3.7.71 \(\int \frac {\sqrt [3]{a+b x^3}}{x^{11} (c+d x^3)} \, dx\) [671]

Optimal. Leaf size=318 \[ -\frac {\sqrt [3]{a+b x^3}}{10 c x^{10}}-\frac {(b c-10 a d) \sqrt [3]{a+b x^3}}{70 a c^2 x^7}+\frac {\left (3 b^2 c^2+5 a b c d-35 a^2 d^2\right ) \sqrt [3]{a+b x^3}}{140 a^2 c^3 x^4}-\frac {\left (9 b^3 c^3+15 a b^2 c^2 d+35 a^2 b c d^2-140 a^3 d^3\right ) \sqrt [3]{a+b x^3}}{140 a^3 c^4 x}+\frac {d^3 \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{13/3}}-\frac {d^3 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 c^{13/3}}+\frac {d^3 \sqrt [3]{b c-a d} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{13/3}} \]

[Out]

-1/10*(b*x^3+a)^(1/3)/c/x^10-1/70*(-10*a*d+b*c)*(b*x^3+a)^(1/3)/a/c^2/x^7+1/140*(-35*a^2*d^2+5*a*b*c*d+3*b^2*c
^2)*(b*x^3+a)^(1/3)/a^2/c^3/x^4-1/140*(-140*a^3*d^3+35*a^2*b*c*d^2+15*a*b^2*c^2*d+9*b^3*c^3)*(b*x^3+a)^(1/3)/a
^3/c^4/x-1/6*d^3*(-a*d+b*c)^(1/3)*ln(d*x^3+c)/c^(13/3)+1/2*d^3*(-a*d+b*c)^(1/3)*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-
(b*x^3+a)^(1/3))/c^(13/3)+1/3*d^3*(-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^(13/3)*3^(1/2)

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Rubi [A]
time = 0.29, antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 6, 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 (-35 a^2 d^2+5 a b c d+3 b^2 c^2\right )}{140 a^2 c^3 x^4}-\frac {\sqrt [3]{a+b x^3} \left (-140 a^3 d^3+35 a^2 b c d^2+15 a b^2 c^2 d+9 b^3 c^3\right )}{140 a^3 c^4 x}+\frac {d^3 \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^{13/3}}-\frac {d^3 \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 c^{13/3}}+\frac {d^3 \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^{13/3}}-\frac {\sqrt [3]{a+b x^3} (b c-10 a d)}{70 a c^2 x^7}-\frac {\sqrt [3]{a+b x^3}}{10 c x^{10}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(1/3)/(x^11*(c + d*x^3)),x]

[Out]

-1/10*(a + b*x^3)^(1/3)/(c*x^10) - ((b*c - 10*a*d)*(a + b*x^3)^(1/3))/(70*a*c^2*x^7) + ((3*b^2*c^2 + 5*a*b*c*d
- 35*a^2*d^2)*(a + b*x^3)^(1/3))/(140*a^2*c^3*x^4) - ((9*b^3*c^3 + 15*a*b^2*c^2*d + 35*a^2*b*c*d^2 - 140*a^3*
d^3)*(a + b*x^3)^(1/3))/(140*a^3*c^4*x) + (d^3*(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^(13/3)) - (d^3*(b*c - a*d)^(1/3)*Log[c + d*x^3])/(6*c^(13/3)) + (d^3*
(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^(13/3))

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^{11} \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^{11} \left (c+d x^3\right )} \, dx}{\sqrt [3]{1+\frac {b x^3}{a}}}\\ &=-\frac {56 a c^4+56 b c^4 x^3-72 a c^3 d x^3-72 b c^3 d x^6+108 a c^2 d^2 x^6+108 b c^2 d^2 x^9-324 a c d^3 x^9-324 b c d^3 x^{12}-28 b c^4 x^3 \, _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 )+28 a c^3 d x^3 \, _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 )+36 b c^3 d x^6 \, _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 )-36 a c^2 d^2 x^6 \, _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 )-54 b c^2 d^2 x^9 \, _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 )+54 a c d^3 x^9 \, _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 )+162 b c d^3 x^{12} \, _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 )-162 a d^4 x^{12} \, _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 )+117 b c^4 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 )-117 a c^3 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 )-99 b c^3 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 )+99 a c^2 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 )+81 b c^2 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 )-81 a c 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 )+297 b c d^3 x^{12} \, _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 )-297 a d^4 x^{12} \, _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 )-54 (b c-a d) x^3 \left (2 c-3 d x^3\right ) \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 )+27 (b c-a d) x^3 \left (c+d x^3\right )^3 \, _4F_3\left (\frac {2}{3},2,2,2;1,1,\frac {5}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )}{560 c^5 x^{10} \left (a+b x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 2.08, size = 419, normalized size = 1.32 \begin {gather*} \frac {\frac {3 \sqrt [3]{c} \sqrt [3]{a+b x^3} \left (-9 b^3 c^3 x^9+3 a b^2 c^2 x^6 \left (c-5 d x^3\right )+a^2 b c x^3 \left (-2 c^2+5 c d x^3-35 d^2 x^6\right )+a^3 \left (-14 c^3+20 c^2 d x^3-35 c d^2 x^6+140 d^3 x^9\right )\right )}{a^3 x^{10}}-70 \sqrt {-6-6 i \sqrt {3}} d^3 \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 )+70 i \left (i+\sqrt {3}\right ) d^3 \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 )+35 \left (1-i \sqrt {3}\right ) d^3 \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 )}{420 c^{13/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(1/3)/(x^11*(c + d*x^3)),x]

[Out]

((3*c^(1/3)*(a + b*x^3)^(1/3)*(-9*b^3*c^3*x^9 + 3*a*b^2*c^2*x^6*(c - 5*d*x^3) + a^2*b*c*x^3*(-2*c^2 + 5*c*d*x^
3 - 35*d^2*x^6) + a^3*(-14*c^3 + 20*c^2*d*x^3 - 35*c*d^2*x^6 + 140*d^3*x^9)))/(a^3*x^10) - 70*Sqrt[-6 - (6*I)*
Sqrt[3]]*d^3*(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))] + (70*I)*(I + Sqrt[3])*d^3*(b*c - a*d)^(1/3)*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqr
t[3])*c^(1/3)*(a + b*x^3)^(1/3)] + 35*(1 - I*Sqrt[3])*d^3*(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)])/(420*
c^(13/3))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x^{11} \left (d \,x^{3}+c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(1/3)/x^11/(d*x^3+c),x)

[Out]

int((b*x^3+a)^(1/3)/x^11/(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^11/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(1/3)/((d*x^3 + c)*x^11), 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^11/(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^{11} \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**11/(d*x**3+c),x)

[Out]

Integral((a + b*x**3)**(1/3)/(x**11*(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^11/(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(1/3)/((d*x^3 + c)*x^11), 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^{11}\,\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^11*(c + d*x^3)),x)

[Out]

int((a + b*x^3)^(1/3)/(x^11*(c + d*x^3)), x)

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