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

Optimal. Leaf size=62 \[ \frac {(a+b x) f^{\frac {c}{(a+b x)^2}}}{b}-\frac {\sqrt {\pi } \sqrt {c} \sqrt {\log (f)} \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right )}{b} \]

[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

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Rubi [A]  time = 0.04, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2206, 2211, 2204} \[ \frac {(a+b x) f^{\frac {c}{(a+b x)^2}}}{b}-\frac {\sqrt {\pi } \sqrt {c} \sqrt {\log (f)} \text {Erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[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 2204

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 2206

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 2211

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*} \int f^{\frac {c}{(a+b x)^2}} \, dx &=\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)) \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 62, normalized size = 1.00 \[ \frac {(a+b x) f^{\frac {c}{(a+b x)^2}}}{b}-\frac {\sqrt {\pi } \sqrt {c} \sqrt {\log (f)} \text {erfi}\left (\frac {\sqrt {c} \sqrt {\log (f)}}{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.45, size = 68, normalized size = 1.10 \[ \frac {\sqrt {\pi } b \sqrt {-\frac {c \log \relax (f)}{b^{2}}} \operatorname {erf}\left (\frac {b \sqrt {-\frac {c \log \relax (f)}{b^{2}}}}{b x + a}\right ) + {\left (b x + a\right )} f^{\frac {c}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{\frac {c}{{\left (b x + a\right )}^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(c/(b*x + a)^2), x)

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maple [A]  time = 0.06, size = 65, normalized size = 1.05 \[ -\frac {\sqrt {\pi }\, c \erf \left (\frac {\sqrt {-c \ln \relax (f )}}{b x +a}\right ) \ln \relax (f )}{\sqrt {-c \ln \relax (f )}\, b}+x \,f^{\frac {c}{\left (b x +a \right )^{2}}}+\frac {a \,f^{\frac {c}{\left (b x +a \right )^{2}}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 2 \, b c \int \frac {f^{\frac {c}{b^{2} x^{2} + 2 \, a b x + a^{2}}} x}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}\,{d x} \log \relax (f) + f^{\frac {c}{b^{2} x^{2} + 2 \, a b x + a^{2}}} x \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 3.64, size = 53, normalized size = 0.85 \[ \frac {f^{\frac {c}{{\left (a+b\,x\right )}^2}}\,\left (a+b\,x\right )}{b}-\frac {c\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {\sqrt {c\,\ln \relax (f)}}{a+b\,x}\right )\,\ln \relax (f)}{b\,\sqrt {c\,\ln \relax (f)}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{\frac {c}{\left (a + b x\right )^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(c/(a + b*x)**2), x)

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