3.1143 \(\int \frac{(a+b x^2)^p (c+d x^2)^q}{x^2} \, dx\)

Optimal. Leaf size=82 \[ -\frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (c+d x^2\right )^q \left (\frac{d x^2}{c}+1\right )^{-q} F_1\left (-\frac{1}{2};-p,-q;\frac{1}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.0686437, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {511, 510} \[ -\frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (c+d x^2\right )^q \left (\frac{d x^2}{c}+1\right )^{-q} F_1\left (-\frac{1}{2};-p,-q;\frac{1}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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])

Rubi steps

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

Mathematica [A]  time = 0.054382, size = 84, normalized size = 1.02 \[ -\frac{\left (a+b x^2\right )^p \left (\frac{a+b x^2}{a}\right )^{-p} \left (c+d x^2\right )^q \left (\frac{c+d x^2}{c}\right )^{-q} F_1\left (-\frac{1}{2};-p,-q;\frac{1}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( b{x}^{2}+a \right ) ^{p} \left ( d{x}^{2}+c \right ) ^{q}}{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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