\(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^3}{(d+e x)^7} \, dx\) [1860]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 35, antiderivative size = 105 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^7} \, dx=\frac {\left (c d^2-a e^2\right )^3}{3 e^4 (d+e x)^3}-\frac {3 c d \left (c d^2-a e^2\right )^2}{2 e^4 (d+e x)^2}+\frac {3 c^2 d^2 \left (c d^2-a e^2\right )}{e^4 (d+e x)}+\frac {c^3 d^3 \log (d+e x)}{e^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {640, 45} \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^7} \, dx=\frac {3 c^2 d^2 \left (c d^2-a e^2\right )}{e^4 (d+e x)}-\frac {3 c d \left (c d^2-a e^2\right )^2}{2 e^4 (d+e x)^2}+\frac {\left (c d^2-a e^2\right )^3}{3 e^4 (d+e x)^3}+\frac {c^3 d^3 \log (d+e x)}{e^4} \]

[In]

Int[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3/(d + e*x)^7,x]

[Out]

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

Rule 45

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])

Rule 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^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]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^7} \, dx=\frac {\frac {\left (c d^2-a e^2\right ) \left (2 a^2 e^4+a c d e^2 (5 d+9 e x)+c^2 d^2 \left (11 d^2+27 d e x+18 e^2 x^2\right )\right )}{(d+e x)^3}+6 c^3 d^3 \log (d+e x)}{6 e^4} \]

[In]

Integrate[(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3/(d + e*x)^7,x]

[Out]

(((c*d^2 - a*e^2)*(2*a^2*e^4 + a*c*d*e^2*(5*d + 9*e*x) + c^2*d^2*(11*d^2 + 27*d*e*x + 18*e^2*x^2)))/(d + e*x)^
3 + 6*c^3*d^3*Log[d + e*x])/(6*e^4)

Maple [A] (verified)

Time = 2.37 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.27

method result size
risch \(\frac {-\frac {3 c^{2} d^{2} \left (e^{2} a -c \,d^{2}\right ) x^{2}}{e^{2}}-\frac {3 c d \left (a^{2} e^{4}+2 a c \,d^{2} e^{2}-3 c^{2} d^{4}\right ) x}{2 e^{3}}-\frac {2 e^{6} a^{3}+3 d^{2} e^{4} a^{2} c +6 d^{4} e^{2} c^{2} a -11 c^{3} d^{6}}{6 e^{4}}}{\left (e x +d \right )^{3}}+\frac {c^{3} d^{3} \ln \left (e x +d \right )}{e^{4}}\) \(133\)
default \(-\frac {3 c^{2} d^{2} \left (e^{2} a -c \,d^{2}\right )}{e^{4} \left (e x +d \right )}-\frac {e^{6} a^{3}-3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a -c^{3} d^{6}}{3 e^{4} \left (e x +d \right )^{3}}-\frac {3 c d \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}{2 e^{4} \left (e x +d \right )^{2}}+\frac {c^{3} d^{3} \ln \left (e x +d \right )}{e^{4}}\) \(139\)
parallelrisch \(\frac {6 \ln \left (e x +d \right ) x^{3} c^{3} d^{3} e^{3}+18 \ln \left (e x +d \right ) x^{2} c^{3} d^{4} e^{2}+18 \ln \left (e x +d \right ) x \,c^{3} d^{5} e -18 x^{2} a \,c^{2} d^{2} e^{4}+18 x^{2} c^{3} d^{4} e^{2}+6 \ln \left (e x +d \right ) c^{3} d^{6}-9 x \,a^{2} c d \,e^{5}-18 x a \,c^{2} d^{3} e^{3}+27 x \,c^{3} d^{5} e -2 e^{6} a^{3}-3 d^{2} e^{4} a^{2} c -6 d^{4} e^{2} c^{2} a +11 c^{3} d^{6}}{6 e^{4} \left (e x +d \right )^{3}}\) \(187\)
norman \(\frac {-\frac {d^{3} \left (2 a^{3} e^{8}+3 a^{2} c \,d^{2} e^{6}+6 a \,c^{2} d^{4} e^{4}-11 c^{3} d^{6} e^{2}\right )}{6 e^{6}}-\frac {\left (a^{3} e^{8}+15 a^{2} c \,d^{2} e^{6}+57 a \,c^{2} d^{4} e^{4}-73 c^{3} d^{6} e^{2}\right ) x^{3}}{3 e^{3}}-\frac {3 d \left (a \,c^{2} d \,e^{4}-e^{2} c^{3} d^{3}\right ) x^{5}}{e}-\frac {3 d \left (a^{2} c \,e^{6}+8 a \,c^{2} d^{2} e^{4}-9 c^{3} d^{4} e^{2}\right ) x^{4}}{2 e^{2}}-\frac {d \left (a^{3} e^{8}+6 a^{2} c \,d^{2} e^{6}+15 a \,c^{2} d^{4} e^{4}-22 c^{3} d^{6} e^{2}\right ) x^{2}}{e^{4}}-\frac {d^{2} \left (a^{3} e^{8}+3 a^{2} c \,d^{2} e^{6}+6 a \,c^{2} d^{4} e^{4}-10 c^{3} d^{6} e^{2}\right ) x}{e^{5}}}{\left (e x +d \right )^{6}}+\frac {c^{3} d^{3} \ln \left (e x +d \right )}{e^{4}}\) \(305\)

[In]

int((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^7,x,method=_RETURNVERBOSE)

[Out]

(-3*c^2*d^2*(a*e^2-c*d^2)/e^2*x^2-3/2*c*d*(a^2*e^4+2*a*c*d^2*e^2-3*c^2*d^4)/e^3*x-1/6*(2*a^3*e^6+3*a^2*c*d^2*e
^4+6*a*c^2*d^4*e^2-11*c^3*d^6)/e^4)/(e*x+d)^3+c^3*d^3*ln(e*x+d)/e^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.85 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^7} \, dx=\frac {11 \, c^{3} d^{6} - 6 \, a c^{2} d^{4} e^{2} - 3 \, a^{2} c d^{2} e^{4} - 2 \, a^{3} e^{6} + 18 \, {\left (c^{3} d^{4} e^{2} - a c^{2} d^{2} e^{4}\right )} x^{2} + 9 \, {\left (3 \, c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} - a^{2} c d e^{5}\right )} x + 6 \, {\left (c^{3} d^{3} e^{3} x^{3} + 3 \, c^{3} d^{4} e^{2} x^{2} + 3 \, c^{3} d^{5} e x + c^{3} d^{6}\right )} \log \left (e x + d\right )}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^7,x, algorithm="fricas")

[Out]

1/6*(11*c^3*d^6 - 6*a*c^2*d^4*e^2 - 3*a^2*c*d^2*e^4 - 2*a^3*e^6 + 18*(c^3*d^4*e^2 - a*c^2*d^2*e^4)*x^2 + 9*(3*
c^3*d^5*e - 2*a*c^2*d^3*e^3 - a^2*c*d*e^5)*x + 6*(c^3*d^3*e^3*x^3 + 3*c^3*d^4*e^2*x^2 + 3*c^3*d^5*e*x + c^3*d^
6)*log(e*x + d))/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)

Sympy [A] (verification not implemented)

Time = 7.19 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.55 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^7} \, dx=\frac {c^{3} d^{3} \log {\left (d + e x \right )}}{e^{4}} + \frac {- 2 a^{3} e^{6} - 3 a^{2} c d^{2} e^{4} - 6 a c^{2} d^{4} e^{2} + 11 c^{3} d^{6} + x^{2} \left (- 18 a c^{2} d^{2} e^{4} + 18 c^{3} d^{4} e^{2}\right ) + x \left (- 9 a^{2} c d e^{5} - 18 a c^{2} d^{3} e^{3} + 27 c^{3} d^{5} e\right )}{6 d^{3} e^{4} + 18 d^{2} e^{5} x + 18 d e^{6} x^{2} + 6 e^{7} x^{3}} \]

[In]

integrate((a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3/(e*x+d)**7,x)

[Out]

c**3*d**3*log(d + e*x)/e**4 + (-2*a**3*e**6 - 3*a**2*c*d**2*e**4 - 6*a*c**2*d**4*e**2 + 11*c**3*d**6 + x**2*(-
18*a*c**2*d**2*e**4 + 18*c**3*d**4*e**2) + x*(-9*a**2*c*d*e**5 - 18*a*c**2*d**3*e**3 + 27*c**3*d**5*e))/(6*d**
3*e**4 + 18*d**2*e**5*x + 18*d*e**6*x**2 + 6*e**7*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^7} \, dx=\frac {c^{3} d^{3} \log \left (e x + d\right )}{e^{4}} + \frac {11 \, c^{3} d^{6} - 6 \, a c^{2} d^{4} e^{2} - 3 \, a^{2} c d^{2} e^{4} - 2 \, a^{3} e^{6} + 18 \, {\left (c^{3} d^{4} e^{2} - a c^{2} d^{2} e^{4}\right )} x^{2} + 9 \, {\left (3 \, c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} - a^{2} c d e^{5}\right )} x}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^7,x, algorithm="maxima")

[Out]

c^3*d^3*log(e*x + d)/e^4 + 1/6*(11*c^3*d^6 - 6*a*c^2*d^4*e^2 - 3*a^2*c*d^2*e^4 - 2*a^3*e^6 + 18*(c^3*d^4*e^2 -
 a*c^2*d^2*e^4)*x^2 + 9*(3*c^3*d^5*e - 2*a*c^2*d^3*e^3 - a^2*c*d*e^5)*x)/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x
+ d^3*e^4)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^7} \, dx=\frac {c^{3} d^{3} \log \left ({\left | e x + d \right |}\right )}{e^{4}} + \frac {18 \, {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x^{2} + 9 \, {\left (3 \, c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} - a^{2} c d e^{4}\right )} x + \frac {11 \, c^{3} d^{6} - 6 \, a c^{2} d^{4} e^{2} - 3 \, a^{2} c d^{2} e^{4} - 2 \, a^{3} e^{6}}{e}}{6 \, {\left (e x + d\right )}^{3} e^{3}} \]

[In]

integrate((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3/(e*x+d)^7,x, algorithm="giac")

[Out]

c^3*d^3*log(abs(e*x + d))/e^4 + 1/6*(18*(c^3*d^4*e - a*c^2*d^2*e^3)*x^2 + 9*(3*c^3*d^5 - 2*a*c^2*d^3*e^2 - a^2
*c*d*e^4)*x + (11*c^3*d^6 - 6*a*c^2*d^4*e^2 - 3*a^2*c*d^2*e^4 - 2*a^3*e^6)/e)/((e*x + d)^3*e^3)

Mupad [B] (verification not implemented)

Time = 10.05 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3}{(d+e x)^7} \, dx=\frac {c^3\,d^3\,\ln \left (d+e\,x\right )}{e^4}-\frac {\frac {2\,a^3\,e^6+3\,a^2\,c\,d^2\,e^4+6\,a\,c^2\,d^4\,e^2-11\,c^3\,d^6}{6\,e^4}+\frac {3\,x\,\left (a^2\,c\,d\,e^4+2\,a\,c^2\,d^3\,e^2-3\,c^3\,d^5\right )}{2\,e^3}+\frac {3\,c^2\,d^2\,x^2\,\left (a\,e^2-c\,d^2\right )}{e^2}}{d^3+3\,d^2\,e\,x+3\,d\,e^2\,x^2+e^3\,x^3} \]

[In]

int((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3/(d + e*x)^7,x)

[Out]

(c^3*d^3*log(d + e*x))/e^4 - ((2*a^3*e^6 - 11*c^3*d^6 + 6*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4)/(6*e^4) + (3*x*(2*a
*c^2*d^3*e^2 - 3*c^3*d^5 + a^2*c*d*e^4))/(2*e^3) + (3*c^2*d^2*x^2*(a*e^2 - c*d^2))/e^2)/(d^3 + e^3*x^3 + 3*d*e
^2*x^2 + 3*d^2*e*x)