### 3.1877 $$\int \frac{(d+e x)^5}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx$$

Optimal. Leaf size=105 $\frac{e^2 x \left (3 c d^2-2 a e^2\right )}{c^3 d^3}-\frac{\left (c d^2-a e^2\right )^3}{c^4 d^4 (a e+c d x)}+\frac{3 e \left (c d^2-a e^2\right )^2 \log (a e+c d x)}{c^4 d^4}+\frac{e^3 x^2}{2 c^2 d^2}$

[Out]

(e^2*(3*c*d^2 - 2*a*e^2)*x)/(c^3*d^3) + (e^3*x^2)/(2*c^2*d^2) - (c*d^2 - a*e^2)^3/(c^4*d^4*(a*e + c*d*x)) + (3
*e*(c*d^2 - a*e^2)^2*Log[a*e + c*d*x])/(c^4*d^4)

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Rubi [A]  time = 0.0928664, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 35, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.057, Rules used = {626, 43} $\frac{e^2 x \left (3 c d^2-2 a e^2\right )}{c^3 d^3}-\frac{\left (c d^2-a e^2\right )^3}{c^4 d^4 (a e+c d x)}+\frac{3 e \left (c d^2-a e^2\right )^2 \log (a e+c d x)}{c^4 d^4}+\frac{e^3 x^2}{2 c^2 d^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^5/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]

[Out]

(e^2*(3*c*d^2 - 2*a*e^2)*x)/(c^3*d^3) + (e^3*x^2)/(2*c^2*d^2) - (c*d^2 - a*e^2)^3/(c^4*d^4*(a*e + c*d*x)) + (3
*e*(c*d^2 - a*e^2)^2*Log[a*e + c*d*x])/(c^4*d^4)

Rule 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx &=\int \frac{(d+e x)^3}{(a e+c d x)^2} \, dx\\ &=\int \left (\frac{3 c d^2 e^2-2 a e^4}{c^3 d^3}+\frac{e^3 x}{c^2 d^2}+\frac{\left (c d^2-a e^2\right )^3}{c^3 d^3 (a e+c d x)^2}+\frac{3 e \left (c d^2-a e^2\right )^2}{c^3 d^3 (a e+c d x)}\right ) \, dx\\ &=\frac{e^2 \left (3 c d^2-2 a e^2\right ) x}{c^3 d^3}+\frac{e^3 x^2}{2 c^2 d^2}-\frac{\left (c d^2-a e^2\right )^3}{c^4 d^4 (a e+c d x)}+\frac{3 e \left (c d^2-a e^2\right )^2 \log (a e+c d x)}{c^4 d^4}\\ \end{align*}

Mathematica [A]  time = 0.0475453, size = 142, normalized size = 1.35 $\frac{-3 a^2 c d^2 e^4+a^3 e^6+3 a c^2 d^4 e^2-c^3 d^6}{c^4 d^4 (a e+c d x)}+\frac{3 \left (a^2 e^5-2 a c d^2 e^3+c^2 d^4 e\right ) \log (a e+c d x)}{c^4 d^4}-\frac{e^2 x \left (2 a e^2-3 c d^2\right )}{c^3 d^3}+\frac{e^3 x^2}{2 c^2 d^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^5/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]

[Out]

-((e^2*(-3*c*d^2 + 2*a*e^2)*x)/(c^3*d^3)) + (e^3*x^2)/(2*c^2*d^2) + (-(c^3*d^6) + 3*a*c^2*d^4*e^2 - 3*a^2*c*d^
2*e^4 + a^3*e^6)/(c^4*d^4*(a*e + c*d*x)) + (3*(c^2*d^4*e - 2*a*c*d^2*e^3 + a^2*e^5)*Log[a*e + c*d*x])/(c^4*d^4
)

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Maple [A]  time = 0.047, size = 184, normalized size = 1.8 \begin{align*}{\frac{{e}^{3}{x}^{2}}{2\,{c}^{2}{d}^{2}}}-2\,{\frac{{e}^{4}ax}{{c}^{3}{d}^{3}}}+3\,{\frac{{e}^{2}x}{{c}^{2}d}}+{\frac{{a}^{3}{e}^{6}}{{c}^{4}{d}^{4} \left ( cdx+ae \right ) }}-3\,{\frac{{a}^{2}{e}^{4}}{{c}^{3}{d}^{2} \left ( cdx+ae \right ) }}+3\,{\frac{a{e}^{2}}{{c}^{2} \left ( cdx+ae \right ) }}-{\frac{{d}^{2}}{c \left ( cdx+ae \right ) }}+3\,{\frac{{e}^{5}\ln \left ( cdx+ae \right ){a}^{2}}{{c}^{4}{d}^{4}}}-6\,{\frac{{e}^{3}\ln \left ( cdx+ae \right ) a}{{c}^{3}{d}^{2}}}+3\,{\frac{e\ln \left ( cdx+ae \right ) }{{c}^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x)

[Out]

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

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Maxima [A]  time = 1.033, size = 193, normalized size = 1.84 \begin{align*} -\frac{c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}}{c^{5} d^{5} x + a c^{4} d^{4} e} + \frac{c d e^{3} x^{2} + 2 \,{\left (3 \, c d^{2} e^{2} - 2 \, a e^{4}\right )} x}{2 \, c^{3} d^{3}} + \frac{3 \,{\left (c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} \log \left (c d x + a e\right )}{c^{4} d^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.82145, size = 408, normalized size = 3.89 \begin{align*} \frac{c^{3} d^{3} e^{3} x^{3} - 2 \, c^{3} d^{6} + 6 \, a c^{2} d^{4} e^{2} - 6 \, a^{2} c d^{2} e^{4} + 2 \, a^{3} e^{6} + 3 \,{\left (2 \, c^{3} d^{4} e^{2} - a c^{2} d^{2} e^{4}\right )} x^{2} + 2 \,{\left (3 \, a c^{2} d^{3} e^{3} - 2 \, a^{2} c d e^{5}\right )} x + 6 \,{\left (a c^{2} d^{4} e^{2} - 2 \, a^{2} c d^{2} e^{4} + a^{3} e^{6} +{\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x\right )} \log \left (c d x + a e\right )}{2 \,{\left (c^{5} d^{5} x + a c^{4} d^{4} e\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.08642, size = 131, normalized size = 1.25 \begin{align*} \frac{a^{3} e^{6} - 3 a^{2} c d^{2} e^{4} + 3 a c^{2} d^{4} e^{2} - c^{3} d^{6}}{a c^{4} d^{4} e + c^{5} d^{5} x} + \frac{e^{3} x^{2}}{2 c^{2} d^{2}} - \frac{x \left (2 a e^{4} - 3 c d^{2} e^{2}\right )}{c^{3} d^{3}} + \frac{3 e \left (a e^{2} - c d^{2}\right )^{2} \log{\left (a e + c d x \right )}}{c^{4} d^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**5/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2,x)

[Out]

(a**3*e**6 - 3*a**2*c*d**2*e**4 + 3*a*c**2*d**4*e**2 - c**3*d**6)/(a*c**4*d**4*e + c**5*d**5*x) + e**3*x**2/(2
*c**2*d**2) - x*(2*a*e**4 - 3*c*d**2*e**2)/(c**3*d**3) + 3*e*(a*e**2 - c*d**2)**2*log(a*e + c*d*x)/(c**4*d**4)

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Giac [B]  time = 1.37179, size = 629, normalized size = 5.99 \begin{align*} \frac{3 \,{\left (c^{5} d^{10} e - 5 \, a c^{4} d^{8} e^{3} + 10 \, a^{2} c^{3} d^{6} e^{5} - 10 \, a^{3} c^{2} d^{4} e^{7} + 5 \, a^{4} c d^{2} e^{9} - a^{5} e^{11}\right )} \arctan \left (\frac{2 \, c d x e + c d^{2} + a e^{2}}{\sqrt{-c^{2} d^{4} + 2 \, a c d^{2} e^{2} - a^{2} e^{4}}}\right )}{{\left (c^{6} d^{8} - 2 \, a c^{5} d^{6} e^{2} + a^{2} c^{4} d^{4} e^{4}\right )} \sqrt{-c^{2} d^{4} + 2 \, a c d^{2} e^{2} - a^{2} e^{4}}} + \frac{{\left (c^{2} d^{2} x^{2} e^{7} + 6 \, c^{2} d^{3} x e^{6} - 4 \, a c d x e^{8}\right )} e^{\left (-4\right )}}{2 \, c^{4} d^{4}} + \frac{3 \,{\left (c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} \log \left (c d x^{2} e + c d^{2} x + a x e^{2} + a d e\right )}{2 \, c^{4} d^{4}} - \frac{c^{5} d^{11} - 5 \, a c^{4} d^{9} e^{2} + 10 \, a^{2} c^{3} d^{7} e^{4} - 10 \, a^{3} c^{2} d^{5} e^{6} + 5 \, a^{4} c d^{3} e^{8} - a^{5} d e^{10} +{\left (c^{5} d^{10} e - 5 \, a c^{4} d^{8} e^{3} + 10 \, a^{2} c^{3} d^{6} e^{5} - 10 \, a^{3} c^{2} d^{4} e^{7} + 5 \, a^{4} c d^{2} e^{9} - a^{5} e^{11}\right )} x}{{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}{\left (c d x^{2} e + c d^{2} x + a x e^{2} + a d e\right )} c^{4} d^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="giac")

[Out]

3*(c^5*d^10*e - 5*a*c^4*d^8*e^3 + 10*a^2*c^3*d^6*e^5 - 10*a^3*c^2*d^4*e^7 + 5*a^4*c*d^2*e^9 - a^5*e^11)*arctan
((2*c*d*x*e + c*d^2 + a*e^2)/sqrt(-c^2*d^4 + 2*a*c*d^2*e^2 - a^2*e^4))/((c^6*d^8 - 2*a*c^5*d^6*e^2 + a^2*c^4*d
^4*e^4)*sqrt(-c^2*d^4 + 2*a*c*d^2*e^2 - a^2*e^4)) + 1/2*(c^2*d^2*x^2*e^7 + 6*c^2*d^3*x*e^6 - 4*a*c*d*x*e^8)*e^
(-4)/(c^4*d^4) + 3/2*(c^2*d^4*e - 2*a*c*d^2*e^3 + a^2*e^5)*log(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e)/(c^4*d^4
) - (c^5*d^11 - 5*a*c^4*d^9*e^2 + 10*a^2*c^3*d^7*e^4 - 10*a^3*c^2*d^5*e^6 + 5*a^4*c*d^3*e^8 - a^5*d*e^10 + (c^
5*d^10*e - 5*a*c^4*d^8*e^3 + 10*a^2*c^3*d^6*e^5 - 10*a^3*c^2*d^4*e^7 + 5*a^4*c*d^2*e^9 - a^5*e^11)*x)/((c^2*d^
4 - 2*a*c*d^2*e^2 + a^2*e^4)*(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e)*c^4*d^4)