3.275 \(\int \frac {a+\frac {b}{x^3}}{c+\frac {d}{x^3}} \, dx\)

Optimal. Leaf size=145 \[ -\frac {(b c-a d) \log \left (c^{2/3} x^2-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3}\right )}{6 c^{4/3} d^{2/3}}+\frac {(b c-a d) \log \left (\sqrt [3]{c} x+\sqrt [3]{d}\right )}{3 c^{4/3} d^{2/3}}-\frac {(b c-a d) \tan ^{-1}\left (\frac {\sqrt [3]{d}-2 \sqrt [3]{c} x}{\sqrt {3} \sqrt [3]{d}}\right )}{\sqrt {3} c^{4/3} d^{2/3}}+\frac {a x}{c} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {374, 388, 200, 31, 634, 617, 204, 628} \[ -\frac {(b c-a d) \log \left (c^{2/3} x^2-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3}\right )}{6 c^{4/3} d^{2/3}}+\frac {(b c-a d) \log \left (\sqrt [3]{c} x+\sqrt [3]{d}\right )}{3 c^{4/3} d^{2/3}}-\frac {(b c-a d) \tan ^{-1}\left (\frac {\sqrt [3]{d}-2 \sqrt [3]{c} x}{\sqrt {3} \sqrt [3]{d}}\right )}{\sqrt {3} c^{4/3} d^{2/3}}+\frac {a x}{c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^3)/(c + d/x^3),x]

[Out]

(a*x)/c - ((b*c - a*d)*ArcTan[(d^(1/3) - 2*c^(1/3)*x)/(Sqrt[3]*d^(1/3))])/(Sqrt[3]*c^(4/3)*d^(2/3)) + ((b*c -
a*d)*Log[d^(1/3) + c^(1/3)*x])/(3*c^(4/3)*d^(2/3)) - ((b*c - a*d)*Log[d^(2/3) - c^(1/3)*d^(1/3)*x + c^(2/3)*x^
2])/(6*c^(4/3)*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 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 374

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

Rule 388

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

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

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Mathematica [A]  time = 0.10, size = 129, normalized size = 0.89 \[ \frac {-(b c-a d) \log \left (c^{2/3} x^2-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3}\right )+2 (b c-a d) \log \left (\sqrt [3]{c} x+\sqrt [3]{d}\right )-2 \sqrt {3} (b c-a d) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{c} x}{\sqrt [3]{d}}}{\sqrt {3}}\right )+6 a \sqrt [3]{c} d^{2/3} x}{6 c^{4/3} d^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^3)/(c + d/x^3),x]

[Out]

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

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fricas [A]  time = 0.66, size = 390, normalized size = 2.69 \[ \left [\frac {6 \, a c d^{2} x - 3 \, \sqrt {\frac {1}{3}} {\left (b c^{2} d - a c d^{2}\right )} \sqrt {\frac {\left (-c d^{2}\right )^{\frac {1}{3}}}{c}} \log \left (\frac {2 \, c d x^{3} + 3 \, \left (-c d^{2}\right )^{\frac {1}{3}} d x - d^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (2 \, c d x^{2} + \left (-c d^{2}\right )^{\frac {2}{3}} x + \left (-c d^{2}\right )^{\frac {1}{3}} d\right )} \sqrt {\frac {\left (-c d^{2}\right )^{\frac {1}{3}}}{c}}}{c x^{3} + d}\right ) - \left (-c d^{2}\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (c d x^{2} - \left (-c d^{2}\right )^{\frac {2}{3}} x - \left (-c d^{2}\right )^{\frac {1}{3}} d\right ) + 2 \, \left (-c d^{2}\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (c d x + \left (-c d^{2}\right )^{\frac {2}{3}}\right )}{6 \, c^{2} d^{2}}, \frac {6 \, a c d^{2} x + 6 \, \sqrt {\frac {1}{3}} {\left (b c^{2} d - a c d^{2}\right )} \sqrt {-\frac {\left (-c d^{2}\right )^{\frac {1}{3}}}{c}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (-c d^{2}\right )^{\frac {2}{3}} x + \left (-c d^{2}\right )^{\frac {1}{3}} d\right )} \sqrt {-\frac {\left (-c d^{2}\right )^{\frac {1}{3}}}{c}}}{d^{2}}\right ) - \left (-c d^{2}\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (c d x^{2} - \left (-c d^{2}\right )^{\frac {2}{3}} x - \left (-c d^{2}\right )^{\frac {1}{3}} d\right ) + 2 \, \left (-c d^{2}\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (c d x + \left (-c d^{2}\right )^{\frac {2}{3}}\right )}{6 \, c^{2} d^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 133, normalized size = 0.92 \[ -\frac {\sqrt {3} {\left (b c - a d\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {d}{c}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {d}{c}\right )^{\frac {1}{3}}}\right )}{3 \, \left (-c^{2} d\right )^{\frac {2}{3}}} - \frac {{\left (b c - a d\right )} \log \left (x^{2} + x \left (-\frac {d}{c}\right )^{\frac {1}{3}} + \left (-\frac {d}{c}\right )^{\frac {2}{3}}\right )}{6 \, \left (-c^{2} d\right )^{\frac {2}{3}}} + \frac {a x}{c} - \frac {{\left (b c - a d\right )} \left (-\frac {d}{c}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {d}{c}\right )^{\frac {1}{3}} \right |}\right )}{3 \, c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 195, normalized size = 1.34 \[ \frac {a x}{c}-\frac {\sqrt {3}\, a d \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {d}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {d}{c}\right )^{\frac {2}{3}} c^{2}}-\frac {a d \ln \left (x +\left (\frac {d}{c}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {d}{c}\right )^{\frac {2}{3}} c^{2}}+\frac {a d \ln \left (x^{2}-\left (\frac {d}{c}\right )^{\frac {1}{3}} x +\left (\frac {d}{c}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {d}{c}\right )^{\frac {2}{3}} c^{2}}+\frac {\sqrt {3}\, b \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {d}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {d}{c}\right )^{\frac {2}{3}} c}+\frac {b \ln \left (x +\left (\frac {d}{c}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {d}{c}\right )^{\frac {2}{3}} c}-\frac {b \ln \left (x^{2}-\left (\frac {d}{c}\right )^{\frac {1}{3}} x +\left (\frac {d}{c}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {d}{c}\right )^{\frac {2}{3}} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.29, size = 128, normalized size = 0.88 \[ \frac {a x}{c} + \frac {\sqrt {3} {\left (b c - a d\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {d}{c}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {d}{c}\right )^{\frac {1}{3}}}\right )}{3 \, c^{2} \left (\frac {d}{c}\right )^{\frac {2}{3}}} - \frac {{\left (b c - a d\right )} \log \left (x^{2} - x \left (\frac {d}{c}\right )^{\frac {1}{3}} + \left (\frac {d}{c}\right )^{\frac {2}{3}}\right )}{6 \, c^{2} \left (\frac {d}{c}\right )^{\frac {2}{3}}} + \frac {{\left (b c - a d\right )} \log \left (x + \left (\frac {d}{c}\right )^{\frac {1}{3}}\right )}{3 \, c^{2} \left (\frac {d}{c}\right )^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.27, size = 123, normalized size = 0.85 \[ \frac {a\,x}{c}-\frac {\ln \left (c^{1/3}\,x+d^{1/3}\right )\,\left (a\,d-b\,c\right )}{3\,c^{4/3}\,d^{2/3}}+\frac {\ln \left (d^{1/3}-2\,c^{1/3}\,x+\sqrt {3}\,d^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (a\,d-b\,c\right )}{3\,c^{4/3}\,d^{2/3}}-\frac {\ln \left (2\,c^{1/3}\,x-d^{1/3}+\sqrt {3}\,d^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (a\,d-b\,c\right )}{3\,c^{4/3}\,d^{2/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^3)/(c + d/x^3),x)

[Out]

(a*x)/c - (log(c^(1/3)*x + d^(1/3))*(a*d - b*c))/(3*c^(4/3)*d^(2/3)) + (log(3^(1/2)*d^(1/3)*1i - 2*c^(1/3)*x +
d^(1/3))*((3^(1/2)*1i)/2 + 1/2)*(a*d - b*c))/(3*c^(4/3)*d^(2/3)) - (log(3^(1/2)*d^(1/3)*1i + 2*c^(1/3)*x - d^
(1/3))*((3^(1/2)*1i)/2 - 1/2)*(a*d - b*c))/(3*c^(4/3)*d^(2/3))

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sympy [A]  time = 0.45, size = 71, normalized size = 0.49 \[ \frac {a x}{c} + \operatorname {RootSum} {\left (27 t^{3} c^{4} d^{2} + a^{3} d^{3} - 3 a^{2} b c d^{2} + 3 a b^{2} c^{2} d - b^{3} c^{3}, \left (t \mapsto t \log {\left (- \frac {3 t c d}{a d - b c} + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x/c + RootSum(27*_t**3*c**4*d**2 + a**3*d**3 - 3*a**2*b*c*d**2 + 3*a*b**2*c**2*d - b**3*c**3, Lambda(_t, _t*
log(-3*_t*c*d/(a*d - b*c) + x)))

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