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

Optimal. Leaf size=89 \[ x \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} (d-e x)^m (d+e x)^m \left (1-\frac {e^2 x^2}{d^2}\right )^{-m} F_1\left (\frac {1}{2};-p,-m;\frac {3}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {519, 430, 429} \[ x \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} (d-e x)^m (d+e x)^m \left (1-\frac {e^2 x^2}{d^2}\right )^{-m} F_1\left (\frac {1}{2};-p,-m;\frac {3}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(d - e*x)^m*(d + e*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 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])

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 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])

Rubi steps

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

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Mathematica [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int (d-e x)^m (d+e x)^m \left (a+c x^2\right )^p \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 1.18, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (c x^{2} + a\right )}^{p} {\left (e x + d\right )}^{m} {\left (-e x + d\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + a\right )}^{p} {\left (e x + d\right )}^{m} {\left (-e x + d\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int \left (c \,x^{2}+a \right )^{p} \left (-e x +d \right )^{m} \left (e x +d \right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + a\right )}^{p} {\left (e x + d\right )}^{m} {\left (-e x + d\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^m\,{\left (d-e\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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