\(\int (d+e x)^m (c d^2 e g-e (c d^2+a e^2) g-c d e^2 g x)^{-1+m} (a d e+(c d^2+a e^2) x+c d e x^2)^{-m} \, dx\) [781]

Optimal result

Integrand size = 73, antiderivative size = 78 \[ \int (d+e x)^m \left (c d^2 e g-e \left (c d^2+a e^2\right ) g-c d e^2 g x\right )^{-1+m} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=-\frac {(d+e x)^m \left (-a e^3 g-c d e^2 g x\right )^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \log (a e+c d x)}{c d e^2 g} \]

[Out]

-(e*x+d)^m*(-c*d*e^2*g*x-a*e^3*g)^m*ln(c*d*x+a*e)/c/d/e^2/g/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {905, 23, 31} \[ \int (d+e x)^m \left (c d^2 e g-e \left (c d^2+a e^2\right ) g-c d e^2 g x\right )^{-1+m} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=-\frac {(d+e x)^m \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{-m} \log (a e+c d x) \left (-a e^3 g-c d e^2 g x\right )^m}{c d e^2 g} \]

[In]

Int[((d + e*x)^m*(c*d^2*e*g - e*(c*d^2 + a*e^2)*g - c*d*e^2*g*x)^(-1 + m))/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*
x^2)^m,x]

[Out]

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

Rule 23

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((c_) + (d_.)*(v_))^(n_), x_Symbol] :> Dist[(a + b*v)^m/(c + d*v)^m, Int[u*
(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] &&  !(IntegerQ[m] || IntegerQ[n
] || GtQ[b/d, 0])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 905

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Dist[(a + b*x + c*x^2)^FracPart[p]/((d + e*x)^FracPart[p]*(a/d + (c*x)/e)^FracPart[p]), Int[(d + e*x)^(m + p)*
(f + g*x)^n*(a/d + (c/e)*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, 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] &&  !IntegerQ[p] &&  !IGtQ[m, 0] &&  !IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \left ((a e+c d x)^m (d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m}\right ) \int (a e+c d x)^{-m} \left (c d^2 e g-e \left (c d^2+a e^2\right ) g-c d e^2 g x\right )^{-1+m} \, dx \\ & = \left ((d+e x)^m \left (c d^2 e g-e \left (c d^2+a e^2\right ) g-c d e^2 g x\right )^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m}\right ) \int \frac {1}{c d^2 e g-e \left (c d^2+a e^2\right ) g-c d e^2 g x} \, dx \\ & = -\frac {(d+e x)^m \left (-a e^3 g-c d e^2 g x\right )^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \log (a e+c d x)}{c d e^2 g} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.82 \[ \int (d+e x)^m \left (c d^2 e g-e \left (c d^2+a e^2\right ) g-c d e^2 g x\right )^{-1+m} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=-\frac {\left (-e^2 g (a e+c d x)\right )^m (d+e x)^m ((a e+c d x) (d+e x))^{-m} \log (a e+c d x)}{c d e^2 g} \]

[In]

Integrate[((d + e*x)^m*(c*d^2*e*g - e*(c*d^2 + a*e^2)*g - c*d*e^2*g*x)^(-1 + m))/(a*d*e + (c*d^2 + a*e^2)*x +
c*d*e*x^2)^m,x]

[Out]

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

Maple [F]

\[\int \left (e x +d \right )^{m} \left (c \,d^{2} e g -e \left (e^{2} a +c \,d^{2}\right ) g -c d \,e^{2} g x \right )^{-1+m} {\left (a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}\right )}^{-m}d x\]

[In]

int((e*x+d)^m*(c*d^2*e*g-e*(a*e^2+c*d^2)*g-c*d*e^2*g*x)^(-1+m)/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m),x)

[Out]

int((e*x+d)^m*(c*d^2*e*g-e*(a*e^2+c*d^2)*g-c*d*e^2*g*x)^(-1+m)/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m),x)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.45 \[ \int (d+e x)^m \left (c d^2 e g-e \left (c d^2+a e^2\right ) g-c d e^2 g x\right )^{-1+m} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=-\frac {\log \left (c d x + a e\right )}{c d e^{2} g \left (-\frac {1}{e^{2} g}\right )^{m}} \]

[In]

integrate((e*x+d)^m*(c*d^2*e*g-e*(a*e^2+c*d^2)*g-c*d*e^2*g*x)^(-1+m)/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m),x,
algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (d+e x)^m \left (c d^2 e g-e \left (c d^2+a e^2\right ) g-c d e^2 g x\right )^{-1+m} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=\text {Timed out} \]

[In]

integrate((e*x+d)**m*(c*d**2*e*g-e*(a*e**2+c*d**2)*g-c*d*e**2*g*x)**(-1+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 = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.41 \[ \int (d+e x)^m \left (c d^2 e g-e \left (c d^2+a e^2\right ) g-c d e^2 g x\right )^{-1+m} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=-\frac {e^{2 \, m - 2} \left (-g\right )^{m} \log \left (c d x + a e\right )}{c d g} \]

[In]

integrate((e*x+d)^m*(c*d^2*e*g-e*(a*e^2+c*d^2)*g-c*d*e^2*g*x)^(-1+m)/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m),x,
algorithm="maxima")

[Out]

-e^(2*m - 2)*(-g)^m*log(c*d*x + a*e)/(c*d*g)

Giac [F]

\[ \int (d+e x)^m \left (c d^2 e g-e \left (c d^2+a e^2\right ) g-c d e^2 g x\right )^{-1+m} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=\int { \frac {{\left (-c d e^{2} g x + c d^{2} e g - {\left (c d^{2} + a e^{2}\right )} e g\right )}^{m - 1} {\left (e x + d\right )}^{m}}{{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{m}} \,d x } \]

[In]

integrate((e*x+d)^m*(c*d^2*e*g-e*(a*e^2+c*d^2)*g-c*d*e^2*g*x)^(-1+m)/((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^m),x,
algorithm="giac")

[Out]

integrate((-c*d*e^2*g*x + c*d^2*e*g - (c*d^2 + a*e^2)*e*g)^(m - 1)*(e*x + d)^m/(c*d*e*x^2 + a*d*e + (c*d^2 + a
*e^2)*x)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (d+e x)^m \left (c d^2 e g-e \left (c d^2+a e^2\right ) g-c d e^2 g x\right )^{-1+m} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{-m} \, dx=\int \frac {{\left (d+e\,x\right )}^m\,{\left (c\,d^2\,e\,g-e\,g\,\left (c\,d^2+a\,e^2\right )-c\,d\,e^2\,g\,x\right )}^{m-1}}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^m} \,d x \]

[In]

int(((d + e*x)^m*(c*d^2*e*g - e*g*(a*e^2 + c*d^2) - c*d*e^2*g*x)^(m - 1))/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x
^2)^m,x)

[Out]

int(((d + e*x)^m*(c*d^2*e*g - e*g*(a*e^2 + c*d^2) - c*d*e^2*g*x)^(m - 1))/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x
^2)^m, x)