3.2901 \(\int \frac{1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )^3} \, dx\)

Optimal. Leaf size=237 \[ -\frac{20 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{11/3} d e^3}+\frac{10 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{27 a^{11/3} d e^3}+\frac{20 b^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{9 \sqrt{3} a^{11/3} d e^3}-\frac{10}{9 a^3 d e^3 (c+d x)^2}+\frac{4}{9 a^2 d e^3 (c+d x)^2 \left (a+b (c+d x)^3\right )}+\frac{1}{6 a d e^3 (c+d x)^2 \left (a+b (c+d x)^3\right )^2} \]

[Out]

-10/(9*a^3*d*e^3*(c + d*x)^2) + 1/(6*a*d*e^3*(c + d*x)^2*(a + b*(c + d*x)^3)^2)
+ 4/(9*a^2*d*e^3*(c + d*x)^2*(a + b*(c + d*x)^3)) + (20*b^(2/3)*ArcTan[(a^(1/3)
- 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(11/3)*d*e^3) - (20*b^(2
/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(27*a^(11/3)*d*e^3) + (10*b^(2/3)*Log[a^(2
/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(27*a^(11/3)*d*e^3)

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Rubi [A]  time = 0.462997, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375 \[ -\frac{20 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{11/3} d e^3}+\frac{10 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{27 a^{11/3} d e^3}+\frac{20 b^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{9 \sqrt{3} a^{11/3} d e^3}-\frac{10}{9 a^3 d e^3 (c+d x)^2}+\frac{4}{9 a^2 d e^3 (c+d x)^2 \left (a+b (c+d x)^3\right )}+\frac{1}{6 a d e^3 (c+d x)^2 \left (a+b (c+d x)^3\right )^2} \]

Antiderivative was successfully verified.

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

[Out]

-10/(9*a^3*d*e^3*(c + d*x)^2) + 1/(6*a*d*e^3*(c + d*x)^2*(a + b*(c + d*x)^3)^2)
+ 4/(9*a^2*d*e^3*(c + d*x)^2*(a + b*(c + d*x)^3)) + (20*b^(2/3)*ArcTan[(a^(1/3)
- 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(11/3)*d*e^3) - (20*b^(2
/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(27*a^(11/3)*d*e^3) + (10*b^(2/3)*Log[a^(2
/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(27*a^(11/3)*d*e^3)

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Rubi in Sympy [A]  time = 53.3033, size = 226, normalized size = 0.95 \[ \frac{1}{6 a d e^{3} \left (a + b \left (c + d x\right )^{3}\right )^{2} \left (c + d x\right )^{2}} + \frac{4}{9 a^{2} d e^{3} \left (a + b \left (c + d x\right )^{3}\right ) \left (c + d x\right )^{2}} - \frac{10}{9 a^{3} d e^{3} \left (c + d x\right )^{2}} - \frac{20 b^{\frac{2}{3}} \log{\left (\sqrt [3]{a} + \sqrt [3]{b} \left (c + d x\right ) \right )}}{27 a^{\frac{11}{3}} d e^{3}} + \frac{10 b^{\frac{2}{3}} \log{\left (a^{\frac{2}{3}} + \sqrt [3]{a} \sqrt [3]{b} \left (- c - d x\right ) + b^{\frac{2}{3}} \left (c + d x\right )^{2} \right )}}{27 a^{\frac{11}{3}} d e^{3}} + \frac{20 \sqrt{3} b^{\frac{2}{3}} \operatorname{atan}{\left (\frac{\sqrt{3} \left (\frac{\sqrt [3]{a}}{3} + \sqrt [3]{b} \left (- \frac{2 c}{3} - \frac{2 d x}{3}\right )\right )}{\sqrt [3]{a}} \right )}}{27 a^{\frac{11}{3}} d e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/(6*a*d*e**3*(a + b*(c + d*x)**3)**2*(c + d*x)**2) + 4/(9*a**2*d*e**3*(a + b*(c
+ d*x)**3)*(c + d*x)**2) - 10/(9*a**3*d*e**3*(c + d*x)**2) - 20*b**(2/3)*log(a*
*(1/3) + b**(1/3)*(c + d*x))/(27*a**(11/3)*d*e**3) + 10*b**(2/3)*log(a**(2/3) +
a**(1/3)*b**(1/3)*(-c - d*x) + b**(2/3)*(c + d*x)**2)/(27*a**(11/3)*d*e**3) + 20
*sqrt(3)*b**(2/3)*atan(sqrt(3)*(a**(1/3)/3 + b**(1/3)*(-2*c/3 - 2*d*x/3))/a**(1/
3))/(27*a**(11/3)*d*e**3)

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Mathematica [A]  time = 0.26636, size = 195, normalized size = 0.82 \[ \frac{20 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )-\frac{9 a^{5/3} b (c+d x)}{\left (a+b (c+d x)^3\right )^2}-\frac{33 a^{2/3} b (c+d x)}{a+b (c+d x)^3}-\frac{27 a^{2/3}}{(c+d x)^2}-40 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )-40 \sqrt{3} b^{2/3} \tan ^{-1}\left (\frac{2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{54 a^{11/3} d e^3} \]

Antiderivative was successfully verified.

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

[Out]

((-27*a^(2/3))/(c + d*x)^2 - (9*a^(5/3)*b*(c + d*x))/(a + b*(c + d*x)^3)^2 - (33
*a^(2/3)*b*(c + d*x))/(a + b*(c + d*x)^3) - 40*Sqrt[3]*b^(2/3)*ArcTan[(-a^(1/3)
+ 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))] - 40*b^(2/3)*Log[a^(1/3) + b^(1/3)*(c
+ d*x)] + 20*b^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)
^2])/(54*a^(11/3)*d*e^3)

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Maple [C]  time = 0.014, size = 446, normalized size = 1.9 \[ -{\frac{11\,{b}^{2}{d}^{3}{x}^{4}}{18\,{e}^{3}{a}^{3} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}}}-{\frac{22\,c{d}^{2}{b}^{2}{x}^{3}}{9\,{e}^{3}{a}^{3} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}}}-{\frac{11\,{b}^{2}{c}^{2}d{x}^{2}}{3\,{e}^{3}{a}^{3} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}}}-{\frac{22\,{b}^{2}x{c}^{3}}{9\,{e}^{3}{a}^{3} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}}}-{\frac{7\,bx}{9\,{e}^{3}{a}^{2} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}}}-{\frac{11\,{b}^{2}{c}^{4}}{18\,{e}^{3}{a}^{3} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}d}}-{\frac{7\,bc}{9\,{e}^{3}{a}^{2} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}d}}-{\frac{20}{27\,{e}^{3}{a}^{3}d}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{3}b{d}^{3}+3\,{{\it \_Z}}^{2}bc{d}^{2}+3\,{\it \_Z}\,b{c}^{2}d+b{c}^{3}+a \right ) }{\frac{\ln \left ( x-{\it \_R} \right ) }{{d}^{2}{{\it \_R}}^{2}+2\,cd{\it \_R}+{c}^{2}}}}-{\frac{1}{2\,{e}^{3}{a}^{3}d \left ( dx+c \right ) ^{2}}} \]

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)^3,x)

[Out]

-11/18/e^3*b^2/a^3/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2*d^3*x^4-22/9/
e^3*b^2/a^3/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2*c*d^2*x^3-11/3/e^3*b
^2/a^3/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2*c^2*d*x^2-22/9/e^3*b^2/a^
3/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2*x*c^3-7/9/e^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)^2*x-11/18/e^3*b^2/a^3/(b*d^3*x^3+3*b*c*d^2*x
^2+3*b*c^2*d*x+b*c^3+a)^2*c^4/d-7/9/e^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)^2*c/d-20/27/e^3/a^3/d*sum(1/(_R^2*d^2+2*_R*c*d+c^2)*ln(x-_R),_R=Root
Of(_Z^3*b*d^3+3*_Z^2*b*c*d^2+3*_Z*b*c^2*d+b*c^3+a))-1/2/a^3/d/e^3/(d*x+c)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ -\frac{20 \, b^{2} d^{6} x^{6} + 120 \, b^{2} c d^{5} x^{5} + 300 \, b^{2} c^{2} d^{4} x^{4} + 20 \, b^{2} c^{6} + 16 \,{\left (25 \, b^{2} c^{3} + 2 \, a b\right )} d^{3} x^{3} + 32 \, a b c^{3} + 12 \,{\left (25 \, b^{2} c^{4} + 8 \, a b c\right )} d^{2} x^{2} + 24 \,{\left (5 \, b^{2} c^{5} + 4 \, a b c^{2}\right )} d x + 9 \, a^{2}}{18 \,{\left (a^{3} b^{2} d^{9} e^{3} x^{8} + 8 \, a^{3} b^{2} c d^{8} e^{3} x^{7} + 28 \, a^{3} b^{2} c^{2} d^{7} e^{3} x^{6} + 2 \,{\left (28 \, a^{3} b^{2} c^{3} + a^{4} b\right )} d^{6} e^{3} x^{5} + 10 \,{\left (7 \, a^{3} b^{2} c^{4} + a^{4} b c\right )} d^{5} e^{3} x^{4} + 4 \,{\left (14 \, a^{3} b^{2} c^{5} + 5 \, a^{4} b c^{2}\right )} d^{4} e^{3} x^{3} +{\left (28 \, a^{3} b^{2} c^{6} + 20 \, a^{4} b c^{3} + a^{5}\right )} d^{3} e^{3} x^{2} + 2 \,{\left (4 \, a^{3} b^{2} c^{7} + 5 \, a^{4} b c^{4} + a^{5} c\right )} d^{2} e^{3} x +{\left (a^{3} b^{2} c^{8} + 2 \, a^{4} b c^{5} + a^{5} c^{2}\right )} d e^{3}\right )}} - \frac{20 \, b \int \frac{1}{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}\,{d x}}{9 \, a^{3} e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/18*(20*b^2*d^6*x^6 + 120*b^2*c*d^5*x^5 + 300*b^2*c^2*d^4*x^4 + 20*b^2*c^6 + 1
6*(25*b^2*c^3 + 2*a*b)*d^3*x^3 + 32*a*b*c^3 + 12*(25*b^2*c^4 + 8*a*b*c)*d^2*x^2
+ 24*(5*b^2*c^5 + 4*a*b*c^2)*d*x + 9*a^2)/(a^3*b^2*d^9*e^3*x^8 + 8*a^3*b^2*c*d^8
*e^3*x^7 + 28*a^3*b^2*c^2*d^7*e^3*x^6 + 2*(28*a^3*b^2*c^3 + a^4*b)*d^6*e^3*x^5 +
10*(7*a^3*b^2*c^4 + a^4*b*c)*d^5*e^3*x^4 + 4*(14*a^3*b^2*c^5 + 5*a^4*b*c^2)*d^4
*e^3*x^3 + (28*a^3*b^2*c^6 + 20*a^4*b*c^3 + a^5)*d^3*e^3*x^2 + 2*(4*a^3*b^2*c^7
+ 5*a^4*b*c^4 + a^5*c)*d^2*e^3*x + (a^3*b^2*c^8 + 2*a^4*b*c^5 + a^5*c^2)*d*e^3)
- 20/9*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^3*e^3)

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Fricas [A]  time = 0.340866, size = 1447, normalized size = 6.11 \[ \text{result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/162*sqrt(3)*(20*sqrt(3)*(b^2*d^8*x^8 + 8*b^2*c*d^7*x^7 + 28*b^2*c^2*d^6*x^6 +
2*(28*b^2*c^3 + a*b)*d^5*x^5 + b^2*c^8 + 10*(7*b^2*c^4 + a*b*c)*d^4*x^4 + 2*a*b
*c^5 + 4*(14*b^2*c^5 + 5*a*b*c^2)*d^3*x^3 + (28*b^2*c^6 + 20*a*b*c^3 + a^2)*d^2*
x^2 + a^2*c^2 + 2*(4*b^2*c^7 + 5*a*b*c^4 + a^2*c)*d*x)*(-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)) - 40*sqrt(3)*(b^2*d^8*x^8 + 8*b^2*c*d^7*x^7 + 28*b^2*c^2*d^6*x^6 +
2*(28*b^2*c^3 + a*b)*d^5*x^5 + b^2*c^8 + 10*(7*b^2*c^4 + a*b*c)*d^4*x^4 + 2*a*b*
c^5 + 4*(14*b^2*c^5 + 5*a*b*c^2)*d^3*x^3 + (28*b^2*c^6 + 20*a*b*c^3 + a^2)*d^2*x
^2 + a^2*c^2 + 2*(4*b^2*c^7 + 5*a*b*c^4 + a^2*c)*d*x)*(-b^2/a^2)^(1/3)*log(b*d*x
+ b*c - a*(-b^2/a^2)^(1/3)) + 120*(b^2*d^8*x^8 + 8*b^2*c*d^7*x^7 + 28*b^2*c^2*d
^6*x^6 + 2*(28*b^2*c^3 + a*b)*d^5*x^5 + b^2*c^8 + 10*(7*b^2*c^4 + a*b*c)*d^4*x^4
+ 2*a*b*c^5 + 4*(14*b^2*c^5 + 5*a*b*c^2)*d^3*x^3 + (28*b^2*c^6 + 20*a*b*c^3 + a
^2)*d^2*x^2 + a^2*c^2 + 2*(4*b^2*c^7 + 5*a*b*c^4 + a^2*c)*d*x)*(-b^2/a^2)^(1/3)*
arctan(1/3*(sqrt(3)*a*(-b^2/a^2)^(1/3) + 2*sqrt(3)*(b*d*x + b*c))/(a*(-b^2/a^2)^
(1/3))) + 3*sqrt(3)*(20*b^2*d^6*x^6 + 120*b^2*c*d^5*x^5 + 300*b^2*c^2*d^4*x^4 +
20*b^2*c^6 + 16*(25*b^2*c^3 + 2*a*b)*d^3*x^3 + 32*a*b*c^3 + 12*(25*b^2*c^4 + 8*a
*b*c)*d^2*x^2 + 24*(5*b^2*c^5 + 4*a*b*c^2)*d*x + 9*a^2))/(a^3*b^2*d^9*e^3*x^8 +
8*a^3*b^2*c*d^8*e^3*x^7 + 28*a^3*b^2*c^2*d^7*e^3*x^6 + 2*(28*a^3*b^2*c^3 + a^4*b
)*d^6*e^3*x^5 + 10*(7*a^3*b^2*c^4 + a^4*b*c)*d^5*e^3*x^4 + 4*(14*a^3*b^2*c^5 + 5
*a^4*b*c^2)*d^4*e^3*x^3 + (28*a^3*b^2*c^6 + 20*a^4*b*c^3 + a^5)*d^3*e^3*x^2 + 2*
(4*a^3*b^2*c^7 + 5*a^4*b*c^4 + a^5*c)*d^2*e^3*x + (a^3*b^2*c^8 + 2*a^4*b*c^5 + a
^5*c^2)*d*e^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

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)**3,x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{1}{{\left ({\left (d x + c\right )}^{3} b + a\right )}^{3}{\left (d e x + c e\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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