\(\int \frac {(c+d x^2) (e+f x^2)^2}{\sqrt {a+b x^2}} \, dx\) [317]

Optimal result

Integrand size = 28, antiderivative size = 198 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\frac {\left (5 a^2 d f^2-6 a b f (2 d e+c f)+8 b^2 e (d e+2 c f)\right ) x \sqrt {a+b x^2}}{16 b^3}-\frac {f (5 a d f-6 b (2 d e+c f)) x^3 \sqrt {a+b x^2}}{24 b^2}+\frac {d f^2 x^5 \sqrt {a+b x^2}}{6 b}+\frac {\left (16 b^3 c e^2-5 a^3 d f^2+6 a^2 b f (2 d e+c f)-8 a b^2 e (d e+2 c f)\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{7/2}} \] Output:

1/16*(5*a^2*d*f^2-6*a*b*f*(c*f+2*d*e)+8*b^2*e*(2*c*f+d*e))*x*(b*x^2+a)^(1/ 
2)/b^3-1/24*f*(5*a*d*f-6*b*(c*f+2*d*e))*x^3*(b*x^2+a)^(1/2)/b^2+1/6*d*f^2* 
x^5*(b*x^2+a)^(1/2)/b+1/16*(16*b^3*c*e^2-5*a^3*d*f^2+6*a^2*b*f*(c*f+2*d*e) 
-8*a*b^2*e*(2*c*f+d*e))*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.87 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b} x \sqrt {a+b x^2} \left (15 a^2 d f^2-2 a b f \left (18 d e+9 c f+5 d f x^2\right )+4 b^2 \left (3 c f \left (4 e+f x^2\right )+2 d \left (3 e^2+3 e f x^2+f^2 x^4\right )\right )\right )+3 \left (-16 b^3 c e^2+5 a^3 d f^2-6 a^2 b f (2 d e+c f)+8 a b^2 e (d e+2 c f)\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{48 b^{7/2}} \] Input:

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

Output:

(Sqrt[b]*x*Sqrt[a + b*x^2]*(15*a^2*d*f^2 - 2*a*b*f*(18*d*e + 9*c*f + 5*d*f 
*x^2) + 4*b^2*(3*c*f*(4*e + f*x^2) + 2*d*(3*e^2 + 3*e*f*x^2 + f^2*x^4))) + 
 3*(-16*b^3*c*e^2 + 5*a^3*d*f^2 - 6*a^2*b*f*(2*d*e + c*f) + 8*a*b^2*e*(d*e 
 + 2*c*f))*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(48*b^(7/2))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {403, 403, 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.

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

\(\Big \downarrow \) 403

\(\displaystyle \frac {\int \frac {\left (f x^2+e\right ) \left ((4 b d e+6 b c f-5 a d f) x^2+(6 b c-a d) e\right )}{\sqrt {b x^2+a}}dx}{6 b}+\frac {d x \sqrt {a+b x^2} \left (e+f x^2\right )^2}{6 b}\)

\(\Big \downarrow \) 403

\(\displaystyle \frac {\frac {\int \frac {\left (4 e (2 d e+9 c f) b^2-2 a f (13 d e+9 c f) b+15 a^2 d f^2\right ) x^2+e \left (5 d f a^2-8 b d e a-6 b c f a+24 b^2 c e\right )}{\sqrt {b x^2+a}}dx}{4 b}+\frac {x \sqrt {a+b x^2} \left (e+f x^2\right ) (-5 a d f+6 b c f+4 b d e)}{4 b}}{6 b}+\frac {d x \sqrt {a+b x^2} \left (e+f x^2\right )^2}{6 b}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\frac {3 \left (-5 a^3 d f^2+6 a^2 b f (c f+2 d e)-8 a b^2 e (2 c f+d e)+16 b^3 c e^2\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}+\frac {x \sqrt {a+b x^2} \left (15 a^2 d f^2-2 a b f (9 c f+13 d e)+4 b^2 e (9 c f+2 d e)\right )}{2 b}}{4 b}+\frac {x \sqrt {a+b x^2} \left (e+f x^2\right ) (-5 a d f+6 b c f+4 b d e)}{4 b}}{6 b}+\frac {d x \sqrt {a+b x^2} \left (e+f x^2\right )^2}{6 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {\frac {3 \left (-5 a^3 d f^2+6 a^2 b f (c f+2 d e)-8 a b^2 e (2 c f+d e)+16 b^3 c e^2\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}+\frac {x \sqrt {a+b x^2} \left (15 a^2 d f^2-2 a b f (9 c f+13 d e)+4 b^2 e (9 c f+2 d e)\right )}{2 b}}{4 b}+\frac {x \sqrt {a+b x^2} \left (e+f x^2\right ) (-5 a d f+6 b c f+4 b d e)}{4 b}}{6 b}+\frac {d x \sqrt {a+b x^2} \left (e+f x^2\right )^2}{6 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {x \sqrt {a+b x^2} \left (15 a^2 d f^2-2 a b f (9 c f+13 d e)+4 b^2 e (9 c f+2 d e)\right )}{2 b}+\frac {3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (-5 a^3 d f^2+6 a^2 b f (c f+2 d e)-8 a b^2 e (2 c f+d e)+16 b^3 c e^2\right )}{2 b^{3/2}}}{4 b}+\frac {x \sqrt {a+b x^2} \left (e+f x^2\right ) (-5 a d f+6 b c f+4 b d e)}{4 b}}{6 b}+\frac {d x \sqrt {a+b x^2} \left (e+f x^2\right )^2}{6 b}\)

Input:

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

Output:

(d*x*Sqrt[a + b*x^2]*(e + f*x^2)^2)/(6*b) + (((4*b*d*e + 6*b*c*f - 5*a*d*f 
)*x*Sqrt[a + b*x^2]*(e + f*x^2))/(4*b) + (((15*a^2*d*f^2 + 4*b^2*e*(2*d*e 
+ 9*c*f) - 2*a*b*f*(13*d*e + 9*c*f))*x*Sqrt[a + b*x^2])/(2*b) + (3*(16*b^3 
*c*e^2 - 5*a^3*d*f^2 + 6*a^2*b*f*(2*d*e + c*f) - 8*a*b^2*e*(d*e + 2*c*f))* 
ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2)))/(4*b))/(6*b)
 

Defintions of rubi rules used

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 403

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

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(-\frac {5 \left (\left (a^{2} \left (a d -\frac {6 b c}{5}\right ) f^{2}-\frac {12 a \left (a d -\frac {4 b c}{3}\right ) b e f}{5}+\frac {8 b^{2} e^{2} \left (a d -2 b c \right )}{5}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )-\sqrt {b \,x^{2}+a}\, \left (\frac {4 \left (\left (\frac {2 x^{2} d}{3}+c \right ) x^{2} f^{2}+4 e \left (\frac {x^{2} d}{2}+c \right ) f +2 d \,e^{2}\right ) b^{\frac {5}{2}}}{5}+a f \left (2 \left (\left (-\frac {x^{2} d}{3}-\frac {3 c}{5}\right ) f -\frac {6 d e}{5}\right ) b^{\frac {3}{2}}+a d f \sqrt {b}\right )\right ) x \right )}{16 b^{\frac {7}{2}}}\)	\(156\)
risch	\(\frac {x \left (8 f^{2} d \,b^{2} x^{4}-10 a b d \,f^{2} x^{2}+12 b^{2} c \,f^{2} x^{2}+24 b^{2} d e f \,x^{2}+15 a^{2} d \,f^{2}-18 a b c \,f^{2}-36 a b d e f +48 b^{2} c e f +24 b^{2} d \,e^{2}\right ) \sqrt {b \,x^{2}+a}}{48 b^{3}}-\frac {\left (5 a^{3} d \,f^{2}-6 a^{2} c \,f^{2} b -12 a^{2} b d e f +16 a c e f \,b^{2}+8 a \,b^{2} d \,e^{2}-16 b^{3} c \,e^{2}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {7}{2}}}\)	\(183\)
default	\(\frac {c \,e^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+f \left (c f +2 d e \right ) \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+e \left (2 c f +d e \right ) \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )+f^{2} d \left (\frac {x^{5} \sqrt {b \,x^{2}+a}}{6 b}-\frac {5 a \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )\)	\(237\)

Input:

int((d*x^2+c)*(f*x^2+e)^2/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-5/16*((a^2*(a*d-6/5*b*c)*f^2-12/5*a*(a*d-4/3*b*c)*b*e*f+8/5*b^2*e^2*(a*d- 
2*b*c))*arctanh((b*x^2+a)^(1/2)/x/b^(1/2))-(b*x^2+a)^(1/2)*(4/5*((2/3*x^2* 
d+c)*x^2*f^2+4*e*(1/2*x^2*d+c)*f+2*d*e^2)*b^(5/2)+a*f*(2*((-1/3*x^2*d-3/5* 
c)*f-6/5*d*e)*b^(3/2)+a*d*f*b^(1/2)))*x)/b^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 396, normalized size of antiderivative = 2.00 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\left [-\frac {3 \, {\left (8 \, {\left (2 \, b^{3} c - a b^{2} d\right )} e^{2} - 4 \, {\left (4 \, a b^{2} c - 3 \, a^{2} b d\right )} e f + {\left (6 \, a^{2} b c - 5 \, a^{3} d\right )} f^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (8 \, b^{3} d f^{2} x^{5} + 2 \, {\left (12 \, b^{3} d e f + {\left (6 \, b^{3} c - 5 \, a b^{2} d\right )} f^{2}\right )} x^{3} + 3 \, {\left (8 \, b^{3} d e^{2} + 4 \, {\left (4 \, b^{3} c - 3 \, a b^{2} d\right )} e f - {\left (6 \, a b^{2} c - 5 \, a^{2} b d\right )} f^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{4}}, -\frac {3 \, {\left (8 \, {\left (2 \, b^{3} c - a b^{2} d\right )} e^{2} - 4 \, {\left (4 \, a b^{2} c - 3 \, a^{2} b d\right )} e f + {\left (6 \, a^{2} b c - 5 \, a^{3} d\right )} f^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, b^{3} d f^{2} x^{5} + 2 \, {\left (12 \, b^{3} d e f + {\left (6 \, b^{3} c - 5 \, a b^{2} d\right )} f^{2}\right )} x^{3} + 3 \, {\left (8 \, b^{3} d e^{2} + 4 \, {\left (4 \, b^{3} c - 3 \, a b^{2} d\right )} e f - {\left (6 \, a b^{2} c - 5 \, a^{2} b d\right )} f^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{4}}\right ] \] Input:

integrate((d*x^2+c)*(f*x^2+e)^2/(b*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

[-1/96*(3*(8*(2*b^3*c - a*b^2*d)*e^2 - 4*(4*a*b^2*c - 3*a^2*b*d)*e*f + (6* 
a^2*b*c - 5*a^3*d)*f^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x 
 - a) - 2*(8*b^3*d*f^2*x^5 + 2*(12*b^3*d*e*f + (6*b^3*c - 5*a*b^2*d)*f^2)* 
x^3 + 3*(8*b^3*d*e^2 + 4*(4*b^3*c - 3*a*b^2*d)*e*f - (6*a*b^2*c - 5*a^2*b* 
d)*f^2)*x)*sqrt(b*x^2 + a))/b^4, -1/48*(3*(8*(2*b^3*c - a*b^2*d)*e^2 - 4*( 
4*a*b^2*c - 3*a^2*b*d)*e*f + (6*a^2*b*c - 5*a^3*d)*f^2)*sqrt(-b)*arctan(sq 
rt(-b)*x/sqrt(b*x^2 + a)) - (8*b^3*d*f^2*x^5 + 2*(12*b^3*d*e*f + (6*b^3*c 
- 5*a*b^2*d)*f^2)*x^3 + 3*(8*b^3*d*e^2 + 4*(4*b^3*c - 3*a*b^2*d)*e*f - (6* 
a*b^2*c - 5*a^2*b*d)*f^2)*x)*sqrt(b*x^2 + a))/b^4]
 

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.26 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {d f^{2} x^{5}}{6 b} + \frac {x^{3} \left (- \frac {5 a d f^{2}}{6 b} + c f^{2} + 2 d e f\right )}{4 b} + \frac {x \left (- \frac {3 a \left (- \frac {5 a d f^{2}}{6 b} + c f^{2} + 2 d e f\right )}{4 b} + 2 c e f + d e^{2}\right )}{2 b}\right ) + \left (- \frac {a \left (- \frac {3 a \left (- \frac {5 a d f^{2}}{6 b} + c f^{2} + 2 d e f\right )}{4 b} + 2 c e f + d e^{2}\right )}{2 b} + c e^{2}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\frac {c e^{2} x + \frac {d f^{2} x^{7}}{7} + \frac {x^{5} \left (c f^{2} + 2 d e f\right )}{5} + \frac {x^{3} \cdot \left (2 c e f + d e^{2}\right )}{3}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:

integrate((d*x**2+c)*(f*x**2+e)**2/(b*x**2+a)**(1/2),x)
 

Output:

Piecewise((sqrt(a + b*x**2)*(d*f**2*x**5/(6*b) + x**3*(-5*a*d*f**2/(6*b) + 
 c*f**2 + 2*d*e*f)/(4*b) + x*(-3*a*(-5*a*d*f**2/(6*b) + c*f**2 + 2*d*e*f)/ 
(4*b) + 2*c*e*f + d*e**2)/(2*b)) + (-a*(-3*a*(-5*a*d*f**2/(6*b) + c*f**2 + 
 2*d*e*f)/(4*b) + 2*c*e*f + d*e**2)/(2*b) + c*e**2)*Piecewise((log(2*sqrt( 
b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), T 
rue)), Ne(b, 0)), ((c*e**2*x + d*f**2*x**7/7 + x**5*(c*f**2 + 2*d*e*f)/5 + 
 x**3*(2*c*e*f + d*e**2)/3)/sqrt(a), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.21 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} d f^{2} x^{5}}{6 \, b} - \frac {5 \, \sqrt {b x^{2} + a} a d f^{2} x^{3}}{24 \, b^{2}} + \frac {5 \, \sqrt {b x^{2} + a} a^{2} d f^{2} x}{16 \, b^{3}} + \frac {{\left (2 \, d e f + c f^{2}\right )} \sqrt {b x^{2} + a} x^{3}}{4 \, b} + \frac {c e^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {5 \, a^{3} d f^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {7}{2}}} - \frac {3 \, {\left (2 \, d e f + c f^{2}\right )} \sqrt {b x^{2} + a} a x}{8 \, b^{2}} + \frac {{\left (d e^{2} + 2 \, c e f\right )} \sqrt {b x^{2} + a} x}{2 \, b} + \frac {3 \, {\left (2 \, d e f + c f^{2}\right )} a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {{\left (d e^{2} + 2 \, c e f\right )} a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} \] Input:

integrate((d*x^2+c)*(f*x^2+e)^2/(b*x^2+a)^(1/2),x, algorithm="maxima")
 

Output:

1/6*sqrt(b*x^2 + a)*d*f^2*x^5/b - 5/24*sqrt(b*x^2 + a)*a*d*f^2*x^3/b^2 + 5 
/16*sqrt(b*x^2 + a)*a^2*d*f^2*x/b^3 + 1/4*(2*d*e*f + c*f^2)*sqrt(b*x^2 + a 
)*x^3/b + c*e^2*arcsinh(b*x/sqrt(a*b))/sqrt(b) - 5/16*a^3*d*f^2*arcsinh(b* 
x/sqrt(a*b))/b^(7/2) - 3/8*(2*d*e*f + c*f^2)*sqrt(b*x^2 + a)*a*x/b^2 + 1/2 
*(d*e^2 + 2*c*e*f)*sqrt(b*x^2 + a)*x/b + 3/8*(2*d*e*f + c*f^2)*a^2*arcsinh 
(b*x/sqrt(a*b))/b^(5/2) - 1/2*(d*e^2 + 2*c*e*f)*a*arcsinh(b*x/sqrt(a*b))/b 
^(3/2)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\frac {1}{48} \, {\left (2 \, {\left (\frac {4 \, d f^{2} x^{2}}{b} + \frac {12 \, b^{4} d e f + 6 \, b^{4} c f^{2} - 5 \, a b^{3} d f^{2}}{b^{5}}\right )} x^{2} + \frac {3 \, {\left (8 \, b^{4} d e^{2} + 16 \, b^{4} c e f - 12 \, a b^{3} d e f - 6 \, a b^{3} c f^{2} + 5 \, a^{2} b^{2} d f^{2}\right )}}{b^{5}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (16 \, b^{3} c e^{2} - 8 \, a b^{2} d e^{2} - 16 \, a b^{2} c e f + 12 \, a^{2} b d e f + 6 \, a^{2} b c f^{2} - 5 \, a^{3} d f^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {7}{2}}} \] Input:

integrate((d*x^2+c)*(f*x^2+e)^2/(b*x^2+a)^(1/2),x, algorithm="giac")
 

Output:

1/48*(2*(4*d*f^2*x^2/b + (12*b^4*d*e*f + 6*b^4*c*f^2 - 5*a*b^3*d*f^2)/b^5) 
*x^2 + 3*(8*b^4*d*e^2 + 16*b^4*c*e*f - 12*a*b^3*d*e*f - 6*a*b^3*c*f^2 + 5* 
a^2*b^2*d*f^2)/b^5)*sqrt(b*x^2 + a)*x - 1/16*(16*b^3*c*e^2 - 8*a*b^2*d*e^2 
 - 16*a*b^2*c*e*f + 12*a^2*b*d*e*f + 6*a^2*b*c*f^2 - 5*a^3*d*f^2)*log(abs( 
-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

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

int(((c + d*x^2)*(e + f*x^2)^2)/(a + b*x^2)^(1/2),x)
 

Output:

int(((c + d*x^2)*(e + f*x^2)^2)/(a + b*x^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.81 \[ \int \frac {\left (c+d x^2\right ) \left (e+f x^2\right )^2}{\sqrt {a+b x^2}} \, dx=\frac {15 \sqrt {b \,x^{2}+a}\, a^{2} b d \,f^{2} x -18 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,f^{2} x -36 \sqrt {b \,x^{2}+a}\, a \,b^{2} d e f x -10 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,f^{2} x^{3}+48 \sqrt {b \,x^{2}+a}\, b^{3} c e f x +12 \sqrt {b \,x^{2}+a}\, b^{3} c \,f^{2} x^{3}+24 \sqrt {b \,x^{2}+a}\, b^{3} d \,e^{2} x +24 \sqrt {b \,x^{2}+a}\, b^{3} d e f \,x^{3}+8 \sqrt {b \,x^{2}+a}\, b^{3} d \,f^{2} x^{5}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} d \,f^{2}+18 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b c \,f^{2}+36 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b d e f -48 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} c e f -24 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} d \,e^{2}+48 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} c \,e^{2}}{48 b^{4}} \] Input:

int((d*x^2+c)*(f*x^2+e)^2/(b*x^2+a)^(1/2),x)
 

Output:

(15*sqrt(a + b*x**2)*a**2*b*d*f**2*x - 18*sqrt(a + b*x**2)*a*b**2*c*f**2*x 
 - 36*sqrt(a + b*x**2)*a*b**2*d*e*f*x - 10*sqrt(a + b*x**2)*a*b**2*d*f**2* 
x**3 + 48*sqrt(a + b*x**2)*b**3*c*e*f*x + 12*sqrt(a + b*x**2)*b**3*c*f**2* 
x**3 + 24*sqrt(a + b*x**2)*b**3*d*e**2*x + 24*sqrt(a + b*x**2)*b**3*d*e*f* 
x**3 + 8*sqrt(a + b*x**2)*b**3*d*f**2*x**5 - 15*sqrt(b)*log((sqrt(a + b*x* 
*2) + sqrt(b)*x)/sqrt(a))*a**3*d*f**2 + 18*sqrt(b)*log((sqrt(a + b*x**2) + 
 sqrt(b)*x)/sqrt(a))*a**2*b*c*f**2 + 36*sqrt(b)*log((sqrt(a + b*x**2) + sq 
rt(b)*x)/sqrt(a))*a**2*b*d*e*f - 48*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b 
)*x)/sqrt(a))*a*b**2*c*e*f - 24*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x) 
/sqrt(a))*a*b**2*d*e**2 + 48*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sq 
rt(a))*b**3*c*e**2)/(48*b**4)