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

Optimal. Leaf size=267 \[ \frac{x \left (a+b x^3\right )^{2/3} (6 b c-5 a d)}{18 c^2 \left (c+d x^3\right ) (b c-a d)}+\frac{a (6 b c-5 a d) \log \left (c+d x^3\right )}{54 c^{8/3} (b c-a d)^{4/3}}-\frac{a (6 b c-5 a d) \log \left (\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{18 c^{8/3} (b c-a d)^{4/3}}+\frac{a (6 b c-5 a d) \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{9 \sqrt{3} c^{8/3} (b c-a d)^{4/3}}-\frac{d x \left (a+b x^3\right )^{5/3}}{6 c \left (c+d x^3\right )^2 (b c-a d)} \]

[Out]

-(d*x*(a + b*x^3)^(5/3))/(6*c*(b*c - a*d)*(c + d*x^3)^2) + ((6*b*c - 5*a*d)*x*(a + b*x^3)^(2/3))/(18*c^2*(b*c
- a*d)*(c + d*x^3)) + (a*(6*b*c - 5*a*d)*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(a + b*x^3)^(1/3)))/Sqrt
[3]])/(9*Sqrt[3]*c^(8/3)*(b*c - a*d)^(4/3)) + (a*(6*b*c - 5*a*d)*Log[c + d*x^3])/(54*c^(8/3)*(b*c - a*d)^(4/3)
) - (a*(6*b*c - 5*a*d)*Log[((b*c - a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)])/(18*c^(8/3)*(b*c - a*d)^(4/3))

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Rubi [A]  time = 0.243439, antiderivative size = 326, normalized size of antiderivative = 1.22, number of steps used = 9, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {382, 378, 377, 200, 31, 634, 617, 204, 628} \[ \frac{x \left (a+b x^3\right )^{2/3} (6 b c-5 a d)}{18 c^2 \left (c+d x^3\right ) (b c-a d)}-\frac{a (6 b c-5 a d) \log \left (\sqrt [3]{c}-\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}\right )}{27 c^{8/3} (b c-a d)^{4/3}}+\frac{a (6 b c-5 a d) \log \left (\frac{x^2 (b c-a d)^{2/3}}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+c^{2/3}\right )}{54 c^{8/3} (b c-a d)^{4/3}}+\frac{a (6 b c-5 a d) \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+\sqrt [3]{c}}{\sqrt{3} \sqrt [3]{c}}\right )}{9 \sqrt{3} c^{8/3} (b c-a d)^{4/3}}-\frac{d x \left (a+b x^3\right )^{5/3}}{6 c \left (c+d x^3\right )^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*x*(a + b*x^3)^(5/3))/(6*c*(b*c - a*d)*(c + d*x^3)^2) + ((6*b*c - 5*a*d)*x*(a + b*x^3)^(2/3))/(18*c^2*(b*c
- a*d)*(c + d*x^3)) + (a*(6*b*c - 5*a*d)*ArcTan[(c^(1/3) + (2*(b*c - a*d)^(1/3)*x)/(a + b*x^3)^(1/3))/(Sqrt[3]
*c^(1/3))])/(9*Sqrt[3]*c^(8/3)*(b*c - a*d)^(4/3)) - (a*(6*b*c - 5*a*d)*Log[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(a
+ b*x^3)^(1/3)])/(27*c^(8/3)*(b*c - a*d)^(4/3)) + (a*(6*b*c - 5*a*d)*Log[c^(2/3) + ((b*c - a*d)^(2/3)*x^2)/(a
+ b*x^3)^(2/3) + (c^(1/3)*(b*c - a*d)^(1/3)*x)/(a + b*x^3)^(1/3)])/(54*c^(8/3)*(b*c - a*d)^(4/3))

Rule 382

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

Rule 378

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

Rule 377

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

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^3\right )^{2/3}}{\left (c+d x^3\right )^3} \, dx &=-\frac{d x \left (a+b x^3\right )^{5/3}}{6 c (b c-a d) \left (c+d x^3\right )^2}+\frac{(6 b c-5 a d) \int \frac{\left (a+b x^3\right )^{2/3}}{\left (c+d x^3\right )^2} \, dx}{6 c (b c-a d)}\\ &=-\frac{d x \left (a+b x^3\right )^{5/3}}{6 c (b c-a d) \left (c+d x^3\right )^2}+\frac{(6 b c-5 a d) x \left (a+b x^3\right )^{2/3}}{18 c^2 (b c-a d) \left (c+d x^3\right )}+\frac{(a (6 b c-5 a d)) \int \frac{1}{\sqrt [3]{a+b x^3} \left (c+d x^3\right )} \, dx}{9 c^2 (b c-a d)}\\ &=-\frac{d x \left (a+b x^3\right )^{5/3}}{6 c (b c-a d) \left (c+d x^3\right )^2}+\frac{(6 b c-5 a d) x \left (a+b x^3\right )^{2/3}}{18 c^2 (b c-a d) \left (c+d x^3\right )}+\frac{(a (6 b c-5 a d)) \operatorname{Subst}\left (\int \frac{1}{c-(b c-a d) x^3} \, dx,x,\frac{x}{\sqrt [3]{a+b x^3}}\right )}{9 c^2 (b c-a d)}\\ &=-\frac{d x \left (a+b x^3\right )^{5/3}}{6 c (b c-a d) \left (c+d x^3\right )^2}+\frac{(6 b c-5 a d) x \left (a+b x^3\right )^{2/3}}{18 c^2 (b c-a d) \left (c+d x^3\right )}+\frac{(a (6 b c-5 a d)) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{c}-\sqrt [3]{b c-a d} x} \, dx,x,\frac{x}{\sqrt [3]{a+b x^3}}\right )}{27 c^{8/3} (b c-a d)}+\frac{(a (6 b c-5 a d)) \operatorname{Subst}\left (\int \frac{2 \sqrt [3]{c}+\sqrt [3]{b c-a d} x}{c^{2/3}+\sqrt [3]{c} \sqrt [3]{b c-a d} x+(b c-a d)^{2/3} x^2} \, dx,x,\frac{x}{\sqrt [3]{a+b x^3}}\right )}{27 c^{8/3} (b c-a d)}\\ &=-\frac{d x \left (a+b x^3\right )^{5/3}}{6 c (b c-a d) \left (c+d x^3\right )^2}+\frac{(6 b c-5 a d) x \left (a+b x^3\right )^{2/3}}{18 c^2 (b c-a d) \left (c+d x^3\right )}-\frac{a (6 b c-5 a d) \log \left (\sqrt [3]{c}-\frac{\sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{27 c^{8/3} (b c-a d)^{4/3}}+\frac{(a (6 b c-5 a d)) \operatorname{Subst}\left (\int \frac{\sqrt [3]{c} \sqrt [3]{b c-a d}+2 (b c-a d)^{2/3} x}{c^{2/3}+\sqrt [3]{c} \sqrt [3]{b c-a d} x+(b c-a d)^{2/3} x^2} \, dx,x,\frac{x}{\sqrt [3]{a+b x^3}}\right )}{54 c^{8/3} (b c-a d)^{4/3}}+\frac{(a (6 b c-5 a d)) \operatorname{Subst}\left (\int \frac{1}{c^{2/3}+\sqrt [3]{c} \sqrt [3]{b c-a d} x+(b c-a d)^{2/3} x^2} \, dx,x,\frac{x}{\sqrt [3]{a+b x^3}}\right )}{18 c^{7/3} (b c-a d)}\\ &=-\frac{d x \left (a+b x^3\right )^{5/3}}{6 c (b c-a d) \left (c+d x^3\right )^2}+\frac{(6 b c-5 a d) x \left (a+b x^3\right )^{2/3}}{18 c^2 (b c-a d) \left (c+d x^3\right )}-\frac{a (6 b c-5 a d) \log \left (\sqrt [3]{c}-\frac{\sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{27 c^{8/3} (b c-a d)^{4/3}}+\frac{a (6 b c-5 a d) \log \left (c^{2/3}+\frac{(b c-a d)^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{54 c^{8/3} (b c-a d)^{4/3}}-\frac{(a (6 b c-5 a d)) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{9 c^{8/3} (b c-a d)^{4/3}}\\ &=-\frac{d x \left (a+b x^3\right )^{5/3}}{6 c (b c-a d) \left (c+d x^3\right )^2}+\frac{(6 b c-5 a d) x \left (a+b x^3\right )^{2/3}}{18 c^2 (b c-a d) \left (c+d x^3\right )}+\frac{a (6 b c-5 a d) \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt{3}}\right )}{9 \sqrt{3} c^{8/3} (b c-a d)^{4/3}}-\frac{a (6 b c-5 a d) \log \left (\sqrt [3]{c}-\frac{\sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{27 c^{8/3} (b c-a d)^{4/3}}+\frac{a (6 b c-5 a d) \log \left (c^{2/3}+\frac{(b c-a d)^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac{\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{54 c^{8/3} (b c-a d)^{4/3}}\\ \end{align*}

Mathematica [C]  time = 0.160942, size = 153, normalized size = 0.57 \[ \frac{x \left (c \left (-a^2 d \left (8 c+5 d x^3\right )+a b \left (6 c^2-5 c d x^3-5 d^2 x^6\right )+3 b^2 c x^3 \left (2 c+d x^3\right )\right )-2 a \left (c+d x^3\right )^2 (5 a d-6 b c) \, _2F_1\left (\frac{1}{3},1;\frac{4}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )\right )}{18 c^3 \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(c*(3*b^2*c*x^3*(2*c + d*x^3) - a^2*d*(8*c + 5*d*x^3) + a*b*(6*c^2 - 5*c*d*x^3 - 5*d^2*x^6)) - 2*a*(-6*b*c
+ 5*a*d)*(c + d*x^3)^2*Hypergeometric2F1[1/3, 1, 4/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))]))/(18*c^3*(b*c - a*d)
*(a + b*x^3)^(1/3)*(c + d*x^3)^2)

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Maple [F]  time = 0.432, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( d{x}^{3}+c \right ) ^{3}} \left ( b{x}^{3}+a \right ) ^{{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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