3.348 \(\int \frac{(a+b x^2)^p}{(c+d x^2)^3} \, dx\)

Optimal. Leaf size=57 \[ \frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{c^3} \]

[Out]

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

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Rubi [A]  time = 0.0251613, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {430, 429} \[ \frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{c^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^p/(c + d*x^2)^3,x]

[Out]

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

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

Mathematica [B]  time = 0.234635, size = 162, normalized size = 2.84 \[ -\frac{3 a c x \left (a+b x^2\right )^p F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{\left (c+d x^2\right )^3 \left (-2 x^2 \left (b c p F_1\left (\frac{3}{2};1-p,3;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )-3 a d F_1\left (\frac{3}{2};-p,4;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )-3 a c F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^2)^p/(c + d*x^2)^3,x]

[Out]

(-3*a*c*x*(a + b*x^2)^p*AppellF1[1/2, -p, 3, 3/2, -((b*x^2)/a), -((d*x^2)/c)])/((c + d*x^2)^3*(-3*a*c*AppellF1
[1/2, -p, 3, 3/2, -((b*x^2)/a), -((d*x^2)/c)] - 2*x^2*(b*c*p*AppellF1[3/2, 1 - p, 3, 5/2, -((b*x^2)/a), -((d*x
^2)/c)] - 3*a*d*AppellF1[3/2, -p, 4, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))

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Maple [F]  time = 0.076, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( b{x}^{2}+a \right ) ^{p}}{ \left ( d{x}^{2}+c \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^p/(d*x^2+c)^3,x)

[Out]

int((b*x^2+a)^p/(d*x^2+c)^3,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 )}^{3}}\,{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)^3,x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^p/(d*x^2 + c)^3, 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}}{d^{3} x^{6} + 3 \, c d^{2} x^{4} + 3 \, c^{2} d x^{2} + c^{3}}, 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)^3,x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^p/(d^3*x^6 + 3*c*d^2*x^4 + 3*c^2*d*x^2 + c^3), 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)**3,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 )}^{3}}\,{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)^3,x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^p/(d*x^2 + c)^3, x)