Integrand size = 26, antiderivative size = 408 \[ \int \frac {(e x)^{1-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\frac {c (e x)^{2-2 p} \left (a+b x^2\right )^{1+p}}{2 a e (1-p) \left (c^2-d^2 x^2\right )^2}-\frac {3 d (e x)^{3-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} (3-2 p),-p,3,\frac {1}{2} (5-2 p),-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^4 e^2 (3-2 p)}-\frac {d^3 (e x)^{5-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} (5-2 p),-p,3,\frac {1}{2} (7-2 p),-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^6 e^4 (5-2 p)}+\frac {3 a^2 d^2 (e x)^{4-2 p} \left (a+b x^2\right )^{-2+p} \operatorname {Hypergeometric2F1}\left (3,2-p,3-p,\frac {\left (b+\frac {a d^2}{c^2}\right ) x^2}{a+b x^2}\right )}{2 c^5 e^3 (2-p)}-\frac {a \left (2 b c^2+a d^2 (1+p)\right ) (e x)^{4-2 p} \left (a+b x^2\right )^{-2+p} \operatorname {Hypergeometric2F1}\left (3,2-p,3-p,\frac {\left (b+\frac {a d^2}{c^2}\right ) x^2}{a+b x^2}\right )}{2 c^5 e^3 (1-p) (2-p)} \] Output:
1/2*c*(e*x)^(2-2*p)*(b*x^2+a)^(p+1)/a/e/(1-p)/(-d^2*x^2+c^2)^2-3*d*(e*x)^( 3-2*p)*(b*x^2+a)^p*AppellF1(3/2-p,3,-p,5/2-p,d^2*x^2/c^2,-b*x^2/a)/c^4/e^2 /(3-2*p)/((1+b*x^2/a)^p)-d^3*(e*x)^(5-2*p)*(b*x^2+a)^p*AppellF1(5/2-p,3,-p ,7/2-p,d^2*x^2/c^2,-b*x^2/a)/c^6/e^4/(5-2*p)/((1+b*x^2/a)^p)+3/2*a^2*d^2*( e*x)^(4-2*p)*(b*x^2+a)^(-2+p)*hypergeom([3, 2-p],[3-p],(b+a*d^2/c^2)*x^2/( b*x^2+a))/c^5/e^3/(2-p)-1/2*a*(2*b*c^2+a*d^2*(p+1))*(e*x)^(4-2*p)*(b*x^2+a )^(-2+p)*hypergeom([3, 2-p],[3-p],(b+a*d^2/c^2)*x^2/(b*x^2+a))/c^5/e^3/(1- p)/(2-p)
\[ \int \frac {(e x)^{1-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int \frac {(e x)^{1-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx \] Input:
Integrate[((e*x)^(1 - 2*p)*(a + b*x^2)^p)/(c + d*x)^3,x]
Output:
Integrate[((e*x)^(1 - 2*p)*(a + b*x^2)^p)/(c + d*x)^3, x]
Time = 0.73 (sec) , antiderivative size = 404, normalized size of antiderivative = 0.99, 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.
| \(\displaystyle \int \frac {(e x)^{1-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx\) |
| \(\Big \downarrow \) 623 |
| \(\displaystyle x^{2 p-1} (e x)^{1-2 p} \int \frac {x^{1-2 p} \left (b x^2+a\right )^p}{(c+d x)^3}dx\) |
| \(\Big \downarrow \) 622 |
| \(\displaystyle x^{2 p-1} (e x)^{1-2 p} \int \left (\frac {c^3 \left (b x^2+a\right )^p x^{1-2 p}}{\left (c^2-d^2 x^2\right )^3}-\frac {3 c^2 d \left (b x^2+a\right )^p x^{2-2 p}}{\left (c^2-d^2 x^2\right )^3}+\frac {3 c d^2 \left (b x^2+a\right )^p x^{3-2 p}}{\left (c^2-d^2 x^2\right )^3}+\frac {d^3 \left (b x^2+a\right )^p x^{4-2 p}}{\left (d^2 x^2-c^2\right )^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle x^{2 p-1} (e x)^{1-2 p} \left (\frac {3 a^2 d^2 x^{4-2 p} \left (a+b x^2\right )^{p-2} \operatorname {Hypergeometric2F1}\left (3,2-p,3-p,\frac {\left (\frac {a d^2}{c^2}+b\right ) x^2}{b x^2+a}\right )}{2 c^5 (2-p)}-\frac {d^3 x^{5-2 p} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2}-p,-p,3,\frac {7}{2}-p,-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^6 (5-2 p)}-\frac {3 d x^{3-2 p} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2}-p,-p,3,\frac {5}{2}-p,-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^4 (3-2 p)}+\frac {c d^2 x^{2-2 p} \left (a+b x^2\right )^{p+1}}{4 \left (c^2-d^2 x^2\right )^2 \left (a d^2+b c^2\right )}+\frac {a x^{2-2 p} \left (a+b x^2\right )^{p-1} \left (a d^2 (p+1)+2 b c^2\right ) \operatorname {Hypergeometric2F1}\left (2,1-p,2-p,\frac {\left (\frac {a d^2}{c^2}+b\right ) x^2}{b x^2+a}\right )}{4 c^3 (1-p) \left (a d^2+b c^2\right )}\right )\) |
Input:
Int[((e*x)^(1 - 2*p)*(a + b*x^2)^p)/(c + d*x)^3,x]
Output:
x^(-1 + 2*p)*(e*x)^(1 - 2*p)*((c*d^2*x^(2 - 2*p)*(a + b*x^2)^(1 + p))/(4*( b*c^2 + a*d^2)*(c^2 - d^2*x^2)^2) - (3*d*x^(3 - 2*p)*(a + b*x^2)^p*AppellF 1[3/2 - p, -p, 3, 5/2 - p, -((b*x^2)/a), (d^2*x^2)/c^2])/(c^4*(3 - 2*p)*(1 + (b*x^2)/a)^p) - (d^3*x^(5 - 2*p)*(a + b*x^2)^p*AppellF1[5/2 - p, -p, 3, 7/2 - p, -((b*x^2)/a), (d^2*x^2)/c^2])/(c^6*(5 - 2*p)*(1 + (b*x^2)/a)^p) + (a*(2*b*c^2 + a*d^2*(1 + p))*x^(2 - 2*p)*(a + b*x^2)^(-1 + p)*Hypergeome tric2F1[2, 1 - p, 2 - p, ((b + (a*d^2)/c^2)*x^2)/(a + b*x^2)])/(4*c^3*(b*c ^2 + a*d^2)*(1 - p)) + (3*a^2*d^2*x^(4 - 2*p)*(a + b*x^2)^(-2 + p)*Hyperge ometric2F1[3, 2 - p, 3 - p, ((b + (a*d^2)/c^2)*x^2)/(a + b*x^2)])/(2*c^5*( 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]
\[\int \frac {\left (e x \right )^{1-2 p} \left (b \,x^{2}+a \right )^{p}}{\left (d x +c \right )^{3}}d x\]
Input:
int((e*x)^(1-2*p)*(b*x^2+a)^p/(d*x+c)^3,x)
Output:
int((e*x)^(1-2*p)*(b*x^2+a)^p/(d*x+c)^3,x)
\[ \int \frac {(e x)^{1-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} \left (e x\right )^{-2 \, p + 1}}{{\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate((e*x)^(1-2*p)*(b*x^2+a)^p/(d*x+c)^3,x, algorithm="fricas")
Output:
integral((b*x^2 + a)^p*(e*x)^(-2*p + 1)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)
Timed out. \[ \int \frac {(e x)^{1-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\text {Timed out} \] Input:
integrate((e*x)**(1-2*p)*(b*x**2+a)**p/(d*x+c)**3,x)
Output:
Timed out
\[ \int \frac {(e x)^{1-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} \left (e x\right )^{-2 \, p + 1}}{{\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate((e*x)^(1-2*p)*(b*x^2+a)^p/(d*x+c)^3,x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^p*(e*x)^(-2*p + 1)/(d*x + c)^3, x)
\[ \int \frac {(e x)^{1-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} \left (e x\right )^{-2 \, p + 1}}{{\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate((e*x)^(1-2*p)*(b*x^2+a)^p/(d*x+c)^3,x, algorithm="giac")
Output:
integrate((b*x^2 + a)^p*(e*x)^(-2*p + 1)/(d*x + c)^3, x)
Timed out. \[ \int \frac {(e x)^{1-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int \frac {{\left (e\,x\right )}^{1-2\,p}\,{\left (b\,x^2+a\right )}^p}{{\left (c+d\,x\right )}^3} \,d x \] Input:
int(((e*x)^(1 - 2*p)*(a + b*x^2)^p)/(c + d*x)^3,x)
Output:
int(((e*x)^(1 - 2*p)*(a + b*x^2)^p)/(c + d*x)^3, x)
\[ \int \frac {(e x)^{1-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\text {too large to display} \] Input:
int((e*x)^(1-2*p)*(b*x^2+a)^p/(d*x+c)^3,x)
Output:
(e*(4*(a + b*x**2)**p*a**2*c*d**4*p**3*x + 4*(a + b*x**2)**p*a**2*c*d**4*p **2*x + 2*(a + b*x**2)**p*a**2*d**5*p**3*x**2 + 2*(a + b*x**2)**p*a**2*d** 5*p**2*x**2 - 10*(a + b*x**2)**p*a*b*c**3*d**2*p**2*x + 2*(a + b*x**2)**p* a*b*c**3*d**2*p*x + (a + b*x**2)**p*a*b*c**2*d**3*p**2*x**2 - 2*(a + b*x** 2)**p*a*b*c**2*d**3*p*x**2 - 6*(a + b*x**2)**p*b**2*c**5*p*x - 9*(a + b*x* *2)**p*b**2*c**4*d*p*x**2 + 3*(a + b*x**2)**p*b**2*c**4*d*x**2 + 16*x**(2* p)*int((a + b*x**2)**p/(2*x**(2*p)*a**2*c**3*d**2*p**2 - x**(2*p)*a**2*c** 3*d**2*p + 6*x**(2*p)*a**2*c**2*d**3*p**2*x - 3*x**(2*p)*a**2*c**2*d**3*p* x + 6*x**(2*p)*a**2*c*d**4*p**2*x**2 - 3*x**(2*p)*a**2*c*d**4*p*x**2 + 2*x **(2*p)*a**2*d**5*p**2*x**3 - x**(2*p)*a**2*d**5*p*x**3 - 2*x**(2*p)*a*b*c **5*p + x**(2*p)*a*b*c**5 - 6*x**(2*p)*a*b*c**4*d*p*x + 3*x**(2*p)*a*b*c** 4*d*x + 2*x**(2*p)*a*b*c**3*d**2*p**2*x**2 - 7*x**(2*p)*a*b*c**3*d**2*p*x* *2 + 3*x**(2*p)*a*b*c**3*d**2*x**2 + 6*x**(2*p)*a*b*c**2*d**3*p**2*x**3 - 5*x**(2*p)*a*b*c**2*d**3*p*x**3 + x**(2*p)*a*b*c**2*d**3*x**3 + 6*x**(2*p) *a*b*c*d**4*p**2*x**4 - 3*x**(2*p)*a*b*c*d**4*p*x**4 + 2*x**(2*p)*a*b*d**5 *p**2*x**5 - x**(2*p)*a*b*d**5*p*x**5 - 2*x**(2*p)*b**2*c**5*p*x**2 + x**( 2*p)*b**2*c**5*x**2 - 6*x**(2*p)*b**2*c**4*d*p*x**3 + 3*x**(2*p)*b**2*c**4 *d*x**3 - 6*x**(2*p)*b**2*c**3*d**2*p*x**4 + 3*x**(2*p)*b**2*c**3*d**2*x** 4 - 2*x**(2*p)*b**2*c**2*d**3*p*x**5 + x**(2*p)*b**2*c**2*d**3*x**5),x)*a* *4*c**4*d**6*p**6 - 12*x**(2*p)*int((a + b*x**2)**p/(2*x**(2*p)*a**2*c*...