Integrand size = 26, antiderivative size = 792 \[ \int \frac {(e x)^{-4-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=-\frac {d^5 \left (2 b c^2+a d^2 (2+p)\right ) (e x)^{2-2 p} \left (a+b x^2\right )^{1+p}}{4 a c^4 \left (b c^2+a d^2\right ) e^6 p \left (c^2-d^2 x^2\right )^2}+\frac {d^3 (e x)^{-2 p} \left (a+b x^2\right )^{1+p}}{2 a c^2 e^4 p \left (c^2-d^2 x^2\right )^2}-\frac {3 d^3 \left (2 b c^2+a d^2 (3+p)\right ) (e x)^{-2 p} \left (a+b x^2\right )^{1+p}}{4 a c^2 \left (b c^2+a d^2\right ) e^4 (1+p) \left (c^2-d^2 x^2\right )^2}+\frac {3 d (e x)^{-2 (1+p)} \left (a+b x^2\right )^{1+p}}{2 a e^2 (1+p) \left (c^2-d^2 x^2\right )^2}-\frac {3 d^3 \left (2 b^2 c^4+a b c^2 d^2 (9+5 p)+a^2 d^4 \left (6+5 p+p^2\right )\right ) (e x)^{-2 p} \left (a+b x^2\right )^{1+p}}{4 a c^4 \left (b c^2+a d^2\right )^2 e^4 (1+p) \left (c^2-d^2 x^2\right )}-\frac {(e x)^{-3-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} (-3-2 p),-p,3,\frac {1}{2} (-1-2 p),-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^3 e (3+2 p)}-\frac {3 d^2 (e x)^{-1-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} (-1-2 p),-p,3,\frac {1}{2} (1-2 p),-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^5 e^3 (1+2 p)}+\frac {3 d^3 \left (6 b^2 c^4+6 a b c^2 d^2 (2+p)+a^2 d^4 \left (6+5 p+p^2\right )\right ) (e x)^{-2 p} \left (a+b x^2\right )^p \operatorname {Hypergeometric2F1}\left (1,-p,1-p,\frac {\left (b+\frac {a d^2}{c^2}\right ) x^2}{a+b x^2}\right )}{4 c^6 \left (b c^2+a d^2\right )^2 e^4 p}-\frac {d^3 \left (2 b^2 c^4+4 a b c^2 d^2 (1+p)+a^2 d^4 \left (2+3 p+p^2\right )\right ) (e x)^{2-2 p} \left (a+b x^2\right )^{-1+p} \operatorname {Hypergeometric2F1}\left (2,1-p,2-p,\frac {\left (b+\frac {a d^2}{c^2}\right ) x^2}{a+b x^2}\right )}{4 c^8 \left (b c^2+a d^2\right ) e^6 (1-p) p} \] Output:
-1/4*d^5*(2*b*c^2+a*d^2*(2+p))*(e*x)^(2-2*p)*(b*x^2+a)^(p+1)/a/c^4/(a*d^2+ b*c^2)/e^6/p/(-d^2*x^2+c^2)^2+1/2*d^3*(b*x^2+a)^(p+1)/a/c^2/e^4/p/((e*x)^( 2*p))/(-d^2*x^2+c^2)^2-3/4*d^3*(2*b*c^2+a*d^2*(3+p))*(b*x^2+a)^(p+1)/a/c^2 /(a*d^2+b*c^2)/e^4/(p+1)/((e*x)^(2*p))/(-d^2*x^2+c^2)^2+3/2*d*(b*x^2+a)^(p +1)/a/e^2/(p+1)/((e*x)^(2*p+2))/(-d^2*x^2+c^2)^2-3/4*d^3*(2*b^2*c^4+a*b*c^ 2*d^2*(9+5*p)+a^2*d^4*(p^2+5*p+6))*(b*x^2+a)^(p+1)/a/c^4/(a*d^2+b*c^2)^2/e ^4/(p+1)/((e*x)^(2*p))/(-d^2*x^2+c^2)-(e*x)^(-3-2*p)*(b*x^2+a)^p*AppellF1( -3/2-p,3,-p,-1/2-p,d^2*x^2/c^2,-b*x^2/a)/c^3/e/(3+2*p)/((1+b*x^2/a)^p)-3*d ^2*(e*x)^(-1-2*p)*(b*x^2+a)^p*AppellF1(-1/2-p,3,-p,1/2-p,d^2*x^2/c^2,-b*x^ 2/a)/c^5/e^3/(1+2*p)/((1+b*x^2/a)^p)+3/4*d^3*(6*b^2*c^4+6*a*b*c^2*d^2*(2+p )+a^2*d^4*(p^2+5*p+6))*(b*x^2+a)^p*hypergeom([1, -p],[1-p],(b+a*d^2/c^2)*x ^2/(b*x^2+a))/c^6/(a*d^2+b*c^2)^2/e^4/p/((e*x)^(2*p))-1/4*d^3*(2*b^2*c^4+4 *a*b*c^2*d^2*(p+1)+a^2*d^4*(p^2+3*p+2))*(e*x)^(2-2*p)*(b*x^2+a)^(-1+p)*hyp ergeom([2, 1-p],[2-p],(b+a*d^2/c^2)*x^2/(b*x^2+a))/c^8/(a*d^2+b*c^2)/e^6/( 1-p)/p
\[ \int \frac {(e x)^{-4-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int \frac {(e x)^{-4-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx \] Input:
Integrate[((e*x)^(-4 - 2*p)*(a + b*x^2)^p)/(c + d*x)^3,x]
Output:
Integrate[((e*x)^(-4 - 2*p)*(a + b*x^2)^p)/(c + d*x)^3, x]
Time = 1.31 (sec) , antiderivative size = 729, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {623, 622, 2009}
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.
\(\displaystyle \int \frac {(e x)^{-2 p-4} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx\) |
\(\Big \downarrow \) 623 |
\(\displaystyle x^{2 (p+2)} (e x)^{-2 (p+2)} \int \frac {x^{-2 (p+2)} \left (b x^2+a\right )^p}{(c+d x)^3}dx\) |
\(\Big \downarrow \) 622 |
\(\displaystyle x^{2 (p+2)} (e x)^{-2 (p+2)} \int \left (\frac {c^3 x^{-2 (p+2)} \left (b x^2+a\right )^p}{\left (c^2-d^2 x^2\right )^3}-\frac {3 c^2 d x^{1-2 (p+2)} \left (b x^2+a\right )^p}{\left (c^2-d^2 x^2\right )^3}+\frac {3 c d^2 x^{2-2 (p+2)} \left (b x^2+a\right )^p}{\left (c^2-d^2 x^2\right )^3}+\frac {d^3 x^{3-2 (p+2)} \left (b x^2+a\right )^p}{\left (d^2 x^2-c^2\right )^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^{2 (p+2)} (e x)^{-2 (p+2)} \left (\frac {3 d x^{-2 (p+1)} \left (a+b x^2\right )^{p+1} \left (a^2 d^4 \left (p^2+5 p+6\right )+a b c^2 d^2 (5 p+9)+2 b^2 c^4\right )}{4 a c^4 (p+1) \left (a d^2+b c^2\right )^2}+\frac {d^3 x^{-2 p} \left (a+b x^2\right )^p \left (a^2 d^4 \left (p^2+3 p+2\right )+4 a b c^2 d^2 (p+1)+2 b^2 c^4\right ) \operatorname {Hypergeometric2F1}\left (1,-p,1-p,\frac {\left (\frac {a d^2}{c^2}+b\right ) x^2}{b x^2+a}\right )}{4 c^6 p \left (a d^2+b c^2\right )^2}+\frac {3 d^3 x^{-2 p} \left (a+b x^2\right )^p \left (a^2 d^4 \left (p^2+5 p+6\right )+6 a b c^2 d^2 (p+2)+6 b^2 c^4\right ) \operatorname {Hypergeometric2F1}\left (1,-p,1-p,\frac {\left (\frac {a d^2}{c^2}+b\right ) x^2}{b x^2+a}\right )}{4 c^6 p \left (a d^2+b c^2\right )^2}-\frac {3 d^2 x^{-2 p-1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (-p-\frac {1}{2},-p,3,\frac {1}{2}-p,-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^5 (2 p+1)}-\frac {x^{-2 p-3} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (-p-\frac {3}{2},-p,3,-p-\frac {1}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^3 (2 p+3)}-\frac {d^5 x^{-2 p} \left (a+b x^2\right )^{p+1}}{4 c^2 \left (c^2-d^2 x^2\right )^2 \left (a d^2+b c^2\right )}-\frac {3 d^3 x^{-2 (p+1)} \left (a+b x^2\right )^{p+1} \left (a d^2 (p+3)+4 b c^2\right )}{4 c^2 \left (c^2-d^2 x^2\right ) \left (a d^2+b c^2\right )^2}-\frac {3 d^3 x^{-2 (p+1)} \left (a+b x^2\right )^{p+1}}{4 \left (c^2-d^2 x^2\right )^2 \left (a d^2+b c^2\right )}-\frac {d^5 x^{-2 p} \left (a+b x^2\right )^{p+1} \left (a d^2 (p+2)+3 b c^2\right )}{4 c^4 \left (c^2-d^2 x^2\right ) \left (a d^2+b c^2\right )^2}\right )\) |
Input:
Int[((e*x)^(-4 - 2*p)*(a + b*x^2)^p)/(c + d*x)^3,x]
Output:
(x^(2*(2 + p))*((3*d*(2*b^2*c^4 + a*b*c^2*d^2*(9 + 5*p) + a^2*d^4*(6 + 5*p + p^2))*(a + b*x^2)^(1 + p))/(4*a*c^4*(b*c^2 + a*d^2)^2*(1 + p)*x^(2*(1 + p))) - (d^5*(a + b*x^2)^(1 + p))/(4*c^2*(b*c^2 + a*d^2)*x^(2*p)*(c^2 - d^ 2*x^2)^2) - (3*d^3*(a + b*x^2)^(1 + p))/(4*(b*c^2 + a*d^2)*x^(2*(1 + p))*( c^2 - d^2*x^2)^2) - (d^5*(3*b*c^2 + a*d^2*(2 + p))*(a + b*x^2)^(1 + p))/(4 *c^4*(b*c^2 + a*d^2)^2*x^(2*p)*(c^2 - d^2*x^2)) - (3*d^3*(4*b*c^2 + a*d^2* (3 + p))*(a + b*x^2)^(1 + p))/(4*c^2*(b*c^2 + a*d^2)^2*x^(2*(1 + p))*(c^2 - d^2*x^2)) - (x^(-3 - 2*p)*(a + b*x^2)^p*AppellF1[-3/2 - p, -p, 3, -1/2 - p, -((b*x^2)/a), (d^2*x^2)/c^2])/(c^3*(3 + 2*p)*(1 + (b*x^2)/a)^p) - (3*d ^2*x^(-1 - 2*p)*(a + b*x^2)^p*AppellF1[-1/2 - p, -p, 3, 1/2 - p, -((b*x^2) /a), (d^2*x^2)/c^2])/(c^5*(1 + 2*p)*(1 + (b*x^2)/a)^p) + (d^3*(2*b^2*c^4 + 4*a*b*c^2*d^2*(1 + p) + a^2*d^4*(2 + 3*p + p^2))*(a + b*x^2)^p*Hypergeome tric2F1[1, -p, 1 - p, ((b + (a*d^2)/c^2)*x^2)/(a + b*x^2)])/(4*c^6*(b*c^2 + a*d^2)^2*p*x^(2*p)) + (3*d^3*(6*b^2*c^4 + 6*a*b*c^2*d^2*(2 + p) + a^2*d^ 4*(6 + 5*p + p^2))*(a + b*x^2)^p*Hypergeometric2F1[1, -p, 1 - p, ((b + (a* d^2)/c^2)*x^2)/(a + b*x^2)])/(4*c^6*(b*c^2 + a*d^2)^2*p*x^(2*p))))/(e*x)^( 2*(2 + p))
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Int[ExpandIntegrand[x^m*(a + b*x^2)^p, (c/(c^2 - d^2*x^2) - d*(x/(c^2 - d^2*x^2)))^(-n), x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[n, -1]
Int[((e_)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*x)^m/x^m Int[x^m*(c + d*x)^n*(a + b*x^2)^p, x], x] / ; FreeQ[{a, b, c, d, e, m, p}, x] && ILtQ[n, 0]
\[\int \frac {\left (e x \right )^{-4-2 p} \left (b \,x^{2}+a \right )^{p}}{\left (d x +c \right )^{3}}d x\]
Input:
int((e*x)^(-4-2*p)*(b*x^2+a)^p/(d*x+c)^3,x)
Output:
int((e*x)^(-4-2*p)*(b*x^2+a)^p/(d*x+c)^3,x)
\[ \int \frac {(e x)^{-4-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} \left (e x\right )^{-2 \, p - 4}}{{\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate((e*x)^(-4-2*p)*(b*x^2+a)^p/(d*x+c)^3,x, algorithm="fricas")
Output:
integral((b*x^2 + a)^p*(e*x)^(-2*p - 4)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)
Timed out. \[ \int \frac {(e x)^{-4-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\text {Timed out} \] Input:
integrate((e*x)**(-4-2*p)*(b*x**2+a)**p/(d*x+c)**3,x)
Output:
Timed out
\[ \int \frac {(e x)^{-4-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} \left (e x\right )^{-2 \, p - 4}}{{\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate((e*x)^(-4-2*p)*(b*x^2+a)^p/(d*x+c)^3,x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^p*(e*x)^(-2*p - 4)/(d*x + c)^3, x)
\[ \int \frac {(e x)^{-4-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} \left (e x\right )^{-2 \, p - 4}}{{\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate((e*x)^(-4-2*p)*(b*x^2+a)^p/(d*x+c)^3,x, algorithm="giac")
Output:
integrate((b*x^2 + a)^p*(e*x)^(-2*p - 4)/(d*x + c)^3, x)
Timed out. \[ \int \frac {(e x)^{-4-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p}{{\left (e\,x\right )}^{2\,p+4}\,{\left (c+d\,x\right )}^3} \,d x \] Input:
int((a + b*x^2)^p/((e*x)^(2*p + 4)*(c + d*x)^3),x)
Output:
int((a + b*x^2)^p/((e*x)^(2*p + 4)*(c + d*x)^3), x)
\[ \int \frac {(e x)^{-4-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\frac {\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} c^{3} x^{4}+3 x^{2 p} c^{2} d \,x^{5}+3 x^{2 p} c \,d^{2} x^{6}+x^{2 p} d^{3} x^{7}}d x}{e^{2 p} e^{4}} \] Input:
int((e*x)^(-4-2*p)*(b*x^2+a)^p/(d*x+c)^3,x)
Output:
int((a + b*x**2)**p/(x**(2*p)*c**3*x**4 + 3*x**(2*p)*c**2*d*x**5 + 3*x**(2 *p)*c*d**2*x**6 + x**(2*p)*d**3*x**7),x)/(e**(2*p)*e**4)