^{2} \log \left (f\right )}{b^{2} c}\right ) \log \left (f\right )^{2}}{\left (c \log \left (f\right )\right )^{\frac {7}{2}} b^{3}}}{2 \, \sqrt {c \log \left (f\right )} b} \]

[In]

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

[Out]

-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*log(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)

none

Time = 0.34 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.67 \[ \int f^{c (a+b x)^2} x^3 \, dx=\frac {\frac {\sqrt {\pi } {\left (2 \, a^{3} c \log \left (f\right ) - 3 \, a\right )} \operatorname {erf}\left (-\sqrt {-c \log \left (f\right )} b {\left (x + \frac {a}{b}\right )}\right )}{\sqrt {-c \log \left (f\right )} b c \log \left (f\right )} + \frac {2 \, {\left (b^{2} c {\left (x + \frac {a}{b}\right )}^{2} \log \left (f\right ) - 3 \, a b c {\left (x + \frac {a}{b}\right )} \log \left (f\right ) + 3 \, a^{2} c \log \left (f\right ) - 1\right )} e^{\left (b^{2} c x^{2} \log \left (f\right ) + 2 \, a b c x \log \left (f\right ) + a^{2} c \log \left (f\right )\right )}}{b c^{2} \log \left (f\right )^{2}}}{4 \, b^{3}} \]

[In]

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

[Out]

1/4*(sqrt(pi)*(2*a^3*c*log(f) - 3*a)*erf(-sqrt(-c*log(f))*b*(x + a/b))/(sqrt(-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 = 0.34 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.84 \[ \int f^{c (a+b x)^2} x^3 \, dx=\frac {f^{b^2\,c\,x^2}\,f^{a^2\,c}\,f^{2\,a\,b\,c\,x}\,x^2}{2\,b^2\,c\,\ln \left (f\right )}-\frac {\sqrt {\pi }\,\mathrm {erfi}\left (\sqrt {c\,\ln \left (f\right )}\,\left (a+b\,x\right )\right )\,\left (\frac {a^3}{b^4}-\frac {3\,a}{2\,b^4\,c\,\ln \left (f\right )}\right )}{2\,\sqrt {c\,\ln \left (f\right )}}+\frac {f^{b^2\,c\,x^2}\,f^{a^2\,c}\,f^{2\,a\,b\,c\,x}\,\left (\frac {a^2\,c\,\ln \left (f\right )}{2}-\frac {1}{2}\right )}{b^4\,c^2\,{\ln \left (f\right )}^2}-\frac {a\,f^{b^2\,c\,x^2}\,f^{a^2\,c}\,f^{2\,a\,b\,c\,x}\,x}{2\,b^3\,c\,\ln \left (f\right )} \]

[In]

int(f^(c*(a + b*x)^2)*x^3,x)

[Out]

(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)*erfi((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))