3.1859 \(\int \frac{(a d e+(c d^2+a e^2) x+c d e x^2)^3}{(d+e x)^6} \, dx\)

Optimal. Leaf size=97 \[ -\frac{3 c^2 d^2 \left (c d^2-a e^2\right ) \log (d+e x)}{e^4}-\frac{3 c d \left (c d^2-a e^2\right )^2}{e^4 (d+e x)}+\frac{\left (c d^2-a e^2\right )^3}{2 e^4 (d+e x)^2}+\frac{c^3 d^3 x}{e^3} \]

[Out]

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

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Rubi [A]  time = 0.0737391, antiderivative size = 97, 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} \[ -\frac{3 c^2 d^2 \left (c d^2-a e^2\right ) \log (d+e x)}{e^4}-\frac{3 c d \left (c d^2-a e^2\right )^2}{e^4 (d+e x)}+\frac{\left (c d^2-a e^2\right )^3}{2 e^4 (d+e x)^2}+\frac{c^3 d^3 x}{e^3} \]

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)^6,x]

[Out]

(c^3*d^3*x)/e^3 + (c*d^2 - a*e^2)^3/(2*e^4*(d + e*x)^2) - (3*c*d*(c*d^2 - a*e^2)^2)/(e^4*(d + e*x)) - (3*c^2*d
^2*(c*d^2 - a*e^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)^6} \, dx &=\int \frac{(a e+c d x)^3}{(d+e x)^3} \, dx\\ &=\int \left (\frac{c^3 d^3}{e^3}+\frac{\left (-c d^2+a e^2\right )^3}{e^3 (d+e x)^3}+\frac{3 c d \left (c d^2-a e^2\right )^2}{e^3 (d+e x)^2}-\frac{3 c^2 d^2 \left (c d^2-a e^2\right )}{e^3 (d+e x)}\right ) \, dx\\ &=\frac{c^3 d^3 x}{e^3}+\frac{\left (c d^2-a e^2\right )^3}{2 e^4 (d+e x)^2}-\frac{3 c d \left (c d^2-a e^2\right )^2}{e^4 (d+e x)}-\frac{3 c^2 d^2 \left (c d^2-a e^2\right ) \log (d+e x)}{e^4}\\ \end{align*}

Mathematica [A]  time = 0.0454638, size = 129, normalized size = 1.33 \[ \frac{-3 a^2 c d e^4 (d+2 e x)-a^3 e^6+3 a c^2 d^3 e^2 (3 d+4 e x)-6 c^2 d^2 (d+e x)^2 \left (c d^2-a e^2\right ) \log (d+e x)+c^3 d^3 \left (-4 d^2 e x-5 d^3+4 d e^2 x^2+2 e^3 x^3\right )}{2 e^4 (d+e x)^2} \]

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)^6,x]

[Out]

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

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Maple [A]  time = 0.048, size = 167, normalized size = 1.7 \begin{align*}{\frac{{c}^{3}{d}^{3}x}{{e}^{3}}}-{\frac{{e}^{2}{a}^{3}}{2\, \left ( ex+d \right ) ^{2}}}+{\frac{3\,{a}^{2}c{d}^{2}}{2\, \left ( ex+d \right ) ^{2}}}-{\frac{3\,a{c}^{2}{d}^{4}}{2\,{e}^{2} \left ( ex+d \right ) ^{2}}}+{\frac{{c}^{3}{d}^{6}}{2\,{e}^{4} \left ( ex+d \right ) ^{2}}}+3\,{\frac{{c}^{2}{d}^{2}\ln \left ( ex+d \right ) a}{{e}^{2}}}-3\,{\frac{{c}^{3}{d}^{4}\ln \left ( ex+d \right ) }{{e}^{4}}}-3\,{\frac{cd{a}^{2}}{ex+d}}+6\,{\frac{{c}^{2}{d}^{3}a}{{e}^{2} \left ( ex+d \right ) }}-3\,{\frac{{c}^{3}{d}^{5}}{{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)^6,x)

[Out]

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

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Maxima [A]  time = 1.09244, size = 192, normalized size = 1.98 \begin{align*} \frac{c^{3} d^{3} x}{e^{3}} - \frac{5 \, c^{3} d^{6} - 9 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6} + 6 \,{\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x}{2 \,{\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} - \frac{3 \,{\left (c^{3} d^{4} - a c^{2} d^{2} e^{2}\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)^6,x, algorithm="maxima")

[Out]

c^3*d^3*x/e^3 - 1/2*(5*c^3*d^6 - 9*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 + a^3*e^6 + 6*(c^3*d^5*e - 2*a*c^2*d^3*e^3
+ a^2*c*d*e^5)*x)/(e^6*x^2 + 2*d*e^5*x + d^2*e^4) - 3*(c^3*d^4 - a*c^2*d^2*e^2)*log(e*x + d)/e^4

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Fricas [B]  time = 1.5291, size = 408, normalized size = 4.21 \begin{align*} \frac{2 \, c^{3} d^{3} e^{3} x^{3} + 4 \, c^{3} d^{4} e^{2} x^{2} - 5 \, c^{3} d^{6} + 9 \, a c^{2} d^{4} e^{2} - 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6} - 2 \,{\left (2 \, c^{3} d^{5} e - 6 \, a c^{2} d^{3} e^{3} + 3 \, a^{2} c d e^{5}\right )} x - 6 \,{\left (c^{3} d^{6} - a c^{2} d^{4} e^{2} +{\left (c^{3} d^{4} e^{2} - a c^{2} d^{2} e^{4}\right )} x^{2} + 2 \,{\left (c^{3} d^{5} e - a c^{2} d^{3} e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \,{\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} 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)^6,x, algorithm="fricas")

[Out]

1/2*(2*c^3*d^3*e^3*x^3 + 4*c^3*d^4*e^2*x^2 - 5*c^3*d^6 + 9*a*c^2*d^4*e^2 - 3*a^2*c*d^2*e^4 - a^3*e^6 - 2*(2*c^
3*d^5*e - 6*a*c^2*d^3*e^3 + 3*a^2*c*d*e^5)*x - 6*(c^3*d^6 - a*c^2*d^4*e^2 + (c^3*d^4*e^2 - a*c^2*d^2*e^4)*x^2
+ 2*(c^3*d^5*e - a*c^2*d^3*e^3)*x)*log(e*x + d))/(e^6*x^2 + 2*d*e^5*x + d^2*e^4)

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Sympy [A]  time = 1.92822, size = 144, normalized size = 1.48 \begin{align*} \frac{c^{3} d^{3} x}{e^{3}} + \frac{3 c^{2} d^{2} \left (a e^{2} - c d^{2}\right ) \log{\left (d + e x \right )}}{e^{4}} - \frac{a^{3} e^{6} + 3 a^{2} c d^{2} e^{4} - 9 a c^{2} d^{4} e^{2} + 5 c^{3} d^{6} + x \left (6 a^{2} c d e^{5} - 12 a c^{2} d^{3} e^{3} + 6 c^{3} d^{5} e\right )}{2 d^{2} e^{4} + 4 d e^{5} x + 2 e^{6} x^{2}} \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)**6,x)

[Out]

c**3*d**3*x/e**3 + 3*c**2*d**2*(a*e**2 - c*d**2)*log(d + e*x)/e**4 - (a**3*e**6 + 3*a**2*c*d**2*e**4 - 9*a*c**
2*d**4*e**2 + 5*c**3*d**6 + x*(6*a**2*c*d*e**5 - 12*a*c**2*d**3*e**3 + 6*c**3*d**5*e))/(2*d**2*e**4 + 4*d*e**5
*x + 2*e**6*x**2)

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Giac [B]  time = 1.20684, size = 352, normalized size = 3.63 \begin{align*} c^{3} d^{3} x e^{\left (-3\right )} - 3 \,{\left (c^{3} d^{4} - a c^{2} d^{2} e^{2}\right )} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) - \frac{{\left (5 \, c^{3} d^{9} - 9 \, a c^{2} d^{7} e^{2} + 3 \, a^{2} c d^{5} e^{4} + a^{3} d^{3} e^{6} + 6 \,{\left (c^{3} d^{5} e^{4} - 2 \, a c^{2} d^{3} e^{6} + a^{2} c d e^{8}\right )} x^{4} +{\left (23 \, c^{3} d^{6} e^{3} - 45 \, a c^{2} d^{4} e^{5} + 21 \, a^{2} c d^{2} e^{7} + a^{3} e^{9}\right )} x^{3} + 3 \,{\left (11 \, c^{3} d^{7} e^{2} - 21 \, a c^{2} d^{5} e^{4} + 9 \, a^{2} c d^{3} e^{6} + a^{3} d e^{8}\right )} x^{2} + 3 \,{\left (7 \, c^{3} d^{8} e - 13 \, a c^{2} d^{6} e^{3} + 5 \, a^{2} c d^{4} e^{5} + a^{3} d^{2} e^{7}\right )} x\right )} e^{\left (-4\right )}}{2 \,{\left (x e + d\right )}^{5}} \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)^6,x, algorithm="giac")

[Out]

c^3*d^3*x*e^(-3) - 3*(c^3*d^4 - a*c^2*d^2*e^2)*e^(-4)*log(abs(x*e + d)) - 1/2*(5*c^3*d^9 - 9*a*c^2*d^7*e^2 + 3
*a^2*c*d^5*e^4 + a^3*d^3*e^6 + 6*(c^3*d^5*e^4 - 2*a*c^2*d^3*e^6 + a^2*c*d*e^8)*x^4 + (23*c^3*d^6*e^3 - 45*a*c^
2*d^4*e^5 + 21*a^2*c*d^2*e^7 + a^3*e^9)*x^3 + 3*(11*c^3*d^7*e^2 - 21*a*c^2*d^5*e^4 + 9*a^2*c*d^3*e^6 + a^3*d*e
^8)*x^2 + 3*(7*c^3*d^8*e - 13*a*c^2*d^6*e^3 + 5*a^2*c*d^4*e^5 + a^3*d^2*e^7)*x)*e^(-4)/(x*e + d)^5