Integrand size = 19, antiderivative size = 176 \[ \int \left (a+b x^2\right )^p \left (c+d x^2\right )^2 \, dx=-\frac {d (3 a d-b c (7+2 p)) x \left (a+b x^2\right )^{1+p}}{b^2 (3+2 p) (5+2 p)}+\frac {d x \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )}{b (5+2 p)}+\frac {\left (3 a^2 d^2-2 a b c d (5+2 p)+b^2 c^2 \left (15+16 p+4 p^2\right )\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{b^2 (3+2 p) (5+2 p)} \]
-d*(3*a*d-b*c*(7+2*p))*x*(b*x^2+a)^(p+1)/b^2/(4*p^2+16*p+15)+d*x*(b*x^2+a) ^(p+1)*(d*x^2+c)/b/(5+2*p)+(3*a^2*d^2-2*a*b*c*d*(5+2*p)+b^2*c^2*(4*p^2+16* p+15))*x*(b*x^2+a)^p*hypergeom([1/2, -p],[3/2],-b*x^2/a)/b^2/(4*p^2+16*p+1 5)/((1+b*x^2/a)^p)
Time = 5.10 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.60 \[ \int \left (a+b x^2\right )^p \left (c+d x^2\right )^2 \, dx=\frac {1}{15} x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (15 c^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )+d x^2 \left (10 c \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )+3 d x^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\right )\right ) \]
(x*(a + b*x^2)^p*(15*c^2*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)] + d *x^2*(10*c*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^2)/a)] + 3*d*x^2*Hyperge ometric2F1[5/2, -p, 7/2, -((b*x^2)/a)])))/(15*(1 + (b*x^2)/a)^p)
Time = 0.30 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {318, 25, 299, 238, 237}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (c+d x^2\right )^2 \left (a+b x^2\right )^p \, dx\) |
\(\Big \downarrow \) 318 |
\(\displaystyle \frac {\int -\left (b x^2+a\right )^p \left (d (3 a d-b c (2 p+7)) x^2+c (a d-b c (2 p+5))\right )dx}{b (2 p+5)}+\frac {d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1}}{b (2 p+5)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1}}{b (2 p+5)}-\frac {\int \left (b x^2+a\right )^p \left (d (3 a d-b c (2 p+7)) x^2+c (a d-b c (2 p+5))\right )dx}{b (2 p+5)}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1}}{b (2 p+5)}-\frac {\frac {d x \left (a+b x^2\right )^{p+1} (3 a d-b c (2 p+7))}{b (2 p+3)}-\frac {\left (3 a^2 d^2-2 a b c d (2 p+5)+b^2 c^2 \left (4 p^2+16 p+15\right )\right ) \int \left (b x^2+a\right )^pdx}{b (2 p+3)}}{b (2 p+5)}\) |
\(\Big \downarrow \) 238 |
\(\displaystyle \frac {d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1}}{b (2 p+5)}-\frac {\frac {d x \left (a+b x^2\right )^{p+1} (3 a d-b c (2 p+7))}{b (2 p+3)}-\frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a^2 d^2-2 a b c d (2 p+5)+b^2 c^2 \left (4 p^2+16 p+15\right )\right ) \int \left (\frac {b x^2}{a}+1\right )^pdx}{b (2 p+3)}}{b (2 p+5)}\) |
\(\Big \downarrow \) 237 |
\(\displaystyle \frac {d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1}}{b (2 p+5)}-\frac {\frac {d x \left (a+b x^2\right )^{p+1} (3 a d-b c (2 p+7))}{b (2 p+3)}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a^2 d^2-2 a b c d (2 p+5)+b^2 c^2 \left (4 p^2+16 p+15\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{b (2 p+3)}}{b (2 p+5)}\) |
(d*x*(a + b*x^2)^(1 + p)*(c + d*x^2))/(b*(5 + 2*p)) - ((d*(3*a*d - b*c*(7 + 2*p))*x*(a + b*x^2)^(1 + p))/(b*(3 + 2*p)) - ((3*a^2*d^2 - 2*a*b*c*d*(5 + 2*p) + b^2*c^2*(15 + 16*p + 4*p^2))*x*(a + b*x^2)^p*Hypergeometric2F1[1/ 2, -p, 3/2, -((b*x^2)/a)])/(b*(3 + 2*p)*(1 + (b*x^2)/a)^p))/(b*(5 + 2*p))
3.4.43.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] && !IntegerQ[2*p ] && GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) ^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(1 + b*(x^2/a))^p, x], x] / ; FreeQ[{a, b, p}, x] && !IntegerQ[2*p] && !GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
\[\int \left (b \,x^{2}+a \right )^{p} \left (d \,x^{2}+c \right )^{2}d x\]
\[ \int \left (a+b x^2\right )^p \left (c+d x^2\right )^2 \, dx=\int { {\left (d x^{2} + c\right )}^{2} {\left (b x^{2} + a\right )}^{p} \,d x } \]
Result contains complex when optimal does not.
Time = 10.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.50 \[ \int \left (a+b x^2\right )^p \left (c+d x^2\right )^2 \, dx=a^{p} c^{2} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {2 a^{p} c d x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + \frac {a^{p} d^{2} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} \]
a**p*c**2*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) + 2*a**p*c* d*x**3*hyper((3/2, -p), (5/2,), b*x**2*exp_polar(I*pi)/a)/3 + a**p*d**2*x* *5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5
\[ \int \left (a+b x^2\right )^p \left (c+d x^2\right )^2 \, dx=\int { {\left (d x^{2} + c\right )}^{2} {\left (b x^{2} + a\right )}^{p} \,d x } \]
\[ \int \left (a+b x^2\right )^p \left (c+d x^2\right )^2 \, dx=\int { {\left (d x^{2} + c\right )}^{2} {\left (b x^{2} + a\right )}^{p} \,d x } \]
Timed out. \[ \int \left (a+b x^2\right )^p \left (c+d x^2\right )^2 \, dx=\int {\left (b\,x^2+a\right )}^p\,{\left (d\,x^2+c\right )}^2 \,d x \]