3.804 \(\int (d+e x)^m (d f-e f x)^m (a+c x^2)^p \, dx\)

Optimal. Leaf size=92 \[ x \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} (d+e x)^m \left (1-\frac{e^2 x^2}{d^2}\right )^{-m} (d f-e f x)^m F_1\left (\frac{1}{2};-p,-m;\frac{3}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right ) \]

[Out]

(x*(d + e*x)^m*(d*f - e*f*x)^m*(a + c*x^2)^p*AppellF1[1/2, -p, -m, 3/2, -((c*x^2)/a), (e^2*x^2)/d^2])/((1 + (c
*x^2)/a)^p*(1 - (e^2*x^2)/d^2)^m)

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Rubi [A]  time = 0.0841984, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {519, 430, 429} \[ x \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} (d+e x)^m \left (1-\frac{e^2 x^2}{d^2}\right )^{-m} (d f-e f x)^m F_1\left (\frac{1}{2};-p,-m;\frac{3}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(d + e*x)^m*(d*f - e*f*x)^m*(a + c*x^2)^p*AppellF1[1/2, -p, -m, 3/2, -((c*x^2)/a), (e^2*x^2)/d^2])/((1 + (c
*x^2)/a)^p*(1 - (e^2*x^2)/d^2)^m)

Rule 519

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_)*((a2_) + (b2_.)*(x_)^(non2_.))^(
p_), x_Symbol] :> Dist[((a1 + b1*x^(n/2))^FracPart[p]*(a2 + b2*x^(n/2))^FracPart[p])/(a1*a2 + b1*b2*x^n)^FracP
art[p], Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, n, p, q}, x] && EqQ[
non2, n/2] && EqQ[a2*b1 + a1*b2, 0] &&  !(EqQ[n, 2] && IGtQ[q, 0])

Rule 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int (d+e x)^m (d f-e f x)^m \left (a+c x^2\right )^p \, dx &=\left ((d+e x)^m (d f-e f x)^m \left (d^2 f-e^2 f x^2\right )^{-m}\right ) \int \left (a+c x^2\right )^p \left (d^2 f-e^2 f x^2\right )^m \, dx\\ &=\left ((d+e x)^m (d f-e f x)^m \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} \left (d^2 f-e^2 f x^2\right )^{-m}\right ) \int \left (1+\frac{c x^2}{a}\right )^p \left (d^2 f-e^2 f x^2\right )^m \, dx\\ &=\left ((d+e x)^m (d f-e f x)^m \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} \left (1-\frac{e^2 x^2}{d^2}\right )^{-m}\right ) \int \left (1+\frac{c x^2}{a}\right )^p \left (1-\frac{e^2 x^2}{d^2}\right )^m \, dx\\ &=x (d+e x)^m (d f-e f x)^m \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} \left (1-\frac{e^2 x^2}{d^2}\right )^{-m} F_1\left (\frac{1}{2};-p,-m;\frac{3}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )\\ \end{align*}

Mathematica [F]  time = 0.0800779, size = 0, normalized size = 0. \[ \int (d+e x)^m (d f-e f x)^m \left (a+c x^2\right )^p \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 1.041, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{m} \left ( -efx+df \right ) ^{m} \left ( c{x}^{2}+a \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (-e f x + d f\right )}^{m}{\left (c x^{2} + a\right )}^{p}{\left (e x + d\right )}^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-e*f*x + d*f)^m*(c*x^2 + a)^p*(e*x + d)^m, x)

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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((e*x+d)**m*(-e*f*x+d*f)**m*(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 (-e f x + d f\right )}^{m}{\left (c x^{2} + a\right )}^{p}{\left (e x + d\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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