3.5.83 \(\int \frac {x^2}{(a+b x^3)^{4/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=167 \[ -\frac {1}{\sqrt [3]{a+b x^3} (b c-a d)}-\frac {\sqrt [3]{d} \log \left (c+d x^3\right )}{6 (b c-a d)^{4/3}}+\frac {\sqrt [3]{d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 (b c-a d)^{4/3}}+\frac {\sqrt [3]{d} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} (b c-a d)^{4/3}} \]

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Rubi [A]  time = 0.17, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {444, 51, 56, 617, 204, 31} \begin {gather*} -\frac {1}{\sqrt [3]{a+b x^3} (b c-a d)}-\frac {\sqrt [3]{d} \log \left (c+d x^3\right )}{6 (b c-a d)^{4/3}}+\frac {\sqrt [3]{d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 (b c-a d)^{4/3}}+\frac {\sqrt [3]{d} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} (b c-a d)^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 56

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, Simp
[Log[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(
1/3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && Ne
gQ[(b*c - a*d)/b]

Rule 204

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

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 617

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]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{4/3} (c+d x)} \, dx,x,x^3\right )\\ &=-\frac {1}{(b c-a d) \sqrt [3]{a+b x^3}}-\frac {d \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a+b x} (c+d x)} \, dx,x,x^3\right )}{3 (b c-a d)}\\ &=-\frac {1}{(b c-a d) \sqrt [3]{a+b x^3}}-\frac {\sqrt [3]{d} \log \left (c+d x^3\right )}{6 (b c-a d)^{4/3}}+\frac {\sqrt [3]{d} \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt [3]{b c-a d}}{\sqrt [3]{d}}+x} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2 (b c-a d)^{4/3}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {(b c-a d)^{2/3}}{d^{2/3}}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{d}}+x^2} \, dx,x,\sqrt [3]{a+b x^3}\right )}{2 (b c-a d)}\\ &=-\frac {1}{(b c-a d) \sqrt [3]{a+b x^3}}-\frac {\sqrt [3]{d} \log \left (c+d x^3\right )}{6 (b c-a d)^{4/3}}+\frac {\sqrt [3]{d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 (b c-a d)^{4/3}}-\frac {\sqrt [3]{d} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}\right )}{(b c-a d)^{4/3}}\\ &=-\frac {1}{(b c-a d) \sqrt [3]{a+b x^3}}+\frac {\sqrt [3]{d} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt [3]{b c-a d}}}{\sqrt {3}}\right )}{\sqrt {3} (b c-a d)^{4/3}}-\frac {\sqrt [3]{d} \log \left (c+d x^3\right )}{6 (b c-a d)^{4/3}}+\frac {\sqrt [3]{d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{2 (b c-a d)^{4/3}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 50, normalized size = 0.30 \begin {gather*} -\frac {\, _2F_1\left (-\frac {1}{3},1;\frac {2}{3};\frac {d \left (b x^3+a\right )}{a d-b c}\right )}{\sqrt [3]{a+b x^3} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Hypergeometric2F1[-1/3, 1, 2/3, (d*(a + b*x^3))/(-(b*c) + a*d)]/((b*c - a*d)*(a + b*x^3)^(1/3)))

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IntegrateAlgebraic [A]  time = 0.24, size = 223, normalized size = 1.34 \begin {gather*} -\frac {\sqrt [3]{d} \log \left (-\sqrt [3]{d} \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+d^{2/3} \left (a+b x^3\right )^{2/3}\right )}{6 (b c-a d)^{4/3}}-\frac {1}{\sqrt [3]{a+b x^3} (b c-a d)}+\frac {\sqrt [3]{d} \log \left (\sqrt [3]{b c-a d}+\sqrt [3]{d} \sqrt [3]{a+b x^3}\right )}{3 (b c-a d)^{4/3}}+\frac {\sqrt [3]{d} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x^3}}{\sqrt {3} \sqrt [3]{b c-a d}}\right )}{\sqrt {3} (b c-a d)^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.67, size = 262, normalized size = 1.57 \begin {gather*} -\frac {2 \, \sqrt {3} {\left (b x^{3} + a\right )} \left (-\frac {d}{b c - a d}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {d}{b c - a d}\right )^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - {\left (b x^{3} + a\right )} \left (-\frac {d}{b c - a d}\right )^{\frac {1}{3}} \log \left (-{\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )} \left (-\frac {d}{b c - a d}\right )^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}} d - {\left (b c - a d\right )} \left (-\frac {d}{b c - a d}\right )^{\frac {1}{3}}\right ) + 2 \, {\left (b x^{3} + a\right )} \left (-\frac {d}{b c - a d}\right )^{\frac {1}{3}} \log \left ({\left (b c - a d\right )} \left (-\frac {d}{b c - a d}\right )^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} d\right ) + 6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{6 \, {\left ({\left (b^{2} c - a b d\right )} x^{3} + a b c - a^{2} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

-1/6*(2*sqrt(3)*(b*x^3 + a)*(-d/(b*c - a*d))^(1/3)*arctan(2/3*sqrt(3)*(b*x^3 + a)^(1/3)*(-d/(b*c - a*d))^(1/3)
+ 1/3*sqrt(3)) - (b*x^3 + a)*(-d/(b*c - a*d))^(1/3)*log(-(b*x^3 + a)^(1/3)*(b*c - a*d)*(-d/(b*c - a*d))^(2/3)
+ (b*x^3 + a)^(2/3)*d - (b*c - a*d)*(-d/(b*c - a*d))^(1/3)) + 2*(b*x^3 + a)*(-d/(b*c - a*d))^(1/3)*log((b*c -
a*d)*(-d/(b*c - a*d))^(2/3) + (b*x^3 + a)^(1/3)*d) + 6*(b*x^3 + a)^(2/3))/((b^2*c - a*b*d)*x^3 + a*b*c - a^2*
d)

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giac [B]  time = 0.26, size = 285, normalized size = 1.71 \begin {gather*} \frac {d \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}} \log \left ({\left | {\left (b x^{3} + a\right )}^{\frac {1}{3}} - \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}} + \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b^{2} c^{2} d - 2 \, \sqrt {3} a b c d^{2} + \sqrt {3} a^{2} d^{3}} - \frac {{\left (-b c d^{2} + a d^{3}\right )}^{\frac {2}{3}} \log \left ({\left (b x^{3} + a\right )}^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-\frac {b c - a d}{d}\right )^{\frac {1}{3}} + \left (-\frac {b c - a d}{d}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )}} - \frac {1}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b c - a d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*d*(-(b*c - a*d)/d)^(2/3)*log(abs((b*x^3 + a)^(1/3) - (-(b*c - a*d)/d)^(1/3)))/(b^2*c^2 - 2*a*b*c*d + a^2*d
^2) + (-b*c*d^2 + a*d^3)^(2/3)*arctan(1/3*sqrt(3)*(2*(b*x^3 + a)^(1/3) + (-(b*c - a*d)/d)^(1/3))/(-(b*c - a*d)
/d)^(1/3))/(sqrt(3)*b^2*c^2*d - 2*sqrt(3)*a*b*c*d^2 + sqrt(3)*a^2*d^3) - 1/6*(-b*c*d^2 + a*d^3)^(2/3)*log((b*x
^3 + a)^(2/3) + (b*x^3 + a)^(1/3)*(-(b*c - a*d)/d)^(1/3) + (-(b*c - a*d)/d)^(2/3))/(b^2*c^2*d - 2*a*b*c*d^2 +
a^2*d^3) - 1/((b*x^3 + a)^(1/3)*(b*c - a*d))

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maple [F]  time = 0.61, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (d \,x^{3}+c \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c positive or negative?

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mupad [B]  time = 4.92, size = 389, normalized size = 2.33 \begin {gather*} \frac {1}{{\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d-b\,c\right )}+\frac {d^{1/3}\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d^4-b\,c\,d^3\right )-\frac {d^{2/3}\,\left (9\,a^4\,d^6-36\,a^3\,b\,c\,d^5+54\,a^2\,b^2\,c^2\,d^4-36\,a\,b^3\,c^3\,d^3+9\,b^4\,c^4\,d^2\right )}{9\,{\left (a\,d-b\,c\right )}^{8/3}}\right )}{3\,{\left (a\,d-b\,c\right )}^{4/3}}-\frac {d^{1/3}\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d^4-b\,c\,d^3\right )-\frac {d^{2/3}\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,\left (9\,a^4\,d^6-36\,a^3\,b\,c\,d^5+54\,a^2\,b^2\,c^2\,d^4-36\,a\,b^3\,c^3\,d^3+9\,b^4\,c^4\,d^2\right )}{9\,{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,{\left (a\,d-b\,c\right )}^{4/3}}+\frac {d^{1/3}\,\ln \left ({\left (b\,x^3+a\right )}^{1/3}\,\left (a\,d^4-b\,c\,d^3\right )-\frac {d^{2/3}\,{\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}^2\,\left (9\,a^4\,d^6-36\,a^3\,b\,c\,d^5+54\,a^2\,b^2\,c^2\,d^4-36\,a\,b^3\,c^3\,d^3+9\,b^4\,c^4\,d^2\right )}{{\left (a\,d-b\,c\right )}^{8/3}}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{{\left (a\,d-b\,c\right )}^{4/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/((a + b*x^3)^(1/3)*(a*d - b*c)) + (d^(1/3)*log((a + b*x^3)^(1/3)*(a*d^4 - b*c*d^3) - (d^(2/3)*(9*a^4*d^6 + 9
*b^4*c^4*d^2 - 36*a*b^3*c^3*d^3 + 54*a^2*b^2*c^2*d^4 - 36*a^3*b*c*d^5))/(9*(a*d - b*c)^(8/3))))/(3*(a*d - b*c)
^(4/3)) - (d^(1/3)*log((a + b*x^3)^(1/3)*(a*d^4 - b*c*d^3) - (d^(2/3)*((3^(1/2)*1i)/2 + 1/2)^2*(9*a^4*d^6 + 9*
b^4*c^4*d^2 - 36*a*b^3*c^3*d^3 + 54*a^2*b^2*c^2*d^4 - 36*a^3*b*c*d^5))/(9*(a*d - b*c)^(8/3)))*((3^(1/2)*1i)/2
+ 1/2))/(3*(a*d - b*c)^(4/3)) + (d^(1/3)*log((a + b*x^3)^(1/3)*(a*d^4 - b*c*d^3) - (d^(2/3)*((3^(1/2)*1i)/6 -
1/6)^2*(9*a^4*d^6 + 9*b^4*c^4*d^2 - 36*a*b^3*c^3*d^3 + 54*a^2*b^2*c^2*d^4 - 36*a^3*b*c*d^5))/(a*d - b*c)^(8/3)
)*((3^(1/2)*1i)/6 - 1/6))/(a*d - b*c)^(4/3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\left (a + b x^{3}\right )^{\frac {4}{3}} \left (c + d x^{3}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/((a + b*x**3)**(4/3)*(c + d*x**3)), x)

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