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

Optimal. Leaf size=169 \[ -\frac{(2 a d+b c) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{18 c^{5/3} d^{4/3}}+\frac{(2 a d+b c) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{4/3}}-\frac{(2 a d+b c) \tan ^{-1}\left (\frac{\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt{3} \sqrt [3]{c}}\right )}{3 \sqrt{3} c^{5/3} d^{4/3}}-\frac{x (b c-a d)}{3 c d \left (c+d x^3\right )} \]

[Out]

-((b*c - a*d)*x)/(3*c*d*(c + d*x^3)) - ((b*c + 2*a*d)*ArcTan[(c^(1/3) - 2*d^(1/3)*x)/(Sqrt[3]*c^(1/3))])/(3*Sq
rt[3]*c^(5/3)*d^(4/3)) + ((b*c + 2*a*d)*Log[c^(1/3) + d^(1/3)*x])/(9*c^(5/3)*d^(4/3)) - ((b*c + 2*a*d)*Log[c^(
2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)*x^2])/(18*c^(5/3)*d^(4/3))

________________________________________________________________________________________

Rubi [A]  time = 0.082311, antiderivative size = 169, normalized size of antiderivative = 1., 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 = {385, 200, 31, 634, 617, 204, 628} \[ -\frac{(2 a d+b c) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{18 c^{5/3} d^{4/3}}+\frac{(2 a d+b c) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{4/3}}-\frac{(2 a d+b c) \tan ^{-1}\left (\frac{\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt{3} \sqrt [3]{c}}\right )}{3 \sqrt{3} c^{5/3} d^{4/3}}-\frac{x (b c-a d)}{3 c d \left (c+d x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)/(c + d*x^3)^2,x]

[Out]

-((b*c - a*d)*x)/(3*c*d*(c + d*x^3)) - ((b*c + 2*a*d)*ArcTan[(c^(1/3) - 2*d^(1/3)*x)/(Sqrt[3]*c^(1/3))])/(3*Sq
rt[3]*c^(5/3)*d^(4/3)) + ((b*c + 2*a*d)*Log[c^(1/3) + d^(1/3)*x])/(9*c^(5/3)*d^(4/3)) - ((b*c + 2*a*d)*Log[c^(
2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)*x^2])/(18*c^(5/3)*d^(4/3))

Rule 385

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

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 31

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

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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{a+b x^3}{\left (c+d x^3\right )^2} \, dx &=-\frac{(b c-a d) x}{3 c d \left (c+d x^3\right )}+\frac{(b c+2 a d) \int \frac{1}{c+d x^3} \, dx}{3 c d}\\ &=-\frac{(b c-a d) x}{3 c d \left (c+d x^3\right )}+\frac{(b c+2 a d) \int \frac{1}{\sqrt [3]{c}+\sqrt [3]{d} x} \, dx}{9 c^{5/3} d}+\frac{(b c+2 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}{9 c^{5/3} d}\\ &=-\frac{(b c-a d) x}{3 c d \left (c+d x^3\right )}+\frac{(b c+2 a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{4/3}}-\frac{(b c+2 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}{18 c^{5/3} d^{4/3}}+\frac{(b c+2 a d) \int \frac{1}{c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2} \, dx}{6 c^{4/3} d}\\ &=-\frac{(b c-a d) x}{3 c d \left (c+d x^3\right )}+\frac{(b c+2 a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{4/3}}-\frac{(b c+2 a d) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{18 c^{5/3} d^{4/3}}+\frac{(b c+2 a d) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{d} x}{\sqrt [3]{c}}\right )}{3 c^{5/3} d^{4/3}}\\ &=-\frac{(b c-a d) x}{3 c d \left (c+d x^3\right )}-\frac{(b c+2 a d) \tan ^{-1}\left (\frac{\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt{3} \sqrt [3]{c}}\right )}{3 \sqrt{3} c^{5/3} d^{4/3}}+\frac{(b c+2 a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{4/3}}-\frac{(b c+2 a d) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{18 c^{5/3} d^{4/3}}\\ \end{align*}

Mathematica [A]  time = 0.0924508, size = 145, normalized size = 0.86 \[ \frac{-(2 a d+b c) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )-\frac{6 c^{2/3} \sqrt [3]{d} x (b c-a d)}{c+d x^3}+2 (2 a d+b c) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )-2 \sqrt{3} (2 a d+b c) \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} x}{\sqrt [3]{c}}}{\sqrt{3}}\right )}{18 c^{5/3} d^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*c^(2/3)*d^(1/3)*(b*c - a*d)*x)/(c + d*x^3) - 2*Sqrt[3]*(b*c + 2*a*d)*ArcTan[(1 - (2*d^(1/3)*x)/c^(1/3))/S
qrt[3]] + 2*(b*c + 2*a*d)*Log[c^(1/3) + d^(1/3)*x] - (b*c + 2*a*d)*Log[c^(2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)*x
^2])/(18*c^(5/3)*d^(4/3))

________________________________________________________________________________________

Maple [A]  time = 0.009, size = 221, normalized size = 1.3 \begin{align*}{\frac{ \left ( ad-bc \right ) x}{3\,cd \left ( d{x}^{3}+c \right ) }}+{\frac{2\,a}{9\,cd}\ln \left ( x+\sqrt [3]{{\frac{c}{d}}} \right ) \left ({\frac{c}{d}} \right ) ^{-{\frac{2}{3}}}}+{\frac{b}{9\,{d}^{2}}\ln \left ( x+\sqrt [3]{{\frac{c}{d}}} \right ) \left ({\frac{c}{d}} \right ) ^{-{\frac{2}{3}}}}-{\frac{a}{9\,cd}\ln \left ({x}^{2}-\sqrt [3]{{\frac{c}{d}}}x+ \left ({\frac{c}{d}} \right ) ^{{\frac{2}{3}}} \right ) \left ({\frac{c}{d}} \right ) ^{-{\frac{2}{3}}}}-{\frac{b}{18\,{d}^{2}}\ln \left ({x}^{2}-\sqrt [3]{{\frac{c}{d}}}x+ \left ({\frac{c}{d}} \right ) ^{{\frac{2}{3}}} \right ) \left ({\frac{c}{d}} \right ) ^{-{\frac{2}{3}}}}+{\frac{2\,\sqrt{3}a}{9\,cd}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{x{\frac{1}{\sqrt [3]{{\frac{c}{d}}}}}}-1 \right ) } \right ) \left ({\frac{c}{d}} \right ) ^{-{\frac{2}{3}}}}+{\frac{\sqrt{3}b}{9\,{d}^{2}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{x{\frac{1}{\sqrt [3]{{\frac{c}{d}}}}}}-1 \right ) } \right ) \left ({\frac{c}{d}} \right ) ^{-{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a*d-b*c)/c/d*x/(d*x^3+c)+2/9/c/d/(c/d)^(2/3)*ln(x+(c/d)^(1/3))*a+1/9/d^2/(c/d)^(2/3)*ln(x+(c/d)^(1/3))*b-
1/9/c/d/(c/d)^(2/3)*ln(x^2-(c/d)^(1/3)*x+(c/d)^(2/3))*a-1/18/d^2/(c/d)^(2/3)*ln(x^2-(c/d)^(1/3)*x+(c/d)^(2/3))
*b+2/9/c/d/(c/d)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(c/d)^(1/3)*x-1))*a+1/9/d^2/(c/d)^(2/3)*3^(1/2)*arctan(1/
3*3^(1/2)*(2/(c/d)^(1/3)*x-1))*b

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.70572, size = 1233, normalized size = 7.3 \begin{align*} \left [\frac{3 \, \sqrt{\frac{1}{3}}{\left (b c^{3} d + 2 \, a c^{2} d^{2} +{\left (b c^{2} d^{2} + 2 \, a c d^{3}\right )} x^{3}\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 ({\left (b c d + 2 \, a d^{2}\right )} x^{3} + b c^{2} + 2 \, a c d\right )} \left (c^{2} d\right )^{\frac{2}{3}} \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 ({\left (b c d + 2 \, a d^{2}\right )} x^{3} + b c^{2} + 2 \, a c d\right )} \left (c^{2} d\right )^{\frac{2}{3}} \log \left (c d x + \left (c^{2} d\right )^{\frac{2}{3}}\right ) - 6 \,{\left (b c^{3} d - a c^{2} d^{2}\right )} x}{18 \,{\left (c^{3} d^{3} x^{3} + c^{4} d^{2}\right )}}, \frac{6 \, \sqrt{\frac{1}{3}}{\left (b c^{3} d + 2 \, a c^{2} d^{2} +{\left (b c^{2} d^{2} + 2 \, a c d^{3}\right )} x^{3}\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 ({\left (b c d + 2 \, a d^{2}\right )} x^{3} + b c^{2} + 2 \, a c d\right )} \left (c^{2} d\right )^{\frac{2}{3}} \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 ({\left (b c d + 2 \, a d^{2}\right )} x^{3} + b c^{2} + 2 \, a c d\right )} \left (c^{2} d\right )^{\frac{2}{3}} \log \left (c d x + \left (c^{2} d\right )^{\frac{2}{3}}\right ) - 6 \,{\left (b c^{3} d - a c^{2} d^{2}\right )} x}{18 \,{\left (c^{3} d^{3} x^{3} + c^{4} d^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/18*(3*sqrt(1/3)*(b*c^3*d + 2*a*c^2*d^2 + (b*c^2*d^2 + 2*a*c*d^3)*x^3)*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)) - ((b*c*d + 2*a*d^2)*x^3 + b*c^2 + 2*a*c*d)*(c^2*d)^(2/3)*log(c*d*x^2 - (c^2*d)^(2/3)*x + (c
^2*d)^(1/3)*c) + 2*((b*c*d + 2*a*d^2)*x^3 + b*c^2 + 2*a*c*d)*(c^2*d)^(2/3)*log(c*d*x + (c^2*d)^(2/3)) - 6*(b*c
^3*d - a*c^2*d^2)*x)/(c^3*d^3*x^3 + c^4*d^2), 1/18*(6*sqrt(1/3)*(b*c^3*d + 2*a*c^2*d^2 + (b*c^2*d^2 + 2*a*c*d^
3)*x^3)*sqrt((c^2*d)^(1/3)/d)*arctan(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
) - ((b*c*d + 2*a*d^2)*x^3 + b*c^2 + 2*a*c*d)*(c^2*d)^(2/3)*log(c*d*x^2 - (c^2*d)^(2/3)*x + (c^2*d)^(1/3)*c) +
2*((b*c*d + 2*a*d^2)*x^3 + b*c^2 + 2*a*c*d)*(c^2*d)^(2/3)*log(c*d*x + (c^2*d)^(2/3)) - 6*(b*c^3*d - a*c^2*d^2
)*x)/(c^3*d^3*x^3 + c^4*d^2)]

________________________________________________________________________________________

Sympy [A]  time = 0.946584, size = 97, normalized size = 0.57 \begin{align*} \frac{x \left (a d - b c\right )}{3 c^{2} d + 3 c d^{2} x^{3}} + \operatorname{RootSum}{\left (729 t^{3} c^{5} d^{4} - 8 a^{3} d^{3} - 12 a^{2} b c d^{2} - 6 a b^{2} c^{2} d - b^{3} c^{3}, \left ( t \mapsto t \log{\left (\frac{9 t c^{2} d}{2 a d + b c} + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(a*d - b*c)/(3*c**2*d + 3*c*d**2*x**3) + RootSum(729*_t**3*c**5*d**4 - 8*a**3*d**3 - 12*a**2*b*c*d**2 - 6*a*
b**2*c**2*d - b**3*c**3, Lambda(_t, _t*log(9*_t*c**2*d/(2*a*d + b*c) + x)))

________________________________________________________________________________________

Giac [A]  time = 1.11013, size = 246, normalized size = 1.46 \begin{align*} -\frac{{\left (b c + 2 \, 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 )}{9 \, c^{2} d} + \frac{\sqrt{3}{\left (\left (-c d^{2}\right )^{\frac{1}{3}} b c + 2 \, \left (-c d^{2}\right )^{\frac{1}{3}} 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 )}{9 \, c^{2} d^{2}} - \frac{b c x - a d x}{3 \,{\left (d x^{3} + c\right )} c d} + \frac{{\left (\left (-c d^{2}\right )^{\frac{1}{3}} b c + 2 \, \left (-c d^{2}\right )^{\frac{1}{3}} 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 )}{18 \, c^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*(b*c + 2*a*d)*(-c/d)^(1/3)*log(abs(x - (-c/d)^(1/3)))/(c^2*d) + 1/9*sqrt(3)*((-c*d^2)^(1/3)*b*c + 2*(-c*d
^2)^(1/3)*a*d)*arctan(1/3*sqrt(3)*(2*x + (-c/d)^(1/3))/(-c/d)^(1/3))/(c^2*d^2) - 1/3*(b*c*x - a*d*x)/((d*x^3 +
c)*c*d) + 1/18*((-c*d^2)^(1/3)*b*c + 2*(-c*d^2)^(1/3)*a*d)*log(x^2 + x*(-c/d)^(1/3) + (-c/d)^(2/3))/(c^2*d^2)