3.12.7 \(\int (d+e x)^{-1-2 p} (c d^2+2 c d e x+c e^2 x^2)^p \, dx\) [1107]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 41, normalized size of antiderivative = 1.00, 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 = {658, 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]

((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 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*} \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 = 30, normalized size = 0.73 \begin {gather*} \frac {(d+e x)^{-2 p} \left (c (d+e x)^2\right )^p \log (d+e x)}{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]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(94\) vs. \(2(43)=86\).
time = 0.66, size = 95, normalized size = 2.32

method	result	size
norman	\(x \ln \left (e x +d \right ) {\mathrm e}^{\left (-1-2 p \right ) \ln \left (e x +d \right )} {\mathrm e}^{p \ln \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )}+\frac {d \ln \left (e x +d \right ) {\mathrm e}^{\left (-1-2 p \right ) \ln \left (e x +d \right )} {\mathrm e}^{p \ln \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )}}{e}\)	\(95\)

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,method=_RETURNVERBOSE)

[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 = 0.29, size = 13, normalized size = 0.32 \begin {gather*} c^{p} e^{\left (-1\right )} \log \left (x e + d\right ) \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]

c^p*e^(-1)*log(x*e + d)

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Fricas [A]
time = 2.61, size = 13, normalized size = 0.32 \begin {gather*} c^{p} e^{\left (-1\right )} \log \left (x e + d\right ) \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]

c^p*e^(-1)*log(x*e + d)

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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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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((c*x^2*e^2 + 2*c*d*x*e + c*d^2)^p*(x*e + d)^(-2*p - 1), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^p}{{\left (d+e\,x\right )}^{2\,p+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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