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

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

[Out]

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

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Rubi [A]  time = 0.026394, 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} \[ \frac{x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^p/(c + d*x^2),x]

[Out]

(x*(a + b*x^2)^p*AppellF1[1/2, -p, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)])/(c*(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}{c+d 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}{c+d x^2} \, 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,1;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{c}\\ \end{align*}

Mathematica [B]  time = 0.178261, size = 162, normalized size = 2.84 \[ -\frac{3 a c x \left (a+b x^2\right )^p F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )}{\left (c+d x^2\right ) \left (2 x^2 \left (a d F_1\left (\frac{3}{2};-p,2;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )-b c p F_1\left (\frac{3}{2};1-p,1;\frac{5}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )-3 a c F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},-\frac{d x^2}{c}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.049, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( b{x}^{2}+a \right ) ^{p}}{d{x}^{2}+c}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x**2)**p/(c + d*x**2), x)

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

[Out]

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