\(\int f^{c (a+b x)^2} x \, dx\) [197]

Optimal result

Integrand size = 13, antiderivative size = 68 \[ \int f^{c (a+b x)^2} x \, dx=\frac {f^{c (a+b x)^2}}{2 b^2 c \log (f)}-\frac {a \sqrt {\pi } \text {erfi}\left (\sqrt {c} (a+b x) \sqrt {\log (f)}\right )}{2 b^2 \sqrt {c} \sqrt {\log (f)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2258, 2235, 2240} \[ \int f^{c (a+b x)^2} x \, dx=\frac {f^{c (a+b x)^2}}{2 b^2 c \log (f)}-\frac {\sqrt {\pi } a \text {erfi}\left (\sqrt {c} \sqrt {\log (f)} (a+b x)\right )}{2 b^2 \sqrt {c} \sqrt {\log (f)}} \]

[In]

Int[f^(c*(a + b*x)^2)*x,x]

[Out]

f^(c*(a + b*x)^2)/(2*b^2*c*Log[f]) - (a*Sqrt[Pi]*Erfi[Sqrt[c]*(a + b*x)*Sqrt[Log[f]]])/(2*b^2*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 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 f^{c (a+b x)^2}}{b}+\frac {f^{c (a+b x)^2} (a+b x)}{b}\right ) \, dx \\ & = \frac {\int f^{c (a+b x)^2} (a+b x) \, dx}{b}-\frac {a \int f^{c (a+b x)^2} \, dx}{b} \\ & = \frac {f^{c (a+b x)^2}}{2 b^2 c \log (f)}-\frac {a \sqrt {\pi } \text {erfi}\left (\sqrt {c} (a+b x) \sqrt {\log (f)}\right )}{2 b^2 \sqrt {c} \sqrt {\log (f)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.93 \[ \int f^{c (a+b x)^2} x \, dx=\frac {f^{c (a+b x)^2}-a \sqrt {c} \sqrt {\pi } \text {erfi}\left (\sqrt {c} (a+b x) \sqrt {\log (f)}\right ) \sqrt {\log (f)}}{2 b^2 c \log (f)} \]

[In]

Integrate[f^(c*(a + b*x)^2)*x,x]

[Out]

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

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.18

method	result	size
risch	\(\frac {f^{b^{2} c \,x^{2}} f^{2 a b c x} f^{a^{2} c}}{2 b^{2} c \ln \left (f \right )}+\frac {a \sqrt {\pi }\, \operatorname {erf}\left (-b \sqrt {-c \ln \left (f \right )}\, x +\frac {a c \ln \left (f \right )}{\sqrt {-c \ln \left (f \right )}}\right )}{2 b^{2} \sqrt {-c \ln \left (f \right )}}\)	\(80\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.06 \[ \int f^{c (a+b x)^2} x \, dx=\frac {\sqrt {\pi } \sqrt {-b^{2} c \log \left (f\right )} a \operatorname {erf}\left (\frac {\sqrt {-b^{2} c \log \left (f\right )} {\left (b x + a\right )}}{b}\right ) + b f^{b^{2} c x^{2} + 2 \, a b c x + a^{2} c}}{2 \, b^{3} c \log \left (f\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(f**(c*(a + b*x)**2)*x, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (54) = 108\).

Time = 0.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.93 \[ \int f^{c (a+b x)^2} x \, dx=-\frac {\frac {\sqrt {\pi } {\left (b^{2} c x + a b c\right )} a c {\left (\operatorname {erf}\left (\sqrt {-\frac {{\left (b^{2} c x + a b c\right )}^{2} \log \left (f\right )}{b^{2} c}}\right ) - 1\right )} \log \left (f\right )^{2}}{\left (c \log \left (f\right )\right )^{\frac {3}{2}} b^{2} \sqrt {-\frac {{\left (b^{2} c x + a b c\right )}^{2} \log \left (f\right )}{b^{2} c}}} - \frac {c f^{\frac {{\left (b^{2} c x + a b c\right )}^{2}}{b^{2} c}} \log \left (f\right )}{\left (c \log \left (f\right )\right )^{\frac {3}{2}} b}}{2 \, \sqrt {c \log \left (f\right )} b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.13 \[ \int f^{c (a+b x)^2} x \, dx=\frac {\frac {\sqrt {\pi } a \operatorname {erf}\left (-\sqrt {-c \log \left (f\right )} b {\left (x + \frac {a}{b}\right )}\right )}{\sqrt {-c \log \left (f\right )} b} + \frac {e^{\left (b^{2} c x^{2} \log \left (f\right ) + 2 \, a b c x \log \left (f\right ) + a^{2} c \log \left (f\right )\right )}}{b c \log \left (f\right )}}{2 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int f^{c (a+b x)^2} x \, dx=\frac {f^{b^2\,c\,x^2}\,f^{a^2\,c}\,f^{2\,a\,b\,c\,x}}{2\,b^2\,c\,\ln \left (f\right )}-\frac {a\,\sqrt {\pi }\,\mathrm {erfi}\left (\sqrt {c\,\ln \left (f\right )}\,\left (a+b\,x\right )\right )}{2\,b^2\,\sqrt {c\,\ln \left (f\right )}} \]

[In]

int(f^(c*(a + b*x)^2)*x,x)

[Out]

(f^(b^2*c*x^2)*f^(a^2*c)*f^(2*a*b*c*x))/(2*b^2*c*log(f)) - (a*pi^(1/2)*erfi((c*log(f))^(1/2)*(a + b*x)))/(2*b^
2*(c*log(f))^(1/2))