3.1.31 \(\int \frac {(c+d x)^2}{\sqrt [3]{a+b x^3}} \, dx\) [31]

Optimal. Leaf size=147 \[ \frac {d^2 \left (a+b x^3\right )^{2/3}}{2 b}+\frac {c^2 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}+\frac {c d x^2 \sqrt [3]{1+\frac {b x^3}{a}} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-\frac {b x^3}{a}\right )}{\sqrt [3]{a+b x^3}}-\frac {c^2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{b}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1900, 267, 1907, 245, 372, 371} \begin {gather*} \frac {c^2 \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {c^2 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}+\frac {c 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}}+\frac {d^2 \left (a+b x^3\right )^{2/3}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2/(a + b*x^3)^(1/3),x]

[Out]

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

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 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 1900

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[Coeff[Pq, x, n - 1], Int[x^(n - 1)*(a + b*x^n)^p, x
], x] + Int[ExpandToSum[Pq - Coeff[Pq, x, n - 1]*x^(n - 1), x]*(a + b*x^n)^p, x] /; FreeQ[{a, b, p}, x] && Pol
yQ[Pq, x] && IGtQ[n, 0] && Expon[Pq, x] == n - 1

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

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Mathematica [A]
time = 10.13, size = 201, normalized size = 1.37 \begin {gather*} \frac {d^2 \left (a+b x^3\right )^{2/3}}{2 b}+\frac {c^2 \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 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 )}{\sqrt [3]{a+b x^3}}-\frac {c^2 \log \left (1-\frac {\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3 \sqrt [3]{b}}+\frac {c^2 \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 \sqrt [3]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2/(a + b*x^3)^(1/3),x]

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (d x +c \right )^{2}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2/(b*x^3+a)^(1/3),x)

[Out]

int((d*x+c)^2/(b*x^3+a)^(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((d*x+c)^2/(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^2 + in
tegrate((d^2*x^2 + 2*c*d*x)/(b*x^3 + a)^(1/3), 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)^2/(b*x^3+a)^(1/3),x, algorithm="fricas")

[Out]

integral((d^2*x^2 + 2*c*d*x + c^2)/(b*x^3 + a)^(1/3), x)

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Sympy [A]
time = 1.58, size = 110, normalized size = 0.75 \begin {gather*} d^{2} \left (\begin {cases} \frac {x^{3}}{3 \sqrt [3]{a}} & \text {for}\: b = 0 \\\frac {\left (a + b x^{3}\right )^{\frac {2}{3}}}{2 b} & \text {otherwise} \end {cases}\right ) + \frac {c^{2} 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 {2 c 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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2/(b*x**3+a)**(1/3),x)

[Out]

d**2*Piecewise((x**3/(3*a**(1/3)), Eq(b, 0)), ((a + b*x**3)**(2/3)/(2*b), True)) + c**2*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)) + 2*c*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))

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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)^2/(b*x^3+a)^(1/3),x, algorithm="giac")

[Out]

integrate((d*x + c)^2/(b*x^3 + a)^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^2/(a + b*x^3)^(1/3),x)

[Out]

int((c + d*x)^2/(a + b*x^3)^(1/3), x)

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