### 3.1108 $$\int (d+e x)^{-1+2 p} (c d^2+2 c d e x+c e^2 x^2)^{-p} \, dx$$

Optimal. Leaf size=43 $\frac{(d+e x)^{2 p} \log (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{-p}}{e}$

[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)

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Rubi [A]  time = 0.0146628, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 36, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.056, Rules used = {644, 31} $\frac{(d+e x)^{2 p} \log (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{-p}}{e}$

Antiderivative was successfully veriﬁed.

[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 644

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]

Rule 31

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

Rubi steps

\begin{align*} \int (d+e x)^{-1+2 p} \left (c d^2+2 c d e x+c e^2 x^2\right )^{-p} \, dx &=\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]  time = 0.0060687, size = 32, normalized size = 0.74 $\frac{(d+e x)^{2 p} \log (d+e x) \left (c (d+e x)^2\right )^{-p}}{e}$

Antiderivative was successfully veriﬁed.

[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)

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Maple [A]  time = 0.077, size = 74, normalized size = 1.7 \begin{align*}{\frac{1}{{{\rm e}^{p\ln \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) }}} \left ( x\ln \left ( ex+d \right ){{\rm e}^{ \left ( -1+2\,p \right ) \ln \left ( ex+d \right ) }}+{\frac{d\ln \left ( ex+d \right ){{\rm e}^{ \left ( -1+2\,p \right ) \ln \left ( ex+d \right ) }}}{e}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.16358, size = 20, normalized size = 0.47 \begin{align*} \frac{\log \left (e x + d\right )}{c^{p} e} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.36749, size = 30, normalized size = 0.7 \begin{align*} \frac{\log \left (e x + d\right )}{c^{p} e} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)