3.757 \(\int \frac {1}{(a+b x^3)^{4/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=179 \[ -\frac {d \log \left (c+d x^3\right )}{6 c^{2/3} (b c-a d)^{4/3}}+\frac {d \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} (b c-a d)^{4/3}}-\frac {d \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} (b c-a d)^{4/3}}+\frac {b x}{a \sqrt [3]{a+b x^3} (b c-a d)} \]

[Out]

b*x/a/(-a*d+b*c)/(b*x^3+a)^(1/3)-1/6*d*ln(d*x^3+c)/c^(2/3)/(-a*d+b*c)^(4/3)+1/2*d*ln((-a*d+b*c)^(1/3)*x/c^(1/3
)-(b*x^3+a)^(1/3))/c^(2/3)/(-a*d+b*c)^(4/3)-1/3*d*arctan(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/3)/(b*x^3+a)^(1/3))*
3^(1/2))/c^(2/3)/(-a*d+b*c)^(4/3)*3^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 238, normalized size of antiderivative = 1.33, number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {382, 377, 200, 31, 634, 617, 204, 628} \[ \frac {d \log \left (\sqrt [3]{c}-\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}\right )}{3 c^{2/3} (b c-a d)^{4/3}}-\frac {d \log \left (\frac {x^2 (b c-a d)^{2/3}}{\left (a+b x^3\right )^{2/3}}+\frac {\sqrt [3]{c} x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+c^{2/3}\right )}{6 c^{2/3} (b c-a d)^{4/3}}-\frac {d \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{a+b x^3}}+\sqrt [3]{c}}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{2/3} (b c-a d)^{4/3}}+\frac {b x}{a \sqrt [3]{a+b x^3} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

(b*x)/(a*(b*c - a*d)*(a + b*x^3)^(1/3)) - (d*ArcTan[(c^(1/3) + (2*(b*c - a*d)^(1/3)*x)/(a + b*x^3)^(1/3))/(Sqr
t[3]*c^(1/3))])/(Sqrt[3]*c^(2/3)*(b*c - a*d)^(4/3)) + (d*Log[c^(1/3) - ((b*c - a*d)^(1/3)*x)/(a + b*x^3)^(1/3)
])/(3*c^(2/3)*(b*c - a*d)^(4/3)) - (d*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^(2/3)*(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 200

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 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 377

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

Rule 382

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

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

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Mathematica [C]  time = 0.12, size = 256, normalized size = 1.43 \[ -\frac {28 c^3 \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-28 c^3 \left (a+b x^3\right )^2+21 c^2 d x^3 \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-21 c^2 d x^3 \left (a+b x^3\right )^2+3 d x^9 (b c-a d)^2 \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+3 c x^6 (b c-a d)^2 \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )}{7 c^3 x^2 \left (a+b x^3\right )^{7/3} (a d-b c)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

-1/7*(-28*c^3*(a + b*x^3)^2 - 21*c^2*d*x^3*(a + b*x^3)^2 + 28*c^3*(a + b*x^3)^2*Hypergeometric2F1[1/3, 1, 4/3,
((b*c - a*d)*x^3)/(c*(a + b*x^3))] + 21*c^2*d*x^3*(a + b*x^3)^2*Hypergeometric2F1[1/3, 1, 4/3, ((b*c - a*d)*x
^3)/(c*(a + b*x^3))] + 3*c*(b*c - a*d)^2*x^6*Hypergeometric2F1[2, 7/3, 10/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))
] + 3*d*(b*c - a*d)^2*x^9*Hypergeometric2F1[2, 7/3, 10/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))])/(c^3*(-(b*c) + a
*d)*x^2*(a + b*x^3)^(7/3))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^3+a)^(4/3)/(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 \[ \int \frac {1}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(4/3)*(d*x^3 + c)), x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (d \,x^{3}+c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^3+a)^(4/3)/(d*x^3+c),x)

[Out]

int(1/(b*x^3+a)^(4/3)/(d*x^3+c),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a)^(4/3)*(d*x^3 + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (b\,x^3+a\right )}^{4/3}\,\left (d\,x^3+c\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^3)^(4/3)*(c + d*x^3)),x)

[Out]

int(1/((a + b*x^3)^(4/3)*(c + d*x^3)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x^{3}\right )^{\frac {4}{3}} \left (c + d x^{3}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**3+a)**(4/3)/(d*x**3+c),x)

[Out]

Integral(1/((a + b*x**3)**(4/3)*(c + d*x**3)), x)

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