3.2887 \(\int \frac{(c e+d e x)^3}{a+b (c+d x)^3} \, dx\)

Optimal. Leaf size=156 \[ \frac{\sqrt [3]{a} e^3 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{4/3} d}-\frac{\sqrt [3]{a} e^3 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{4/3} d}+\frac{\sqrt [3]{a} e^3 \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} b^{4/3} d}+\frac{e^3 x}{b} \]

[Out]

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

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Rubi [A]  time = 0.129492, antiderivative size = 156, 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, 321, 200, 31, 634, 617, 204, 628} \[ \frac{\sqrt [3]{a} e^3 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{4/3} d}-\frac{\sqrt [3]{a} e^3 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{4/3} d}+\frac{\sqrt [3]{a} e^3 \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} b^{4/3} d}+\frac{e^3 x}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*x)/b + (a^(1/3)*e^3*ArcTan[(a^(1/3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*b^(4/3)*d) - (a^(
1/3)*e^3*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(3*b^(4/3)*d) + (a^(1/3)*e^3*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x
) + b^(2/3)*(c + d*x)^2])/(6*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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 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 e+d e x)^3}{a+b (c+d x)^3} \, dx &=\frac{e^3 \operatorname{Subst}\left (\int \frac{x^3}{a+b x^3} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 x}{b}-\frac{\left (a e^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^3} \, dx,x,c+d x\right )}{b d}\\ &=\frac{e^3 x}{b}-\frac{\left (\sqrt [3]{a} e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{3 b d}-\frac{\left (\sqrt [3]{a} e^3\right ) \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 b d}\\ &=\frac{e^3 x}{b}-\frac{\sqrt [3]{a} e^3 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{4/3} d}+\frac{\left (\sqrt [3]{a} e^3\right ) \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 b^{4/3} d}-\frac{\left (a^{2/3} e^3\right ) \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 b d}\\ &=\frac{e^3 x}{b}-\frac{\sqrt [3]{a} e^3 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{4/3} d}+\frac{\sqrt [3]{a} e^3 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{4/3} d}-\frac{\left (\sqrt [3]{a} e^3\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )}{b^{4/3} d}\\ &=\frac{e^3 x}{b}+\frac{\sqrt [3]{a} e^3 \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{\sqrt{3} b^{4/3} d}-\frac{\sqrt [3]{a} e^3 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 b^{4/3} d}+\frac{\sqrt [3]{a} e^3 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 b^{4/3} d}\\ \end{align*}

Mathematica [A]  time = 0.0192573, size = 145, normalized size = 0.93 \[ \frac{e^3 \left (\sqrt [3]{a} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )-2 \sqrt [3]{a} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )-2 \sqrt{3} \sqrt [3]{a} \tan ^{-1}\left (\frac{2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )+6 \sqrt [3]{b} c+6 \sqrt [3]{b} d x\right )}{6 b^{4/3} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.002, size = 84, normalized size = 0.5 \begin{align*}{\frac{{e}^{3}x}{b}}-{\frac{{e}^{3}a}{3\,{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*e*x+c*e)^3/(a+b*(d*x+c)^3),x)

[Out]

e^3*x/b-1/3*e^3/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))*a

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\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 )} a e^{3}}{b} + \frac{e^{3} x}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*e^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 + e^3*x/b

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Fricas [A]  time = 1.71419, size = 343, normalized size = 2.2 \begin{align*} \frac{6 \, d e^{3} x + 2 \, \sqrt{3} e^{3} \left (-\frac{a}{b}\right )^{\frac{1}{3}} \arctan \left (\frac{2 \, \sqrt{3}{\left (b d x + b c\right )} \left (-\frac{a}{b}\right )^{\frac{2}{3}} - \sqrt{3} a}{3 \, a}\right ) - e^{3} \left (-\frac{a}{b}\right )^{\frac{1}{3}} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} +{\left (d x + c\right )} \left (-\frac{a}{b}\right )^{\frac{1}{3}} + \left (-\frac{a}{b}\right )^{\frac{2}{3}}\right ) + 2 \, e^{3} \left (-\frac{a}{b}\right )^{\frac{1}{3}} \log \left (d x + c - \left (-\frac{a}{b}\right )^{\frac{1}{3}}\right )}{6 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.488936, size = 44, normalized size = 0.28 \begin{align*} \frac{e^{3} \operatorname{RootSum}{\left (27 t^{3} b^{4} + a, \left ( t \mapsto t \log{\left (x + \frac{- 3 t b e^{3} + c e^{3}}{d e^{3}} \right )} \right )\right )}}{d} + \frac{e^{3} x}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15404, size = 298, normalized size = 1.91 \begin{align*} \frac{1}{3} \, \sqrt{3} \left (-\frac{a e^{9}}{b^{4} d^{3}}\right )^{\frac{1}{3}} \arctan \left (-\frac{b d^{4} x + b c d^{3} - \left (-a b^{2}\right )^{\frac{1}{3}} d^{3}}{\sqrt{3} b d^{4} x + \sqrt{3} b c d^{3} + \sqrt{3} \left (-a b^{2}\right )^{\frac{1}{3}} d^{3}}\right ) - \frac{1}{6} \, \left (-\frac{a e^{9}}{b^{4} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left (\sqrt{3} b d^{4} x + \sqrt{3} b c d^{3} + \sqrt{3} \left (-a b^{2}\right )^{\frac{1}{3}} d^{3}\right )}^{2} +{\left (b d^{4} x + b c d^{3} - \left (-a b^{2}\right )^{\frac{1}{3}} d^{3}\right )}^{2}\right ) + \frac{1}{3} \, \left (-\frac{a e^{9}}{b^{4} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left | -b^{2} d^{4} x - b^{2} c d^{3} + \left (-a b^{2}\right )^{\frac{1}{3}} b d^{3} \right |}\right ) + \frac{x e^{3}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*(-a*e^9/(b^4*d^3))^(1/3)*arctan(-(b*d^4*x + b*c*d^3 - (-a*b^2)^(1/3)*d^3)/(sqrt(3)*b*d^4*x + sqrt(
3)*b*c*d^3 + sqrt(3)*(-a*b^2)^(1/3)*d^3)) - 1/6*(-a*e^9/(b^4*d^3))^(1/3)*log((sqrt(3)*b*d^4*x + sqrt(3)*b*c*d^
3 + sqrt(3)*(-a*b^2)^(1/3)*d^3)^2 + (b*d^4*x + b*c*d^3 - (-a*b^2)^(1/3)*d^3)^2) + 1/3*(-a*e^9/(b^4*d^3))^(1/3)
*log(abs(-b^2*d^4*x - b^2*c*d^3 + (-a*b^2)^(1/3)*b*d^3)) + x*e^3/b