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

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

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Rubi [A]  time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 31

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

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]

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*}

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Mathematica [A]  time = 0.01, size = 32, normalized size = 0.74 \begin {gather*} \frac {(d+e x)^{2 p} \log (d+e x) \left (c (d+e x)^2\right )^{-p}}{e} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.18, size = 32, normalized size = 0.74 \begin {gather*} \frac {(d+e x)^{2 p} \log (d+e x) \left (c (d+e x)^2\right )^{-p}}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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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fricas [A]  time = 0.41, size = 15, normalized size = 0.35 \begin {gather*} \frac {\log \left (e x + d\right )}{c^{p} e} \end {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.10, size = 74, normalized size = 1.72 \begin {gather*} \left (x \,{\mathrm e}^{\left (2 p -1\right ) \ln \left (e x +d \right )} \ln \left (e x +d \right )+\frac {d \,{\mathrm e}^{\left (2 p -1\right ) \ln \left (e x +d \right )} \ln \left (e x +d \right )}{e}\right ) {\mathrm e}^{-p \ln \left (c \,e^{2} x^{2}+2 c d e x +c \,d^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))

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maxima [A]  time = 1.44, size = 15, normalized size = 0.35 \begin {gather*} \frac {\log \left (e x + d\right )}{c^{p} e} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (c \left (d + e x\right )^{2}\right )^{- p} \left (d + e x\right )^{2 p - 1}\, dx \end {gather*}

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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