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

Optimal. Leaf size=42 \[ -\frac {c (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{p-1}}{e (1-2 p)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {642, 609} \begin {gather*} -\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)} \end {gather*}

Antiderivative was successfully verified.

[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 609

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 642

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*} \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^2} \, dx &=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]  time = 0.02, size = 30, normalized size = 0.71 \begin {gather*} \frac {c (d+e x) \left (c (d+e x)^2\right )^{p-1}}{e (2 p-1)} \end {gather*}

Antiderivative was successfully verified.

[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*p))

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.16, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^p}{(d+e x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 49, normalized size = 1.17 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 41, normalized size = 0.98 \begin {gather*} \frac {\left (c \,e^{2} x^{2}+2 c d e x +c \,d^{2}\right )^{p}}{\left (e x +d \right ) \left (2 p -1\right ) e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.49, size = 34, normalized size = 0.81 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

mupad [B]  time = 0.46, size = 42, normalized size = 1.00 \begin {gather*} \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 )} \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,x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \text {NaN} & \text {for}\: d = 0 \wedge e = 0 \wedge p = \frac {1}{2} \\0^{p} \tilde {\infty } x & \text {for}\: d = - e x \\\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((nan, Eq(d, 0) & Eq(e, 0) & Eq(p, 1/2)), (0**p*zoo*x, Eq(d, -e*x)), (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))

________________________________________________________________________________________