### 3.1107 $$\int (d+e x)^{-1-2 p} (c d^2+2 c d e x+c e^2 x^2)^p \, dx$$

Optimal. Leaf size=41 $\frac{(d+e x)^{-2 p} \log (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^p}{e}$

[Out]

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

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Rubi [A]  time = 0.0147421, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 34, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.059, 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]

((c*d^2 + 2*c*d*e*x + c*e^2*x^2)^p*Log[d + e*x])/(e*(d + e*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 = 0.0070629, size = 30, normalized size = 0.73 $\frac{(d+e x)^{-2 p} \log (d+e x) \left (c (d+e x)^2\right )^p}{e}$

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]

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

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Maple [B]  time = 0.079, size = 95, normalized size = 2.3 \begin{align*} x\ln \left ( ex+d \right ){{\rm e}^{ \left ( -1-2\,p \right ) \ln \left ( ex+d \right ) }}{{\rm e}^{p\ln \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) }}+{\frac{d\ln \left ( ex+d \right ){{\rm e}^{ \left ( -1-2\,p \right ) \ln \left ( ex+d \right ) }}{{\rm e}^{p\ln \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) }}}{e}} \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))*exp(p*ln(c*e^2*x^2+2*c*d*e*x+c*d^2))+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 = 1.34949, size = 18, normalized size = 0.44 \begin{align*} \frac{c^{p} \log \left (e x + d\right )}{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]

c^p*log(e*x + d)/e

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Fricas [A]  time = 2.40496, size = 27, normalized size = 0.66 \begin{align*} \frac{c^{p} \log \left (e x + d\right )}{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]

c^p*log(e*x + d)/e

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \left (d + e x\right )^{2}\right )^{p} \left (d + e x\right )^{- 2 p - 1}\, dx \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]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p}{\left (e x + d\right )}^{-2 \, p - 1}\,{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((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^p*(e*x + d)^(-2*p - 1), x)