\(\int (a e+c d x)^n (d+e x)^m (a d e+(c d^2+a e^2) x+c d e x^2)^{-m} \, dx\) [780]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 47, antiderivative size = 65 \[ \int (a e+c d x)^n (d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=\frac {(a e+c d x)^n (d+e x)^{-1+m} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{1-m}}{c d (1-m+n)} \]

[Out]

(c*d*x+a*e)^n*(e*x+d)^(-1+m)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1-m)/c/d/(1-m+n)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {872} \[ \int (a e+c d x)^n (d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=\frac {(d+e x)^{m-1} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{1-m} (a e+c d x)^n}{c d (-m+n+1)} \]

[In]

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

[Out]

((a*e + c*d*x)^n*(d + e*x)^(-1 + m)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(1 - m))/(c*d*(1 - m + n))

Rule 872

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Simp[(-e)*(d + e*x)^(m - 1)*(f + g*x)^n*((a + b*x + c*x^2)^(p + 1)/(c*(m - n - 1))), x] /; FreeQ[{a, b, c, d,
e, f, g, m, n, p}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !Intege
rQ[p] && EqQ[m + p, 0] && EqQ[c*e*f + c*d*g - b*e*g, 0] && NeQ[m - n - 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(a e+c d x)^n (d+e x)^{-1+m} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{1-m}}{c d (1-m+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.80 \[ \int (a e+c d x)^n (d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=\frac {(a e+c d x)^{1+n} (d+e x)^m ((a e+c d x) (d+e x))^{-m}}{c d (1-m+n)} \]

[In]

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

[Out]

((a*e + c*d*x)^(1 + n)*(d + e*x)^m)/(c*d*(1 - m + n)*((a*e + c*d*x)*(d + e*x))^m)

Maple [A] (verified)

Time = 3.88 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.98

method result size
gosper \(-\frac {\left (e x +d \right )^{m} \left (c d x +a e \right )^{1+n} \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{-m}}{c d \left (-1+m -n \right )}\) \(64\)
parallelrisch \(-\frac {\left (x \left (e x +d \right )^{m} \left (c d x +a e \right )^{n} c d e m +\left (e x +d \right )^{m} \left (c d x +a e \right )^{n} a \,e^{2} m \right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{-m}}{m c d e \left (-1+m -n \right )}\) \(98\)
risch \(-\frac {\left (c d x +a e \right )^{n} \left (c d x +a e \right ) \left (c d x +a e \right )^{-m} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i \left (c d x +a e \right ) \left (e x +d \right )\right ) m \left (-\operatorname {csgn}\left (i \left (c d x +a e \right ) \left (e x +d \right )\right )+\operatorname {csgn}\left (i \left (c d x +a e \right )\right )\right ) \left (-\operatorname {csgn}\left (i \left (c d x +a e \right ) \left (e x +d \right )\right )+\operatorname {csgn}\left (i \left (e x +d \right )\right )\right )}{2}}}{c d \left (-1+m -n \right )}\) \(131\)

[In]

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

[Out]

-1/c/d/(-1+m-n)*(e*x+d)^m*(c*d*x+a*e)^(1+n)/((c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^m)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02 \[ \int (a e+c d x)^n (d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=-\frac {{\left (c d x + a e\right )} {\left (c d x + a e\right )}^{n} {\left (e x + d\right )}^{m} e^{\left (-m \log \left (c d x + a e\right ) - m \log \left (e x + d\right )\right )}}{c d m - c d n - c d} \]

[In]

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

[Out]

-(c*d*x + a*e)*(c*d*x + a*e)^n*(e*x + d)^m*e^(-m*log(c*d*x + a*e) - m*log(e*x + d))/(c*d*m - c*d*n - c*d)

Sympy [F(-1)]

Timed out. \[ \int (a e+c d x)^n (d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.75 \[ \int (a e+c d x)^n (d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=-\frac {{\left (c d x + a e\right )} e^{\left (-m \log \left (c d x + a e\right ) + n \log \left (c d x + a e\right )\right )}}{c d {\left (m - n - 1\right )}} \]

[In]

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

[Out]

-(c*d*x + a*e)*e^(-m*log(c*d*x + a*e) + n*log(c*d*x + a*e))/(c*d*(m - n - 1))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.63 \[ \int (a e+c d x)^n (d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=-\frac {{\left (c d x + a e\right )}^{n} {\left (e x + d\right )}^{m} c d x e^{\left (-m \log \left (c d x + a e\right ) - m \log \left (e x + d\right )\right )} + {\left (c d x + a e\right )}^{n} {\left (e x + d\right )}^{m} a e e^{\left (-m \log \left (c d x + a e\right ) - m \log \left (e x + d\right )\right )}}{c d m - c d n - c d} \]

[In]

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

[Out]

-((c*d*x + a*e)^n*(e*x + d)^m*c*d*x*e^(-m*log(c*d*x + a*e) - m*log(e*x + d)) + (c*d*x + a*e)^n*(e*x + d)^m*a*e
*e^(-m*log(c*d*x + a*e) - m*log(e*x + d)))/(c*d*m - c*d*n - c*d)

Mupad [B] (verification not implemented)

Time = 12.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97 \[ \int (a e+c d x)^n (d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=\frac {{\left (a\,e+c\,d\,x\right )}^{n+1}\,{\left (d+e\,x\right )}^m}{c\,d\,{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^m\,\left (n-m+1\right )} \]

[In]

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

[Out]

((a*e + c*d*x)^(n + 1)*(d + e*x)^m)/(c*d*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^m*(n - m + 1))