3.7.70 \(\int \frac {\sqrt [3]{a+b x^3}}{x^8 (c+d x^3)} \, dx\) [670]

Optimal. Leaf size=258 \[ -\frac {\sqrt [3]{a+b x^3}}{7 c x^7}-\frac {(b c-7 a d) \sqrt [3]{a+b x^3}}{28 a c^2 x^4}+\frac {\left (3 b^2 c^2+7 a b c d-28 a^2 d^2\right ) \sqrt [3]{a+b x^3}}{28 a^2 c^3 x}-\frac {d^2 \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{10/3}}+\frac {d^2 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 c^{10/3}}-\frac {d^2 \sqrt [3]{b c-a d} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{10/3}} \]

[Out]

-1/7*(b*x^3+a)^(1/3)/c/x^7-1/28*(-7*a*d+b*c)*(b*x^3+a)^(1/3)/a/c^2/x^4+1/28*(-28*a^2*d^2+7*a*b*c*d+3*b^2*c^2)*
(b*x^3+a)^(1/3)/a^2/c^3/x+1/6*d^2*(-a*d+b*c)^(1/3)*ln(d*x^3+c)/c^(10/3)-1/2*d^2*(-a*d+b*c)^(1/3)*ln((-a*d+b*c)
^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(10/3)-1/3*d^2*(-a*d+b*c)^(1/3)*arctan(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/3)
/(b*x^3+a)^(1/3))*3^(1/2))/c^(10/3)*3^(1/2)

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Rubi [A]
time = 0.20, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {486, 597, 12, 503} \begin {gather*} \frac {\sqrt [3]{a+b x^3} \left (-28 a^2 d^2+7 a b c d+3 b^2 c^2\right )}{28 a^2 c^3 x}-\frac {d^2 \sqrt [3]{b c-a d} \text {ArcTan}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{10/3}}+\frac {d^2 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 c^{10/3}}-\frac {d^2 \sqrt [3]{b c-a d} \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{10/3}}-\frac {\sqrt [3]{a+b x^3} (b c-7 a d)}{28 a c^2 x^4}-\frac {\sqrt [3]{a+b x^3}}{7 c x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(1/3)/(x^8*(c + d*x^3)),x]

[Out]

-1/7*(a + b*x^3)^(1/3)/(c*x^7) - ((b*c - 7*a*d)*(a + b*x^3)^(1/3))/(28*a*c^2*x^4) + ((3*b^2*c^2 + 7*a*b*c*d -
28*a^2*d^2)*(a + b*x^3)^(1/3))/(28*a^2*c^3*x) - (d^2*(b*c - a*d)^(1/3)*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^
(1/3)*(a + b*x^3)^(1/3)))/Sqrt[3]])/(Sqrt[3]*c^(10/3)) + (d^2*(b*c - a*d)^(1/3)*Log[c + d*x^3])/(6*c^(10/3)) -
(d^2*(b*c - a*d)^(1/3)*Log[((b*c - a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)])/(2*c^(10/3))

Rule 12

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

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 503

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

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 {\sqrt [3]{a+b x^3}}{x^8 \left (c+d x^3\right )} \, dx &=\frac {\sqrt [3]{a+b x^3} \int \frac {\sqrt [3]{1+\frac {b x^3}{a}}}{x^8 \left (c+d x^3\right )} \, dx}{\sqrt [3]{1+\frac {b x^3}{a}}}\\ &=-\frac {8 a c^3+8 b c^3 x^3-12 a c^2 d x^3-12 b c^2 d x^6+36 a c d^2 x^6+36 b c d^2 x^9-2 (b c-a d) x^3 \left (2 c^2-3 c d x^3+9 d^2 x^6\right ) \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+15 b c^3 x^3 \, _2F_1\left (\frac {2}{3},2;\frac {5}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-15 a c^2 d x^3 \, _2F_1\left (\frac {2}{3},2;\frac {5}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-12 b c^2 d x^6 \, _2F_1\left (\frac {2}{3},2;\frac {5}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+12 a c d^2 x^6 \, _2F_1\left (\frac {2}{3},2;\frac {5}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-27 b c d^2 x^9 \, _2F_1\left (\frac {2}{3},2;\frac {5}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )+27 a d^3 x^9 \, _2F_1\left (\frac {2}{3},2;\frac {5}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-9 (b c-a d) x^3 \left (c+d x^3\right )^2 \, _3F_2\left (\frac {2}{3},2,2;1,\frac {5}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )}{56 c^4 x^7 \left (a+b x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.87, size = 373, normalized size = 1.45 \begin {gather*} \frac {-\frac {3 \sqrt [3]{c} \sqrt [3]{a+b x^3} \left (-3 b^2 c^2 x^6+a b c x^3 \left (c-7 d x^3\right )+a^2 \left (4 c^2-7 c d x^3+28 d^2 x^6\right )\right )}{a^2 x^7}+14 \sqrt {-6-6 i \sqrt {3}} d^2 \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )+14 \left (1-i \sqrt {3}\right ) d^2 \sqrt [3]{b c-a d} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )+7 i \left (i+\sqrt {3}\right ) d^2 \sqrt [3]{b c-a d} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{84 c^{10/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(1/3)/(x^8*(c + d*x^3)),x]

[Out]

((-3*c^(1/3)*(a + b*x^3)^(1/3)*(-3*b^2*c^2*x^6 + a*b*c*x^3*(c - 7*d*x^3) + a^2*(4*c^2 - 7*c*d*x^3 + 28*d^2*x^6
)))/(a^2*x^7) + 14*Sqrt[-6 - (6*I)*Sqrt[3]]*d^2*(b*c - a*d)^(1/3)*ArcTan[(3*(b*c - a*d)^(1/3)*x)/(Sqrt[3]*(b*c
- a*d)^(1/3)*x - (3*I + Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3))] + 14*(1 - I*Sqrt[3])*d^2*(b*c - a*d)^(1/3)*Log[2
*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)] + (7*I)*(I + Sqrt[3])*d^2*(b*c - a*d)^(1/3)*
Log[2*(b*c - a*d)^(2/3)*x^2 + (-1 - I*Sqrt[3])*c^(1/3)*(b*c - a*d)^(1/3)*x*(a + b*x^3)^(1/3) + I*(I + Sqrt[3])
*c^(2/3)*(a + b*x^3)^(2/3)])/(84*c^(10/3))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x^{8} \left (d \,x^{3}+c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(1/3)/x^8/(d*x^3+c),x)

[Out]

int((b*x^3+a)^(1/3)/x^8/(d*x^3+c),x)

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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)^(1/3)/x^8/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(1/3)/((d*x^3 + c)*x^8), x)

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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)^(1/3)/x^8/(d*x^3+c),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [3]{a + b x^{3}}}{x^{8} \left (c + d x^{3}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(1/3)/x**8/(d*x**3+c),x)

[Out]

Integral((a + b*x**3)**(1/3)/(x**8*(c + d*x**3)), x)

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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)^(1/3)/x^8/(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(1/3)/((d*x^3 + c)*x^8), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (b\,x^3+a\right )}^{1/3}}{x^8\,\left (d\,x^3+c\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(1/3)/(x^8*(c + d*x^3)),x)

[Out]

int((a + b*x^3)^(1/3)/(x^8*(c + d*x^3)), x)

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