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

Optimal. Leaf size=144 \[ \frac {(b c-a d) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{2/3} d^{4/3}}-\frac {(b c-a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{2/3} d^{4/3}}+\frac {(b c-a d) \tan ^{-1}\left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{2/3} d^{4/3}}+\frac {b x}{d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {388, 200, 31, 634, 617, 204, 628} \[ \frac {(b c-a d) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{2/3} d^{4/3}}-\frac {(b c-a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{2/3} d^{4/3}}+\frac {(b c-a d) \tan ^{-1}\left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{2/3} d^{4/3}}+\frac {b x}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)/(c + d*x^3),x]

[Out]

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

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Mathematica [A]  time = 0.12, size = 128, normalized size = 0.89 \[ \frac {(b c-a d) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )-2 (b c-a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )+2 \sqrt {3} (b c-a d) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} x}{\sqrt [3]{c}}}{\sqrt {3}}\right )+6 b c^{2/3} \sqrt [3]{d} x}{6 c^{2/3} d^{4/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)/(c + d*x^3),x]

[Out]

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

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fricas [A]  time = 0.45, size = 369, normalized size = 2.56 \[ \left [\frac {6 \, b c^{2} d x - 3 \, \sqrt {\frac {1}{3}} {\left (b c^{2} d - a c d^{2}\right )} \sqrt {-\frac {\left (c^{2} d\right )^{\frac {1}{3}}}{d}} \log \left (\frac {2 \, c d x^{3} - 3 \, \left (c^{2} d\right )^{\frac {1}{3}} c x - c^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, c d x^{2} + \left (c^{2} d\right )^{\frac {2}{3}} x - \left (c^{2} d\right )^{\frac {1}{3}} c\right )} \sqrt {-\frac {\left (c^{2} d\right )^{\frac {1}{3}}}{d}}}{d x^{3} + c}\right ) + \left (c^{2} d\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (c d x^{2} - \left (c^{2} d\right )^{\frac {2}{3}} x + \left (c^{2} d\right )^{\frac {1}{3}} c\right ) - 2 \, \left (c^{2} d\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (c d x + \left (c^{2} d\right )^{\frac {2}{3}}\right )}{6 \, c^{2} d^{2}}, \frac {6 \, b c^{2} d x - 6 \, \sqrt {\frac {1}{3}} {\left (b c^{2} d - a c d^{2}\right )} \sqrt {\frac {\left (c^{2} d\right )^{\frac {1}{3}}}{d}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (c^{2} d\right )^{\frac {2}{3}} x - \left (c^{2} d\right )^{\frac {1}{3}} c\right )} \sqrt {\frac {\left (c^{2} d\right )^{\frac {1}{3}}}{d}}}{c^{2}}\right ) + \left (c^{2} d\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (c d x^{2} - \left (c^{2} d\right )^{\frac {2}{3}} x + \left (c^{2} d\right )^{\frac {1}{3}} c\right ) - 2 \, \left (c^{2} d\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (c d x + \left (c^{2} d\right )^{\frac {2}{3}}\right )}{6 \, c^{2} d^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 133, normalized size = 0.92 \[ \frac {\sqrt {3} {\left (b c - a d\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{3 \, \left (-c d^{2}\right )^{\frac {2}{3}}} + \frac {{\left (b c - a d\right )} \log \left (x^{2} + x \left (-\frac {c}{d}\right )^{\frac {1}{3}} + \left (-\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \, \left (-c d^{2}\right )^{\frac {2}{3}}} + \frac {b x}{d} + \frac {{\left (b c - a d\right )} \left (-\frac {c}{d}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {c}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 195, normalized size = 1.35 \[ \frac {\sqrt {3}\, a \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {c}{d}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {c}{d}\right )^{\frac {2}{3}} d}+\frac {a \ln \left (x +\left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {c}{d}\right )^{\frac {2}{3}} d}-\frac {a \ln \left (x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{3}} x +\left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {c}{d}\right )^{\frac {2}{3}} d}-\frac {\sqrt {3}\, b c \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {c}{d}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {c}{d}\right )^{\frac {2}{3}} d^{2}}-\frac {b c \ln \left (x +\left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {c}{d}\right )^{\frac {2}{3}} d^{2}}+\frac {b c \ln \left (x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{3}} x +\left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {c}{d}\right )^{\frac {2}{3}} d^{2}}+\frac {b x}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.28, size = 128, normalized size = 0.89 \[ \frac {b x}{d} - \frac {\sqrt {3} {\left (b c - a d\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{3 \, d^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}}} + \frac {{\left (b c - a d\right )} \log \left (x^{2} - x \left (\frac {c}{d}\right )^{\frac {1}{3}} + \left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \, d^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}}} - \frac {{\left (b c - a d\right )} \log \left (x + \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 \, d^{2} \left (\frac {c}{d}\right )^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.38, size = 123, normalized size = 0.85 \[ \frac {b\,x}{d}+\frac {\ln \left (d^{1/3}\,x+c^{1/3}\right )\,\left (a\,d-b\,c\right )}{3\,c^{2/3}\,d^{4/3}}-\frac {\ln \left (c^{1/3}-2\,d^{1/3}\,x+\sqrt {3}\,c^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (a\,d-b\,c\right )}{3\,c^{2/3}\,d^{4/3}}+\frac {\ln \left (2\,d^{1/3}\,x-c^{1/3}+\sqrt {3}\,c^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (a\,d-b\,c\right )}{3\,c^{2/3}\,d^{4/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)/(c + d*x^3),x)

[Out]

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

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sympy [A]  time = 0.42, size = 71, normalized size = 0.49 \[ \frac {b x}{d} + \operatorname {RootSum} {\left (27 t^{3} c^{2} d^{4} - a^{3} d^{3} + 3 a^{2} b c d^{2} - 3 a b^{2} c^{2} d + b^{3} c^{3}, \left (t \mapsto t \log {\left (\frac {3 t c d}{a d - b c} + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x/d + RootSum(27*_t**3*c**2*d**4 - 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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