\(\int \frac {(a+b x^2)^p}{(c+d x)^4} \, dx\) [59]

Optimal result

Integrand size = 17, antiderivative size = 291 \[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^4} \, dx=-\frac {\left (a+b x^2\right )^p}{3 d (c+d x)^3}+\frac {c p \left (a+b x^2\right )^p}{d (2-p) \left (c^2-d^2 x^2\right )^2}-\frac {2 b p x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},1-p,3,\frac {5}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{3 a c^4}-\frac {2 b d^2 p x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},1-p,3,\frac {7}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{15 a c^6}-\frac {2 b^2 c \left (3 a d^2-b c^2 (1-2 p)\right ) \left (a+b x^2\right )^p \operatorname {Hypergeometric2F1}\left (3,p,1+p,\frac {d^2 \left (a+b x^2\right )}{b c^2+a d^2}\right )}{3 d \left (b c^2+a d^2\right )^3 (2-p)} \] Output:

-1/3*(b*x^2+a)^p/d/(d*x+c)^3+c*p*(b*x^2+a)^p/d/(2-p)/(-d^2*x^2+c^2)^2-2/3* 
b*p*x^3*(b*x^2+a)^p*AppellF1(3/2,3,1-p,5/2,d^2*x^2/c^2,-b*x^2/a)/a/c^4/((1 
+b*x^2/a)^p)-2/15*b*d^2*p*x^5*(b*x^2+a)^p*AppellF1(5/2,3,1-p,7/2,d^2*x^2/c 
^2,-b*x^2/a)/a/c^6/((1+b*x^2/a)^p)-2/3*b^2*c*(3*a*d^2-b*c^2*(1-2*p))*(b*x^ 
2+a)^p*hypergeom([3, p],[p+1],d^2*(b*x^2+a)/(a*d^2+b*c^2))/d/(a*d^2+b*c^2) 
^3/(2-p)
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.48 \[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^4} \, dx=\frac {\left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x\right )}{c+d x}\right )^{-p} \left (a+b x^2\right )^p \operatorname {AppellF1}\left (3-2 p,-p,-p,4-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x}\right )}{d (-3+2 p) (c+d x)^3} \] Input:

Integrate[(a + b*x^2)^p/(c + d*x)^4,x]
 

Output:

((a + b*x^2)^p*AppellF1[3 - 2*p, -p, -p, 4 - 2*p, (c - Sqrt[-(a/b)]*d)/(c 
+ d*x), (c + Sqrt[-(a/b)]*d)/(c + d*x)])/(d*(-3 + 2*p)*((d*(-Sqrt[-(a/b)] 
+ x))/(c + d*x))^p*((d*(Sqrt[-(a/b)] + x))/(c + d*x))^p*(c + d*x)^3)
 

Rubi [A] (warning: unable to verify)

Time = 1.12 (sec) , antiderivative size = 390, normalized size of antiderivative = 1.34, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {505, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^p}{(c+d x)^4} \, dx\)

\(\Big \downarrow \) 505

\(\displaystyle \int \left (\frac {6 c^2 d^2 x^2 \left (a+b x^2\right )^p}{\left (c^2-d^2 x^2\right )^4}+\frac {d^4 x^4 \left (a+b x^2\right )^p}{\left (d^2 x^2-c^2\right )^4}-\frac {4 c d^3 x^3 \left (a+b x^2\right )^p}{\left (c^2-d^2 x^2\right )^4}+\frac {c^4 \left (a+b x^2\right )^p}{\left (c^2-d^2 x^2\right )^4}-\frac {4 c^3 d x \left (a+b x^2\right )^p}{\left (c^2-d^2 x^2\right )^4}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {d^4 x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},-p,4,\frac {7}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{5 c^8}+\frac {2 d^2 x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,4,\frac {5}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^6}+\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,4,\frac {3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^4}-\frac {2 b^3 c^3 d \left (a+b x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (4,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{(p+1) \left (a d^2+b c^2\right )^4}+\frac {2 b^2 c d \left (a+b x^2\right )^{p+1} \left (3 a d^2+b c^2 (p+1)\right ) \operatorname {Hypergeometric2F1}\left (3,p+1,p+2,\frac {d^2 \left (b x^2+a\right )}{b c^2+a d^2}\right )}{3 (p+1) \left (a d^2+b c^2\right )^4}-\frac {2 c^3 d \left (a+b x^2\right )^{p+1}}{3 \left (c^2-d^2 x^2\right )^3 \left (a d^2+b c^2\right )}\)

Input:

Int[(a + b*x^2)^p/(c + d*x)^4,x]
 

Output:

(-2*c^3*d*(a + b*x^2)^(1 + p))/(3*(b*c^2 + a*d^2)*(c^2 - d^2*x^2)^3) + (x* 
(a + b*x^2)^p*AppellF1[1/2, -p, 4, 3/2, -((b*x^2)/a), (d^2*x^2)/c^2])/(c^4 
*(1 + (b*x^2)/a)^p) + (2*d^2*x^3*(a + b*x^2)^p*AppellF1[3/2, -p, 4, 5/2, - 
((b*x^2)/a), (d^2*x^2)/c^2])/(c^6*(1 + (b*x^2)/a)^p) + (d^4*x^5*(a + b*x^2 
)^p*AppellF1[5/2, -p, 4, 7/2, -((b*x^2)/a), (d^2*x^2)/c^2])/(5*c^8*(1 + (b 
*x^2)/a)^p) + (2*b^2*c*d*(3*a*d^2 + b*c^2*(1 + p))*(a + b*x^2)^(1 + p)*Hyp 
ergeometric2F1[3, 1 + p, 2 + p, (d^2*(a + b*x^2))/(b*c^2 + a*d^2)])/(3*(b* 
c^2 + a*d^2)^4*(1 + p)) - (2*b^3*c^3*d*(a + b*x^2)^(1 + p)*Hypergeometric2 
F1[4, 1 + p, 2 + p, (d^2*(a + b*x^2))/(b*c^2 + a*d^2)])/((b*c^2 + a*d^2)^4 
*(1 + p))
 

Defintions of rubi rules used

rule 505

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[E 
xpandIntegrand[(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, p}, x] && ILtQ[n, -1] && PosQ[a/b]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

\[\int \frac {\left (b \,x^{2}+a \right )^{p}}{\left (d x +c \right )^{4}}d x\]

Input:

int((b*x^2+a)^p/(d*x+c)^4,x)
 

Output:

int((b*x^2+a)^p/(d*x+c)^4,x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^4} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (d x + c\right )}^{4}} \,d x } \] Input:

integrate((b*x^2+a)^p/(d*x+c)^4,x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^p/(d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x 
+ c^4), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^4} \, dx=\text {Timed out} \] Input:

integrate((b*x**2+a)**p/(d*x+c)**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^4} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (d x + c\right )}^{4}} \,d x } \] Input:

integrate((b*x^2+a)^p/(d*x+c)^4,x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^p/(d*x + c)^4, x)
 

Giac [F]

\[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^4} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p}}{{\left (d x + c\right )}^{4}} \,d x } \] Input:

integrate((b*x^2+a)^p/(d*x+c)^4,x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^p/(d*x + c)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^4} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p}{{\left (c+d\,x\right )}^4} \,d x \] Input:

int((a + b*x^2)^p/(c + d*x)^4,x)
 

Output:

int((a + b*x^2)^p/(c + d*x)^4, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^p}{(c+d x)^4} \, dx=\int \frac {\left (b \,x^{2}+a \right )^{p}}{d^{4} x^{4}+4 c \,d^{3} x^{3}+6 c^{2} d^{2} x^{2}+4 c^{3} d x +c^{4}}d x \] Input:

int((b*x^2+a)^p/(d*x+c)^4,x)
 

Output:

int((a + b*x**2)**p/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 
+ d**4*x**4),x)