\(\int f^{\frac {c}{(a+b x)^2}} \, dx\) [228]

Optimal result

Integrand size = 11, antiderivative size = 62 \[ \int f^{\frac {c}{(a+b x)^2}} \, dx=\frac {f^{\frac {c}{(a+b x)^2}} (a+b x)}{b}-\frac {\sqrt {c} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right ) \sqrt {\log (f)}}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2237, 2242, 2235} \[ \int f^{\frac {c}{(a+b x)^2}} \, dx=\frac {(a+b x) f^{\frac {c}{(a+b x)^2}}}{b}-\frac {\sqrt {\pi } \sqrt {c} \sqrt {\log (f)} \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right )}{b} \]

[In]

Int[f^(c/(a + b*x)^2),x]

[Out]

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

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 2237

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

Rule 2242

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[1/(d*(m + 1))
, Subst[Int[F^(a + b*x^2), x], x, (c + d*x)^(m + 1)], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && EqQ[n, 2*(m + 1
)]

Rubi steps \begin{align*} \text {integral}& = \frac {f^{\frac {c}{(a+b x)^2}} (a+b x)}{b}+(2 c \log (f)) \int \frac {f^{\frac {c}{(a+b x)^2}}}{(a+b x)^2} \, dx \\ & = \frac {f^{\frac {c}{(a+b x)^2}} (a+b x)}{b}-\frac {(2 c \log (f)) \text {Subst}\left (\int f^{c x^2} \, dx,x,\frac {1}{a+b x}\right )}{b} \\ & = \frac {f^{\frac {c}{(a+b x)^2}} (a+b x)}{b}-\frac {\sqrt {c} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right ) \sqrt {\log (f)}}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00 \[ \int f^{\frac {c}{(a+b x)^2}} \, dx=\frac {f^{\frac {c}{(a+b x)^2}} (a+b x)}{b}-\frac {\sqrt {c} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right ) \sqrt {\log (f)}}{b} \]

[In]

Integrate[f^(c/(a + b*x)^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.05

method	result	size
risch	\(f^{\frac {c}{\left (b x +a \right )^{2}}} x +\frac {f^{\frac {c}{\left (b x +a \right )^{2}}} a}{b}-\frac {\ln \left (f \right ) c \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-c \ln \left (f \right )}}{b x +a}\right )}{b \sqrt {-c \ln \left (f \right )}}\)	\(65\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.10 \[ \int f^{\frac {c}{(a+b x)^2}} \, dx=\frac {\sqrt {\pi } b \sqrt {-\frac {c \log \left (f\right )}{b^{2}}} \operatorname {erf}\left (\frac {b \sqrt {-\frac {c \log \left (f\right )}{b^{2}}}}{b x + a}\right ) + {\left (b x + a\right )} f^{\frac {c}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{b} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(f**(c/(a + b*x)**2), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(f^(c/(b*x + a)^2), x)

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.85 \[ \int f^{\frac {c}{(a+b x)^2}} \, dx=\frac {f^{\frac {c}{{\left (a+b\,x\right )}^2}}\,\left (a+b\,x\right )}{b}-\frac {c\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {\sqrt {c\,\ln \left (f\right )}}{a+b\,x}\right )\,\ln \left (f\right )}{b\,\sqrt {c\,\ln \left (f\right )}} \]

[In]

int(f^(c/(a + b*x)^2),x)

[Out]

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