3.1493 \(\int (-a e+c d x) (d+e x)^{-3-2 p} (a+c x^2)^p \, dx\)

Optimal. Leaf size=31 \[ \frac{\left (a+c x^2\right )^{p+1} (d+e x)^{-2 (p+1)}}{2 (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0188862, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.033, Rules used = {803} \[ \frac{\left (a+c x^2\right )^{p+1} (d+e x)^{-2 (p+1)}}{2 (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 803

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] /; FreeQ[{a, c, d, e, f, g, m, p}, x]
&& NeQ[c*d^2 + a*e^2, 0] && EqQ[Simplify[m + 2*p + 3], 0] && EqQ[c*d*f + a*e*g, 0]

Rubi steps

\begin{align*} \int (-a e+c d x) (d+e x)^{-3-2 p} \left (a+c x^2\right )^p \, dx &=\frac{(d+e x)^{-2 (1+p)} \left (a+c x^2\right )^{1+p}}{2 (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.0648407, size = 30, normalized size = 0.97 \[ \frac{\left (a+c x^2\right )^{p+1} (d+e x)^{-2 p-2}}{2 p+2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 30, normalized size = 1. \begin{align*}{\frac{ \left ( c{x}^{2}+a \right ) ^{1+p} \left ( ex+d \right ) ^{-2\,p-2}}{2+2\,p}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.39777, size = 78, normalized size = 2.52 \begin{align*} \frac{{\left (c x^{2} + a\right )} e^{\left (p \log \left (c x^{2} + a\right ) - 2 \, p \log \left (e x + d\right )\right )}}{2 \,{\left (e^{2}{\left (p + 1\right )} x^{2} + 2 \, d e{\left (p + 1\right )} x + d^{2}{\left (p + 1\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.915, size = 111, normalized size = 3.58 \begin{align*} \frac{{\left (c e x^{3} + c d x^{2} + a e x + a d\right )}{\left (c x^{2} + a\right )}^{p}{\left (e x + d\right )}^{-2 \, p - 3}}{2 \,{\left (p + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c d x - a e\right )}{\left (c x^{2} + a\right )}^{p}{\left (e x + d\right )}^{-2 \, p - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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