\(\int \frac {A+B x^2+C x^4+D x^6}{(a+b x^2)^{3/2}} \, dx\) [234]

Optimal result

Integrand size = 29, antiderivative size = 137 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {\left (\frac {A}{a}-\frac {b^2 B-a b C+a^2 D}{b^3}\right ) x}{\sqrt {a+b x^2}}+\frac {(4 b C-7 a D) x \sqrt {a+b x^2}}{8 b^3}+\frac {D x^3 \sqrt {a+b x^2}}{4 b^2}+\frac {\left (8 b^2 B-12 a b C+15 a^2 D\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{7/2}} \] Output:

(A/a-(B*b^2-C*a*b+D*a^2)/b^3)*x/(b*x^2+a)^(1/2)+1/8*(4*C*b-7*D*a)*x*(b*x^2 
+a)^(1/2)/b^3+1/4*D*x^3*(b*x^2+a)^(1/2)/b^2+1/8*(8*B*b^2-12*C*a*b+15*D*a^2 
)*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {b} x \left (8 A b^3+a \left (-15 a^2 D+a b \left (12 C-5 D x^2\right )+b^2 \left (-8 B+4 C x^2+2 D x^4\right )\right )\right )}{a \sqrt {a+b x^2}}+\left (-8 b^2 B+12 a b C-15 a^2 D\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{7/2}} \] Input:

Integrate[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^(3/2),x]
 

Output:

((Sqrt[b]*x*(8*A*b^3 + a*(-15*a^2*D + a*b*(12*C - 5*D*x^2) + b^2*(-8*B + 4 
*C*x^2 + 2*D*x^4))))/(a*Sqrt[a + b*x^2]) + (-8*b^2*B + 12*a*b*C - 15*a^2*D 
)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(8*b^(7/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2345, 25, 1473, 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.

\(\displaystyle \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 2345

\(\displaystyle \frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{a \sqrt {a+b x^2}}-\frac {\int -\frac {\frac {a D x^4}{b}+\frac {a (b C-a D) x^2}{b^2}+\frac {a \left (D a^2-b C a+b^2 B\right )}{b^3}}{\sqrt {b x^2+a}}dx}{a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\frac {a D x^4}{b}+\frac {a (b C-a D) x^2}{b^2}+\frac {a \left (D a^2-b C a+b^2 B\right )}{b^3}}{\sqrt {b x^2+a}}dx}{a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{a \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 1473

\(\displaystyle \frac {\frac {\int \frac {a \left (b^2 \left (4 C-\frac {7 a D}{b}\right ) x^2+4 \left (D a^2-b C a+b^2 B\right )\right )}{b^2 \sqrt {b x^2+a}}dx}{4 b}+\frac {a D x^3 \sqrt {a+b x^2}}{4 b^2}}{a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{a \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a \int \frac {b (4 b C-7 a D) x^2+4 \left (D a^2-b C a+b^2 B\right )}{\sqrt {b x^2+a}}dx}{4 b^3}+\frac {a D x^3 \sqrt {a+b x^2}}{4 b^2}}{a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{a \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {a \left (\frac {1}{2} \left (15 a^2 D-12 a b C+8 b^2 B\right ) \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2} (4 b C-7 a D)\right )}{4 b^3}+\frac {a D x^3 \sqrt {a+b x^2}}{4 b^2}}{a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{a \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {a \left (\frac {1}{2} \left (15 a^2 D-12 a b C+8 b^2 B\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2} (4 b C-7 a D)\right )}{4 b^3}+\frac {a D x^3 \sqrt {a+b x^2}}{4 b^2}}{a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{a \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{a \sqrt {a+b x^2}}+\frac {\frac {a \left (\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (15 a^2 D-12 a b C+8 b^2 B\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2} (4 b C-7 a D)\right )}{4 b^3}+\frac {a D x^3 \sqrt {a+b x^2}}{4 b^2}}{a}\)

Input:

Int[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^(3/2),x]
 

Output:

((A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*x)/(a*Sqrt[a + b*x^2]) + ((a*D*x^3* 
Sqrt[a + b*x^2])/(4*b^2) + (a*(((4*b*C - 7*a*D)*x*Sqrt[a + b*x^2])/2 + ((8 
*b^2*B - 12*a*b*C + 15*a^2*D)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqr 
t[b])))/(4*b^3))/a
 

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 1473

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> Simp[c^p*x^(4*p - 1)*((d + e*x^2)^(q + 1)/(e*(4*p + 2*q + 1))) 
, x] + Simp[1/(e*(4*p + 2*q + 1))   Int[(d + e*x^2)^q*ExpandToSum[e*(4*p + 
2*q + 1)*(a + b*x^2 + c*x^4)^p - d*c^p*(4*p - 1)*x^(4*p - 2) - e*c^p*(4*p + 
 2*q + 1)*x^(4*p), x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] &&  !LtQ[q, -1]
 

rule 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.83

method	result	size
pseudoelliptic	\(\frac {\sqrt {b \,x^{2}+a}\, a \left (B \,b^{2}-\frac {3}{2} C a b +\frac {15}{8} D a^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+\left (-\frac {15 a^{3} D}{8}+\frac {3 \left (-\frac {5 D x^{2}}{12}+C \right ) b \,a^{2}}{2}-\left (-\frac {1}{4} D x^{4}-\frac {1}{2} C \,x^{2}+B \right ) b^{2} a +b^{3} A \right ) \sqrt {b}\, x}{\sqrt {b \,x^{2}+a}\, a \,b^{\frac {7}{2}}}\)	\(114\)
default	\(\frac {A x}{a \sqrt {b \,x^{2}+a}}+C \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )+D \left (\frac {x^{5}}{4 b \sqrt {b \,x^{2}+a}}-\frac {5 a \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )}{4 b}\right )+B \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )\)	\(203\)

Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/(b*x^2+a)^(1/2)*((b*x^2+a)^(1/2)*a*(B*b^2-3/2*C*a*b+15/8*D*a^2)*arctanh( 
(b*x^2+a)^(1/2)/x/b^(1/2))+(-15/8*a^3*D+3/2*(-5/12*D*x^2+C)*b*a^2-(-1/4*D* 
x^4-1/2*C*x^2+B)*b^2*a+b^3*A)*b^(1/2)*x)/a/b^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.57 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/2}} \, dx=\left [\frac {{\left (15 \, D a^{4} - 12 \, C a^{3} b + 8 \, B a^{2} b^{2} + {\left (15 \, D a^{3} b - 12 \, C a^{2} b^{2} + 8 \, B a b^{3}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, D a b^{3} x^{5} - {\left (5 \, D a^{2} b^{2} - 4 \, C a b^{3}\right )} x^{3} - {\left (15 \, D a^{3} b - 12 \, C a^{2} b^{2} + 8 \, B a b^{3} - 8 \, A b^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, {\left (a b^{5} x^{2} + a^{2} b^{4}\right )}}, -\frac {{\left (15 \, D a^{4} - 12 \, C a^{3} b + 8 \, B a^{2} b^{2} + {\left (15 \, D a^{3} b - 12 \, C a^{2} b^{2} + 8 \, B a b^{3}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, D a b^{3} x^{5} - {\left (5 \, D a^{2} b^{2} - 4 \, C a b^{3}\right )} x^{3} - {\left (15 \, D a^{3} b - 12 \, C a^{2} b^{2} + 8 \, B a b^{3} - 8 \, A b^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, {\left (a b^{5} x^{2} + a^{2} b^{4}\right )}}\right ] \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="fricas")
 

Output:

[1/16*((15*D*a^4 - 12*C*a^3*b + 8*B*a^2*b^2 + (15*D*a^3*b - 12*C*a^2*b^2 + 
 8*B*a*b^3)*x^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 
 2*(2*D*a*b^3*x^5 - (5*D*a^2*b^2 - 4*C*a*b^3)*x^3 - (15*D*a^3*b - 12*C*a^2 
*b^2 + 8*B*a*b^3 - 8*A*b^4)*x)*sqrt(b*x^2 + a))/(a*b^5*x^2 + a^2*b^4), -1/ 
8*((15*D*a^4 - 12*C*a^3*b + 8*B*a^2*b^2 + (15*D*a^3*b - 12*C*a^2*b^2 + 8*B 
*a*b^3)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (2*D*a*b^3*x^5 
- (5*D*a^2*b^2 - 4*C*a*b^3)*x^3 - (15*D*a^3*b - 12*C*a^2*b^2 + 8*B*a*b^3 - 
 8*A*b^4)*x)*sqrt(b*x^2 + a))/(a*b^5*x^2 + a^2*b^4)]
 

Sympy [A] (verification not implemented)

Time = 8.51 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.74 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {A x}{a^{\frac {3}{2}} \sqrt {1 + \frac {b x^{2}}{a}}} + B \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{b^{\frac {3}{2}}} - \frac {x}{\sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + C \left (\frac {3 \sqrt {a} x}{2 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 b^{\frac {5}{2}}} + \frac {x^{3}}{2 \sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + D \left (- \frac {15 a^{\frac {3}{2}} x}{8 b^{3} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {5 \sqrt {a} x^{3}}{8 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {15 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 b^{\frac {7}{2}}} + \frac {x^{5}}{4 \sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \] Input:

integrate((D*x**6+C*x**4+B*x**2+A)/(b*x**2+a)**(3/2),x)
 

Output:

A*x/(a**(3/2)*sqrt(1 + b*x**2/a)) + B*(asinh(sqrt(b)*x/sqrt(a))/b**(3/2) - 
 x/(sqrt(a)*b*sqrt(1 + b*x**2/a))) + C*(3*sqrt(a)*x/(2*b**2*sqrt(1 + b*x** 
2/a)) - 3*a*asinh(sqrt(b)*x/sqrt(a))/(2*b**(5/2)) + x**3/(2*sqrt(a)*b*sqrt 
(1 + b*x**2/a))) + D*(-15*a**(3/2)*x/(8*b**3*sqrt(1 + b*x**2/a)) - 5*sqrt( 
a)*x**3/(8*b**2*sqrt(1 + b*x**2/a)) + 15*a**2*asinh(sqrt(b)*x/sqrt(a))/(8* 
b**(7/2)) + x**5/(4*sqrt(a)*b*sqrt(1 + b*x**2/a)))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.25 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {D x^{5}}{4 \, \sqrt {b x^{2} + a} b} - \frac {5 \, D a x^{3}}{8 \, \sqrt {b x^{2} + a} b^{2}} + \frac {C x^{3}}{2 \, \sqrt {b x^{2} + a} b} + \frac {A x}{\sqrt {b x^{2} + a} a} - \frac {15 \, D a^{2} x}{8 \, \sqrt {b x^{2} + a} b^{3}} + \frac {3 \, C a x}{2 \, \sqrt {b x^{2} + a} b^{2}} - \frac {B x}{\sqrt {b x^{2} + a} b} + \frac {15 \, D a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {7}{2}}} - \frac {3 \, C a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} + \frac {B \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="maxima")
 

Output:

1/4*D*x^5/(sqrt(b*x^2 + a)*b) - 5/8*D*a*x^3/(sqrt(b*x^2 + a)*b^2) + 1/2*C* 
x^3/(sqrt(b*x^2 + a)*b) + A*x/(sqrt(b*x^2 + a)*a) - 15/8*D*a^2*x/(sqrt(b*x 
^2 + a)*b^3) + 3/2*C*a*x/(sqrt(b*x^2 + a)*b^2) - B*x/(sqrt(b*x^2 + a)*b) + 
 15/8*D*a^2*arcsinh(b*x/sqrt(a*b))/b^(7/2) - 3/2*C*a*arcsinh(b*x/sqrt(a*b) 
)/b^(5/2) + B*arcsinh(b*x/sqrt(a*b))/b^(3/2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (\frac {2 \, D x^{2}}{b} - \frac {5 \, D a^{2} b^{3} - 4 \, C a b^{4}}{a b^{5}}\right )} x^{2} - \frac {15 \, D a^{3} b^{2} - 12 \, C a^{2} b^{3} + 8 \, B a b^{4} - 8 \, A b^{5}}{a b^{5}}\right )} x}{8 \, \sqrt {b x^{2} + a}} - \frac {{\left (15 \, D a^{2} - 12 \, C a b + 8 \, B b^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {7}{2}}} \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="giac")
 

Output:

1/8*((2*D*x^2/b - (5*D*a^2*b^3 - 4*C*a*b^4)/(a*b^5))*x^2 - (15*D*a^3*b^2 - 
 12*C*a^2*b^3 + 8*B*a*b^4 - 8*A*b^5)/(a*b^5))*x/sqrt(b*x^2 + a) - 1/8*(15* 
D*a^2 - 12*C*a*b + 8*B*b^2)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {A+B\,x^2+C\,x^4+x^6\,D}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \] Input:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(3/2),x)
 

Output:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.28 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {-15 \sqrt {b \,x^{2}+a}\, a^{2} b d x +12 \sqrt {b \,x^{2}+a}\, a \,b^{2} c x -5 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{3}+4 \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{3}+2 \sqrt {b \,x^{2}+a}\, b^{3} d \,x^{5}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} d -12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b c +15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b d \,x^{2}+8 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{3}-12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} c \,x^{2}+8 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{4} x^{2}-10 \sqrt {b}\, a^{3} d +9 \sqrt {b}\, a^{2} b c -10 \sqrt {b}\, a^{2} b d \,x^{2}+9 \sqrt {b}\, a \,b^{2} c \,x^{2}}{8 b^{4} \left (b \,x^{2}+a \right )} \] Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(3/2),x)
 

Output:

( - 15*sqrt(a + b*x**2)*a**2*b*d*x + 12*sqrt(a + b*x**2)*a*b**2*c*x - 5*sq 
rt(a + b*x**2)*a*b**2*d*x**3 + 4*sqrt(a + b*x**2)*b**3*c*x**3 + 2*sqrt(a + 
 b*x**2)*b**3*d*x**5 + 15*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt( 
a))*a**3*d - 12*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b 
*c + 15*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b*d*x**2 
+ 8*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**3 - 12*sqrt(b 
)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**2*c*x**2 + 8*sqrt(b)*lo 
g((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*b**4*x**2 - 10*sqrt(b)*a**3*d + 
9*sqrt(b)*a**2*b*c - 10*sqrt(b)*a**2*b*d*x**2 + 9*sqrt(b)*a*b**2*c*x**2)/( 
8*b**4*(a + b*x**2))