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

Optimal. Leaf size=111 \[ -\frac {(3 b c-a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{6 b^{4/3}}+\frac {(3 b c-a d) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{4/3}}+\frac {d x \left (a+b x^3\right )^{2/3}}{3 b} \]

[Out]

1/3*d*x*(b*x^3+a)^(2/3)/b-1/6*(-a*d+3*b*c)*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(4/3)+1/9*(-a*d+3*b*c)*arctan(1/3*
(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/b^(4/3)*3^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 111, 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 = {388, 239} \[ -\frac {(3 b c-a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{6 b^{4/3}}+\frac {(3 b c-a d) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} b^{4/3}}+\frac {d x \left (a+b x^3\right )^{2/3}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)/(a + b*x^3)^(1/3),x]

[Out]

(d*x*(a + b*x^3)^(2/3))/(3*b) + ((3*b*c - a*d)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[
3]*b^(4/3)) - ((3*b*c - a*d)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(6*b^(4/3))

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 141, normalized size = 1.27 \[ \frac {\frac {(3 b c-a d) \left (\log \left (\frac {b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )-2 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )\right )}{6 \sqrt [3]{b}}+d x \left (a+b x^3\right )^{2/3}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x*(a + b*x^3)^(2/3) + ((3*b*c - a*d)*(2*Sqrt[3]*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]] - 2*L
og[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
)]))/(6*b^(1/3)))/(3*b)

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fricas [A]  time = 0.65, size = 362, normalized size = 3.26 \[ \left [\frac {6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b d x - 3 \, \sqrt {\frac {1}{3}} {\left (3 \, b^{2} c - a b d\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \log \left (3 \, b x^{3} - 3 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {2}{3}} x^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (b^{\frac {4}{3}} x^{3} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x^{2} - 2 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{\frac {2}{3}} x\right )} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} + 2 \, a\right ) - 2 \, {\left (3 \, b c - a d\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + {\left (3 \, b c - a d\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{18 \, b^{2}}, \frac {6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b d x - 2 \, {\left (3 \, b c - a d\right )} b^{\frac {2}{3}} \log \left (-\frac {b^{\frac {1}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + {\left (3 \, b c - a d\right )} b^{\frac {2}{3}} \log \left (\frac {b^{\frac {2}{3}} x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) - \frac {6 \, \sqrt {\frac {1}{3}} {\left (3 \, b^{2} c - a b d\right )} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (b^{\frac {1}{3}} x + 2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}\right )}}{b^{\frac {1}{3}} x}\right )}{b^{\frac {1}{3}}}}{18 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/18*(6*(b*x^3 + a)^(2/3)*b*d*x - 3*sqrt(1/3)*(3*b^2*c - a*b*d)*sqrt(-1/b^(2/3))*log(3*b*x^3 - 3*(b*x^3 + a)^
(1/3)*b^(2/3)*x^2 - 3*sqrt(1/3)*(b^(4/3)*x^3 + (b*x^3 + a)^(1/3)*b*x^2 - 2*(b*x^3 + a)^(2/3)*b^(2/3)*x)*sqrt(-
1/b^(2/3)) + 2*a) - 2*(3*b*c - a*d)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) + (3*b*c - a*d)*b^(2/3)*lo
g((b^(2/3)*x^2 + (b*x^3 + a)^(1/3)*b^(1/3)*x + (b*x^3 + a)^(2/3))/x^2))/b^2, 1/18*(6*(b*x^3 + a)^(2/3)*b*d*x -
2*(3*b*c - a*d)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) + (3*b*c - a*d)*b^(2/3)*log((b^(2/3)*x^2 + (b
*x^3 + a)^(1/3)*b^(1/3)*x + (b*x^3 + a)^(2/3))/x^2) - 6*sqrt(1/3)*(3*b^2*c - a*b*d)*arctan(sqrt(1/3)*(b^(1/3)*
x + 2*(b*x^3 + a)^(1/3))/(b^(1/3)*x))/b^(1/3))/b^2]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.40, size = 0, normalized size = 0.00 \[ \int \frac {d \,x^{3}+c}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.20, size = 244, normalized size = 2.20 \[ -\frac {1}{6} \, {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {\log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {1}{3}}} + \frac {2 \, \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {1}{3}}}\right )} c + \frac {1}{18} \, {\left (\frac {2 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {4}{3}}} - \frac {a \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {4}{3}}} + \frac {2 \, a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {4}{3}}} - \frac {6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} a}{{\left (b^{2} - \frac {{\left (b x^{3} + a\right )} b}{x^{3}}\right )} x^{2}}\right )} d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^3+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 + 1/18
*(2*sqrt(3)*a*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/b^(1/3))/b^(4/3) - a*log(b^(2/3) + (b*x^3 +
a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(4/3) + 2*a*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(4/3) - 6*(b*
x^3 + a)^(2/3)*a/((b^2 - (b*x^3 + a)*b/x^3)*x^2))*d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^3)/(a + b*x^3)^(1/3),x)

[Out]

int((c + d*x^3)/(a + b*x^3)^(1/3), x)

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sympy [C]  time = 4.44, size = 78, normalized size = 0.70 \[ \frac {c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {4}{3}\right )} + \frac {d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {7}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**3+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**4*gamma(4/3)
*hyper((1/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(1/3)*gamma(7/3))

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