3.73 \(\int \frac{(c+d x^3)^2}{(a+b x^3)^{4/3}} \, dx\)

Optimal. Leaf size=159 \[ -\frac{d x \left (a+b x^3\right )^{2/3} (3 b c-4 a d)}{3 a b^2}-\frac{d (3 b c-2 a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{3 b^{7/3}}+\frac{2 d (3 b c-2 a d) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{7/3}}+\frac{x \left (c+d x^3\right ) (b c-a d)}{a b \sqrt [3]{a+b x^3}} \]

[Out]

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

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Rubi [A]  time = 0.102334, antiderivative size = 159, 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 = {413, 388, 239} \[ -\frac{d x \left (a+b x^3\right )^{2/3} (3 b c-4 a d)}{3 a b^2}-\frac{d (3 b c-2 a d) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{3 b^{7/3}}+\frac{2 d (3 b c-2 a d) \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{7/3}}+\frac{x \left (c+d x^3\right ) (b c-a d)}{a b \sqrt [3]{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)^2/(a + b*x^3)^(4/3),x]

[Out]

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

Rule 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[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}{\left (a+b x^3\right )^{4/3}} \, dx &=\frac{(b c-a d) x \left (c+d x^3\right )}{a b \sqrt [3]{a+b x^3}}+\frac{\int \frac{a c d-d (3 b c-4 a d) x^3}{\sqrt [3]{a+b x^3}} \, dx}{a b}\\ &=-\frac{d (3 b c-4 a d) x \left (a+b x^3\right )^{2/3}}{3 a b^2}+\frac{(b c-a d) x \left (c+d x^3\right )}{a b \sqrt [3]{a+b x^3}}+\frac{(2 d (3 b c-2 a d)) \int \frac{1}{\sqrt [3]{a+b x^3}} \, dx}{3 b^2}\\ &=-\frac{d (3 b c-4 a d) x \left (a+b x^3\right )^{2/3}}{3 a b^2}+\frac{(b c-a d) x \left (c+d x^3\right )}{a b \sqrt [3]{a+b x^3}}+\frac{2 d (3 b c-2 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^{7/3}}-\frac{d (3 b c-2 a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{3 b^{7/3}}\\ \end{align*}

Mathematica [A]  time = 5.14927, size = 168, normalized size = 1.06 \[ \frac{x \left (a+b x^3\right )^{2/3} \left (\frac{3 (b c-a d)^2}{a \left (a+b x^3\right )}+d^2\right )}{3 b^2}+\frac{d (3 b c-2 a d) \left (\log \left (\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )-2 \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )\right )}{9 b^{7/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^3)^2/(a + b*x^3)^(4/3),x]

[Out]

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

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Maple [F]  time = 0.416, size = 0, normalized size = 0. \begin{align*} \int{ \left ( d{x}^{3}+c \right ) ^{2} \left ( b{x}^{3}+a \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^3+c)^2/(b*x^3+a)^(4/3),x)

[Out]

int((d*x^3+c)^2/(b*x^3+a)^(4/3),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)^(4/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.73604, size = 1538, normalized size = 9.67 \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)^(4/3),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d x^{3}\right )^{2}}{\left (a + b x^{3}\right )^{\frac{4}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**3+c)**2/(b*x**3+a)**(4/3),x)

[Out]

Integral((c + d*x**3)**2/(a + b*x**3)**(4/3), x)

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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{4}{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)^(4/3),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)^2/(b*x^3 + a)^(4/3), x)