3.2892 \(\int \frac{1}{(c e+d e x)^3 (a+b (c+d x)^3)} \, dx\)

Optimal. Leaf size=168 \[ -\frac{b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{5/3} d e^3}+\frac{b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{5/3} d e^3}+\frac{b^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{5/3} d e^3}-\frac{1}{2 a d e^3 (c+d x)^2} \]

[Out]

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

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Rubi [A]  time = 0.126265, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {372, 325, 200, 31, 634, 617, 204, 628} \[ -\frac{b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{5/3} d e^3}+\frac{b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{5/3} d e^3}+\frac{b^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{5/3} d e^3}-\frac{1}{2 a d e^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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{1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^3 \left (a+b x^3\right )} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac{1}{2 a d e^3 (c+d x)^2}-\frac{b \operatorname{Subst}\left (\int \frac{1}{a+b x^3} \, dx,x,c+d x\right )}{a d e^3}\\ &=-\frac{1}{2 a d e^3 (c+d x)^2}-\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{3 a^{5/3} d e^3}-\frac{b \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 )}{3 a^{5/3} d e^3}\\ &=-\frac{1}{2 a d e^3 (c+d x)^2}-\frac{b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{5/3} d e^3}+\frac{b^{2/3} \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 )}{6 a^{5/3} d e^3}-\frac{b \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 )}{2 a^{4/3} d e^3}\\ &=-\frac{1}{2 a d e^3 (c+d x)^2}-\frac{b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{5/3} d e^3}+\frac{b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{5/3} d e^3}-\frac{b^{2/3} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )}{a^{5/3} d e^3}\\ &=-\frac{1}{2 a d e^3 (c+d x)^2}+\frac{b^{2/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{\sqrt{3} a^{5/3} d e^3}-\frac{b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{5/3} d e^3}+\frac{b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{5/3} d e^3}\\ \end{align*}

Mathematica [A]  time = 0.0363882, size = 142, normalized size = 0.85 \[ \frac{b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )-\frac{3 a^{2/3}}{(c+d x)^2}-2 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )-2 \sqrt{3} b^{2/3} \tan ^{-1}\left (\frac{2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{6 a^{5/3} d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.005, size = 93, normalized size = 0.6 \begin{align*} -{\frac{1}{3\,{e}^{3}da}\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}}}}-{\frac{1}{2\,{e}^{3}da \left ( dx+c \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/e^3/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))/a
-1/2/a/d/e^3/(d*x+c)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(a*d^3*e^3*x^2 + 2*a*c*d^2*e^3*x + a*c^2*d*e^3) - b*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)/(a*e^3)

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Fricas [A]  time = 1.60144, size = 556, normalized size = 3.31 \begin{align*} \frac{2 \, \sqrt{3}{\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} \arctan \left (\frac{2 \, \sqrt{3}{\left (a d x + a c\right )} \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{2}{3}} - \sqrt{3} b}{3 \, b}\right ) -{\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} \log \left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + a^{2} \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{2}{3}} +{\left (a b d x + a b c\right )} \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}}\right ) + 2 \,{\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} \log \left (b d x + b c - a \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}}\right ) - 3}{6 \,{\left (a d^{3} e^{3} x^{2} + 2 \, a c d^{2} e^{3} x + a c^{2} d e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.29902, size = 75, normalized size = 0.45 \begin{align*} - \frac{1}{2 a c^{2} d e^{3} + 4 a c d^{2} e^{3} x + 2 a d^{3} e^{3} x^{2}} + \frac{\operatorname{RootSum}{\left (27 t^{3} a^{5} + b^{2}, \left ( t \mapsto t \log{\left (x + \frac{- 3 t a^{2} + b c}{b d} \right )} \right )\right )}}{d e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(2*a*c**2*d*e**3 + 4*a*c*d**2*e**3*x + 2*a*d**3*e**3*x**2) + RootSum(27*_t**3*a**5 + b**2, Lambda(_t, _t*lo
g(x + (-3*_t*a**2 + b*c)/(b*d))))/(d*e**3)

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Giac [A]  time = 1.19555, size = 271, normalized size = 1.61 \begin{align*} \frac{1}{3} \, \sqrt{3} \left (-\frac{b^{2} e^{\left (-9\right )}}{a^{5} 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{b^{2} e^{\left (-9\right )}}{a^{5} 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{b^{2} e^{\left (-9\right )}}{a^{5} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left | -a b d x e^{3} - a b c e^{3} + \left (-a b^{2}\right )^{\frac{1}{3}} a e^{3} \right |}\right ) - \frac{e^{\left (-3\right )}}{2 \,{\left (d x + c\right )}^{2} a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*(-b^2*e^(-9)/(a^5*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))) - 1/6*(-b^2*e^(-9)/(a^5*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) + 1/3*(-b^2*e^(-9)/(a^5*d^3))^(1/3)*log(abs(-a*b*d*x*e^3 - a
*b*c*e^3 + (-a*b^2)^(1/3)*a*e^3)) - 1/2*e^(-3)/((d*x + c)^2*a*d)