3.1.39 \(\int \frac {c+d x}{(a+b x^3)^{2/3}} \, dx\) [39]

Optimal. Leaf size=121 \[ -\frac {d \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}+\frac {c x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}-\frac {d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3}} \]

[Out]

c*x*(1+b*x^3/a)^(2/3)*hypergeom([1/3, 2/3],[4/3],-b*x^3/a)/(b*x^3+a)^(2/3)-1/2*d*ln(b^(1/3)*x-(b*x^3+a)^(1/3))
/b^(2/3)-1/3*d*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/b^(2/3)*3^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1907, 252, 251, 337} \begin {gather*} -\frac {d \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}-\frac {d \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3}}+\frac {c x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + b*x^3)^(2/3),x]

[Out]

-((d*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(2/3))) + (c*x*(1 + (b*x^3)/a)^(2/3)*Hy
pergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(a + b*x^3)^(2/3) - (d*Log[b^(1/3)*x - (a + b*x^3)^(1/3)])/(2*b^
(2/3))

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 337

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

Rule 1907

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n])

Rubi steps

\begin {align*} \int \frac {c+d x}{\left (a+b x^3\right )^{2/3}} \, dx &=\int \left (\frac {c}{\left (a+b x^3\right )^{2/3}}+\frac {d x}{\left (a+b x^3\right )^{2/3}}\right ) \, dx\\ &=c \int \frac {1}{\left (a+b x^3\right )^{2/3}} \, dx+d \int \frac {x}{\left (a+b x^3\right )^{2/3}} \, dx\\ &=d \text {Subst}\left (\int \frac {x}{1-b x^3} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )+\frac {\left (c \left (1+\frac {b x^3}{a}\right )^{2/3}\right ) \int \frac {1}{\left (1+\frac {b x^3}{a}\right )^{2/3}} \, dx}{\left (a+b x^3\right )^{2/3}}\\ &=\frac {c x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}+\frac {d \text {Subst}\left (\int \frac {1}{1-\sqrt [3]{b} x} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{3 \sqrt [3]{b}}-\frac {d \text {Subst}\left (\int \frac {1-\sqrt [3]{b} x}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{3 \sqrt [3]{b}}\\ &=\frac {c x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}-\frac {d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}+\frac {d \text {Subst}\left (\int \frac {\sqrt [3]{b}+2 b^{2/3} x}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{6 b^{2/3}}-\frac {d \text {Subst}\left (\int \frac {1}{1+\sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{2 \sqrt [3]{b}}\\ &=\frac {c x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}-\frac {d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}+\frac {d \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{6 b^{2/3}}+\frac {d \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{b^{2/3}}\\ &=-\frac {d \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}+\frac {c x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}}-\frac {d \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}+\frac {d \log \left (1+\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{6 b^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 10.04, size = 78, normalized size = 0.64 \begin {gather*} \frac {x \left (2 c \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )+d x \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {b x^3}{a+b x^3}\right )\right )}{2 \left (a+b x^3\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)/(a + b*x^3)^(2/3),x]

[Out]

(x*(2*c*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)] + d*x*Hypergeometric2F1[2/3, 1, 5
/3, (b*x^3)/(a + b*x^3)]))/(2*(a + b*x^3)^(2/3))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {d x +c}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)/(b*x^3+a)^(2/3),x)

[Out]

int((d*x+c)/(b*x^3+a)^(2/3),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((d*x+c)/(b*x^3+a)^(2/3),x, algorithm="maxima")

[Out]

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

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Fricas [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((d*x+c)/(b*x^3+a)^(2/3),x, algorithm="fricas")

[Out]

integral((d*x + c)/(b*x^3 + a)^(2/3), x)

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Sympy [C] Result contains complex when optimal does not.
time = 1.13, size = 78, normalized size = 0.64 \begin {gather*} \frac {c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {4}{3}\right )} + \frac {d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {5}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(b*x**3+a)**(2/3),x)

[Out]

c*x*gamma(1/3)*hyper((1/3, 2/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(4/3)) + d*x**2*gamma(2/3)
*hyper((2/3, 2/3), (5/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(5/3))

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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((d*x+c)/(b*x^3+a)^(2/3),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {c+d\,x}{{\left (b\,x^3+a\right )}^{2/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/(a + b*x^3)^(2/3),x)

[Out]

int((c + d*x)/(a + b*x^3)^(2/3), x)

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