3.39 \(\int \frac{c+d x}{(a+b x^3)^{2/3}} \, dx\)

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

[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))

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Rubi [A]  time = 0.114944, antiderivative size = 172, normalized size of antiderivative = 1.42, number of steps used = 11, number of rules used = 10, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.588, Rules used = {1893, 246, 245, 331, 292, 31, 634, 617, 204, 628} \[ -\frac{d \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 b^{2/3}}+\frac{d \log \left (\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )}{6 b^{2/3}}-\frac{d \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} b^{2/3}}+\frac{c x \left (\frac{b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{\left (a+b x^3\right )^{2/3}} \]

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[1 - (b^(1/3)*x)/(a + b*x^3)^(1/3)])/(
3*b^(2/3)) + (d*Log[1 + (b^(2/3)*x^2)/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)])/(6*b^(2/3))

Rule 1893

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])

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[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 245

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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 292

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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*}

Mathematica [A]  time = 0.031943, size = 78, normalized size = 0.64 \[ \frac{x \left (2 c \left (\frac{b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{4}{3};-\frac{b x^3}{a}\right )+d x \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{b x^3}{b x^3+a}\right )\right )}{2 \left (a+b x^3\right )^{2/3}} \]

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. \begin{align*} \int{(dx+c) \left ( b{x}^{3}+a \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{d x + c}{{\left (b x^{3} + a\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{d x + c}{{\left (b x^{3} + a\right )}^{\frac{2}{3}}}, x\right ) \end{align*}

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]  time = 1.89121, size = 78, normalized size = 0.64 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{d x + c}{{\left (b x^{3} + a\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

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)