3.7.3 \(\int \frac {(a+b x^3)^{2/3}}{x^2 (a d-b d x^3)} \, dx\) [603]

Optimal. Leaf size=483 \[ -\frac {\left (a+b x^3\right )^{2/3}}{a d x}+\frac {2^{2/3} \sqrt [3]{b} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} a^{2/3} d}+\frac {\sqrt [3]{b} \tan ^{-1}\left (\frac {1+\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3} a^{2/3} d}+\frac {b x^2 \sqrt [3]{1+\frac {b x^3}{a}} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-\frac {b x^3}{a}\right )}{2 a d \sqrt [3]{a+b x^3}}+\frac {\sqrt [3]{b} \log \left (\frac {\left (\sqrt [3]{a}-\sqrt [3]{b} x\right )^2 \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{a}\right )}{6 \sqrt [3]{2} a^{2/3} d}+\frac {\sqrt [3]{b} \log \left (1+\frac {2^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )^2}{\left (a+b x^3\right )^{2/3}}-\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}\right )}{3 \sqrt [3]{2} a^{2/3} d}-\frac {2^{2/3} \sqrt [3]{b} \log \left (1+\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}\right )}{3 a^{2/3} d}-\frac {\sqrt [3]{b} \log \left (\frac {\sqrt [3]{b} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a}}-\frac {2^{2/3} \sqrt [3]{b} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}\right )}{2 \sqrt [3]{2} a^{2/3} d} \]

[Out]

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

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Rubi [A]
time = 0.29, antiderivative size = 483, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {486, 598, 372, 371, 502, 2174, 206, 31, 648, 631, 210, 642} \begin {gather*} \frac {2^{2/3} \sqrt [3]{b} \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^{2/3} d}+\frac {\sqrt [3]{b} \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 )}{\sqrt [3]{2} \sqrt {3} a^{2/3} d}+\frac {\sqrt [3]{b} \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 \sqrt [3]{2} a^{2/3} d}-\frac {2^{2/3} \sqrt [3]{b} \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^{2/3} d}-\frac {\sqrt [3]{b} \log \left (\frac {\sqrt [3]{b} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a}}-\frac {2^{2/3} \sqrt [3]{b} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}\right )}{2 \sqrt [3]{2} a^{2/3} d}+\frac {\sqrt [3]{b} \log \left (\frac {\left (\sqrt [3]{a}-\sqrt [3]{b} x\right )^2 \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{a}\right )}{6 \sqrt [3]{2} a^{2/3} d}+\frac {b x^2 \sqrt [3]{\frac {b x^3}{a}+1} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-\frac {b x^3}{a}\right )}{2 a d \sqrt [3]{a+b x^3}}-\frac {\left (a+b x^3\right )^{2/3}}{a d x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 206

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 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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 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 502

Int[(x_)/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[b/a, 3]}, Dist[-q^2/(3
*d), Int[1/((1 - q*x)*(a + b*x^3)^(1/3)), x], x] + Dist[q/d, Subst[Int[1/(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 598

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 0]

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]

Rule 2174

Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)), x_Symbol] :> Simp[Sqrt[3]*(ArcTan[(1 - 2^(1/3)*Rt[b,
3]*((c - d*x)/(d*(a + b*x^3)^(1/3))))/Sqrt[3]]/(2^(4/3)*Rt[b, 3]*c)), x] + (Simp[Log[(c + d*x)^2*(c - d*x)]/(2
^(7/3)*Rt[b, 3]*c), x] - Simp[(3*Log[Rt[b, 3]*(c - d*x) + 2^(2/3)*d*(a + b*x^3)^(1/3)])/(2^(7/3)*Rt[b, 3]*c),
x]) /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^3 + a*d^3, 0]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 20.07, size = 136, normalized size = 0.28 \begin {gather*} \frac {15 a b x^3 \sqrt [3]{1+\frac {b x^3}{a}} F_1\left (\frac {2}{3};\frac {1}{3},1;\frac {5}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )-2 \left (5 a \left (a+b x^3\right )+b^2 x^6 \sqrt [3]{1+\frac {b x^3}{a}} F_1\left (\frac {5}{3};\frac {1}{3},1;\frac {8}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )\right )}{10 a^2 d x \sqrt [3]{a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a*b*x^3*(1 + (b*x^3)/a)^(1/3)*AppellF1[2/3, 1/3, 1, 5/3, -((b*x^3)/a), (b*x^3)/a] - 2*(5*a*(a + b*x^3) + b
^2*x^6*(1 + (b*x^3)/a)^(1/3)*AppellF1[5/3, 1/3, 1, 8/3, -((b*x^3)/a), (b*x^3)/a]))/(10*a^2*d*x*(a + b*x^3)^(1/
3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

-integrate((b*x^3 + a)^(2/3)/((b*d*x^3 - a*d)*x^2), 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)^(2/3)/x^2/(-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 {\left (a + b x^{3}\right )^{\frac {2}{3}}}{- a x^{2} + b x^{5}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral((a + b*x**3)**(2/3)/(-a*x**2 + b*x**5), 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)^(2/3)/x^2/(-b*d*x^3+a*d),x, algorithm="giac")

[Out]

integrate(-(b*x^3 + a)^(2/3)/((b*d*x^3 - a*d)*x^2), 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 )}^{2/3}}{x^2\,\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^2*(a*d - b*d*x^3)),x)

[Out]

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

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