3.111 \(\int f^{a+b x^3} \, dx\)

Optimal. Leaf size=32 \[ -\frac {x f^a \Gamma \left (\frac {1}{3},-b x^3 \log (f)\right )}{3 \sqrt [3]{-b x^3 \log (f)}} \]

[Out]

-1/3*f^a*x*GAMMA(1/3,-b*x^3*ln(f))/(-b*x^3*ln(f))^(1/3)

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Rubi [A]  time = 0.00, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2208} \[ -\frac {x f^a \text {Gamma}\left (\frac {1}{3},-b x^3 \log (f)\right )}{3 \sqrt [3]{-b x^3 \log (f)}} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^3),x]

[Out]

-(f^a*x*Gamma[1/3, -(b*x^3*Log[f])])/(3*(-(b*x^3*Log[f]))^(1/3))

Rule 2208

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

Rubi steps

\begin {align*} \int f^{a+b x^3} \, dx &=-\frac {f^a x \Gamma \left (\frac {1}{3},-b x^3 \log (f)\right )}{3 \sqrt [3]{-b x^3 \log (f)}}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 32, normalized size = 1.00 \[ -\frac {x f^a \Gamma \left (\frac {1}{3},-b x^3 \log (f)\right )}{3 \sqrt [3]{-b x^3 \log (f)}} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^3),x]

[Out]

-1/3*(f^a*x*Gamma[1/3, -(b*x^3*Log[f])])/(-(b*x^3*Log[f]))^(1/3)

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fricas [A]  time = 0.42, size = 29, normalized size = 0.91 \[ \frac {\left (-b \log \relax (f)\right )^{\frac {2}{3}} f^{a} \Gamma \left (\frac {1}{3}, -b x^{3} \log \relax (f)\right )}{3 \, b \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-b*log(f))^(2/3)*f^a*gamma(1/3, -b*x^3*log(f))/(b*log(f))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{b x^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(b*x^3 + a), x)

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maple [B]  time = 0.03, size = 78, normalized size = 2.44 \[ \frac {\left (-\frac {\left (-b \right )^{\frac {1}{3}} x \Gamma \left (\frac {1}{3}, -b \,x^{3} \ln \relax (f )\right ) \ln \relax (f )^{\frac {1}{3}}}{\left (-b \,x^{3} \ln \relax (f )\right )^{\frac {1}{3}}}+\frac {2 \left (-b \right )^{\frac {1}{3}} \pi \sqrt {3}\, x \ln \relax (f )^{\frac {1}{3}}}{3 \Gamma \left (\frac {2}{3}\right ) \left (-b \,x^{3} \ln \relax (f )\right )^{\frac {1}{3}}}\right ) f^{a}}{3 \left (-b \right )^{\frac {1}{3}} \ln \relax (f )^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^3+a),x)

[Out]

1/3*f^a/(-b)^(1/3)/ln(f)^(1/3)*(2/3*x*(-b)^(1/3)*ln(f)^(1/3)*Pi*3^(1/2)/GAMMA(2/3)/(-b*x^3*ln(f))^(1/3)-x*(-b)
^(1/3)*ln(f)^(1/3)/(-b*x^3*ln(f))^(1/3)*GAMMA(1/3,-b*x^3*ln(f)))

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maxima [A]  time = 1.25, size = 26, normalized size = 0.81 \[ -\frac {f^{a} x \Gamma \left (\frac {1}{3}, -b x^{3} \log \relax (f)\right )}{3 \, \left (-b x^{3} \log \relax (f)\right )^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*f^a*x*gamma(1/3, -b*x^3*log(f))/(-b*x^3*log(f))^(1/3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int f^{b\,x^3+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^3),x)

[Out]

int(f^(a + b*x^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + b x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(a + b*x**3), x)

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