3.2892 \(\int \frac{1}{(c e+d e x)^2 \left (a+b (c+d x)^3\right )^2} \, dx\)

Optimal. Leaf size=204 \[ -\frac{2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 a^{7/3} d e^2}+\frac{4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{7/3} d e^2}+\frac{4 \sqrt [3]{b} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{7/3} d e^2}-\frac{4}{3 a^2 d e^2 (c+d x)}+\frac{1}{3 a d e^2 (c+d x) \left (a+b (c+d x)^3\right )} \]

[Out]

-4/(3*a^2*d*e^2*(c + d*x)) + 1/(3*a*d*e^2*(c + d*x)*(a + b*(c + d*x)^3)) + (4*b^
(1/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(7
/3)*d*e^2) + (4*b^(1/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(9*a^(7/3)*d*e^2) - (2
*b^(1/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(9*a^(7
/3)*d*e^2)

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Rubi [A]  time = 0.390851, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375 \[ -\frac{2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 a^{7/3} d e^2}+\frac{4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{7/3} d e^2}+\frac{4 \sqrt [3]{b} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{7/3} d e^2}-\frac{4}{3 a^2 d e^2 (c+d x)}+\frac{1}{3 a d e^2 (c+d x) \left (a+b (c+d x)^3\right )} \]

Antiderivative was successfully verified.

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

[Out]

-4/(3*a^2*d*e^2*(c + d*x)) + 1/(3*a*d*e^2*(c + d*x)*(a + b*(c + d*x)^3)) + (4*b^
(1/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(7
/3)*d*e^2) + (4*b^(1/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(9*a^(7/3)*d*e^2) - (2
*b^(1/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(9*a^(7
/3)*d*e^2)

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Rubi in Sympy [A]  time = 47.089, size = 192, normalized size = 0.94 \[ \frac{1}{3 a d e^{2} \left (a + b \left (c + d x\right )^{3}\right ) \left (c + d x\right )} - \frac{4}{3 a^{2} d e^{2} \left (c + d x\right )} + \frac{4 \sqrt [3]{b} \log{\left (\sqrt [3]{a} + \sqrt [3]{b} \left (c + d x\right ) \right )}}{9 a^{\frac{7}{3}} d e^{2}} - \frac{2 \sqrt [3]{b} \log{\left (a^{\frac{2}{3}} + \sqrt [3]{a} \sqrt [3]{b} \left (- c - d x\right ) + b^{\frac{2}{3}} \left (c + d x\right )^{2} \right )}}{9 a^{\frac{7}{3}} d e^{2}} + \frac{4 \sqrt{3} \sqrt [3]{b} \operatorname{atan}{\left (\frac{\sqrt{3} \left (\frac{\sqrt [3]{a}}{3} + \sqrt [3]{b} \left (- \frac{2 c}{3} - \frac{2 d x}{3}\right )\right )}{\sqrt [3]{a}} \right )}}{9 a^{\frac{7}{3}} d e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/(3*a*d*e**2*(a + b*(c + d*x)**3)*(c + d*x)) - 4/(3*a**2*d*e**2*(c + d*x)) + 4*
b**(1/3)*log(a**(1/3) + b**(1/3)*(c + d*x))/(9*a**(7/3)*d*e**2) - 2*b**(1/3)*log
(a**(2/3) + a**(1/3)*b**(1/3)*(-c - d*x) + b**(2/3)*(c + d*x)**2)/(9*a**(7/3)*d*
e**2) + 4*sqrt(3)*b**(1/3)*atan(sqrt(3)*(a**(1/3)/3 + b**(1/3)*(-2*c/3 - 2*d*x/3
))/a**(1/3))/(9*a**(7/3)*d*e**2)

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Mathematica [A]  time = 0.176949, size = 171, normalized size = 0.84 \[ \frac{-2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )-\frac{3 \sqrt [3]{a} b (c+d x)^2}{a+b (c+d x)^3}+4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )-4 \sqrt{3} \sqrt [3]{b} \tan ^{-1}\left (\frac{2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )-\frac{9 \sqrt [3]{a}}{c+d x}}{9 a^{7/3} d e^2} \]

Antiderivative was successfully verified.

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

[Out]

((-9*a^(1/3))/(c + d*x) - (3*a^(1/3)*b*(c + d*x)^2)/(a + b*(c + d*x)^3) - 4*Sqrt
[3]*b^(1/3)*ArcTan[(-a^(1/3) + 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))] + 4*b^(1/
3)*Log[a^(1/3) + b^(1/3)*(c + d*x)] - 2*b^(1/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c
+ d*x) + b^(2/3)*(c + d*x)^2])/(9*a^(7/3)*d*e^2)

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Maple [C]  time = 0.013, size = 242, normalized size = 1.2 \[ -{\frac{b{x}^{2}d}{3\,{e}^{2}{a}^{2} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) }}-{\frac{2\,bcx}{3\,{e}^{2}{a}^{2} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) }}-{\frac{b{c}^{2}}{3\,{e}^{2}{a}^{2} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) d}}-{\frac{4}{9\,{e}^{2}{a}^{2}d}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{3}b{d}^{3}+3\,{{\it \_Z}}^{2}bc{d}^{2}+3\,{\it \_Z}\,b{c}^{2}d+b{c}^{3}+a \right ) }{\frac{ \left ({\it \_R}\,d+c \right ) \ln \left ( x-{\it \_R} \right ) }{{d}^{2}{{\it \_R}}^{2}+2\,cd{\it \_R}+{c}^{2}}}}-{\frac{1}{{e}^{2}{a}^{2}d \left ( dx+c \right ) }} \]

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)^2,x)

[Out]

-1/3/e^2*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)*x^2*d-2/3/e^2*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)*c*x-1/3/e^2*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)*c^2/d-4/9/e^2/a^2/d*sum((_R*d+c)/(_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))-1/a^
2/d/e^2/(d*x+c)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ -\frac{4 \, b d^{3} x^{3} + 12 \, b c d^{2} x^{2} + 12 \, b c^{2} d x + 4 \, b c^{3} + 3 \, a}{3 \,{\left (a^{2} b d^{5} e^{2} x^{4} + 4 \, a^{2} b c d^{4} e^{2} x^{3} + 6 \, a^{2} b c^{2} d^{3} e^{2} x^{2} +{\left (4 \, a^{2} b c^{3} + a^{3}\right )} d^{2} e^{2} x +{\left (a^{2} b c^{4} + a^{3} c\right )} d e^{2}\right )}} - \frac{4 \, b \int \frac{d x + c}{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}\,{d x}}{3 \, a^{2} e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/3*(4*b*d^3*x^3 + 12*b*c*d^2*x^2 + 12*b*c^2*d*x + 4*b*c^3 + 3*a)/(a^2*b*d^5*e^
2*x^4 + 4*a^2*b*c*d^4*e^2*x^3 + 6*a^2*b*c^2*d^3*e^2*x^2 + (4*a^2*b*c^3 + a^3)*d^
2*e^2*x + (a^2*b*c^4 + a^3*c)*d*e^2) - 4/3*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^2*e^2)

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Fricas [A]  time = 0.231904, size = 575, normalized size = 2.82 \[ -\frac{\sqrt{3}{\left (2 \, \sqrt{3}{\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + b c^{4} +{\left (4 \, b c^{3} + a\right )} d x + a c\right )} \left (\frac{b}{a}\right )^{\frac{1}{3}} \log \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} -{\left (a d x + a c\right )} \left (\frac{b}{a}\right )^{\frac{2}{3}} + a \left (\frac{b}{a}\right )^{\frac{1}{3}}\right ) - 4 \, \sqrt{3}{\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + b c^{4} +{\left (4 \, b c^{3} + a\right )} d x + a c\right )} \left (\frac{b}{a}\right )^{\frac{1}{3}} \log \left (b d x + b c + a \left (\frac{b}{a}\right )^{\frac{2}{3}}\right ) - 12 \,{\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + b c^{4} +{\left (4 \, b c^{3} + a\right )} d x + a c\right )} \left (\frac{b}{a}\right )^{\frac{1}{3}} \arctan \left (\frac{\sqrt{3} a \left (\frac{b}{a}\right )^{\frac{2}{3}} - 2 \, \sqrt{3}{\left (b d x + b c\right )}}{3 \, a \left (\frac{b}{a}\right )^{\frac{2}{3}}}\right ) + 3 \, \sqrt{3}{\left (4 \, b d^{3} x^{3} + 12 \, b c d^{2} x^{2} + 12 \, b c^{2} d x + 4 \, b c^{3} + 3 \, a\right )}\right )}}{27 \,{\left (a^{2} b d^{5} e^{2} x^{4} + 4 \, a^{2} b c d^{4} e^{2} x^{3} + 6 \, a^{2} b c^{2} d^{3} e^{2} x^{2} +{\left (4 \, a^{2} b c^{3} + a^{3}\right )} d^{2} e^{2} x +{\left (a^{2} b c^{4} + a^{3} c\right )} d e^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/27*sqrt(3)*(2*sqrt(3)*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + b*c^4 +
(4*b*c^3 + a)*d*x + a*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)) - 4*sqrt(3)*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b
*c^2*d^2*x^2 + b*c^4 + (4*b*c^3 + a)*d*x + a*c)*(b/a)^(1/3)*log(b*d*x + b*c + a*
(b/a)^(2/3)) - 12*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + b*c^4 + (4*b*c^
3 + a)*d*x + a*c)*(b/a)^(1/3)*arctan(1/3*(sqrt(3)*a*(b/a)^(2/3) - 2*sqrt(3)*(b*d
*x + b*c))/(a*(b/a)^(2/3))) + 3*sqrt(3)*(4*b*d^3*x^3 + 12*b*c*d^2*x^2 + 12*b*c^2
*d*x + 4*b*c^3 + 3*a))/(a^2*b*d^5*e^2*x^4 + 4*a^2*b*c*d^4*e^2*x^3 + 6*a^2*b*c^2*
d^3*e^2*x^2 + (4*a^2*b*c^3 + a^3)*d^2*e^2*x + (a^2*b*c^4 + a^3*c)*d*e^2)

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Sympy [A]  time = 33.8171, size = 196, normalized size = 0.96 \[ - \frac{3 a + 4 b c^{3} + 12 b c^{2} d x + 12 b c d^{2} x^{2} + 4 b d^{3} x^{3}}{3 a^{3} c d e^{2} + 3 a^{2} b c^{4} d e^{2} + 18 a^{2} b c^{2} d^{3} e^{2} x^{2} + 12 a^{2} b c d^{4} e^{2} x^{3} + 3 a^{2} b d^{5} e^{2} x^{4} + x \left (3 a^{3} d^{2} e^{2} + 12 a^{2} b c^{3} d^{2} e^{2}\right )} + \frac{\operatorname{RootSum}{\left (729 t^{3} a^{7} - 64 b, \left ( t \mapsto t \log{\left (x + \frac{81 t^{2} a^{5} + 16 b c}{16 b d} \right )} \right )\right )}}{d e^{2}} \]

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)**2,x)

[Out]

-(3*a + 4*b*c**3 + 12*b*c**2*d*x + 12*b*c*d**2*x**2 + 4*b*d**3*x**3)/(3*a**3*c*d
*e**2 + 3*a**2*b*c**4*d*e**2 + 18*a**2*b*c**2*d**3*e**2*x**2 + 12*a**2*b*c*d**4*
e**2*x**3 + 3*a**2*b*d**5*e**2*x**4 + x*(3*a**3*d**2*e**2 + 12*a**2*b*c**3*d**2*
e**2)) + RootSum(729*_t**3*a**7 - 64*b, Lambda(_t, _t*log(x + (81*_t**2*a**5 + 1
6*b*c)/(16*b*d))))/(d*e**2)

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GIAC/XCAS [A]  time = 0.235175, size = 363, normalized size = 1.78 \[ \frac{4 \, \left (\frac{b}{a d^{3}}\right )^{\frac{1}{3}} e^{\left (-2\right )}{\rm ln}\left ({\left | -\left (\frac{b}{a d^{3}}\right )^{\frac{1}{3}} e^{\left (-2\right )} - \frac{e^{\left (-1\right )}}{{\left (d x e + c e\right )} d} \right |}\right )}{9 \, a^{2}} - \frac{4 \, \sqrt{3} \left (a^{2} b\right )^{\frac{1}{3}} \arctan \left (\frac{\sqrt{3}{\left (\left (\frac{b}{a d^{3}}\right )^{\frac{1}{3}} e^{\left (-2\right )} - \frac{2 \, e^{\left (-1\right )}}{{\left (d x e + c e\right )} d}\right )} e^{2}}{3 \, \left (\frac{b}{a d^{3}}\right )^{\frac{1}{3}}}\right ) e^{\left (-2\right )}}{9 \, a^{3} d} - \frac{2 \, \left (a^{2} b\right )^{\frac{1}{3}} e^{\left (-2\right )}{\rm ln}\left (\left (\frac{b}{a d^{3}}\right )^{\frac{2}{3}} e^{\left (-4\right )} - \frac{\left (\frac{b}{a d^{3}}\right )^{\frac{1}{3}} e^{\left (-3\right )}}{{\left (d x e + c e\right )} d} + \frac{e^{\left (-2\right )}}{{\left (d x e + c e\right )}^{2} d^{2}}\right )}{9 \, a^{3} d} - \frac{e^{\left (-1\right )}}{{\left (d x e + c e\right )} a^{2} d} - \frac{b e^{\left (-1\right )}}{3 \,{\left (d x e + c e\right )} a^{2}{\left (b + \frac{a e^{3}}{{\left (d x e + c e\right )}^{3}}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

4/9*(b/(a*d^3))^(1/3)*e^(-2)*ln(abs(-(b/(a*d^3))^(1/3)*e^(-2) - e^(-1)/((d*x*e +
c*e)*d)))/a^2 - 4/9*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^3*d) - 2/
9*(a^2*b)^(1/3)*e^(-2)*ln((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^3*d) - e^(-1)/((d*x*e + c*e)*
a^2*d) - 1/3*b*e^(-1)/((d*x*e + c*e)*a^2*(b + a*e^3/(d*x*e + c*e)^3)*d)