3.6.98 \(\int \frac {(a+b x^3)^{2/3}}{x^9 (a d-b d x^3)} \, dx\) [598]

Optimal. Leaf size=209 \[ -\frac {\left (a+b x^3\right )^{2/3}}{8 a d x^8}-\frac {b \left (a+b x^3\right )^{2/3}}{4 a^2 d x^5}-\frac {5 b^2 \left (a+b x^3\right )^{2/3}}{8 a^3 d x^2}+\frac {2^{2/3} b^{8/3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} a^3 d}+\frac {b^{8/3} \log \left (a d-b d x^3\right )}{3 \sqrt [3]{2} a^3 d}-\frac {b^{8/3} \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} a^3 d} \]

[Out]

-1/8*(b*x^3+a)^(2/3)/a/d/x^8-1/4*b*(b*x^3+a)^(2/3)/a^2/d/x^5-5/8*b^2*(b*x^3+a)^(2/3)/a^3/d/x^2+1/6*b^(8/3)*ln(
-b*d*x^3+a*d)*2^(2/3)/a^3/d-1/2*b^(8/3)*ln(2^(1/3)*b^(1/3)*x-(b*x^3+a)^(1/3))*2^(2/3)/a^3/d+1/3*2^(2/3)*b^(8/3
)*arctan(1/3*(1+2*2^(1/3)*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/a^3/d*3^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.16, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {486, 597, 12, 384} \begin {gather*} \frac {2^{2/3} b^{8/3} \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} a^3 d}+\frac {b^{8/3} \log \left (a d-b d x^3\right )}{3 \sqrt [3]{2} a^3 d}-\frac {b^{8/3} \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} a^3 d}-\frac {5 b^2 \left (a+b x^3\right )^{2/3}}{8 a^3 d x^2}-\frac {b \left (a+b x^3\right )^{2/3}}{4 a^2 d x^5}-\frac {\left (a+b x^3\right )^{2/3}}{8 a d x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(2/3)/(x^9*(a*d - b*d*x^3)),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 486

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

Rule 597

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*g*(m + 1))), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^3\right )^{2/3}}{x^9 \left (a d-b d x^3\right )} \, dx &=\frac {\left (a+b x^3\right )^{2/3} \int \frac {\left (1+\frac {b x^3}{a}\right )^{2/3}}{x^9 \left (a d-b d x^3\right )} \, dx}{\left (1+\frac {b x^3}{a}\right )^{2/3}}\\ &=-\frac {5 a^3+11 a^2 b x^3+15 a b^2 x^6+9 b^3 x^9-4 b x^3 \left (5 a^2+6 a b x^3+9 b^2 x^6\right ) \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {2 b x^3}{a+b x^3}\right )+42 a^2 b x^3 \, _2F_1\left (\frac {1}{3},2;\frac {4}{3};\frac {2 b x^3}{a+b x^3}\right )+12 a b^2 x^6 \, _2F_1\left (\frac {1}{3},2;\frac {4}{3};\frac {2 b x^3}{a+b x^3}\right )-54 b^3 x^9 \, _2F_1\left (\frac {1}{3},2;\frac {4}{3};\frac {2 b x^3}{a+b x^3}\right )-18 b x^3 \left (a-b x^3\right )^2 \, _3F_2\left (\frac {1}{3},2,2;1,\frac {4}{3};\frac {2 b x^3}{a+b x^3}\right )}{40 a^3 d x^8 \sqrt [3]{a+b x^3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.43, size = 206, normalized size = 0.99 \begin {gather*} \frac {-\frac {3 \left (a+b x^3\right )^{2/3} \left (a^2+2 a b x^3+5 b^2 x^6\right )}{x^8}+8\ 2^{2/3} \sqrt {3} b^{8/3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}}\right )-8\ 2^{2/3} b^{8/3} \log \left (-2 \sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}\right )+4\ 2^{2/3} b^{8/3} \log \left (2 b^{2/3} x^2+2^{2/3} \sqrt [3]{b} x \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}\right )}{24 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(2/3)/(x^9*(a*d - b*d*x^3)),x]

[Out]

((-3*(a + b*x^3)^(2/3)*(a^2 + 2*a*b*x^3 + 5*b^2*x^6))/x^8 + 8*2^(2/3)*Sqrt[3]*b^(8/3)*ArcTan[(Sqrt[3]*b^(1/3)*
x)/(b^(1/3)*x + 2^(2/3)*(a + b*x^3)^(1/3))] - 8*2^(2/3)*b^(8/3)*Log[-2*b^(1/3)*x + 2^(2/3)*(a + b*x^3)^(1/3)]
+ 4*2^(2/3)*b^(8/3)*Log[2*b^(2/3)*x^2 + 2^(2/3)*b^(1/3)*x*(a + b*x^3)^(1/3) + 2^(1/3)*(a + b*x^3)^(2/3)])/(24*
a^3*d)

________________________________________________________________________________________

Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{9} \left (-b d \,x^{3}+a d \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(2/3)/x^9/(-b*d*x^3+a*d),x)

[Out]

int((b*x^3+a)^(2/3)/x^9/(-b*d*x^3+a*d),x)

________________________________________________________________________________________

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((b*x^3+a)^(2/3)/x^9/(-b*d*x^3+a*d),x, algorithm="maxima")

[Out]

-integrate((b*x^3 + a)^(2/3)/((b*d*x^3 - a*d)*x^9), x)

________________________________________________________________________________________

Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(2/3)/x^9/(-b*d*x^3+a*d),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\left (a + b x^{3}\right )^{\frac {2}{3}}}{- a x^{9} + b x^{12}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(2/3)/x**9/(-b*d*x**3+a*d),x)

[Out]

-Integral((a + b*x**3)**(2/3)/(-a*x**9 + b*x**12), x)/d

________________________________________________________________________________________

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((b*x^3+a)^(2/3)/x^9/(-b*d*x^3+a*d),x, algorithm="giac")

[Out]

integrate(-(b*x^3 + a)^(2/3)/((b*d*x^3 - a*d)*x^9), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (b\,x^3+a\right )}^{2/3}}{x^9\,\left (a\,d-b\,d\,x^3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(2/3)/(x^9*(a*d - b*d*x^3)),x)

[Out]

int((a + b*x^3)^(2/3)/(x^9*(a*d - b*d*x^3)), x)

________________________________________________________________________________________