\(\int \frac {(c d^2+2 c d e x+c e^2 x^2)^p}{(d+e x)^2} \, dx\) [1105]

Optimal result

Integrand size = 30, antiderivative size = 42 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^2} \, dx=-\frac {c (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{-1+p}}{e (1-2 p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {656, 623} \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^2} \, dx=-\frac {c (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{p-1}}{e (1-2 p)} \]

[In]

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

[Out]

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

Rule 623

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1)
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rule 656

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^m/c^(m/2), Int[(a +
b*x + c*x^2)^(p + m/2), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && EqQ[
2*c*d - b*e, 0] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = c \int \left (c d^2+2 c d e x+c e^2 x^2\right )^{-1+p} \, dx \\ & = -\frac {c (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{-1+p}}{e (1-2 p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.76 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^2} \, dx=\frac {c (d+e x) \left (c (d+e x)^2\right )^{-1+p}}{e (1+2 (-1+p))} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.79 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.71

method	result	size
risch	\(\frac {\left (c \left (e x +d \right )^{2}\right )^{p}}{\left (2 p -1\right ) e \left (e x +d \right )}\)	\(30\)
parallelrisch	\(\frac {{\left (c \left (x^{2} e^{2}+2 d e x +d^{2}\right )\right )}^{p}}{e \left (e x +d \right ) \left (2 p -1\right )}\)	\(39\)
gosper	\(\frac {\left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{p}}{\left (e x +d \right ) \left (2 p -1\right ) e}\)	\(41\)
norman	\(\frac {{\mathrm e}^{p \ln \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )}}{\left (2 p -1\right ) e \left (e x +d \right )}\)	\(43\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.17 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^2} \, dx=\frac {{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p}}{2 \, d e p - d e + {\left (2 \, e^{2} p - e^{2}\right )} x} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^2} \, dx=\begin {cases} \frac {x \sqrt {c d^{2}}}{d^{2}} & \text {for}\: e = 0 \wedge p = \frac {1}{2} \\\frac {x \left (c d^{2}\right )^{p}}{d^{2}} & \text {for}\: e = 0 \\\int \frac {\sqrt {c \left (d + e x\right )^{2}}}{\left (d + e x\right )^{2}}\, dx & \text {for}\: p = \frac {1}{2} \\\frac {\left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 d e p - d e + 2 e^{2} p x - e^{2} x} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*sqrt(c*d**2)/d**2, Eq(e, 0) & Eq(p, 1/2)), (x*(c*d**2)**p/d**2, Eq(e, 0)), (Integral(sqrt(c*(d +
e*x)**2)/(d + e*x)**2, x), Eq(p, 1/2)), ((c*d**2 + 2*c*d*e*x + c*e**2*x**2)**p/(2*d*e*p - d*e + 2*e**2*p*x - e
**2*x), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.81 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^2} \, dx=\frac {{\left (e x + d\right )}^{2 \, p} c^{p}}{e^{2} {\left (2 \, p - 1\right )} x + d e {\left (2 \, p - 1\right )}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.97 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^2} \, dx=\frac {{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^p}{e^2\,\left (2\,p-1\right )\,\left (x+\frac {d}{e}\right )} \]

[In]

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

[Out]

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