3.2891 \(\int \frac {1}{(c e+d e x)^2 (a+b (c+d x)^3)} \, dx\)

Optimal. Leaf size=166 \[ -\frac {\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{4/3} d e^2}+\frac {\sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{4/3} d e^2}+\frac {\sqrt [3]{b} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{4/3} d e^2}-\frac {1}{a d e^2 (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 166, normalized size of antiderivative = 1.00, 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, 292, 31, 634, 617, 204, 628} \[ -\frac {\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{4/3} d e^2}+\frac {\sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{4/3} d e^2}+\frac {\sqrt [3]{b} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{4/3} d e^2}-\frac {1}{a d e^2 (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*e + d*e*x)^2*(a + b*(c + d*x)^3)),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, 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 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(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 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 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 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 {1}{(c e+d e x)^2 \left (a+b (c+d x)^3\right )} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b x^3\right )} \, dx,x,c+d x\right )}{d e^2}\\ &=-\frac {1}{a d e^2 (c+d x)}-\frac {b \operatorname {Subst}\left (\int \frac {x}{a+b x^3} \, dx,x,c+d x\right )}{a d e^2}\\ &=-\frac {1}{a d e^2 (c+d x)}+\frac {b^{2/3} \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{3 a^{4/3} d e^2}-\frac {b^{2/3} \operatorname {Subst}\left (\int \frac {\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^{4/3} d e^2}\\ &=-\frac {1}{a d e^2 (c+d x)}+\frac {\sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{4/3} d e^2}-\frac {\sqrt [3]{b} \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^{4/3} d e^2}-\frac {b^{2/3} \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 d e^2}\\ &=-\frac {1}{a d e^2 (c+d x)}+\frac {\sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{4/3} d e^2}-\frac {\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{4/3} d e^2}-\frac {\sqrt [3]{b} \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^{4/3} d e^2}\\ &=-\frac {1}{a d e^2 (c+d x)}+\frac {\sqrt [3]{b} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} a^{4/3} d e^2}+\frac {\sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{4/3} d e^2}-\frac {\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{4/3} d e^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 143, normalized size = 0.86 \[ \frac {-\sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )+2 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )-2 \sqrt {3} \sqrt [3]{b} \tan ^{-1}\left (\frac {2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt {3} \sqrt [3]{a}}\right )-\frac {6 \sqrt [3]{a}}{c+d x}}{6 a^{4/3} d e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*e + d*e*x)^2*(a + b*(c + d*x)^3)),x]

[Out]

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

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fricas [A]  time = 1.10, size = 156, normalized size = 0.94 \[ -\frac {2 \, \sqrt {3} {\left (d x + c\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (d x + c\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + {\left (d x + c\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} - {\left (a d x + a c\right )} \left (\frac {b}{a}\right )^{\frac {2}{3}} + a \left (\frac {b}{a}\right )^{\frac {1}{3}}\right ) - 2 \, {\left (d x + c\right )} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b d x + b c + a \left (\frac {b}{a}\right )^{\frac {2}{3}}\right ) + 6}{6 \, {\left (a d^{2} e^{2} x + a c d e^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)^2/(a+b*(d*x+c)^3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.21, size = 226, normalized size = 1.36 \[ \frac {\left (\frac {b}{a d^{3}}\right )^{\frac {1}{3}} e^{\left (-2\right )} \log \left ({\left | -\left (\frac {b}{a d^{3}}\right )^{\frac {1}{3}} e^{\left (-2\right )} - \frac {e^{\left (-1\right )}}{{\left (d x e + c e\right )} d} \right |}\right )}{3 \, a} - \frac {\sqrt {3} \left (a^{2} b\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (\frac {b}{a d^{3}}\right )^{\frac {1}{3}} e^{\left (-2\right )} - \frac {2 \, e^{\left (-1\right )}}{{\left (d x e + c e\right )} d}\right )} e^{2}}{3 \, \left (\frac {b}{a d^{3}}\right )^{\frac {1}{3}}}\right ) e^{\left (-2\right )}}{3 \, a^{2} d} - \frac {\left (a^{2} b\right )^{\frac {1}{3}} e^{\left (-2\right )} \log \left (\left (\frac {b}{a d^{3}}\right )^{\frac {2}{3}} e^{\left (-4\right )} - \frac {\left (\frac {b}{a d^{3}}\right )^{\frac {1}{3}} e^{\left (-3\right )}}{{\left (d x e + c e\right )} d} + \frac {e^{\left (-2\right )}}{{\left (d x e + c e\right )}^{2} d^{2}}\right )}{6 \, a^{2} d} - \frac {e^{\left (-1\right )}}{{\left (d x e + c e\right )} a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)^2/(a+b*(d*x+c)^3),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.00, size = 98, normalized size = 0.59 \[ -\frac {\left (\RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right ) d +c \right ) \ln \left (-\RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )+x \right )}{3 a d \,e^{2} \left (d^{2} \RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )^{2}+2 c d \RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )+c^{2}\right )}-\frac {1}{\left (d x +c \right ) a d \,e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/d/e^2/(d*x+c)-1/3/e^2/d*sum((_R*d+c)/(_R^2*d^2+2*_R*c*d+c^2)*ln(-_R+x),_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 = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{a d^{2} e^{2} x + a c d e^{2}} - \frac {-\frac {1}{6} \, {\left (2 \, \sqrt {3} \left (-\frac {1}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac {2}{3}}\right )}}{3 \, \left (-a^{2} b\right )^{\frac {2}{3}}}\right ) + \left (-\frac {1}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac {2}{3}}\right )}^{2} + 3 \, \left (-a^{2} b\right )^{\frac {4}{3}}\right ) - 2 \, \left (-\frac {1}{a b^{2} d^{3}}\right )^{\frac {1}{3}} \log \left ({\left | a b d x + a b c + \left (-a^{2} b\right )^{\frac {2}{3}} \right |}\right )\right )} b}{a e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)^2/(a+b*(d*x+c)^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.59, size = 154, normalized size = 0.93 \[ \frac {b^{1/3}\,\ln \left (b^{1/3}\,c+a^{1/3}+b^{1/3}\,d\,x\right )}{3\,a^{4/3}\,d\,e^2}-\frac {1}{a\,d\,\left (c\,e^2+d\,e^2\,x\right )}-\frac {b^{1/3}\,\ln \left (2\,b^{1/3}\,c-a^{1/3}+2\,b^{1/3}\,d\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{3\,a^{4/3}\,d\,e^2}+\frac {b^{1/3}\,\ln \left (2\,b^{1/3}\,c-a^{1/3}+2\,b^{1/3}\,d\,x-\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )}{a^{4/3}\,d\,e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^(1/3)*log(b^(1/3)*c + a^(1/3) + b^(1/3)*d*x))/(3*a^(4/3)*d*e^2) - 1/(a*d*(c*e^2 + d*e^2*x)) - (b^(1/3)*log(
3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*c - a^(1/3) + 2*b^(1/3)*d*x)*((3^(1/2)*1i)/2 + 1/2))/(3*a^(4/3)*d*e^2) + (b^(1/
3)*log(2*b^(1/3)*c - 3^(1/2)*a^(1/3)*1i - a^(1/3) + 2*b^(1/3)*d*x)*((3^(1/2)*1i)/6 - 1/6))/(a^(4/3)*d*e^2)

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sympy [A]  time = 0.99, size = 54, normalized size = 0.33 \[ - \frac {1}{a c d e^{2} + a d^{2} e^{2} x} + \frac {\operatorname {RootSum} {\left (27 t^{3} a^{4} - b, \left (t \mapsto t \log {\left (x + \frac {9 t^{2} a^{3} + b c}{b d} \right )} \right )\right )}}{d e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)**2/(a+b*(d*x+c)**3),x)

[Out]

-1/(a*c*d*e**2 + a*d**2*e**2*x) + RootSum(27*_t**3*a**4 - b, Lambda(_t, _t*log(x + (9*_t**2*a**3 + b*c)/(b*d))
))/(d*e**2)

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