\(\int \frac {(c d^2+2 c d e x+c e^2 x^2)^p}{(d+e x)^3} \, dx\) [1106]

Optimal result

Integrand size = 30, antiderivative size = 39 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^3} \, dx=-\frac {c \left (c d^2+2 c d e x+c e^2 x^2\right )^{-1+p}}{2 e (1-p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, 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 = {657, 643} \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^3} \, dx=-\frac {c \left (c d^2+2 c d e x+c e^2 x^2\right )^{p-1}}{2 e (1-p)} \]

[In]

Int[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^p/(d + e*x)^3,x]

[Out]

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

Rule 643

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

Rule 657

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^(m - 1)/c^((m - 1)/2
), Int[(d + e*x)*(a + b*x + c*x^2)^(p + (m - 1)/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 - 1)/2]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.64 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^3} \, dx=\frac {c \left (c (d+e x)^2\right )^{-1+p}}{e (-2+2 p)} \]

[In]

Integrate[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^p/(d + e*x)^3,x]

[Out]

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

Maple [A] (verified)

Time = 3.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {\left (c \left (e x +d \right )^{2}\right )^{p}}{2 \left (p -1\right ) e \left (e x +d \right )^{2}}\)	\(29\)
parallelrisch	\(\frac {{\left (c \left (x^{2} e^{2}+2 d e x +d^{2}\right )\right )}^{p}}{2 \left (e x +d \right )^{2} \left (p -1\right ) e}\)	\(38\)
gosper	\(\frac {\left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{p}}{2 \left (e x +d \right )^{2} \left (p -1\right ) e}\)	\(40\)
norman	\(\frac {{\mathrm e}^{p \ln \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )}}{2 \left (p -1\right ) e \left (e x +d \right )^{2}}\)	\(42\)

[In]

int((c*e^2*x^2+2*c*d*e*x+c*d^2)^p/(e*x+d)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.79 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^3} \, dx=\frac {{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p}}{2 \, {\left (d^{2} e p - d^{2} e + {\left (e^{3} p - e^{3}\right )} x^{2} + 2 \, {\left (d e^{2} p - d e^{2}\right )} x\right )}} \]

[In]

integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)^p/(e*x+d)^3,x, algorithm="fricas")

[Out]

1/2*(c*e^2*x^2 + 2*c*d*e*x + c*d^2)^p/(d^2*e*p - d^2*e + (e^3*p - e^3)*x^2 + 2*(d*e^2*p - d*e^2)*x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (34) = 68\).

Time = 0.47 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.56 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^3} \, dx=\begin {cases} \frac {c x}{d} & \text {for}\: e = 0 \wedge p = 1 \\\frac {x \left (c d^{2}\right )^{p}}{d^{3}} & \text {for}\: e = 0 \\\frac {c \log {\left (\frac {d}{e} + x \right )}}{e} & \text {for}\: p = 1 \\\frac {\left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{p}}{2 d^{2} e p - 2 d^{2} e + 4 d e^{2} p x - 4 d e^{2} x + 2 e^{3} p x^{2} - 2 e^{3} x^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate((c*e**2*x**2+2*c*d*e*x+c*d**2)**p/(e*x+d)**3,x)

[Out]

Piecewise((c*x/d, Eq(e, 0) & Eq(p, 1)), (x*(c*d**2)**p/d**3, Eq(e, 0)), (c*log(d/e + x)/e, Eq(p, 1)), ((c*d**2
+ 2*c*d*e*x + c*e**2*x**2)**p/(2*d**2*e*p - 2*d**2*e + 4*d*e**2*p*x - 4*d*e**2*x + 2*e**3*p*x**2 - 2*e**3*x**
2), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.15 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^3} \, dx=\frac {{\left (e x + d\right )}^{2 \, p} c^{p}}{2 \, {\left (e^{3} {\left (p - 1\right )} x^{2} + 2 \, d e^{2} {\left (p - 1\right )} x + d^{2} e {\left (p - 1\right )}\right )}} \]

[In]

integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)^p/(e*x+d)^3,x, algorithm="maxima")

[Out]

1/2*(e*x + d)^(2*p)*c^p/(e^3*(p - 1)*x^2 + 2*d*e^2*(p - 1)*x + d^2*e*(p - 1))

Giac [F]

\[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^3} \, dx=\int { \frac {{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{p}}{{\left (e x + d\right )}^{3}} \,d x } \]

[In]

integrate((c*e^2*x^2+2*c*d*e*x+c*d^2)^p/(e*x+d)^3,x, algorithm="giac")

[Out]

integrate((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^p/(e*x + d)^3, x)

Mupad [B] (verification not implemented)

Time = 10.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.33 \[ \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^3} \, dx=\frac {{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^p}{2\,e^3\,\left (p-1\right )\,\left (x^2+\frac {d^2}{e^2}+\frac {2\,d\,x}{e}\right )} \]

[In]

int((c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^p/(d + e*x)^3,x)

[Out]

(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^p/(2*e^3*(p - 1)*(x^2 + d^2/e^2 + (2*d*x)/e))