\(\int (d+e x)^{-1+2 p} (c d^2+2 c d e x+c e^2 x^2)^{-p} \, dx\) [1108]

Optimal result

Integrand size = 36, antiderivative size = 43 \[ \int (d+e x)^{-1+2 p} \left (c d^2+2 c d e x+c e^2 x^2\right )^{-p} \, dx=\frac {(d+e x)^{2 p} \left (c d^2+2 c d e x+c e^2 x^2\right )^{-p} \log (d+e x)}{e} \]

[Out]

(e*x+d)^(2*p)*ln(e*x+d)/e/((c*e^2*x^2+2*c*d*e*x+c*d^2)^p)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {658, 31} \[ \int (d+e x)^{-1+2 p} \left (c d^2+2 c d e x+c e^2 x^2\right )^{-p} \, dx=\frac {(d+e x)^{2 p} \log (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{-p}}{e} \]

[In]

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

[Out]

((d + e*x)^(2*p)*Log[d + e*x])/(e*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^p)

Rule 31

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

Rule 658

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^p/(d
+ e*x)^(2*p), Int[(d + e*x)^(m + 2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !
IntegerQ[p] && EqQ[2*c*d - b*e, 0] &&  !IntegerQ[m]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.74 \[ \int (d+e x)^{-1+2 p} \left (c d^2+2 c d e x+c e^2 x^2\right )^{-p} \, dx=\frac {(d+e x)^{2 p} \left (c (d+e x)^2\right )^{-p} \log (d+e x)}{e} \]

[In]

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

[Out]

((d + e*x)^(2*p)*Log[d + e*x])/(e*(c*(d + e*x)^2)^p)

Maple [A] (verified)

Time = 2.51 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.72

method	result	size
norman	\(\left (x \ln \left (e x +d \right ) {\mathrm e}^{\left (2 p -1\right ) \ln \left (e x +d \right )}+\frac {d \ln \left (e x +d \right ) {\mathrm e}^{\left (2 p -1\right ) \ln \left (e x +d \right )}}{e}\right ) {\mathrm e}^{-p \ln \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )}\)	\(74\)

[In]

int((e*x+d)^(2*p-1)/((c*e^2*x^2+2*c*d*e*x+c*d^2)^p),x,method=_RETURNVERBOSE)

[Out]

(x*ln(e*x+d)*exp((2*p-1)*ln(e*x+d))+d/e*ln(e*x+d)*exp((2*p-1)*ln(e*x+d)))/exp(p*ln(c*e^2*x^2+2*c*d*e*x+c*d^2))

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.35 \[ \int (d+e x)^{-1+2 p} \left (c d^2+2 c d e x+c e^2 x^2\right )^{-p} \, dx=\frac {\log \left (e x + d\right )}{c^{p} e} \]

[In]

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

[Out]

log(e*x + d)/(c^p*e)

Sympy [F]

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

[In]

integrate((e*x+d)**(-1+2*p)/((c*e**2*x**2+2*c*d*e*x+c*d**2)**p),x)

[Out]

Integral((d + e*x)**(2*p - 1)/(c*(d + e*x)**2)**p, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.35 \[ \int (d+e x)^{-1+2 p} \left (c d^2+2 c d e x+c e^2 x^2\right )^{-p} \, dx=\frac {\log \left (e x + d\right )}{c^{p} e} \]

[In]

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

[Out]

log(e*x + d)/(c^p*e)

Giac [F]

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

[In]

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

[Out]

integrate((e*x + d)^(2*p - 1)/(c*e^2*x^2 + 2*c*d*e*x + c*d^2)^p, x)

Mupad [F(-1)]

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

[In]

int((d + e*x)^(2*p - 1)/(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^p,x)

[Out]

int((d + e*x)^(2*p - 1)/(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^p, x)