3.6.85 \(\int \frac {\sqrt [3]{a+b x^3}}{x^6 (a d-b d x^3)} \, dx\) [585]

Optimal. Leaf size=523 \[ -\frac {\sqrt [3]{a+b x^3}}{5 a d x^5}-\frac {3 b \sqrt [3]{a+b x^3}}{5 a^2 d x^2}-\frac {\sqrt [3]{2} b^{5/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} a^{7/3} d}-\frac {b^{5/3} \tan ^{-1}\left (\frac {1+\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3} a^{7/3} d}+\frac {2 b^2 x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{5 a^2 d \left (a+b x^3\right )^{2/3}}-\frac {b^{5/3} \log \left (2^{2/3}-\frac {\sqrt [3]{a}+\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3\ 2^{2/3} a^{7/3} d}+\frac {b^{5/3} \log \left (1+\frac {2^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )^2}{\left (a+b x^3\right )^{2/3}}-\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}\right )}{3\ 2^{2/3} a^{7/3} d}-\frac {\sqrt [3]{2} b^{5/3} \log \left (1+\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}\right )}{3 a^{7/3} d}+\frac {b^{5/3} \log \left (2 \sqrt [3]{2}+\frac {\left (\sqrt [3]{a}+\sqrt [3]{b} x\right )^2}{\left (a+b x^3\right )^{2/3}}+\frac {2^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}\right )}{6\ 2^{2/3} a^{7/3} d} \]

[Out]

-1/5*(b*x^3+a)^(1/3)/a/d/x^5-3/5*b*(b*x^3+a)^(1/3)/a^2/d/x^2+2/5*b^2*x*(1+b*x^3/a)^(2/3)*hypergeom([1/3, 2/3],
[4/3],-b*x^3/a)/a^2/d/(b*x^3+a)^(2/3)-1/6*b^(5/3)*ln(2^(2/3)+(-a^(1/3)-b^(1/3)*x)/(b*x^3+a)^(1/3))*2^(1/3)/a^(
7/3)/d+1/6*b^(5/3)*ln(1+2^(2/3)*(a^(1/3)+b^(1/3)*x)^2/(b*x^3+a)^(2/3)-2^(1/3)*(a^(1/3)+b^(1/3)*x)/(b*x^3+a)^(1
/3))*2^(1/3)/a^(7/3)/d-1/3*2^(1/3)*b^(5/3)*ln(1+2^(1/3)*(a^(1/3)+b^(1/3)*x)/(b*x^3+a)^(1/3))/a^(7/3)/d+1/12*b^
(5/3)*ln(2*2^(1/3)+(a^(1/3)+b^(1/3)*x)^2/(b*x^3+a)^(2/3)+2^(2/3)*(a^(1/3)+b^(1/3)*x)/(b*x^3+a)^(1/3))*2^(1/3)/
a^(7/3)/d-1/3*2^(1/3)*b^(5/3)*arctan(1/3*(1-2*2^(1/3)*(a^(1/3)+b^(1/3)*x)/(b*x^3+a)^(1/3))*3^(1/2))/a^(7/3)/d*
3^(1/2)-1/6*b^(5/3)*arctan(1/3*(1+2^(1/3)*(a^(1/3)+b^(1/3)*x)/(b*x^3+a)^(1/3))*3^(1/2))*2^(1/3)/a^(7/3)/d*3^(1
/2)

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Rubi [A]
time = 0.38, antiderivative size = 523, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 14, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {486, 597, 544, 252, 251, 421, 420, 493, 298, 31, 648, 631, 210, 642} \begin {gather*} -\frac {\sqrt [3]{2} b^{5/3} \text {ArcTan}\left (\frac {1-\frac {2 \sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} a^{7/3} d}-\frac {b^{5/3} \text {ArcTan}\left (\frac {\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3} a^{7/3} d}-\frac {b^{5/3} \log \left (2^{2/3}-\frac {\sqrt [3]{a}+\sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}\right )}{3\ 2^{2/3} a^{7/3} d}+\frac {b^{5/3} \log \left (\frac {2^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )^2}{\left (a+b x^3\right )^{2/3}}-\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+1\right )}{3\ 2^{2/3} a^{7/3} d}-\frac {\sqrt [3]{2} b^{5/3} \log \left (\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+1\right )}{3 a^{7/3} d}+\frac {b^{5/3} \log \left (\frac {\left (\sqrt [3]{a}+\sqrt [3]{b} x\right )^2}{\left (a+b x^3\right )^{2/3}}+\frac {2^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+2 \sqrt [3]{2}\right )}{6\ 2^{2/3} a^{7/3} d}+\frac {2 b^2 x \left (\frac {b x^3}{a}+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{5 a^2 d \left (a+b x^3\right )^{2/3}}-\frac {3 b \sqrt [3]{a+b x^3}}{5 a^2 d x^2}-\frac {\sqrt [3]{a+b x^3}}{5 a d x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(1/3)/(x^6*(a*d - b*d*x^3)),x]

[Out]

-1/5*(a + b*x^3)^(1/3)/(a*d*x^5) - (3*b*(a + b*x^3)^(1/3))/(5*a^2*d*x^2) - (2^(1/3)*b^(5/3)*ArcTan[(1 - (2*2^(
1/3)*(a^(1/3) + b^(1/3)*x))/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*a^(7/3)*d) - (b^(5/3)*ArcTan[(1 + (2^(1/3)*(
a^(1/3) + b^(1/3)*x))/(a + b*x^3)^(1/3))/Sqrt[3]])/(2^(2/3)*Sqrt[3]*a^(7/3)*d) + (2*b^2*x*(1 + (b*x^3)/a)^(2/3
)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(5*a^2*d*(a + b*x^3)^(2/3)) - (b^(5/3)*Log[2^(2/3) - (a^(1/3
) + b^(1/3)*x)/(a + b*x^3)^(1/3)])/(3*2^(2/3)*a^(7/3)*d) + (b^(5/3)*Log[1 + (2^(2/3)*(a^(1/3) + b^(1/3)*x)^2)/
(a + b*x^3)^(2/3) - (2^(1/3)*(a^(1/3) + b^(1/3)*x))/(a + b*x^3)^(1/3)])/(3*2^(2/3)*a^(7/3)*d) - (2^(1/3)*b^(5/
3)*Log[1 + (2^(1/3)*(a^(1/3) + b^(1/3)*x))/(a + b*x^3)^(1/3)])/(3*a^(7/3)*d) + (b^(5/3)*Log[2*2^(1/3) + (a^(1/
3) + b^(1/3)*x)^2/(a + b*x^3)^(2/3) + (2^(2/3)*(a^(1/3) + b^(1/3)*x))/(a + b*x^3)^(1/3)])/(6*2^(2/3)*a^(7/3)*d
)

Rule 31

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

Rule 210

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

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(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 420

Int[((a_) + (b_.)*(x_)^3)^(1/3)/((c_) + (d_.)*(x_)^3), x_Symbol] :> With[{q = Rt[b/a, 3]}, Dist[9*(a/(c*q)), S
ubst[Int[x/((4 - a*x^3)*(1 + 2*a*x^3)), x], x, (1 + q*x)/(a + b*x^3)^(1/3)], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[b*c - a*d, 0] && EqQ[b*c + a*d, 0]

Rule 421

Int[1/(((a_) + (b_.)*(x_)^3)^(2/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x^3)^
(2/3), x], x] - Dist[d/(b*c - a*d), Int[(a + b*x^3)^(1/3)/(c + d*x^3), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ
[b*c - a*d, 0] && EqQ[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 493

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

Rule 544

Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Dist[f/d,
Int[(a + b*x^n)^p, x], x] + Dist[(d*e - c*f)/d, Int[(a + b*x^n)^p/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, p, n}, 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]

Rule 631

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 642

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]

Rule 648

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]

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{a+b x^3}}{x^6 \left (a d-b d x^3\right )} \, dx &=\frac {\sqrt [3]{a+b x^3} \int \frac {\sqrt [3]{1+\frac {b x^3}{a}}}{x^6 \left (a d-b d x^3\right )} \, dx}{\sqrt [3]{1+\frac {b x^3}{a}}}\\ &=-\frac {\sqrt [3]{a+b x^3} F_1\left (-\frac {5}{3};-\frac {1}{3},1;-\frac {2}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )}{5 a d x^5 \sqrt [3]{1+\frac {b x^3}{a}}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 20.13, size = 243, normalized size = 0.46 \begin {gather*} \frac {-\frac {4 \left (a^2+4 a b x^3+3 b^2 x^6\right )}{a^2 x^5}+\frac {3 b^3 x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )}{a^3}+\frac {112 b^2 x F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )}{\left (a-b x^3\right ) \left (4 a F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )+b x^3 \left (3 F_1\left (\frac {4}{3};\frac {2}{3},2;\frac {7}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )-2 F_1\left (\frac {4}{3};\frac {5}{3},1;\frac {7}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )\right )\right )}}{20 d \left (a+b x^3\right )^{2/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3)^(1/3)/(x^6*(a*d - b*d*x^3)),x]

[Out]

((-4*(a^2 + 4*a*b*x^3 + 3*b^2*x^6))/(a^2*x^5) + (3*b^3*x^4*(1 + (b*x^3)/a)^(2/3)*AppellF1[4/3, 2/3, 1, 7/3, -(
(b*x^3)/a), (b*x^3)/a])/a^3 + (112*b^2*x*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), (b*x^3)/a])/((a - b*x^3)*(4*
a*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), (b*x^3)/a] + b*x^3*(3*AppellF1[4/3, 2/3, 2, 7/3, -((b*x^3)/a), (b*x
^3)/a] - 2*AppellF1[4/3, 5/3, 1, 7/3, -((b*x^3)/a), (b*x^3)/a]))))/(20*d*(a + b*x^3)^(2/3))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x^{6} \left (-b d \,x^{3}+a d \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(1/3)/x^6/(-b*d*x^3+a*d),x)

[Out]

int((b*x^3+a)^(1/3)/x^6/(-b*d*x^3+a*d),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^6/(-b*d*x^3+a*d),x, algorithm="maxima")

[Out]

-integrate((b*x^3 + a)^(1/3)/((b*d*x^3 - a*d)*x^6), 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^6/(-b*d*x^3+a*d),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\sqrt [3]{a + b x^{3}}}{- a x^{6} + b x^{9}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(1/3)/x**6/(-b*d*x**3+a*d),x)

[Out]

-Integral((a + b*x**3)**(1/3)/(-a*x**6 + b*x**9), x)/d

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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^6/(-b*d*x^3+a*d),x, algorithm="giac")

[Out]

integrate(-(b*x^3 + a)^(1/3)/((b*d*x^3 - a*d)*x^6), 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^6\,\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)^(1/3)/(x^6*(a*d - b*d*x^3)),x)

[Out]

int((a + b*x^3)^(1/3)/(x^6*(a*d - b*d*x^3)), x)

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