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

Optimal. Leaf size=41 \[ \frac {(a+b x) f^{\frac {c}{a+b x}}}{b}-\frac {c \log (f) \text {Ei}\left (\frac {c \log (f)}{a+b x}\right )}{b} \]

[Out]

f^(c/(b*x+a))*(b*x+a)/b-c*Ei(c*ln(f)/(b*x+a))*ln(f)/b

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Rubi [A]  time = 0.03, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2206, 2210} \[ \frac {(a+b x) f^{\frac {c}{a+b x}}}{b}-\frac {c \log (f) \text {Ei}\left (\frac {c \log (f)}{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[f^(c/(a + b*x)),x]

[Out]

(f^(c/(a + b*x))*(a + b*x))/b - (c*ExpIntegralEi[(c*Log[f])/(a + b*x)]*Log[f])/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 2210

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 41, normalized size = 1.00 \[ \frac {(a+b x) f^{\frac {c}{a+b x}}}{b}-\frac {c \log (f) \text {Ei}\left (\frac {c \log (f)}{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(c/(a + b*x)),x]

[Out]

(f^(c/(a + b*x))*(a + b*x))/b - (c*ExpIntegralEi[(c*Log[f])/(a + b*x)]*Log[f])/b

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fricas [A]  time = 0.43, size = 40, normalized size = 0.98 \[ -\frac {c {\rm Ei}\left (\frac {c \log \relax (f)}{b x + a}\right ) \log \relax (f) - {\left (b x + a\right )} f^{\frac {c}{b x + a}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c*Ei(c*log(f)/(b*x + a))*log(f) - (b*x + a)*f^(c/(b*x + a)))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(c/(b*x + a)), x)

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maple [A]  time = 0.10, size = 52, normalized size = 1.27 \[ \frac {c \Ei \left (1, -\frac {c \ln \relax (f )}{b x +a}\right ) \ln \relax (f )}{b}+x \,f^{\frac {c}{b x +a}}+\frac {a \,f^{\frac {c}{b x +a}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

f^(1/(b*x+a)*c)*x+1/b*f^(1/(b*x+a)*c)*a+c/b*ln(f)*Ei(1,-1/(b*x+a)*c*ln(f))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.55, size = 50, normalized size = 1.22 \[ f^{\frac {c}{a+b\,x}}\,x+\frac {a\,f^{\frac {c}{a+b\,x}}}{b}-\frac {c\,\mathrm {ei}\left (\frac {c\,\ln \relax (f)}{a+b\,x}\right )\,\ln \relax (f)}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(c/(a + b*x)),x)

[Out]

f^(c/(a + b*x))*x + (a*f^(c/(a + b*x)))/b - (c*ei((c*log(f))/(a + b*x))*log(f))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(c/(a + b*x)), x)

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