3.2869 \(\int \frac{(c+d x)^3}{(a+b (c+d x)^3)^2} \, dx\)

Optimal. Leaf size=170 \[ \frac{\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{2/3} b^{4/3} d}-\frac{\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{2/3} b^{4/3} d}-\frac{\tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{2/3} b^{4/3} d}-\frac{c+d x}{3 b d \left (a+b (c+d x)^3\right )} \]

[Out]

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

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Rubi [A]  time = 0.14064, antiderivative size = 170, normalized size of antiderivative = 1., 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 = {372, 288, 200, 31, 634, 617, 204, 628} \[ \frac{\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{2/3} b^{4/3} d}-\frac{\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{2/3} b^{4/3} d}-\frac{\tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{2/3} b^{4/3} d}-\frac{c+d x}{3 b d \left (a+b (c+d x)^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 372

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m), Subst[Int[x^m
*(a + b*x^n)^p, x], x, v], x] /; FreeQ[{a, b, m, n, p}, x] && LinearPairQ[u, v, x]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, 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 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{(c+d x)^3}{\left (a+b (c+d x)^3\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^3}{\left (a+b x^3\right )^2} \, dx,x,c+d x\right )}{d}\\ &=-\frac{c+d x}{3 b d \left (a+b (c+d x)^3\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{a+b x^3} \, dx,x,c+d x\right )}{3 b d}\\ &=-\frac{c+d x}{3 b d \left (a+b (c+d x)^3\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{9 a^{2/3} b d}+\frac{\operatorname{Subst}\left (\int \frac{2 \sqrt [3]{a}-\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,c+d x\right )}{9 a^{2/3} b d}\\ &=-\frac{c+d x}{3 b d \left (a+b (c+d x)^3\right )}+\frac{\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{2/3} b^{4/3} d}-\frac{\operatorname{Subst}\left (\int \frac{-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,c+d x\right )}{18 a^{2/3} b^{4/3} d}+\frac{\operatorname{Subst}\left (\int \frac{1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,c+d x\right )}{6 \sqrt [3]{a} b d}\\ &=-\frac{c+d x}{3 b d \left (a+b (c+d x)^3\right )}+\frac{\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{2/3} b^{4/3} d}-\frac{\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{2/3} b^{4/3} d}+\frac{\operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )}{3 a^{2/3} b^{4/3} d}\\ &=-\frac{c+d x}{3 b d \left (a+b (c+d x)^3\right )}-\frac{\tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{3 \sqrt{3} a^{2/3} b^{4/3} d}+\frac{\log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{2/3} b^{4/3} d}-\frac{\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{2/3} b^{4/3} d}\\ \end{align*}

Mathematica [A]  time = 0.0651127, size = 151, normalized size = 0.89 \[ \frac{-\frac{\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{a^{2/3}}+\frac{2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{a^{2/3}}+\frac{2 \sqrt{3} \tan ^{-1}\left (\frac{2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{a^{2/3}}-\frac{6 \sqrt [3]{b} (c+d x)}{a+b (c+d x)^3}}{18 b^{4/3} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.01, size = 124, normalized size = 0.7 \begin{align*}{\frac{1}{b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a} \left ( -{\frac{x}{3\,b}}-{\frac{c}{3\,bd}} \right ) }+{\frac{1}{9\,{b}^{2}d}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{3}b{d}^{3}+3\,{{\it \_Z}}^{2}bc{d}^{2}+3\,{\it \_Z}\,b{c}^{2}d+b{c}^{3}+a \right ) }{\frac{\ln \left ( x-{\it \_R} \right ) }{{d}^{2}{{\it \_R}}^{2}+2\,cd{\it \_R}+{c}^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/3*x/b-1/3*c/d/b)/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)+1/9/b^2/d*sum(1/(_R^2*d^2+2*_R*c*d+c^2)*ln(
x-_R),_R=RootOf(_Z^3*b*d^3+3*_Z^2*b*c*d^2+3*_Z*b*c^2*d+b*c^3+a))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{d x + c}{3 \,{\left (b^{2} d^{4} x^{3} + 3 \, b^{2} c d^{3} x^{2} + 3 \, b^{2} c^{2} d^{2} x +{\left (b^{2} c^{3} + a b\right )} d\right )}} + \frac{\frac{1}{3} \, \sqrt{3} \left (\frac{1}{a^{2} b d^{3}}\right )^{\frac{1}{3}} \arctan \left (-\frac{b d x + b c + \left (a b^{2}\right )^{\frac{1}{3}}}{\sqrt{3} b d x + \sqrt{3} b c - \sqrt{3} \left (a b^{2}\right )^{\frac{1}{3}}}\right ) - \frac{1}{6} \, \left (\frac{1}{a^{2} b d^{3}}\right )^{\frac{1}{3}} \log \left ({\left (\sqrt{3} b d x + \sqrt{3} b c - \sqrt{3} \left (a b^{2}\right )^{\frac{1}{3}}\right )}^{2} +{\left (b d x + b c + \left (a b^{2}\right )^{\frac{1}{3}}\right )}^{2}\right ) + \frac{1}{3} \, \left (\frac{1}{a^{2} b d^{3}}\right )^{\frac{1}{3}} \log \left ({\left | b d x + b c + \left (a b^{2}\right )^{\frac{1}{3}} \right |}\right )}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.66629, size = 1851, normalized size = 10.89 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.10418, size = 90, normalized size = 0.53 \begin{align*} - \frac{c + d x}{3 a b d + 3 b^{2} c^{3} d + 9 b^{2} c^{2} d^{2} x + 9 b^{2} c d^{3} x^{2} + 3 b^{2} d^{4} x^{3}} + \frac{\operatorname{RootSum}{\left (729 t^{3} a^{2} b^{4} - 1, \left ( t \mapsto t \log{\left (x + \frac{9 t a b + c}{d} \right )} \right )\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c + d*x)/(3*a*b*d + 3*b**2*c**3*d + 9*b**2*c**2*d**2*x + 9*b**2*c*d**3*x**2 + 3*b**2*d**4*x**3) + RootSum(72
9*_t**3*a**2*b**4 - 1, Lambda(_t, _t*log(x + (9*_t*a*b + c)/d)))/d

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Giac [A]  time = 1.17302, size = 288, normalized size = 1.69 \begin{align*} \frac{1}{9} \, \sqrt{3} \left (\frac{1}{a^{2} b^{4} d^{3}}\right )^{\frac{1}{3}} \arctan \left (-\frac{b d x + b c + \left (a b^{2}\right )^{\frac{1}{3}}}{\sqrt{3} b d x + \sqrt{3} b c - \sqrt{3} \left (a b^{2}\right )^{\frac{1}{3}}}\right ) - \frac{1}{18} \, \left (\frac{1}{a^{2} b^{4} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left (\sqrt{3} b d x + \sqrt{3} b c - \sqrt{3} \left (a b^{2}\right )^{\frac{1}{3}}\right )}^{2} +{\left (b d x + b c + \left (a b^{2}\right )^{\frac{1}{3}}\right )}^{2}\right ) + \frac{1}{9} \, \left (\frac{1}{a^{2} b^{4} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left | 3 \, b^{2} d x + 3 \, b^{2} c + 3 \, \left (a b^{2}\right )^{\frac{1}{3}} b \right |}\right ) - \frac{d x + c}{3 \,{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*sqrt(3)*(1/(a^2*b^4*d^3))^(1/3)*arctan(-(b*d*x + b*c + (a*b^2)^(1/3))/(sqrt(3)*b*d*x + sqrt(3)*b*c - sqrt(
3)*(a*b^2)^(1/3))) - 1/18*(1/(a^2*b^4*d^3))^(1/3)*log((sqrt(3)*b*d*x + sqrt(3)*b*c - sqrt(3)*(a*b^2)^(1/3))^2
+ (b*d*x + b*c + (a*b^2)^(1/3))^2) + 1/9*(1/(a^2*b^4*d^3))^(1/3)*log(abs(3*b^2*d*x + 3*b^2*c + 3*(a*b^2)^(1/3)
*b)) - 1/3*(d*x + c)/((b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)*b*d)