Integrand size = 20, antiderivative size = 127 \[ \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^2} \, dx=-\frac {d^2 \left (a+b x^2\right )^{1+p}}{a x}+\frac {\left (a e^2+b d^2 (1+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{a}-\frac {d e \left (a+b x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )}{a (1+p)} \]
-d^2*(b*x^2+a)^(p+1)/a/x+(a*e^2+b*d^2*(1+2*p))*x*(b*x^2+a)^p*hypergeom([1/ 2, -p],[3/2],-b*x^2/a)/a/((1+b*x^2/a)^p)-d*e*(b*x^2+a)^(p+1)*hypergeom([1, p+1],[2+p],1+b*x^2/a)/a/(p+1)
Time = 0.19 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.06 \[ \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^2} \, dx=-\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (a d^2 (1+p) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^2}{a}\right )+e x \left (-a e (1+p) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )+d \left (a+b x^2\right ) \left (1+\frac {b x^2}{a}\right )^p \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^2}{a}\right )\right )\right )}{a (1+p) x} \]
-(((a + b*x^2)^p*(a*d^2*(1 + p)*Hypergeometric2F1[-1/2, -p, 1/2, -((b*x^2) /a)] + e*x*(-(a*e*(1 + p)*x*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)]) + d*(a + b*x^2)*(1 + (b*x^2)/a)^p*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a])))/(a*(1 + p)*x*(1 + (b*x^2)/a)^p))
Time = 0.25 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {543, 27, 243, 75, 359, 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.
\(\displaystyle \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^2} \, dx\) |
\(\Big \downarrow \) 543 |
\(\displaystyle \int \frac {\left (b x^2+a\right )^p \left (d^2+e^2 x^2\right )}{x^2}dx+\int \frac {2 d e \left (b x^2+a\right )^p}{x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {\left (b x^2+a\right )^p \left (d^2+e^2 x^2\right )}{x^2}dx+2 d e \int \frac {\left (b x^2+a\right )^p}{x}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \int \frac {\left (b x^2+a\right )^p \left (d^2+e^2 x^2\right )}{x^2}dx+d e \int \frac {\left (b x^2+a\right )^p}{x^2}dx^2\) |
\(\Big \downarrow \) 75 |
\(\displaystyle \int \frac {\left (b x^2+a\right )^p \left (d^2+e^2 x^2\right )}{x^2}dx-\frac {d e \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a (p+1)}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {\left (a e^2+b d^2 (2 p+1)\right ) \int \left (b x^2+a\right )^pdx}{a}-\frac {d^2 \left (a+b x^2\right )^{p+1}}{a x}-\frac {d e \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a (p+1)}\) |
\(\Big \downarrow \) 238 |
\(\displaystyle \frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a e^2+b d^2 (2 p+1)\right ) \int \left (\frac {b x^2}{a}+1\right )^pdx}{a}-\frac {d^2 \left (a+b x^2\right )^{p+1}}{a x}-\frac {d e \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a (p+1)}\) |
\(\Big \downarrow \) 237 |
\(\displaystyle \frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a e^2+b d^2 (2 p+1)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{a}-\frac {d^2 \left (a+b x^2\right )^{p+1}}{a x}-\frac {d e \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^2}{a}+1\right )}{a (p+1)}\) |
-((d^2*(a + b*x^2)^(1 + p))/(a*x)) + ((a*e^2 + b*d^2*(1 + 2*p))*x*(a + b*x ^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(a*(1 + (b*x^2)/a)^p) - (d*e*(a + b*x^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2) /a])/(a*(1 + p))
3.4.97.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x )^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[n] && (IntegerQ[m] || GtQ[-d/(b*c), 0])
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[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Module[{k}, Int[x^m*Sum[Binomial[n, 2*k]*c^(n - 2*k)*d^(2*k)*x^(2*k), {k, 0, n/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Binomial[n, 2*k + 1]*c^ (n - 2*k - 1)*d^(2*k + 1)*x^(2*k), {k, 0, (n - 1)/2}]*(a + b*x^2)^p, x]] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[n, 1] && IntegerQ[m] && !IntegerQ[2*p] && !(EqQ[m, 1] && EqQ[b*c^2 + a*d^2, 0])
\[\int \frac {\left (e x +d \right )^{2} \left (b \,x^{2}+a \right )^{p}}{x^{2}}d x\]
\[ \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p}}{x^{2}} \,d x } \]
Result contains complex when optimal does not.
Time = 4.99 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.75 \[ \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^2} \, dx=- \frac {a^{p} d^{2} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{x} + a^{p} e^{2} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} - \frac {b^{p} d e x^{2 p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{\Gamma \left (1 - p\right )} \]
-a**p*d**2*hyper((-1/2, -p), (1/2,), b*x**2*exp_polar(I*pi)/a)/x + a**p*e* *2*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) - b**p*d*e*x**(2*p )*gamma(-p)*hyper((-p, -p), (1 - p,), a*exp_polar(I*pi)/(b*x**2))/gamma(1 - p)
\[ \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p}}{x^{2}} \,d x } \]
\[ \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^2} \, dx=\int { \frac {{\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p}}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^2 \left (a+b x^2\right )^p}{x^2} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^2}{x^2} \,d x \]