3.996 \(\int (a+\frac {b}{x^2})^p (c+\frac {d}{x^2})^q (e x)^{5/2} \, dx\)

Optimal. Leaf size=91 \[ \frac {2 (e x)^{7/2} \left (a+\frac {b}{x^2}\right )^p \left (\frac {b}{a x^2}+1\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {d}{c x^2}+1\right )^{-q} F_1\left (-\frac {7}{4};-p,-q;-\frac {3}{4};-\frac {b}{a x^2},-\frac {d}{c x^2}\right )}{7 e} \]

[Out]

2/7*(a+b/x^2)^p*(c+d/x^2)^q*(e*x)^(7/2)*AppellF1(-7/4,-p,-q,-3/4,-b/a/x^2,-d/c/x^2)/e/((1+b/a/x^2)^p)/((1+d/c/
x^2)^q)

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Rubi [A]  time = 0.12, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {496, 511, 510} \[ \frac {2 (e x)^{7/2} \left (a+\frac {b}{x^2}\right )^p \left (\frac {b}{a x^2}+1\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {d}{c x^2}+1\right )^{-q} F_1\left (-\frac {7}{4};-p,-q;-\frac {3}{4};-\frac {b}{a x^2},-\frac {d}{c x^2}\right )}{7 e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^p*(c + d/x^2)^q*(e*x)^(5/2),x]

[Out]

(2*(a + b/x^2)^p*(c + d/x^2)^q*(e*x)^(7/2)*AppellF1[-7/4, -p, -q, -3/4, -(b/(a*x^2)), -(d/(c*x^2))])/(7*e*(1 +
b/(a*x^2))^p*(1 + d/(c*x^2))^q)

Rule 496

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{g = Deno
minator[m]}, -Dist[g/e, Subst[Int[((a + b/(e^n*x^(g*n)))^p*(c + d/(e^n*x^(g*n)))^q)/x^(g*(m + 1) + 1), x], x,
1/(e*x)^(1/g)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && ILtQ[n, 0] && FractionQ[m]

Rule 510

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

Rule 511

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

Rubi steps

\begin {align*} \int \left (a+\frac {b}{x^2}\right )^p \left (c+\frac {d}{x^2}\right )^q (e x)^{5/2} \, dx &=-\frac {2 \operatorname {Subst}\left (\int \frac {\left (a+b e^2 x^4\right )^p \left (c+d e^2 x^4\right )^q}{x^8} \, dx,x,\frac {1}{\sqrt {e x}}\right )}{e}\\ &=-\frac {\left (2 \left (a+\frac {b}{x^2}\right )^p \left (1+\frac {b}{a x^2}\right )^{-p}\right ) \operatorname {Subst}\left (\int \frac {\left (1+\frac {b e^2 x^4}{a}\right )^p \left (c+d e^2 x^4\right )^q}{x^8} \, dx,x,\frac {1}{\sqrt {e x}}\right )}{e}\\ &=-\frac {\left (2 \left (a+\frac {b}{x^2}\right )^p \left (1+\frac {b}{a x^2}\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (1+\frac {d}{c x^2}\right )^{-q}\right ) \operatorname {Subst}\left (\int \frac {\left (1+\frac {b e^2 x^4}{a}\right )^p \left (1+\frac {d e^2 x^4}{c}\right )^q}{x^8} \, dx,x,\frac {1}{\sqrt {e x}}\right )}{e}\\ &=\frac {2 \left (a+\frac {b}{x^2}\right )^p \left (1+\frac {b}{a x^2}\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (1+\frac {d}{c x^2}\right )^{-q} (e x)^{7/2} F_1\left (-\frac {7}{4};-p,-q;-\frac {3}{4};-\frac {b}{a x^2},-\frac {d}{c x^2}\right )}{7 e}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 111, normalized size = 1.22 \[ -\frac {2 x (e x)^{5/2} \left (a+\frac {b}{x^2}\right )^p \left (\frac {a x^2}{b}+1\right )^{-p} \left (c+\frac {d}{x^2}\right )^q \left (\frac {c x^2}{d}+1\right )^{-q} F_1\left (-p-q+\frac {7}{4};-p,-q;-p-q+\frac {11}{4};-\frac {a x^2}{b},-\frac {c x^2}{d}\right )}{4 p+4 q-7} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b/x^2)^p*(c + d/x^2)^q*(e*x)^(5/2),x]

[Out]

(-2*(a + b/x^2)^p*(c + d/x^2)^q*x*(e*x)^(5/2)*AppellF1[7/4 - p - q, -p, -q, 11/4 - p - q, -((a*x^2)/b), -((c*x
^2)/d)])/((-7 + 4*p + 4*q)*(1 + (a*x^2)/b)^p*(1 + (c*x^2)/d)^q)

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fricas [F]  time = 1.06, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {e x} e^{2} x^{2} \left (\frac {a x^{2} + b}{x^{2}}\right )^{p} \left (\frac {c x^{2} + d}{x^{2}}\right )^{q}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)^p*(c+d/x^2)^q*(e*x)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(e*x)*e^2*x^2*((a*x^2 + b)/x^2)^p*((c*x^2 + d)/x^2)^q, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)^p*(c+d/x^2)^q*(e*x)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{1,[0,6,1,1,0]%%%} / %%%{1,[0,0,0,0,3]%%%} Error: Bad Argument V
alue

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{\frac {5}{2}} \left (a +\frac {b}{x^{2}}\right )^{p} \left (c +\frac {d}{x^{2}}\right )^{q}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)^p*(c+d/x^2)^q*(e*x)^(5/2),x)

[Out]

int((a+b/x^2)^p*(c+d/x^2)^q*(e*x)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e x\right )^{\frac {5}{2}} {\left (a + \frac {b}{x^{2}}\right )}^{p} {\left (c + \frac {d}{x^{2}}\right )}^{q}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)^p*(c+d/x^2)^q*(e*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((e*x)^(5/2)*(a + b/x^2)^p*(c + d/x^2)^q, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (e\,x\right )}^{5/2}\,{\left (a+\frac {b}{x^2}\right )}^p\,{\left (c+\frac {d}{x^2}\right )}^q \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(5/2)*(a + b/x^2)^p*(c + d/x^2)^q,x)

[Out]

int((e*x)^(5/2)*(a + b/x^2)^p*(c + d/x^2)^q, 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((a+b/x**2)**p*(c+d/x**2)**q*(e*x)**(5/2),x)

[Out]

Timed out

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