\(\int \frac {(c+d x^2) (e+f x^2)^2}{(a+b x^2)^{5/2}} \, dx\) [359]

Optimal result

Integrand size = 28, antiderivative size = 165 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {(b c-a d) (b e-a f)^2 x}{3 a b^3 \left (a+b x^2\right )^{3/2}}+\frac {(b e-a f) \left (2 b^2 c e-7 a^2 d f+a b (d e+4 c f)\right ) x}{3 a^2 b^3 \sqrt {a+b x^2}}+\frac {d f^2 x \sqrt {a+b x^2}}{2 b^3}+\frac {f (4 b d e+2 b c f-5 a d f) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{7/2}} \] Output:

1/3*(-a*d+b*c)*(-a*f+b*e)^2*x/a/b^3/(b*x^2+a)^(3/2)+1/3*(-a*f+b*e)*(2*b^2* 
c*e-7*a^2*d*f+a*b*(4*c*f+d*e))*x/a^2/b^3/(b*x^2+a)^(1/2)+1/2*d*f^2*x*(b*x^ 
2+a)^(1/2)/b^3+1/2*f*(-5*a*d*f+2*b*c*f+4*b*d*e)*arctanh(b^(1/2)*x/(b*x^2+a 
)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.01 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {x \left (15 a^4 d f^2+4 b^4 c e^2 x^2+2 a b^3 e \left (3 c e+d e x^2+2 c f x^2\right )+a^2 b^2 f x^2 \left (-16 d e-8 c f+3 d f x^2\right )+2 a^3 b f \left (-6 d e-3 c f+10 d f x^2\right )\right )}{6 a^2 b^3 \left (a+b x^2\right )^{3/2}}-\frac {f (4 b d e+2 b c f-5 a d f) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{2 b^{7/2}} \] Input:

Integrate[((c + d*x^2)*(e + f*x^2)^2)/(a + b*x^2)^(5/2),x]
 

Output:

(x*(15*a^4*d*f^2 + 4*b^4*c*e^2*x^2 + 2*a*b^3*e*(3*c*e + d*e*x^2 + 2*c*f*x^ 
2) + a^2*b^2*f*x^2*(-16*d*e - 8*c*f + 3*d*f*x^2) + 2*a^3*b*f*(-6*d*e - 3*c 
*f + 10*d*f*x^2)))/(6*a^2*b^3*(a + b*x^2)^(3/2)) - (f*(4*b*d*e + 2*b*c*f - 
 5*a*d*f)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(2*b^(7/2))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.24, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {401, 25, 401, 27, 299, 224, 219}

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.

\(\displaystyle \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\left (a+b x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 401

\(\displaystyle \frac {x \left (e+f x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/2}}-\frac {\int -\frac {\left (f x^2+e\right ) \left ((2 b c+a d) e-(2 b c-5 a d) f x^2\right )}{\left (b x^2+a\right )^{3/2}}dx}{3 a b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\left (f x^2+e\right ) \left ((2 b c+a d) e-(2 b c-5 a d) f x^2\right )}{\left (b x^2+a\right )^{3/2}}dx}{3 a b}+\frac {x \left (e+f x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 401

\(\displaystyle \frac {\frac {x \left (e+f x^2\right ) (b e (a d+2 b c)+a f (2 b c-5 a d))}{a b \sqrt {a+b x^2}}-\frac {\int \frac {f \left (\left (-15 d f a^2+2 b (d e+3 c f) a+4 b^2 c e\right ) x^2+a (2 b c-5 a d) e\right )}{\sqrt {b x^2+a}}dx}{a b}}{3 a b}+\frac {x \left (e+f x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {x \left (e+f x^2\right ) (b e (a d+2 b c)+a f (2 b c-5 a d))}{a b \sqrt {a+b x^2}}-\frac {f \int \frac {\left (-15 d f a^2+2 b (d e+3 c f) a+4 b^2 c e\right ) x^2+a (2 b c-5 a d) e}{\sqrt {b x^2+a}}dx}{a b}}{3 a b}+\frac {x \left (e+f x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {x \left (e+f x^2\right ) (b e (a d+2 b c)+a f (2 b c-5 a d))}{a b \sqrt {a+b x^2}}-\frac {f \left (\frac {x \sqrt {a+b x^2} \left (-15 a^2 d f+2 a b (3 c f+d e)+4 b^2 c e\right )}{2 b}-\frac {3 a^2 (-5 a d f+2 b c f+4 b d e) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}\right )}{a b}}{3 a b}+\frac {x \left (e+f x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {x \left (e+f x^2\right ) (b e (a d+2 b c)+a f (2 b c-5 a d))}{a b \sqrt {a+b x^2}}-\frac {f \left (\frac {x \sqrt {a+b x^2} \left (-15 a^2 d f+2 a b (3 c f+d e)+4 b^2 c e\right )}{2 b}-\frac {3 a^2 (-5 a d f+2 b c f+4 b d e) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}\right )}{a b}}{3 a b}+\frac {x \left (e+f x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {x \left (e+f x^2\right ) (b e (a d+2 b c)+a f (2 b c-5 a d))}{a b \sqrt {a+b x^2}}-\frac {f \left (\frac {x \sqrt {a+b x^2} \left (-15 a^2 d f+2 a b (3 c f+d e)+4 b^2 c e\right )}{2 b}-\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (-5 a d f+2 b c f+4 b d e)}{2 b^{3/2}}\right )}{a b}}{3 a b}+\frac {x \left (e+f x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/2}}\)

Input:

Int[((c + d*x^2)*(e + f*x^2)^2)/(a + b*x^2)^(5/2),x]
 

Output:

((b*c - a*d)*x*(e + f*x^2)^2)/(3*a*b*(a + b*x^2)^(3/2)) + (((b*(2*b*c + a* 
d)*e + a*(2*b*c - 5*a*d)*f)*x*(e + f*x^2))/(a*b*Sqrt[a + b*x^2]) - (f*(((4 
*b^2*c*e - 15*a^2*d*f + 2*a*b*(d*e + 3*c*f))*x*Sqrt[a + b*x^2])/(2*b) - (3 
*a^2*(4*b*d*e + 2*b*c*f - 5*a*d*f)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/( 
2*b^(3/2))))/(a*b))/(3*a*b)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 401

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
q/(a*b*2*(p + 1))), x] + Simp[1/(a*b*2*(p + 1))   Int[(a + b*x^2)^(p + 1)*( 
c + d*x^2)^(q - 1)*Simp[c*(b*e*2*(p + 1) + b*e - a*f) + d*(b*e*2*(p + 1) + 
(b*e - a*f)*(2*q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && L 
tQ[p, -1] && GtQ[q, 0]
 

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.02

method	result	size
pseudoelliptic	\(\frac {-\frac {5 a^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}} \left (a d f -\frac {2 b \left (c f +2 d e \right )}{5}\right ) f \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )}{2}+\frac {2 \left (-\frac {3 a^{3} \left (\left (-\frac {10 x^{2} d}{3}+c \right ) f +2 d e \right ) f \,b^{\frac {3}{2}}}{2}-2 \left (\left (-\frac {3 x^{2} d}{8}+c \right ) f +2 d e \right ) x^{2} f \,a^{2} b^{\frac {5}{2}}+\frac {3 a \left (\frac {2 c f \,x^{2}}{3}+e \left (\frac {x^{2} d}{3}+c \right )\right ) e \,b^{\frac {7}{2}}}{2}+b^{\frac {9}{2}} e^{2} c \,x^{2}+\frac {15 a^{4} d \,f^{2} \sqrt {b}}{4}\right ) x}{3}}{b^{\frac {7}{2}} \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2}}\)	\(169\)
default	\(c \,e^{2} \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )+f \left (c f +2 d e \right ) \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )+e \left (2 c f +d e \right ) \left (-\frac {x}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {a \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )}{2 b}\right )+f^{2} d \left (\frac {x^{5}}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {5 a \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )}{2 b}\right )\)	\(256\)
risch	\(\frac {d \,f^{2} x \sqrt {b \,x^{2}+a}}{2 b^{3}}-\frac {\frac {f \left (5 a d f -2 b c f -4 b d e \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}-\frac {\left (a^{3} d \,f^{2}-a^{2} c \,f^{2} b -2 a^{2} b d e f +2 a c e f \,b^{2}+a \,b^{2} d \,e^{2}-b^{3} c \,e^{2}\right ) \left (-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{3 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{3 a \left (x -\frac {\sqrt {-a b}}{b}\right )}\right )}{2 b a}-\frac {\left (a^{3} d \,f^{2}-a^{2} c \,f^{2} b -2 a^{2} b d e f +2 a c e f \,b^{2}+a \,b^{2} d \,e^{2}-b^{3} c \,e^{2}\right ) \left (\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{3 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{3 a \left (x +\frac {\sqrt {-a b}}{b}\right )}\right )}{2 b a}-\frac {\left (5 a^{3} d \,f^{2}-3 a^{2} c \,f^{2} b -6 a^{2} b d e f +2 a c e f \,b^{2}+a \,b^{2} d \,e^{2}+b^{3} c \,e^{2}\right ) \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{2 a^{2} b \left (x -\frac {\sqrt {-a b}}{b}\right )}-\frac {\left (5 a^{3} d \,f^{2}-3 a^{2} c \,f^{2} b -6 a^{2} b d e f +2 a c e f \,b^{2}+a \,b^{2} d \,e^{2}+b^{3} c \,e^{2}\right ) \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{2 a^{2} b \left (x +\frac {\sqrt {-a b}}{b}\right )}}{2 b^{3}}\)	\(671\)

Input:

int((d*x^2+c)*(f*x^2+e)^2/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/3/(b*x^2+a)^(3/2)/b^(7/2)*(-15/4*a^2*(b*x^2+a)^(3/2)*(a*d*f-2/5*b*(c*f+2 
*d*e))*f*arctanh((b*x^2+a)^(1/2)/x/b^(1/2))+(-3/2*a^3*((-10/3*x^2*d+c)*f+2 
*d*e)*f*b^(3/2)-2*((-3/8*x^2*d+c)*f+2*d*e)*x^2*f*a^2*b^(5/2)+3/2*a*(2/3*c* 
f*x^2+e*(1/3*x^2*d+c))*e*b^(7/2)+b^(9/2)*e^2*c*x^2+15/4*a^4*d*f^2*b^(1/2)) 
*x)/a^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (145) = 290\).

Time = 0.18 (sec) , antiderivative size = 616, normalized size of antiderivative = 3.73 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\left (a+b x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left (4 \, a^{4} b d e f + {\left (4 \, a^{2} b^{3} d e f + {\left (2 \, a^{2} b^{3} c - 5 \, a^{3} b^{2} d\right )} f^{2}\right )} x^{4} + {\left (2 \, a^{4} b c - 5 \, a^{5} d\right )} f^{2} + 2 \, {\left (4 \, a^{3} b^{2} d e f + {\left (2 \, a^{3} b^{2} c - 5 \, a^{4} b d\right )} f^{2}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (3 \, a^{2} b^{3} d f^{2} x^{5} + 2 \, {\left ({\left (2 \, b^{5} c + a b^{4} d\right )} e^{2} + 2 \, {\left (a b^{4} c - 4 \, a^{2} b^{3} d\right )} e f - 2 \, {\left (2 \, a^{2} b^{3} c - 5 \, a^{3} b^{2} d\right )} f^{2}\right )} x^{3} + 3 \, {\left (2 \, a b^{4} c e^{2} - 4 \, a^{3} b^{2} d e f - {\left (2 \, a^{3} b^{2} c - 5 \, a^{4} b d\right )} f^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{12 \, {\left (a^{2} b^{6} x^{4} + 2 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}}, -\frac {3 \, {\left (4 \, a^{4} b d e f + {\left (4 \, a^{2} b^{3} d e f + {\left (2 \, a^{2} b^{3} c - 5 \, a^{3} b^{2} d\right )} f^{2}\right )} x^{4} + {\left (2 \, a^{4} b c - 5 \, a^{5} d\right )} f^{2} + 2 \, {\left (4 \, a^{3} b^{2} d e f + {\left (2 \, a^{3} b^{2} c - 5 \, a^{4} b d\right )} f^{2}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (3 \, a^{2} b^{3} d f^{2} x^{5} + 2 \, {\left ({\left (2 \, b^{5} c + a b^{4} d\right )} e^{2} + 2 \, {\left (a b^{4} c - 4 \, a^{2} b^{3} d\right )} e f - 2 \, {\left (2 \, a^{2} b^{3} c - 5 \, a^{3} b^{2} d\right )} f^{2}\right )} x^{3} + 3 \, {\left (2 \, a b^{4} c e^{2} - 4 \, a^{3} b^{2} d e f - {\left (2 \, a^{3} b^{2} c - 5 \, a^{4} b d\right )} f^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{2} b^{6} x^{4} + 2 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}}\right ] \] Input:

integrate((d*x^2+c)*(f*x^2+e)^2/(b*x^2+a)^(5/2),x, algorithm="fricas")
 

Output:

[-1/12*(3*(4*a^4*b*d*e*f + (4*a^2*b^3*d*e*f + (2*a^2*b^3*c - 5*a^3*b^2*d)* 
f^2)*x^4 + (2*a^4*b*c - 5*a^5*d)*f^2 + 2*(4*a^3*b^2*d*e*f + (2*a^3*b^2*c - 
 5*a^4*b*d)*f^2)*x^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - 
 a) - 2*(3*a^2*b^3*d*f^2*x^5 + 2*((2*b^5*c + a*b^4*d)*e^2 + 2*(a*b^4*c - 4 
*a^2*b^3*d)*e*f - 2*(2*a^2*b^3*c - 5*a^3*b^2*d)*f^2)*x^3 + 3*(2*a*b^4*c*e^ 
2 - 4*a^3*b^2*d*e*f - (2*a^3*b^2*c - 5*a^4*b*d)*f^2)*x)*sqrt(b*x^2 + a))/( 
a^2*b^6*x^4 + 2*a^3*b^5*x^2 + a^4*b^4), -1/6*(3*(4*a^4*b*d*e*f + (4*a^2*b^ 
3*d*e*f + (2*a^2*b^3*c - 5*a^3*b^2*d)*f^2)*x^4 + (2*a^4*b*c - 5*a^5*d)*f^2 
 + 2*(4*a^3*b^2*d*e*f + (2*a^3*b^2*c - 5*a^4*b*d)*f^2)*x^2)*sqrt(-b)*arcta 
n(sqrt(-b)*x/sqrt(b*x^2 + a)) - (3*a^2*b^3*d*f^2*x^5 + 2*((2*b^5*c + a*b^4 
*d)*e^2 + 2*(a*b^4*c - 4*a^2*b^3*d)*e*f - 2*(2*a^2*b^3*c - 5*a^3*b^2*d)*f^ 
2)*x^3 + 3*(2*a*b^4*c*e^2 - 4*a^3*b^2*d*e*f - (2*a^3*b^2*c - 5*a^4*b*d)*f^ 
2)*x)*sqrt(b*x^2 + a))/(a^2*b^6*x^4 + 2*a^3*b^5*x^2 + a^4*b^4)]
 

Sympy [F]

\[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\left (a+b x^2\right )^{5/2}} \, dx=\int \frac {\left (c + d x^{2}\right ) \left (e + f x^{2}\right )^{2}}{\left (a + b x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((d*x**2+c)*(f*x**2+e)**2/(b*x**2+a)**(5/2),x)
 

Output:

Integral((c + d*x**2)*(e + f*x**2)**2/(a + b*x**2)**(5/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (145) = 290\).

Time = 0.04 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.79 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {d f^{2} x^{5}}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {5 \, a d f^{2} x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )}}{6 \, b} - \frac {1}{3} \, {\left (2 \, d e f + c f^{2}\right )} x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} + \frac {2 \, c e^{2} x}{3 \, \sqrt {b x^{2} + a} a^{2}} + \frac {c e^{2} x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} + \frac {5 \, a d f^{2} x}{6 \, \sqrt {b x^{2} + a} b^{3}} - \frac {5 \, a d f^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {7}{2}}} - \frac {{\left (2 \, d e f + c f^{2}\right )} x}{3 \, \sqrt {b x^{2} + a} b^{2}} - \frac {{\left (d e^{2} + 2 \, c e f\right )} x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {{\left (d e^{2} + 2 \, c e f\right )} x}{3 \, \sqrt {b x^{2} + a} a b} + \frac {{\left (2 \, d e f + c f^{2}\right )} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} \] Input:

integrate((d*x^2+c)*(f*x^2+e)^2/(b*x^2+a)^(5/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/2*d*f^2*x^5/((b*x^2 + a)^(3/2)*b) + 5/6*a*d*f^2*x*(3*x^2/((b*x^2 + a)^(3 
/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2))/b - 1/3*(2*d*e*f + c*f^2)*x*(3*x^2/( 
(b*x^2 + a)^(3/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2)) + 2/3*c*e^2*x/(sqrt(b* 
x^2 + a)*a^2) + 1/3*c*e^2*x/((b*x^2 + a)^(3/2)*a) + 5/6*a*d*f^2*x/(sqrt(b* 
x^2 + a)*b^3) - 5/2*a*d*f^2*arcsinh(b*x/sqrt(a*b))/b^(7/2) - 1/3*(2*d*e*f 
+ c*f^2)*x/(sqrt(b*x^2 + a)*b^2) - 1/3*(d*e^2 + 2*c*e*f)*x/((b*x^2 + a)^(3 
/2)*b) + 1/3*(d*e^2 + 2*c*e*f)*x/(sqrt(b*x^2 + a)*a*b) + (2*d*e*f + c*f^2) 
*arcsinh(b*x/sqrt(a*b))/b^(5/2)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.21 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {{\left ({\left (\frac {3 \, d f^{2} x^{2}}{b} + \frac {2 \, {\left (2 \, b^{6} c e^{2} + a b^{5} d e^{2} + 2 \, a b^{5} c e f - 8 \, a^{2} b^{4} d e f - 4 \, a^{2} b^{4} c f^{2} + 10 \, a^{3} b^{3} d f^{2}\right )}}{a^{2} b^{5}}\right )} x^{2} + \frac {3 \, {\left (2 \, a b^{5} c e^{2} - 4 \, a^{3} b^{3} d e f - 2 \, a^{3} b^{3} c f^{2} + 5 \, a^{4} b^{2} d f^{2}\right )}}{a^{2} b^{5}}\right )} x}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} - \frac {{\left (4 \, b d e f + 2 \, b c f^{2} - 5 \, a d f^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {7}{2}}} \] Input:

integrate((d*x^2+c)*(f*x^2+e)^2/(b*x^2+a)^(5/2),x, algorithm="giac")
 

Output:

1/6*((3*d*f^2*x^2/b + 2*(2*b^6*c*e^2 + a*b^5*d*e^2 + 2*a*b^5*c*e*f - 8*a^2 
*b^4*d*e*f - 4*a^2*b^4*c*f^2 + 10*a^3*b^3*d*f^2)/(a^2*b^5))*x^2 + 3*(2*a*b 
^5*c*e^2 - 4*a^3*b^3*d*e*f - 2*a^3*b^3*c*f^2 + 5*a^4*b^2*d*f^2)/(a^2*b^5)) 
*x/(b*x^2 + a)^(3/2) - 1/2*(4*b*d*e*f + 2*b*c*f^2 - 5*a*d*f^2)*log(abs(-sq 
rt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\left (a+b x^2\right )^{5/2}} \, dx=\int \frac {\left (d\,x^2+c\right )\,{\left (f\,x^2+e\right )}^2}{{\left (b\,x^2+a\right )}^{5/2}} \,d x \] Input:

int(((c + d*x^2)*(e + f*x^2)^2)/(a + b*x^2)^(5/2),x)
 

Output:

int(((c + d*x^2)*(e + f*x^2)^2)/(a + b*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 738, normalized size of antiderivative = 4.47 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

int((d*x^2+c)*(f*x^2+e)^2/(b*x^2+a)^(5/2),x)
 

Output:

(30*sqrt(a + b*x**2)*a**4*b*d*f**2*x - 12*sqrt(a + b*x**2)*a**3*b**2*c*f** 
2*x - 24*sqrt(a + b*x**2)*a**3*b**2*d*e*f*x + 40*sqrt(a + b*x**2)*a**3*b** 
2*d*f**2*x**3 - 16*sqrt(a + b*x**2)*a**2*b**3*c*f**2*x**3 - 32*sqrt(a + b* 
x**2)*a**2*b**3*d*e*f*x**3 + 6*sqrt(a + b*x**2)*a**2*b**3*d*f**2*x**5 + 12 
*sqrt(a + b*x**2)*a*b**4*c*e**2*x + 8*sqrt(a + b*x**2)*a*b**4*c*e*f*x**3 + 
 4*sqrt(a + b*x**2)*a*b**4*d*e**2*x**3 + 8*sqrt(a + b*x**2)*b**5*c*e**2*x* 
*3 - 30*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**5*d*f**2 + 
12*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**4*b*c*f**2 + 24* 
sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**4*b*d*e*f - 60*sqrt 
(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**4*b*d*f**2*x**2 + 24*sq 
rt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*b**2*c*f**2*x**2 + 
48*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*b**2*d*e*f*x** 
2 - 30*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*b**2*d*f** 
2*x**4 + 12*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b**3* 
c*f**2*x**4 + 24*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2* 
b**3*d*e*f*x**4 - 5*sqrt(b)*a**5*d*f**2 - 10*sqrt(b)*a**4*b*d*f**2*x**2 + 
8*sqrt(b)*a**3*b**2*c*e*f + 4*sqrt(b)*a**3*b**2*d*e**2 - 5*sqrt(b)*a**3*b* 
*2*d*f**2*x**4 - 8*sqrt(b)*a**2*b**3*c*e**2 + 16*sqrt(b)*a**2*b**3*c*e*f*x 
**2 + 8*sqrt(b)*a**2*b**3*d*e**2*x**2 - 16*sqrt(b)*a*b**4*c*e**2*x**2 + 8* 
sqrt(b)*a*b**4*c*e*f*x**4 + 4*sqrt(b)*a*b**4*d*e**2*x**4 - 8*sqrt(b)*b*...