Integrand size = 28, antiderivative size = 231 \[ \int \frac {(e x)^{-1-2 p} \left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=-\frac {d (2 b c-a d (2+p)) (e x)^{2-2 p} \left (a+b x^2\right )^{1+p}}{4 a c^2 (b c-a d) e^3 p \left (c+d x^2\right )^2}-\frac {(e x)^{-2 p} \left (a+b x^2\right )^{1+p}}{2 a c e p \left (c+d x^2\right )^2}+\frac {\left (2 b^2 c^2-4 a b c d (1+p)+a^2 d^2 \left (2+3 p+p^2\right )\right ) (e x)^{2-2 p} \left (a+b x^2\right )^{-1+p} \operatorname {Hypergeometric2F1}\left (2,1-p,2-p,\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}\right )}{4 c^4 (b c-a d) e^3 (1-p) p} \] Output:
-1/4*d*(2*b*c-a*d*(2+p))*(e*x)^(2-2*p)*(b*x^2+a)^(p+1)/a/c^2/(-a*d+b*c)/e^ 3/p/(d*x^2+c)^2-1/2*(b*x^2+a)^(p+1)/a/c/e/p/((e*x)^(2*p))/(d*x^2+c)^2+1/4* (2*b^2*c^2-4*a*b*c*d*(p+1)+a^2*d^2*(p^2+3*p+2))*(e*x)^(2-2*p)*(b*x^2+a)^(- 1+p)*hypergeom([2, -p+1],[2-p],(-a*d+b*c)*x^2/c/(b*x^2+a))/c^4/(-a*d+b*c)/ e^3/(-p+1)/p
Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
Time = 0.38 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.83 \[ \int \frac {(e x)^{-1-2 p} \left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=-\frac {x (e x)^{-1-2 p} \left (a+b x^2\right )^p \left (2 d p x^2 \left (2 c+d (1+p) x^2\right ) \Phi \left (\frac {(b c-a d) x^2}{c \left (a+b x^2\right )},1,1-p\right )-d^2 (-1+p) p x^4 \Phi \left (\frac {(b c-a d) x^2}{c \left (a+b x^2\right )},1,2-p\right )-\left (2 c^2+4 c d (1+p) x^2+d^2 \left (2+3 p+p^2\right ) x^4\right ) \Phi \left (\frac {(b c-a d) x^2}{c \left (a+b x^2\right )},1,-p\right )\right )}{4 c^3 \left (c+d x^2\right )^2} \] Input:
Integrate[((e*x)^(-1 - 2*p)*(a + b*x^2)^p)/(c + d*x^2)^3,x]
Output:
-1/4*(x*(e*x)^(-1 - 2*p)*(a + b*x^2)^p*(2*d*p*x^2*(2*c + d*(1 + p)*x^2)*Hu rwitzLerchPhi[((b*c - a*d)*x^2)/(c*(a + b*x^2)), 1, 1 - p] - d^2*(-1 + p)* p*x^4*HurwitzLerchPhi[((b*c - a*d)*x^2)/(c*(a + b*x^2)), 1, 2 - p] - (2*c^ 2 + 4*c*d*(1 + p)*x^2 + d^2*(2 + 3*p + p^2)*x^4)*HurwitzLerchPhi[((b*c - a *d)*x^2)/(c*(a + b*x^2)), 1, -p]))/(c^3*(c + d*x^2)^2)
Time = 0.35 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {393, 114, 25, 168, 25, 27, 141}
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.
| \(\displaystyle \int \frac {(e x)^{-2 p-1} \left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 393 |
| \(\displaystyle \frac {\left (x^2\right )^{p+1} (e x)^{-2 p-1} \int \frac {\left (x^2\right )^{-p-1} \left (b x^2+a\right )^p}{\left (d x^2+c\right )^3}dx^2}{2 x}\) |
| \(\Big \downarrow \) 114 |
| \(\displaystyle \frac {\left (x^2\right )^{p+1} (e x)^{-2 p-1} \left (-\frac {\int -\frac {\left (x^2\right )^{-p-1} \left (b x^2+a\right )^p \left (-b d x^2+2 b c-a d (p+2)\right )}{\left (d x^2+c\right )^2}dx^2}{2 c (b c-a d)}-\frac {d \left (x^2\right )^{-p} \left (a+b x^2\right )^{p+1}}{2 c \left (c+d x^2\right )^2 (b c-a d)}\right )}{2 x}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\left (x^2\right )^{p+1} (e x)^{-2 p-1} \left (\frac {\int \frac {\left (x^2\right )^{-p-1} \left (b x^2+a\right )^p \left (-b d x^2+2 b c-a d (p+2)\right )}{\left (d x^2+c\right )^2}dx^2}{2 c (b c-a d)}-\frac {d \left (x^2\right )^{-p} \left (a+b x^2\right )^{p+1}}{2 c \left (c+d x^2\right )^2 (b c-a d)}\right )}{2 x}\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle \frac {\left (x^2\right )^{p+1} (e x)^{-2 p-1} \left (\frac {-\frac {\int -\frac {\left (2 b^2 c^2-4 a b d (p+1) c+a^2 d^2 \left (p^2+3 p+2\right )\right ) \left (x^2\right )^{-p-1} \left (b x^2+a\right )^p}{d x^2+c}dx^2}{c (b c-a d)}-\frac {d \left (x^2\right )^{-p} \left (a+b x^2\right )^{p+1} (3 b c-a d (p+2))}{c \left (c+d x^2\right ) (b c-a d)}}{2 c (b c-a d)}-\frac {d \left (x^2\right )^{-p} \left (a+b x^2\right )^{p+1}}{2 c \left (c+d x^2\right )^2 (b c-a d)}\right )}{2 x}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\left (x^2\right )^{p+1} (e x)^{-2 p-1} \left (\frac {\frac {\int \frac {\left (2 b^2 c^2-4 a b d (p+1) c+a^2 d^2 \left (p^2+3 p+2\right )\right ) \left (x^2\right )^{-p-1} \left (b x^2+a\right )^p}{d x^2+c}dx^2}{c (b c-a d)}-\frac {d \left (x^2\right )^{-p} \left (a+b x^2\right )^{p+1} (3 b c-a d (p+2))}{c \left (c+d x^2\right ) (b c-a d)}}{2 c (b c-a d)}-\frac {d \left (x^2\right )^{-p} \left (a+b x^2\right )^{p+1}}{2 c \left (c+d x^2\right )^2 (b c-a d)}\right )}{2 x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (x^2\right )^{p+1} (e x)^{-2 p-1} \left (\frac {\frac {\left (a^2 d^2 \left (p^2+3 p+2\right )-4 a b c d (p+1)+2 b^2 c^2\right ) \int \frac {\left (x^2\right )^{-p-1} \left (b x^2+a\right )^p}{d x^2+c}dx^2}{c (b c-a d)}-\frac {d \left (x^2\right )^{-p} \left (a+b x^2\right )^{p+1} (3 b c-a d (p+2))}{c \left (c+d x^2\right ) (b c-a d)}}{2 c (b c-a d)}-\frac {d \left (x^2\right )^{-p} \left (a+b x^2\right )^{p+1}}{2 c \left (c+d x^2\right )^2 (b c-a d)}\right )}{2 x}\) |
| \(\Big \downarrow \) 141 |
| \(\displaystyle \frac {\left (x^2\right )^{p+1} (e x)^{-2 p-1} \left (\frac {-\frac {\left (x^2\right )^{-p} \left (a+b x^2\right )^p \left (a^2 d^2 \left (p^2+3 p+2\right )-4 a b c d (p+1)+2 b^2 c^2\right ) \operatorname {Hypergeometric2F1}\left (1,-p,1-p,\frac {(b c-a d) x^2}{c \left (b x^2+a\right )}\right )}{c^2 p (b c-a d)}-\frac {d \left (x^2\right )^{-p} \left (a+b x^2\right )^{p+1} (3 b c-a d (p+2))}{c \left (c+d x^2\right ) (b c-a d)}}{2 c (b c-a d)}-\frac {d \left (x^2\right )^{-p} \left (a+b x^2\right )^{p+1}}{2 c \left (c+d x^2\right )^2 (b c-a d)}\right )}{2 x}\) |
Input:
Int[((e*x)^(-1 - 2*p)*(a + b*x^2)^p)/(c + d*x^2)^3,x]
Output:
((e*x)^(-1 - 2*p)*(x^2)^(1 + p)*(-1/2*(d*(a + b*x^2)^(1 + p))/(c*(b*c - a* d)*(x^2)^p*(c + d*x^2)^2) + (-((d*(3*b*c - a*d*(2 + p))*(a + b*x^2)^(1 + p ))/(c*(b*c - a*d)*(x^2)^p*(c + d*x^2))) - ((2*b^2*c^2 - 4*a*b*c*d*(1 + p) + a^2*d^2*(2 + 3*p + p^2))*(a + b*x^2)^p*Hypergeometric2F1[1, -p, 1 - p, ( (b*c - a*d)*x^2)/(c*(a + b*x^2))])/(c^2*(b*c - a*d)*p*(x^2)^p))/(2*c*(b*c - a*d))))/(2*x)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/((m + 1)*(b*e - a*f)^( n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f ))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] || !Su mSimplerQ[p, 1]) && !ILtQ[m, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.), x_Symbol] :> Simp[(e*x)^m/(2*x*(x^2)^(Simplify[(m + 1)/2] - 1)) Subs t[Int[x^(Simplify[(m + 1)/2] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ[Simp lify[m + 2*p]] && !IntegerQ[m]
\[\int \frac {\left (e x \right )^{-1-2 p} \left (b \,x^{2}+a \right )^{p}}{\left (x^{2} d +c \right )^{3}}d x\]
Input:
int((e*x)^(-1-2*p)*(b*x^2+a)^p/(d*x^2+c)^3,x)
Output:
int((e*x)^(-1-2*p)*(b*x^2+a)^p/(d*x^2+c)^3,x)
\[ \int \frac {(e x)^{-1-2 p} \left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} \left (e x\right )^{-2 \, p - 1}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \] Input:
integrate((e*x)^(-1-2*p)*(b*x^2+a)^p/(d*x^2+c)^3,x, algorithm="fricas")
Output:
integral((b*x^2 + a)^p*(e*x)^(-2*p - 1)/(d^3*x^6 + 3*c*d^2*x^4 + 3*c^2*d*x ^2 + c^3), x)
Timed out. \[ \int \frac {(e x)^{-1-2 p} \left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=\text {Timed out} \] Input:
integrate((e*x)**(-1-2*p)*(b*x**2+a)**p/(d*x**2+c)**3,x)
Output:
Timed out
\[ \int \frac {(e x)^{-1-2 p} \left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} \left (e x\right )^{-2 \, p - 1}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \] Input:
integrate((e*x)^(-1-2*p)*(b*x^2+a)^p/(d*x^2+c)^3,x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^p*(e*x)^(-2*p - 1)/(d*x^2 + c)^3, x)
\[ \int \frac {(e x)^{-1-2 p} \left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} \left (e x\right )^{-2 \, p - 1}}{{\left (d x^{2} + c\right )}^{3}} \,d x } \] Input:
integrate((e*x)^(-1-2*p)*(b*x^2+a)^p/(d*x^2+c)^3,x, algorithm="giac")
Output:
integrate((b*x^2 + a)^p*(e*x)^(-2*p - 1)/(d*x^2 + c)^3, x)
Timed out. \[ \int \frac {(e x)^{-1-2 p} \left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p}{{\left (e\,x\right )}^{2\,p+1}\,{\left (d\,x^2+c\right )}^3} \,d x \] Input:
int((a + b*x^2)^p/((e*x)^(2*p + 1)*(c + d*x^2)^3),x)
Output:
int((a + b*x^2)^p/((e*x)^(2*p + 1)*(c + d*x^2)^3), x)
\[ \int \frac {(e x)^{-1-2 p} \left (a+b x^2\right )^p}{\left (c+d x^2\right )^3} \, dx=\text {too large to display} \] Input:
int((e*x)^(-1-2*p)*(b*x^2+a)^p/(d*x^2+c)^3,x)
Output:
( - (a + b*x**2)**p*a**5*d**4*p**5 - 6*(a + b*x**2)**p*a**5*d**4*p**4 - 13 *(a + b*x**2)**p*a**5*d**4*p**3 - 12*(a + b*x**2)**p*a**5*d**4*p**2 - 4*(a + b*x**2)**p*a**5*d**4*p + 12*(a + b*x**2)**p*a**4*b*c*d**3*p**4 + 48*(a + b*x**2)**p*a**4*b*c*d**3*p**3 + 60*(a + b*x**2)**p*a**4*b*c*d**3*p**2 + 24*(a + b*x**2)**p*a**4*b*c*d**3*p + 2*(a + b*x**2)**p*a**4*b*d**4*p**4*x* *2 + 10*(a + b*x**2)**p*a**4*b*d**4*p**3*x**2 + 16*(a + b*x**2)**p*a**4*b* d**4*p**2*x**2 + 8*(a + b*x**2)**p*a**4*b*d**4*p*x**2 - 48*(a + b*x**2)**p *a**3*b**2*c**2*d**2*p**3 - 108*(a + b*x**2)**p*a**3*b**2*c**2*d**2*p**2 - 60*(a + b*x**2)**p*a**3*b**2*c**2*d**2*p - 16*(a + b*x**2)**p*a**3*b**2*c *d**3*p**3*x**2 - 56*(a + b*x**2)**p*a**3*b**2*c*d**3*p**2*x**2 - 56*(a + b*x**2)**p*a**3*b**2*c*d**3*p*x**2 - 16*(a + b*x**2)**p*a**3*b**2*c*d**3*x **2 - 2*(a + b*x**2)**p*a**3*b**2*d**4*p**3*x**4 - 10*(a + b*x**2)**p*a**3 *b**2*d**4*p**2*x**4 - 16*(a + b*x**2)**p*a**3*b**2*d**4*p*x**4 - 8*(a + b *x**2)**p*a**3*b**2*d**4*x**4 + 72*(a + b*x**2)**p*a**2*b**3*c**3*d*p**2 + 72*(a + b*x**2)**p*a**2*b**3*c**3*d*p - 4*(a + b*x**2)**p*a**2*b**3*c**2* d**2*p**3*x**2 + 40*(a + b*x**2)**p*a**2*b**3*c**2*d**2*p**2*x**2 + 96*(a + b*x**2)**p*a**2*b**3*c**2*d**2*p*x**2 + 48*(a + b*x**2)**p*a**2*b**3*c** 2*d**2*x**2 + 12*(a + b*x**2)**p*a**2*b**3*c*d**3*p**2*x**4 + 36*(a + b*x* *2)**p*a**2*b**3*c*d**3*p*x**4 + 24*(a + b*x**2)**p*a**2*b**3*c*d**3*x**4 - 36*(a + b*x**2)**p*a*b**4*c**4*p + 16*(a + b*x**2)**p*a*b**4*c**3*d*p...