\(\int \frac {(d+e x)^m}{c d^2+2 c d e x+c e^2 x^2} \, dx\) [1091]

Optimal result

Integrand size = 30, antiderivative size = 24 \[ \int \frac {(d+e x)^m}{c d^2+2 c d e x+c e^2 x^2} \, dx=-\frac {(d+e x)^{-1+m}}{c e (1-m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {27, 12, 32} \[ \int \frac {(d+e x)^m}{c d^2+2 c d e x+c e^2 x^2} \, dx=-\frac {(d+e x)^{m-1}}{c e (1-m)} \]

[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 12

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

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 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*} \text {integral}& = \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] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^m}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {(d+e x)^{-1+m}}{c e (-1+m)} \]

[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))

Maple [A] (verified)

Time = 2.42 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92

method	result	size
gosper	\(\frac {\left (e x +d \right )^{-1+m}}{c e \left (-1+m \right )}\)	\(22\)
risch	\(\frac {\left (e x +d \right )^{m}}{\left (e x +d \right ) c e \left (-1+m \right )}\)	\(27\)
parallelrisch	\(\frac {\left (e x +d \right )^{m}}{\left (e x +d \right ) c e \left (-1+m \right )}\)	\(27\)
norman	\(\frac {{\mathrm e}^{m \ln \left (e x +d \right )}}{c e \left (-1+m \right ) \left (e x +d \right )}\)	\(29\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {(d+e x)^m}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {{\left (e x + d\right )}^{m}}{c d e m - c d e + {\left (c e^{2} m - c e^{2}\right )} x} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (15) = 30\).

Time = 0.35 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33 \[ \int \frac {(d+e x)^m}{c d^2+2 c d e x+c e^2 x^2} \, dx=\begin {cases} \frac {x}{c d} & \text {for}\: e = 0 \wedge m = 1 \\\frac {d^{m} x}{c d^{2}} & \text {for}\: e = 0 \\\frac {\log {\left (\frac {d}{e} + x \right )}}{c e} & \text {for}\: m = 1 \\\frac {\left (d + e x\right )^{m}}{c d e m - c d e + c e^{2} m x - c e^{2} x} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x/(c*d), Eq(e, 0) & Eq(m, 1)), (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))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {(d+e x)^m}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {{\left (e x + d\right )}^{m}}{c e^{2} {\left (m - 1\right )} x + c d e {\left (m - 1\right )}} \]

[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))

Giac [F]

\[ \int \frac {(d+e x)^m}{c d^2+2 c d e x+c e^2 x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{c e^{2} x^{2} + 2 \, c d e x + c d^{2}} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 9.71 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {(d+e x)^m}{c d^2+2 c d e x+c e^2 x^2} \, dx=\frac {{\left (d+e\,x\right )}^m}{c\,e^2\,\left (x+\frac {d}{e}\right )\,\left (m-1\right )} \]

[In]

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

[Out]

(d + e*x)^m/(c*e^2*(x + d/e)*(m - 1))