3.74 \(\int \frac{(c+d x^3)^2}{(a+b x^3)^{7/3}} \, dx\)

Optimal. Leaf size=152 \[ \frac{x (b c-a d) (4 a d+3 b c)}{4 a^2 b^2 \sqrt [3]{a+b x^3}}-\frac{d^2 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 b^{7/3}}+\frac{d^2 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} b^{7/3}}+\frac{x \left (c+d x^3\right ) (b c-a d)}{4 a b \left (a+b x^3\right )^{4/3}} \]

[Out]

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

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Rubi [A]  time = 0.0676022, antiderivative size = 152, 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, 385, 239} \[ \frac{x (b c-a d) (4 a d+3 b c)}{4 a^2 b^2 \sqrt [3]{a+b x^3}}-\frac{d^2 \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 b^{7/3}}+\frac{d^2 \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} b^{7/3}}+\frac{x \left (c+d x^3\right ) (b c-a d)}{4 a b \left (a+b x^3\right )^{4/3}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^3)^2/(a + b*x^3)^(7/3),x]

[Out]

((b*c - a*d)*(3*b*c + 4*a*d)*x)/(4*a^2*b^2*(a + b*x^3)^(1/3)) + ((b*c - a*d)*x*(c + d*x^3))/(4*a*b*(a + b*x^3)
^(4/3)) + (d^2*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(7/3)) - (d^2*Log[-(b^(1/3)*x
) + (a + b*x^3)^(1/3)])/(2*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 385

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

Mathematica [A]  time = 5.2114, size = 180, normalized size = 1.18 \[ \frac{x \left (\left (a+b x^3\right ) \left (-5 a^2 d^2+2 a b c d+3 b^2 c^2\right )+a (b c-a d)^2\right )}{4 a^2 b^2 \left (a+b x^3\right )^{4/3}}+\frac{d^2 \left (\log \left (\frac{b^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1\right )-2 \log \left (1-\frac{\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )\right )}{6 b^{7/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^3)^2/(a + b*x^3)^(7/3),x]

[Out]

(x*(a*(b*c - a*d)^2 + (3*b^2*c^2 + 2*a*b*c*d - 5*a^2*d^2)*(a + b*x^3)))/(4*a^2*b^2*(a + b*x^3)^(4/3)) + (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)] + L
og[1 + (b^(2/3)*x^2)/(a + b*x^3)^(2/3) + (b^(1/3)*x)/(a + b*x^3)^(1/3)]))/(6*b^(7/3))

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Maple [F]  time = 0.376, size = 0, normalized size = 0. \begin{align*} \int{ \left ( d{x}^{3}+c \right ) ^{2} \left ( b{x}^{3}+a \right ) ^{-{\frac{7}{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)^(7/3),x)

[Out]

int((d*x^3+c)^2/(b*x^3+a)^(7/3),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)^(7/3),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.76132, size = 1617, normalized size = 10.64 \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)^(7/3),x, algorithm="fricas")

[Out]

[1/12*(6*sqrt(1/3)*(a^2*b^3*d^2*x^6 + 2*a^3*b^2*d^2*x^3 + a^4*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) - 4*(a^2*b^2*d^2*x^6 + 2*a^3*b*d^2*x^3 + a^4*d^2)*(-b)^(2/3)*log(((-b)^(1
/3)*x + (b*x^3 + a)^(1/3))/x) + 2*(a^2*b^2*d^2*x^6 + 2*a^3*b*d^2*x^3 + a^4*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^4*c^2 + 2*a*b^3*c*d - 5*a^2*b^2*d^2)*x^4
+ 4*(a*b^3*c^2 - a^3*b*d^2)*x)*(b*x^3 + a)^(2/3))/(a^2*b^5*x^6 + 2*a^3*b^4*x^3 + a^4*b^3), -1/12*(12*sqrt(1/3
)*(a^2*b^3*d^2*x^6 + 2*a^3*b^2*d^2*x^3 + a^4*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) + 4*(a^2*b^2*d^2*x^6 + 2*a^3*b*d^2*x^3 + a^4*d^2)*(-b)^(2/3)*log(((-b)
^(1/3)*x + (b*x^3 + a)^(1/3))/x) - 2*(a^2*b^2*d^2*x^6 + 2*a^3*b*d^2*x^3 + a^4*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^4*c^2 + 2*a*b^3*c*d - 5*a^2*b^2*d^2)*
x^4 + 4*(a*b^3*c^2 - a^3*b*d^2)*x)*(b*x^3 + a)^(2/3))/(a^2*b^5*x^6 + 2*a^3*b^4*x^3 + a^4*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{7}{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)**(7/3),x)

[Out]

Integral((c + d*x**3)**2/(a + b*x**3)**(7/3), 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{7}{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)^(7/3),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)^2/(b*x^3 + a)^(7/3), x)