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

Optimal. Leaf size=175 \[ -\frac{\left (2 a^2 d^2-6 a b c d+9 b^2 c^2\right ) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{18 b^{7/3}}+\frac{\left (2 a^2 d^2-6 a b c d+9 b^2 c^2\right ) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{9 \sqrt{3} b^{7/3}}+\frac{d x \left (a+b x^3\right )^{2/3} (9 b c-4 a d)}{18 b^2}+\frac{d x \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )}{6 b} \]

[Out]

(d*(9*b*c - 4*a*d)*x*(a + b*x^3)^(2/3))/(18*b^2) + (d*x*(a + b*x^3)^(2/3)*(c + d*x^3))/(6*b) + ((9*b^2*c^2 - 6
*a*b*c*d + 2*a^2*d^2)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(9*Sqrt[3]*b^(7/3)) - ((9*b^2*c^2
- 6*a*b*c*d + 2*a^2*d^2)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(18*b^(7/3))

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Rubi [A]  time = 0.096277, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {416, 388, 239} \[ -\frac{\left (2 a^2 d^2-6 a b c d+9 b^2 c^2\right ) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{18 b^{7/3}}+\frac{\left (2 a^2 d^2-6 a b c d+9 b^2 c^2\right ) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{9 \sqrt{3} b^{7/3}}+\frac{d x \left (a+b x^3\right )^{2/3} (9 b c-4 a d)}{18 b^2}+\frac{d x \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )}{6 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(9*b*c - 4*a*d)*x*(a + b*x^3)^(2/3))/(18*b^2) + (d*x*(a + b*x^3)^(2/3)*(c + d*x^3))/(6*b) + ((9*b^2*c^2 - 6
*a*b*c*d + 2*a^2*d^2)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(9*Sqrt[3]*b^(7/3)) - ((9*b^2*c^2
- 6*a*b*c*d + 2*a^2*d^2)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(18*b^(7/3))

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, 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]

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]

Rubi steps

\begin{align*} \int \frac{\left (c+d x^3\right )^2}{\sqrt [3]{a+b x^3}} \, dx &=\frac{d x \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )}{6 b}+\frac{\int \frac{c (6 b c-a d)+d (9 b c-4 a d) x^3}{\sqrt [3]{a+b x^3}} \, dx}{6 b}\\ &=\frac{d (9 b c-4 a d) x \left (a+b x^3\right )^{2/3}}{18 b^2}+\frac{d x \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )}{6 b}+\frac{\left (9 b^2 c^2-6 a b c d+2 a^2 d^2\right ) \int \frac{1}{\sqrt [3]{a+b x^3}} \, dx}{9 b^2}\\ &=\frac{d (9 b c-4 a d) x \left (a+b x^3\right )^{2/3}}{18 b^2}+\frac{d x \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )}{6 b}+\frac{\left (9 b^2 c^2-6 a b c d+2 a^2 d^2\right ) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt{3}}\right )}{9 \sqrt{3} b^{7/3}}-\frac{\left (9 b^2 c^2-6 a b c d+2 a^2 d^2\right ) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{18 b^{7/3}}\\ \end{align*}

Mathematica [A]  time = 5.13845, size = 172, normalized size = 0.98 \[ \frac{\left (2 a^2 d^2-6 a b c d+9 b^2 c^2\right ) \left (\log \left (\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )-2 \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )\right )+3 \sqrt [3]{b} d x \left (a+b x^3\right )^{2/3} \left (3 b \left (4 c+d x^3\right )-4 a d\right )}{54 b^{7/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^(1/3)*d*x*(a + b*x^3)^(2/3)*(-4*a*d + 3*b*(4*c + d*x^3)) + (9*b^2*c^2 - 6*a*b*c*d + 2*a^2*d^2)*(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)]))/(54*b^(7/3))

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Maple [F]  time = 0.215, size = 0, normalized size = 0. \begin{align*} \int{ \left ( d{x}^{3}+c \right ) ^{2}{\frac{1}{\sqrt [3]{b{x}^{3}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.7547, size = 1341, normalized size = 7.66 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/54*(3*sqrt(1/3)*(9*b^3*c^2 - 6*a*b^2*c*d + 2*a^2*b*d^2)*sqrt((-b)^(1/3)/b)*log(3*b*x^3 - 3*(b*x^3 + a)^(1/3
)*(-b)^(2/3)*x^2 - 3*sqrt(1/3)*((-b)^(1/3)*b*x^3 - (b*x^3 + a)^(1/3)*b*x^2 + 2*(b*x^3 + a)^(2/3)*(-b)^(2/3)*x)
*sqrt((-b)^(1/3)/b) + 2*a) - 2*(9*b^2*c^2 - 6*a*b*c*d + 2*a^2*d^2)*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^
(1/3))/x) + (9*b^2*c^2 - 6*a*b*c*d + 2*a^2*d^2)*(-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) + 3*(3*b^2*d^2*x^4 + 4*(3*b^2*c*d - a*b*d^2)*x)*(b*x^3 + a)^(2/3))/b^3, -1/54*(6*s
qrt(1/3)*(9*b^3*c^2 - 6*a*b^2*c*d + 2*a^2*b*d^2)*sqrt(-(-b)^(1/3)/b)*arctan(-sqrt(1/3)*((-b)^(1/3)*x - 2*(b*x^
3 + a)^(1/3))*sqrt(-(-b)^(1/3)/b)/x) + 2*(9*b^2*c^2 - 6*a*b*c*d + 2*a^2*d^2)*(-b)^(2/3)*log(((-b)^(1/3)*x + (b
*x^3 + a)^(1/3))/x) - (9*b^2*c^2 - 6*a*b*c*d + 2*a^2*d^2)*(-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) - 3*(3*b^2*d^2*x^4 + 4*(3*b^2*c*d - a*b*d^2)*x)*(b*x^3 + a)^(2/3))/b^3]

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Sympy [C]  time = 3.90301, size = 126, normalized size = 0.72 \begin{align*} \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^{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 )} + \frac{d^{2} x^{7} \Gamma \left (\frac{7}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{7}{3} \\ \frac{10}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac{10}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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**4*gam
ma(4/3)*hyper((1/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(1/3)*gamma(7/3)) + d**2*x**7*gamma(7/3)*hyp
er((1/3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(1/3)*gamma(10/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{3} + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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