3.7.1 \(\int \frac {x^4 (a+b x^3)^{2/3}}{a d-b d x^3} \, dx\) [601]

Optimal. Leaf size=485 \[ -\frac {x^2 \left (a+b x^3\right )^{2/3}}{4 b d}+\frac {2^{2/3} a^{4/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^{5/3} d}+\frac {a^{4/3} \tan ^{-1}\left (\frac {1+\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3} b^{5/3} d}-\frac {3 a x^2 \sqrt [3]{1+\frac {b x^3}{a}} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-\frac {b x^3}{a}\right )}{4 b d \sqrt [3]{a+b x^3}}+\frac {a^{4/3} \log \left (\frac {\left (\sqrt [3]{a}-\sqrt [3]{b} x\right )^2 \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{a}\right )}{6 \sqrt [3]{2} b^{5/3} d}+\frac {a^{4/3} \log \left (1+\frac {2^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )^2}{\left (a+b x^3\right )^{2/3}}-\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}\right )}{3 \sqrt [3]{2} b^{5/3} d}-\frac {2^{2/3} a^{4/3} \log \left (1+\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}\right )}{3 b^{5/3} d}-\frac {a^{4/3} \log \left (\frac {\sqrt [3]{b} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a}}-\frac {2^{2/3} \sqrt [3]{b} \sqrt [3]{a+b x^3}}{\sqrt [3]{a}}\right )}{2 \sqrt [3]{2} b^{5/3} d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.29, antiderivative size = 485, 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 = {489, 598, 372, 371, 502, 2174, 206, 31, 648, 631, 210, 642} \begin {gather*} \frac {2^{2/3} a^{4/3} \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} b^{5/3} d}+\frac {a^{4/3} \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 )}{\sqrt [3]{2} \sqrt {3} b^{5/3} d}+\frac {a^{4/3} \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 \sqrt [3]{2} b^{5/3} d}-\frac {2^{2/3} a^{4/3} \log \left (\frac {\sqrt [3]{2} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{a+b x^3}}+1\right )}{3 b^{5/3} d}-\frac {a^{4/3} \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} b^{5/3} d}+\frac {a^{4/3} \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} b^{5/3} d}-\frac {3 a 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 )}{4 b d \sqrt [3]{a+b x^3}}-\frac {x^2 \left (a+b x^3\right )^{2/3}}{4 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*x^3)^(2/3))/(a*d - b*d*x^3),x]

[Out]

-1/4*(x^2*(a + b*x^3)^(2/3))/(b*d) + (2^(2/3)*a^(4/3)*ArcTan[(1 - (2*2^(1/3)*(a^(1/3) + b^(1/3)*x))/(a + b*x^3
)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(5/3)*d) + (a^(4/3)*ArcTan[(1 + (2^(1/3)*(a^(1/3) + b^(1/3)*x))/(a + b*x^3)^(1/3
))/Sqrt[3]])/(2^(1/3)*Sqrt[3]*b^(5/3)*d) - (3*a*x^2*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, -((
b*x^3)/a)])/(4*b*d*(a + b*x^3)^(1/3)) + (a^(4/3)*Log[((a^(1/3) - b^(1/3)*x)^2*(a^(1/3) + b^(1/3)*x))/a])/(6*2^
(1/3)*b^(5/3)*d) + (a^(4/3)*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)*b^(5/3)*d) - (2^(2/3)*a^(4/3)*Log[1 + (2^(1/3)*(a^(1/3) + b^(1/3)*x
))/(a + b*x^3)^(1/3)])/(3*b^(5/3)*d) - (a^(4/3)*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)*b^(5/3)*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 489

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(n - 1
)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*(m + n*(p + q) + 1))), x] - Dist[e^n/(b*(m + n*(p +
q) + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[a*c*(m - n + 1) + (a*d*(m - n + 1) - n*q*(b
*c - a*d))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 0] &&
GtQ[m - n + 1, 0] && 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 {x^4 \left (a+b x^3\right )^{2/3}}{a d-b d x^3} \, dx &=\frac {\left (a+b x^3\right )^{2/3} \int \frac {x^4 \left (1+\frac {b x^3}{a}\right )^{2/3}}{a d-b d x^3} \, dx}{\left (1+\frac {b x^3}{a}\right )^{2/3}}\\ &=\frac {x^5 \left (a+b x^3\right )^{2/3} F_1\left (\frac {5}{3};-\frac {2}{3},1;\frac {8}{3};-\frac {b x^3}{a},\frac {b x^3}{a}\right )}{5 a d \left (1+\frac {b x^3}{a}\right )^{2/3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 6.70, size = 127, normalized size = 0.26 \begin {gather*} \frac {x^2 \left (-5 \left (a+b x^3\right )+5 a \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 )+6 b x^3 \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 )}{20 b d \sqrt [3]{a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*x^3)^(2/3))/(a*d - b*d*x^3),x]

[Out]

(x^2*(-5*(a + b*x^3) + 5*a*(1 + (b*x^3)/a)^(1/3)*AppellF1[2/3, 1/3, 1, 5/3, -((b*x^3)/a), (b*x^3)/a] + 6*b*x^3
*(1 + (b*x^3)/a)^(1/3)*AppellF1[5/3, 1/3, 1, 8/3, -((b*x^3)/a), (b*x^3)/a]))/(20*b*d*(a + b*x^3)^(1/3))

________________________________________________________________________________________

Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {x^{4} \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{-b d \,x^{3}+a d}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x)

[Out]

int(x^4*(b*x^3+a)^(2/3)/(-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(x^4*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x, algorithm="maxima")

[Out]

-integrate((b*x^3 + a)^(2/3)*x^4/(b*d*x^3 - a*d), 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(x^4*(b*x^3+a)^(2/3)/(-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 {x^{4} \left (a + b x^{3}\right )^{\frac {2}{3}}}{- a + b x^{3}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b*x**3+a)**(2/3)/(-b*d*x**3+a*d),x)

[Out]

-Integral(x**4*(a + b*x**3)**(2/3)/(-a + b*x**3), 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(x^4*(b*x^3+a)^(2/3)/(-b*d*x^3+a*d),x, algorithm="giac")

[Out]

integrate(-(b*x^3 + a)^(2/3)*x^4/(b*d*x^3 - a*d), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4\,{\left (b\,x^3+a\right )}^{2/3}}{a\,d-b\,d\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*x^3)^(2/3))/(a*d - b*d*x^3),x)

[Out]

int((x^4*(a + b*x^3)^(2/3))/(a*d - b*d*x^3), x)

________________________________________________________________________________________