3.1858 \(\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^3}{(d+e x)^5} \, dx\)

Optimal. Leaf size=94 \[ -\frac{c^2 d^2 x \left (2 c d^2-3 a e^2\right )}{e^3}+\frac{\left (c d^2-a e^2\right )^3}{e^4 (d+e x)}+\frac{3 c d \left (c d^2-a e^2\right )^2 \log (d+e x)}{e^4}+\frac{c^3 d^3 x^2}{2 e^2} \]

[Out]

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

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Rubi [A]  time = 0.0814272, antiderivative size = 94, 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} \[ -\frac{c^2 d^2 x \left (2 c d^2-3 a e^2\right )}{e^3}+\frac{\left (c d^2-a e^2\right )^3}{e^4 (d+e x)}+\frac{3 c d \left (c d^2-a e^2\right )^2 \log (d+e x)}{e^4}+\frac{c^3 d^3 x^2}{2 e^2} \]

Antiderivative was successfully verified.

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3/(d + e*x)^5,x]

[Out]

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

Mathematica [A]  time = 0.0412388, size = 129, normalized size = 1.37 \[ \frac{6 a^2 c d^2 e^4-2 a^3 e^6+6 a c^2 d^2 e^2 \left (-d^2+d e x+e^2 x^2\right )+6 c d (d+e x) \left (c d^2-a e^2\right )^2 \log (d+e x)+c^3 d^3 \left (-4 d^2 e x+2 d^3-3 d e^2 x^2+e^3 x^3\right )}{2 e^4 (d+e x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3/(d + e*x)^5,x]

[Out]

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

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Maple [A]  time = 0.046, size = 156, normalized size = 1.7 \begin{align*}{\frac{{c}^{3}{d}^{3}{x}^{2}}{2\,{e}^{2}}}+3\,{\frac{a{c}^{2}{d}^{2}x}{e}}-2\,{\frac{{c}^{3}{d}^{4}x}{{e}^{3}}}+3\,dc\ln \left ( ex+d \right ){a}^{2}-6\,{\frac{{c}^{2}{d}^{3}\ln \left ( ex+d \right ) a}{{e}^{2}}}+3\,{\frac{{c}^{3}{d}^{5}\ln \left ( ex+d \right ) }{{e}^{4}}}-{\frac{{e}^{2}{a}^{3}}{ex+d}}+3\,{\frac{{a}^{2}c{d}^{2}}{ex+d}}-3\,{\frac{a{c}^{2}{d}^{4}}{{e}^{2} \left ( ex+d \right ) }}+{\frac{{c}^{3}{d}^{6}}{{e}^{4} \left ( ex+d \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^5,x)

[Out]

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

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Maxima [A]  time = 1.06743, size = 184, normalized size = 1.96 \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}}{e^{5} x + d e^{4}} + \frac{c^{3} d^{3} e x^{2} - 2 \,{\left (2 \, c^{3} d^{4} - 3 \, a c^{2} d^{2} e^{2}\right )} x}{2 \, e^{3}} + \frac{3 \,{\left (c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} \log \left (e x + d\right )}{e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [B]  time = 1.62843, size = 386, normalized size = 4.11 \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 (c^{3} d^{4} e^{2} - 2 \, a c^{2} d^{2} e^{4}\right )} x^{2} - 2 \,{\left (2 \, c^{3} d^{5} e - 3 \, a c^{2} d^{3} e^{3}\right )} x + 6 \,{\left (c^{3} d^{6} - 2 \, a c^{2} d^{4} e^{2} + a^{2} c d^{2} e^{4} +{\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 (e x + d\right )}{2 \,{\left (e^{5} x + d e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [A]  time = 1.09834, size = 119, normalized size = 1.27 \begin{align*} \frac{c^{3} d^{3} x^{2}}{2 e^{2}} + \frac{3 c d \left (a e^{2} - c d^{2}\right )^{2} \log{\left (d + e x \right )}}{e^{4}} - \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}}{d e^{4} + e^{5} x} + \frac{x \left (3 a c^{2} d^{2} e^{2} - 2 c^{3} d^{4}\right )}{e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3/(e*x+d)**5,x)

[Out]

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

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Giac [A]  time = 1.2441, size = 240, normalized size = 2.55 \begin{align*} \frac{1}{2} \,{\left (c^{3} d^{3} - \frac{6 \,{\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} e^{\left (-1\right )}}{x e + d}\right )}{\left (x e + d\right )}^{2} e^{\left (-4\right )} - 3 \,{\left (c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} e^{\left (-4\right )} \log \left (\frac{{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) +{\left (\frac{c^{3} d^{6} e^{20}}{x e + d} - \frac{3 \, a c^{2} d^{4} e^{22}}{x e + d} + \frac{3 \, a^{2} c d^{2} e^{24}}{x e + d} - \frac{a^{3} e^{26}}{x e + d}\right )} e^{\left (-24\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^5,x, algorithm="giac")

[Out]

1/2*(c^3*d^3 - 6*(c^3*d^4*e - a*c^2*d^2*e^3)*e^(-1)/(x*e + d))*(x*e + d)^2*e^(-4) - 3*(c^3*d^5 - 2*a*c^2*d^3*e
^2 + a^2*c*d*e^4)*e^(-4)*log(abs(x*e + d)*e^(-1)/(x*e + d)^2) + (c^3*d^6*e^20/(x*e + d) - 3*a*c^2*d^4*e^22/(x*
e + d) + 3*a^2*c*d^2*e^24/(x*e + d) - a^3*e^26/(x*e + d))*e^(-24)