3.27.100 \(\int \frac {x^m}{\sqrt [3]{a+b x^{3 (1+m)}}} \, dx\) [2700]

Optimal. Leaf size=97 \[ \frac {\tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x^{1+m}}{\sqrt [3]{a+b x^{3 (1+m)}}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b} (1+m)}-\frac {\log \left (\sqrt [3]{b} x^{1+m}-\sqrt [3]{a+b x^{3 (1+m)}}\right )}{2 \sqrt [3]{b} (1+m)} \]

[Out]

-1/2*ln(b^(1/3)*x^(1+m)-(a+b*x^(3+3*m))^(1/3))/b^(1/3)/(1+m)+1/3*arctan(1/3*(1+2*b^(1/3)*x^(1+m)/(a+b*x^(3+3*m
))^(1/3))*3^(1/2))/b^(1/3)/(1+m)*3^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {352, 245} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b} x^{m+1}}{\sqrt [3]{a+b x^{3 (m+1)}}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b} (m+1)}-\frac {\log \left (\sqrt [3]{b} x^{m+1}-\sqrt [3]{a+b x^{3 (m+1)}}\right )}{2 \sqrt [3]{b} (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^m/(a + b*x^(3*(1 + m)))^(1/3),x]

[Out]

ArcTan[(1 + (2*b^(1/3)*x^(1 + m))/(a + b*x^(3*(1 + m)))^(1/3))/Sqrt[3]]/(Sqrt[3]*b^(1/3)*(1 + m)) - Log[b^(1/3
)*x^(1 + m) - (a + b*x^(3*(1 + m)))^(1/3)]/(2*b^(1/3)*(1 + m))

Rule 245

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 352

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/(m + 1), Subst[Int[(a + b*x^Simplify[n/(m +
1)])^p, x], x, x^(m + 1)], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[n/(m + 1)]] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {x^m}{\sqrt [3]{a+b x^{3 (1+m)}}} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{\sqrt [3]{a+b x^3}} \, dx,x,x^{1+m}\right )}{1+m}\\ &=\frac {\tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x^{1+m}}{\sqrt [3]{a+b x^{3 (1+m)}}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b} (1+m)}-\frac {\log \left (\sqrt [3]{b} x^{1+m}-\sqrt [3]{a+b x^{3 (1+m)}}\right )}{2 \sqrt [3]{b} (1+m)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.06, size = 67, normalized size = 0.69 \begin {gather*} \frac {x^{1+m} \sqrt [3]{1+\frac {b x^{3+3 m}}{a}} \, _2F_1\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};-\frac {b x^{3+3 m}}{a}\right )}{(1+m) \sqrt [3]{a+b x^{3+3 m}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^m/(a + b*x^(3*(1 + m)))^(1/3),x]

[Out]

(x^(1 + m)*(1 + (b*x^(3 + 3*m))/a)^(1/3)*Hypergeometric2F1[1/3, 1/3, 4/3, -((b*x^(3 + 3*m))/a)])/((1 + m)*(a +
b*x^(3 + 3*m))^(1/3))

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {x^{m}}{\left (a +b \,x^{3+3 m}\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(a+b*x^(3+3*m))^(1/3),x)

[Out]

int(x^m/(a+b*x^(3+3*m))^(1/3),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(a+b*x^(3+3*m))^(1/3),x, algorithm="maxima")

[Out]

integrate(x^m/(b*x^(3*m + 3) + a)^(1/3), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(a+b*x^(3+3*m))^(1/3),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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Sympy [C] Result contains complex when optimal does not.
time = 1.38, size = 117, normalized size = 1.21 \begin {gather*} \frac {x x^{m} \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {m}{3 m + 3} + 1 + \frac {1}{3 m + 3} \end {matrix}\middle | {\frac {b x^{3} x^{3 m} e^{i \pi }}{a}} \right )}}{3 a^{\frac {m}{3 m + 3}} a^{\frac {1}{3 m + 3}} m \Gamma \left (\frac {m}{3 m + 3} + 1 + \frac {1}{3 m + 3}\right ) + 3 a^{\frac {m}{3 m + 3}} a^{\frac {1}{3 m + 3}} \Gamma \left (\frac {m}{3 m + 3} + 1 + \frac {1}{3 m + 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m/(a+b*x**(3+3*m))**(1/3),x)

[Out]

x*x**m*gamma(1/3)*hyper((1/3, 1/3), (m/(3*m + 3) + 1 + 1/(3*m + 3),), b*x**3*x**(3*m)*exp_polar(I*pi)/a)/(3*a*
*(m/(3*m + 3))*a**(1/(3*m + 3))*m*gamma(m/(3*m + 3) + 1 + 1/(3*m + 3)) + 3*a**(m/(3*m + 3))*a**(1/(3*m + 3))*g
amma(m/(3*m + 3) + 1 + 1/(3*m + 3)))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(a+b*x^(3+3*m))^(1/3),x, algorithm="giac")

[Out]

integrate(x^m/(b*x^(3*m + 3) + a)^(1/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^m}{{\left (a+b\,x^{3\,m+3}\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(a + b*x^(3*m + 3))^(1/3),x)

[Out]

int(x^m/(a + b*x^(3*m + 3))^(1/3), x)

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