Integrand size = 26, antiderivative size = 590 \[ \int \frac {(e x)^{-1-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\frac {3 d^2 (e x)^{2-2 p} \left (a+b x^2\right )^{1+p}}{2 a c e^3 (1-p) \left (c^2-d^2 x^2\right )^2}+\frac {d^2 \left (2 b c^2+a d^2 (2+p)\right ) (e x)^{2-2 p} \left (a+b x^2\right )^{1+p}}{4 a c \left (b c^2+a d^2\right ) e^3 p \left (c^2-d^2 x^2\right )^2}-\frac {c (e x)^{-2 p} \left (a+b x^2\right )^{1+p}}{2 a e p \left (c^2-d^2 x^2\right )^2}-\frac {3 d (e x)^{1-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} (1-2 p),-p,3,\frac {1}{2} (3-2 p),-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^4 e^2 (1-2 p)}-\frac {d^3 (e x)^{3-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} (3-2 p),-p,3,\frac {1}{2} (5-2 p),-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^6 e^4 (3-2 p)}+\frac {\left (2 b^2 c^4+4 a b c^2 d^2 (1+p)+a^2 d^4 \left (2+3 p+p^2\right )\right ) (e x)^{2-2 p} \left (a+b x^2\right )^{-1+p} \operatorname {Hypergeometric2F1}\left (2,1-p,2-p,\frac {\left (b+\frac {a d^2}{c^2}\right ) x^2}{a+b x^2}\right )}{4 c^5 \left (b c^2+a d^2\right ) e^3 (1-p) p}-\frac {3 a d^2 \left (2 b c^2+a d^2 (1+p)\right ) (e x)^{4-2 p} \left (a+b x^2\right )^{-2+p} \operatorname {Hypergeometric2F1}\left (3,2-p,3-p,\frac {\left (b+\frac {a d^2}{c^2}\right ) x^2}{a+b x^2}\right )}{2 c^7 e^5 (1-p) (2-p)} \] Output:
3/2*d^2*(e*x)^(2-2*p)*(b*x^2+a)^(p+1)/a/c/e^3/(1-p)/(-d^2*x^2+c^2)^2+1/4*d ^2*(2*b*c^2+a*d^2*(2+p))*(e*x)^(2-2*p)*(b*x^2+a)^(p+1)/a/c/(a*d^2+b*c^2)/e ^3/p/(-d^2*x^2+c^2)^2-1/2*c*(b*x^2+a)^(p+1)/a/e/p/((e*x)^(2*p))/(-d^2*x^2+ c^2)^2-3*d*(e*x)^(1-2*p)*(b*x^2+a)^p*AppellF1(1/2-p,3,-p,3/2-p,d^2*x^2/c^2 ,-b*x^2/a)/c^4/e^2/(1-2*p)/((1+b*x^2/a)^p)-d^3*(e*x)^(3-2*p)*(b*x^2+a)^p*A ppellF1(3/2-p,3,-p,5/2-p,d^2*x^2/c^2,-b*x^2/a)/c^6/e^4/(3-2*p)/((1+b*x^2/a )^p)+1/4*(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)*hypergeom([2, 1-p],[2-p],(b+a*d^2/c^2)*x^2/(b*x^2+a))/c^ 5/(a*d^2+b*c^2)/e^3/(1-p)/p-3/2*a*d^2*(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^7 /e^5/(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 = 0.97 (sec) , antiderivative size = 562, normalized size of antiderivative = 0.95, 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.
| \(\displaystyle \int \frac {(e x)^{-2 p-1} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx\) |
| \(\Big \downarrow \) 623 |
| \(\displaystyle x^{2 p+1} (e x)^{-2 p-1} \int \frac {x^{-2 p-1} \left (b x^2+a\right )^p}{(c+d x)^3}dx\) |
| \(\Big \downarrow \) 622 |
| \(\displaystyle x^{2 p+1} (e x)^{-2 p-1} \int \left (\frac {c^3 \left (b x^2+a\right )^p x^{-2 p-1}}{\left (c^2-d^2 x^2\right )^3}+\frac {3 c d^2 \left (b x^2+a\right )^p x^{1-2 p}}{\left (c^2-d^2 x^2\right )^3}+\frac {d^3 \left (b x^2+a\right )^p x^{2-2 p}}{\left (d^2 x^2-c^2\right )^3}-\frac {3 c^2 d \left (b x^2+a\right )^p x^{-2 p}}{\left (c^2-d^2 x^2\right )^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle x^{2 p+1} (e x)^{-2 p-1} \left (-\frac {x^{-2 p} \left (a+b x^2\right )^p \left (a^2 d^4 \left (p^2+3 p+2\right )+4 a b c^2 d^2 (p+1)+2 b^2 c^4\right ) \operatorname {Hypergeometric2F1}\left (1,-p,1-p,\frac {\left (\frac {a d^2}{c^2}+b\right ) x^2}{b x^2+a}\right )}{4 c^3 p \left (a d^2+b c^2\right )^2}-\frac {d^3 x^{3-2 p} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2}-p,-p,3,\frac {5}{2}-p,-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^6 (3-2 p)}-\frac {3 d x^{1-2 p} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2}-p,-p,3,\frac {3}{2}-p,-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^4 (1-2 p)}+\frac {d^2 x^{-2 p} \left (a+b x^2\right )^{p+1} \left (a d^2 (p+2)+3 b c^2\right )}{4 c \left (c^2-d^2 x^2\right ) \left (a d^2+b c^2\right )^2}+\frac {c d^2 x^{-2 p} \left (a+b x^2\right )^{p+1}}{4 \left (c^2-d^2 x^2\right )^2 \left (a d^2+b c^2\right )}+\frac {3 d^4 x^{2-2 p} \left (a+b x^2\right )^{p+1}}{4 c \left (c^2-d^2 x^2\right )^2 \left (a d^2+b c^2\right )}+\frac {3 a d^2 x^{2-2 p} \left (a+b x^2\right )^{p-1} \left (a d^2 (p+1)+2 b c^2\right ) \operatorname {Hypergeometric2F1}\left (2,1-p,2-p,\frac {\left (\frac {a d^2}{c^2}+b\right ) x^2}{b x^2+a}\right )}{4 c^5 (1-p) \left (a d^2+b c^2\right )}\right )\) |
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)*((3*d^4*x^(2 - 2*p)*(a + b*x^2)^(1 + p))/(4*c *(b*c^2 + a*d^2)*(c^2 - d^2*x^2)^2) + (c*d^2*(a + b*x^2)^(1 + p))/(4*(b*c^ 2 + a*d^2)*x^(2*p)*(c^2 - d^2*x^2)^2) + (d^2*(3*b*c^2 + a*d^2*(2 + p))*(a + b*x^2)^(1 + p))/(4*c*(b*c^2 + a*d^2)^2*x^(2*p)*(c^2 - d^2*x^2)) - (3*d*x ^(1 - 2*p)*(a + b*x^2)^p*AppellF1[1/2 - p, -p, 3, 3/2 - p, -((b*x^2)/a), ( d^2*x^2)/c^2])/(c^4*(1 - 2*p)*(1 + (b*x^2)/a)^p) - (d^3*x^(3 - 2*p)*(a + b *x^2)^p*AppellF1[3/2 - p, -p, 3, 5/2 - p, -((b*x^2)/a), (d^2*x^2)/c^2])/(c ^6*(3 - 2*p)*(1 + (b*x^2)/a)^p) - ((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*Hypergeometric2F1[1, -p, 1 - p, ((b + (a*d^2)/c^2)*x^2)/(a + b*x^2)])/(4*c^3*(b*c^2 + a*d^2)^2*p*x^(2*p)) + (3* a*d^2*(2*b*c^2 + a*d^2*(1 + p))*x^(2 - 2*p)*(a + b*x^2)^(-1 + p)*Hypergeom etric2F1[2, 1 - p, 2 - p, ((b + (a*d^2)/c^2)*x^2)/(a + b*x^2)])/(4*c^5*(b* c^2 + a*d^2)*(1 - 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. \[ \int \frac {(e x)^{-1-2 p} \left (a+b x^2\right )^p}{(c+d x)^3} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p}{{\left (e\,x\right )}^{2\,p+1}\,{\left (c+d\,x\right )}^3} \,d x \] Input:
int((a + b*x^2)^p/((e*x)^(2*p + 1)*(c + d*x)^3),x)
Output:
int((a + b*x^2)^p/((e*x)^(2*p + 1)*(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:
( - 6*(a + b*x**2)**p*a**2*c**2*d**3*p**2 + 3*(a + b*x**2)**p*a**2*c**2*d* *3*p - 2*(a + b*x**2)**p*a**2*c*d**4*p**3*x - 2*(a + b*x**2)**p*a**2*c*d** 4*p**2*x - (a + b*x**2)**p*a**2*d**5*p**3*x**2 - (a + b*x**2)**p*a**2*d**5 *p**2*x**2 + 6*(a + b*x**2)**p*a*b*c**4*d*p - 3*(a + b*x**2)**p*a*b*c**4*d + 8*(a + b*x**2)**p*a*b*c**3*d**2*p**2*x + 14*(a + b*x**2)**p*a*b*c**3*d* *2*p*x - 6*(a + b*x**2)**p*a*b*c**3*d**2*x + 4*(a + b*x**2)**p*a*b*c**2*d* *3*p**2*x**2 + 7*(a + b*x**2)**p*a*b*c**2*d**3*p*x**2 - 3*(a + b*x**2)**p* a*b*c**2*d**3*x**2 - 6*(a + b*x**2)**p*b**2*c**5*p*x - 3*(a + b*x**2)**p*b **2*c**4*d*p*x**2 - 8*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 ...