3.32 \(\int \frac {c+d x}{\sqrt [3]{a+b x^3}} \, dx\)

Optimal. Leaf size=124 \[ -\frac {c \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}+\frac {c \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {d x^2 \sqrt [3]{\frac {b x^3}{a}+1} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-\frac {b x^3}{a}\right )}{2 \sqrt [3]{a+b x^3}} \]

[Out]

1/2*d*x^2*(1+b*x^3/a)^(1/3)*hypergeom([1/3, 2/3],[5/3],-b*x^3/a)/(b*x^3+a)^(1/3)-1/2*c*ln(-b^(1/3)*x+(b*x^3+a)
^(1/3))/b^(1/3)+1/3*c*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/b^(1/3)*3^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 124, 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 = {1893, 239, 365, 364} \[ -\frac {c \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}+\frac {c \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {d x^2 \sqrt [3]{\frac {b x^3}{a}+1} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-\frac {b x^3}{a}\right )}{2 \sqrt [3]{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + b*x^3)^(1/3),x]

[Out]

(c*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(1/3)) + (d*x^2*(1 + (b*x^3)/a)^(1/3)*Hyp
ergeometric2F1[1/3, 2/3, 5/3, -((b*x^3)/a)])/(2*(a + b*x^3)^(1/3)) - (c*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])
/(2*b^(1/3))

Rule 239

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 364

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

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

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])

Rubi steps

\begin {align*} \int \frac {c+d x}{\sqrt [3]{a+b x^3}} \, dx &=\int \left (\frac {c}{\sqrt [3]{a+b x^3}}+\frac {d x}{\sqrt [3]{a+b x^3}}\right ) \, dx\\ &=c \int \frac {1}{\sqrt [3]{a+b x^3}} \, dx+d \int \frac {x}{\sqrt [3]{a+b x^3}} \, dx\\ &=\frac {c \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {c \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b}}+\frac {\left (d \sqrt [3]{1+\frac {b x^3}{a}}\right ) \int \frac {x}{\sqrt [3]{1+\frac {b x^3}{a}}} \, dx}{\sqrt [3]{a+b x^3}}\\ &=\frac {c \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {d x^2 \sqrt [3]{1+\frac {b x^3}{a}} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-\frac {b x^3}{a}\right )}{2 \sqrt [3]{a+b x^3}}-\frac {c \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 163, normalized size = 1.31 \[ \frac {1}{6} \left (\frac {c \left (\log \left (\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )-2 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )\right )}{\sqrt [3]{b}}+\frac {3 d x^2 \sqrt [3]{\frac {b x^3}{a}+1} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-\frac {b x^3}{a}\right )}{\sqrt [3]{a+b x^3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*d*x^2*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, -((b*x^3)/a)])/(a + b*x^3)^(1/3) + (c*(2*Sqrt
[3]*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]] - 2*Log[1 - (b^(1/3)*x)/(a + b*x^3)^(1/3)] + Log[1 +
(b^(2/3)*x^2)/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)]))/b^(1/3))/6

________________________________________________________________________________________

fricas [F]  time = 35.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d x + c}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x + c)/(b*x^3 + a)^(1/3), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x + c}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {d x +c}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{6} \, {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {\log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {1}{3}}} + \frac {2 \, \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {1}{3}}}\right )} c + d \int \frac {x}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*sqrt(3)*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(1/3) - log(b^(2/3) + (b*x^3 +
a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(1/3) + 2*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(1/3))*c + d*in
tegrate(x/(b*x^3 + a)^(1/3), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/(a + b*x^3)^(1/3),x)

[Out]

int((c + d*x)/(a + b*x^3)^(1/3), x)

________________________________________________________________________________________

sympy [C]  time = 2.27, size = 78, normalized size = 0.63 \[ \frac {c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {4}{3}\right )} + \frac {d x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {5}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________