3.9.4 \(\int (d+e x)^m (d f-e f x)^m (a+c x^2)^p \, dx\) [804]

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

[Out]

x*(e*x+d)^m*(-e*f*x+d*f)^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 = 0.05, antiderivative size = 92, normalized size of antiderivative = 1.00, 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 = {533, 441, 440} \begin {gather*} 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 ) \end {gather*}

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 440

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 441

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 533

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)^FracPa
rt[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 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*}

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Mathematica [F]
time = 0.06, size = 0, normalized size = 0.00 \begin {gather*} \int (d+e x)^m (d f-e f x)^m \left (a+c x^2\right )^p \, dx \end {gather*}

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 = 0.04, size = 0, normalized size = 0.00 \[\int \left (e x +d \right )^{m} \left (-e f x +d f \right )^{m} \left (c \,x^{2}+a \right )^{p}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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