\(\int \frac {(a+b x^2)^p}{(c+d x^2)^3} \, dx\) [348]

Optimal result

Integrand size = 19, antiderivative size = 57 \[ \int \frac {\left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=\frac {x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,3,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {441, 440} \[ \int \frac {\left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,3,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c^3} \]

[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 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])

Rubi steps \begin{align*} \text {integral}& = \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] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(162\) vs. \(2(57)=114\).

Time = 0.26 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.84 \[ \int \frac {\left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=-\frac {3 a c x \left (a+b x^2\right )^p \operatorname {AppellF1}\left (\frac {1}{2},-p,3,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{\left (c+d x^2\right )^3 \left (-3 a c \operatorname {AppellF1}\left (\frac {1}{2},-p,3,\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )-2 x^2 \left (b c p \operatorname {AppellF1}\left (\frac {3}{2},1-p,3,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )-3 a d \operatorname {AppellF1}\left (\frac {3}{2},-p,4,\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )} \]

[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)])))

Maple [F]

\[\int \frac {\left (b \,x^{2}+a \right )^{p}}{\left (d \,x^{2}+c \right )^{3}}d x\]

[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)

Fricas [F]

\[ \int \frac {\left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {\left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p}{{\left (d\,x^2+c\right )}^3} \,d x \]

[In]

int((a + b*x^2)^p/(c + d*x^2)^3,x)

[Out]

int((a + b*x^2)^p/(c + d*x^2)^3, x)