### 3.1091 $$\int \frac{(d+e x)^m}{c d^2+2 c d e x+c e^2 x^2} \, dx$$

Optimal. Leaf size=24 $-\frac{(d+e x)^{m-1}}{c e (1-m)}$

[Out]

-((d + e*x)^(-1 + m)/(c*e*(1 - m)))

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Rubi [A]  time = 0.0115782, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.1, Rules used = {27, 12, 32} $-\frac{(d+e x)^{m-1}}{c e (1-m)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^m/(c*d^2 + 2*c*d*e*x + c*e^2*x^2),x]

[Out]

-((d + e*x)^(-1 + m)/(c*e*(1 - m)))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int \frac{(d+e x)^m}{c d^2+2 c d e x+c e^2 x^2} \, dx &=\int \frac{(d+e x)^{-2+m}}{c} \, dx\\ &=\frac{\int (d+e x)^{-2+m} \, dx}{c}\\ &=-\frac{(d+e x)^{-1+m}}{c e (1-m)}\\ \end{align*}

Mathematica [A]  time = 0.0140982, size = 21, normalized size = 0.88 $\frac{(d+e x)^{m-1}}{c e (m-1)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^m/(c*d^2 + 2*c*d*e*x + c*e^2*x^2),x]

[Out]

(d + e*x)^(-1 + m)/(c*e*(-1 + m))

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Maple [A]  time = 0.041, size = 22, normalized size = 0.9 \begin{align*}{\frac{ \left ( ex+d \right ) ^{-1+m}}{ce \left ( -1+m \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*x+d)^(-1+m)/c/e/(-1+m)

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Maxima [A]  time = 1.18196, size = 36, normalized size = 1.5 \begin{align*} \frac{{\left (e x + d\right )}^{m}}{c e^{2}{\left (m - 1\right )} x + c d e{\left (m - 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*x + d)^m/(c*e^2*(m - 1)*x + c*d*e*(m - 1))

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Fricas [A]  time = 2.33927, size = 72, normalized size = 3. \begin{align*} \frac{{\left (e x + d\right )}^{m}}{c d e m - c d e +{\left (c e^{2} m - c e^{2}\right )} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*x + d)^m/(c*d*e*m - c*d*e + (c*e^2*m - c*e^2)*x)

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Sympy [A]  time = 1.18284, size = 63, normalized size = 2.62 \begin{align*} \begin{cases} \text{NaN} & \text{for}\: d = 0 \wedge e = 0 \wedge m = 1 \\0^{m} \tilde{\infty } x & \text{for}\: d = - e x \\\frac{d^{m} x}{c d^{2}} & \text{for}\: e = 0 \\\frac{\log{\left (\frac{d}{e} + x \right )}}{c e} & \text{for}\: m = 1 \\\frac{\left (d + e x\right )^{m}}{c d e m - c d e + c e^{2} m x - c e^{2} x} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m/(c*e**2*x**2+2*c*d*e*x+c*d**2),x)

[Out]

Piecewise((nan, Eq(d, 0) & Eq(e, 0) & Eq(m, 1)), (0**m*zoo*x, Eq(d, -e*x)), (d**m*x/(c*d**2), Eq(e, 0)), (log(
d/e + x)/(c*e), Eq(m, 1)), ((d + e*x)**m/(c*d*e*m - c*d*e + c*e**2*m*x - c*e**2*x), True))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)^m/(c*e^2*x^2 + 2*c*d*e*x + c*d^2), x)