3.5.78 \(\int \frac {1}{x^5 (a+b x^3)^{2/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=215 \[ \frac {\sqrt [3]{a+b x^3} (4 a d+3 b c)}{4 a^2 c^2 x}+\frac {d^2 \log \left (c+d x^3\right )}{6 c^{7/3} (b c-a d)^{2/3}}-\frac {d^2 \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{7/3} (b c-a d)^{2/3}}-\frac {d^2 \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{7/3} (b c-a d)^{2/3}}-\frac {\sqrt [3]{a+b x^3}}{4 a c x^4} \]

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Rubi [A]  time = 0.30, antiderivative size = 269, normalized size of antiderivative = 1.25, number of steps used = 9, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {494, 461, 292, 31, 634, 617, 204, 628} \begin {gather*} \frac {\sqrt [3]{a+b x^3} (a d+b c)}{a^2 c^2 x}-\frac {\left (a+b x^3\right )^{4/3}}{4 a^2 c x^4}-\frac {d^2 \log \left (\sqrt [3]{c}-\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}\right )}{3 c^{7/3} (b c-a d)^{2/3}}+\frac {d^2 \log \left (\frac {x^2 (b c-a d)^{2/3}}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+c^{2/3}\right )}{6 c^{7/3} (b c-a d)^{2/3}}-\frac {d^2 \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+\sqrt [3]{c}}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{7/3} (b c-a d)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^3)^(2/3)*(c + d*x^3)),x]

[Out]

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

Rule 31

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

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 292

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 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 494

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[(k*a^(p + (m + 1)/n))/n, Subst[Int[(x^((k*(m + 1))/n - 1)*(c - (b*c - a*d)*x^k)^q)/(1 - b*x^k)^(p
+ q + (m + 1)/n + 1), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 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]

Rule 628

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 634

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 {1}{x^5 \left (a+b x^3\right )^{2/3} \left (c+d x^3\right )} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-b x^3\right )^2}{x^5 \left (c-(b c-a d) x^3\right )} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{a^2}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{c x^5}+\frac {-b c-a d}{c^2 x^2}+\frac {a^2 d^2 x}{c^2 \left (c-(b c-a d) x^3\right )}\right ) \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{a^2}\\ &=\frac {(b c+a d) \sqrt [3]{a+b x^3}}{a^2 c^2 x}-\frac {\left (a+b x^3\right )^{4/3}}{4 a^2 c x^4}+\frac {d^2 \operatorname {Subst}\left (\int \frac {x}{c-(b c-a d) x^3} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{c^2}\\ &=\frac {(b c+a d) \sqrt [3]{a+b x^3}}{a^2 c^2 x}-\frac {\left (a+b x^3\right )^{4/3}}{4 a^2 c x^4}+\frac {d^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{c}-\sqrt [3]{b c-a d} x} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{3 c^{7/3} \sqrt [3]{b c-a d}}-\frac {d^2 \operatorname {Subst}\left (\int \frac {\sqrt [3]{c}-\sqrt [3]{b c-a d} x}{c^{2/3}+\sqrt [3]{c} \sqrt [3]{b c-a d} x+(b c-a d)^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{3 c^{7/3} \sqrt [3]{b c-a d}}\\ &=\frac {(b c+a d) \sqrt [3]{a+b x^3}}{a^2 c^2 x}-\frac {\left (a+b x^3\right )^{4/3}}{4 a^2 c x^4}-\frac {d^2 \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{3 c^{7/3} (b c-a d)^{2/3}}+\frac {d^2 \operatorname {Subst}\left (\int \frac {\sqrt [3]{c} \sqrt [3]{b c-a d}+2 (b c-a d)^{2/3} x}{c^{2/3}+\sqrt [3]{c} \sqrt [3]{b c-a d} x+(b c-a d)^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{6 c^{7/3} (b c-a d)^{2/3}}-\frac {d^2 \operatorname {Subst}\left (\int \frac {1}{c^{2/3}+\sqrt [3]{c} \sqrt [3]{b c-a d} x+(b c-a d)^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{a+b x^3}}\right )}{2 c^2 \sqrt [3]{b c-a d}}\\ &=\frac {(b c+a d) \sqrt [3]{a+b x^3}}{a^2 c^2 x}-\frac {\left (a+b x^3\right )^{4/3}}{4 a^2 c x^4}-\frac {d^2 \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{3 c^{7/3} (b c-a d)^{2/3}}+\frac {d^2 \log \left (c^{2/3}+\frac {(b c-a d)^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{6 c^{7/3} (b c-a d)^{2/3}}+\frac {d^2 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{c^{7/3} (b c-a d)^{2/3}}\\ &=\frac {(b c+a d) \sqrt [3]{a+b x^3}}{a^2 c^2 x}-\frac {\left (a+b x^3\right )^{4/3}}{4 a^2 c x^4}-\frac {d^2 \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^{7/3} (b c-a d)^{2/3}}-\frac {d^2 \log \left (\sqrt [3]{c}-\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{3 c^{7/3} (b c-a d)^{2/3}}+\frac {d^2 \log \left (c^{2/3}+\frac {(b c-a d)^{2/3} x^2}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{c} \sqrt [3]{b c-a d} x}{\sqrt [3]{a+b x^3}}\right )}{6 c^{7/3} (b c-a d)^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 1.89, size = 267, normalized size = 1.24 \begin {gather*} -\frac {-81 x^3 \left (c+d x^3\right )^2 (b c-a d) \, _4F_3\left (\frac {2}{3},2,2,2;1,1,\frac {8}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+216 d x^6 \left (c+d x^3\right ) (a d-b c) \, _3F_2\left (\frac {2}{3},2,2;1,\frac {8}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-5 \left (2 c \left (a+b x^3\right ) \left (c^2+10 c d x^3+9 d^2 x^6\right )+\left (a \left (-8 c^3+17 c^2 d x^3+46 c d^2 x^6+9 d^3 x^9\right )+3 b c x^3 \left (-3 c^2+2 c d x^3+9 d^2 x^6\right )\right ) \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )\right )}{120 c^4 x^4 \left (a+b x^3\right )^{5/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(x^5*(a + b*x^3)^(2/3)*(c + d*x^3)),x]

[Out]

-1/120*(-5*(2*c*(a + b*x^3)*(c^2 + 10*c*d*x^3 + 9*d^2*x^6) + (3*b*c*x^3*(-3*c^2 + 2*c*d*x^3 + 9*d^2*x^6) + a*(
-8*c^3 + 17*c^2*d*x^3 + 46*c*d^2*x^6 + 9*d^3*x^9))*Hypergeometric2F1[2/3, 1, 5/3, ((b*c - a*d)*x^3)/(c*(a + b*
x^3))]) + 216*d*(-(b*c) + a*d)*x^6*(c + d*x^3)*HypergeometricPFQ[{2/3, 2, 2}, {1, 8/3}, ((b*c - a*d)*x^3)/(c*(
a + b*x^3))] - 81*(b*c - a*d)*x^3*(c + d*x^3)^2*HypergeometricPFQ[{2/3, 2, 2, 2}, {1, 1, 8/3}, ((b*c - a*d)*x^
3)/(c*(a + b*x^3))])/(c^4*x^4*(a + b*x^3)^(5/3))

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IntegrateAlgebraic [C]  time = 2.41, size = 377, normalized size = 1.75 \begin {gather*} \frac {\sqrt [3]{a+b x^3} \left (-a c+4 a d x^3+3 b c x^3\right )}{4 a^2 c^2 x^4}+\frac {\left (d^2-i \sqrt {3} d^2\right ) \log \left (2 x \sqrt [3]{b c-a d}+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{6 c^{7/3} (b c-a d)^{2/3}}+\frac {\sqrt {\frac {1}{6} \left (-1-i \sqrt {3}\right )} d^2 \tan ^{-1}\left (\frac {3 x \sqrt [3]{b c-a d}}{\sqrt {3} x \sqrt [3]{b c-a d}-\sqrt {3} \sqrt [3]{c} \sqrt [3]{a+b x^3}-3 i \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{c^{7/3} (b c-a d)^{2/3}}+\frac {i \left (\sqrt {3} d^2+i d^2\right ) \log \left (\left (\sqrt {3}+i\right ) c^{2/3} \left (a+b x^3\right )^{2/3}+\sqrt [3]{c} \left (-\sqrt {3} x+i x\right ) \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}-2 i x^2 (b c-a d)^{2/3}\right )}{12 c^{7/3} (b c-a d)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^5*(a + b*x^3)^(2/3)*(c + d*x^3)),x]

[Out]

((a + b*x^3)^(1/3)*(-(a*c) + 3*b*c*x^3 + 4*a*d*x^3))/(4*a^2*c^2*x^4) + (Sqrt[(-1 - I*Sqrt[3])/6]*d^2*ArcTan[(3
*(b*c - a*d)^(1/3)*x)/(Sqrt[3]*(b*c - a*d)^(1/3)*x - (3*I)*c^(1/3)*(a + b*x^3)^(1/3) - Sqrt[3]*c^(1/3)*(a + b*
x^3)^(1/3))])/(c^(7/3)*(b*c - a*d)^(2/3)) + ((d^2 - I*Sqrt[3]*d^2)*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])
*c^(1/3)*(a + b*x^3)^(1/3)])/(6*c^(7/3)*(b*c - a*d)^(2/3)) + ((I/12)*(I*d^2 + Sqrt[3]*d^2)*Log[(-2*I)*(b*c - a
*d)^(2/3)*x^2 + c^(1/3)*(b*c - a*d)^(1/3)*(I*x - Sqrt[3]*x)*(a + b*x^3)^(1/3) + (I + Sqrt[3])*c^(2/3)*(a + b*x
^3)^(2/3)])/(c^(7/3)*(b*c - a*d)^(2/3))

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fricas [F(-1)]  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(1/x^5/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )} x^{5}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(2/3)*(d*x^3 + c)*x^5), x)

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maple [F]  time = 0.61, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (d \,x^{3}+c \right ) x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(b*x^3+a)^(2/3)/(d*x^3+c),x)

[Out]

int(1/x^5/(b*x^3+a)^(2/3)/(d*x^3+c),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )} x^{5}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^3+a)^(2/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a)^(2/3)*(d*x^3 + c)*x^5), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^5\,{\left (b\,x^3+a\right )}^{2/3}\,\left (d\,x^3+c\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a + b*x^3)^(2/3)*(c + d*x^3)),x)

[Out]

int(1/(x^5*(a + b*x^3)^(2/3)*(c + d*x^3)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{5} \left (a + b x^{3}\right )^{\frac {2}{3}} \left (c + d x^{3}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(b*x**3+a)**(2/3)/(d*x**3+c),x)

[Out]

Integral(1/(x**5*(a + b*x**3)**(2/3)*(c + d*x**3)), x)

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