\(\int \frac {A+B x^2+C x^4+D x^6}{(a+b x^2)^{5/2}} \, dx\) [137]

Optimal result

Integrand size = 29, antiderivative size = 149 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {\left (\frac {A}{a}-\frac {b^2 B-a b C+a^2 D}{b^3}\right ) x}{3 \left (a+b x^2\right )^{3/2}}+\frac {\left (2 A b^3+a \left (b^2 B-4 a b C+7 a^2 D\right )\right ) x}{3 a^2 b^3 \sqrt {a+b x^2}}+\frac {D x \sqrt {a+b x^2}}{2 b^3}+\frac {(2 b C-5 a D) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {x \left (15 a^4 D+4 A b^4 x^2+2 a b^3 \left (3 A+B x^2\right )+a^2 b^2 x^2 \left (-8 C+3 D x^2\right )+a^3 b \left (-6 C+20 D x^2\right )\right )}{6 a^2 b^3 \left (a+b x^2\right )^{3/2}}+\frac {(-2 b C+5 a D) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{2 b^{7/2}} \] Input:

Integrate[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^(5/2),x]
 

Output:

(x*(15*a^4*D + 4*A*b^4*x^2 + 2*a*b^3*(3*A + B*x^2) + a^2*b^2*x^2*(-8*C + 3 
*D*x^2) + a^3*b*(-6*C + 20*D*x^2)))/(6*a^2*b^3*(a + b*x^2)^(3/2)) + ((-2*b 
*C + 5*a*D)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(2*b^(7/2))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2345, 25, 1471, 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.

\(\displaystyle \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 2345

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1471

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {3 a \left (\frac {1}{2} (2 b C-5 a D) \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} D x \sqrt {a+b x^2}\right )}{b^3}+\frac {x \left (\frac {7 a^2 D-4 a b C+b^2 B}{b^3}+\frac {2 A}{a}\right )}{\sqrt {a+b x^2}}}{3 a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{3 a \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {3 a \left (\frac {1}{2} (2 b C-5 a D) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} D x \sqrt {a+b x^2}\right )}{b^3}+\frac {x \left (\frac {7 a^2 D-4 a b C+b^2 B}{b^3}+\frac {2 A}{a}\right )}{\sqrt {a+b x^2}}}{3 a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{3 a \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {x \left (\frac {7 a^2 D-4 a b C+b^2 B}{b^3}+\frac {2 A}{a}\right )}{\sqrt {a+b x^2}}+\frac {3 a \left (\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (2 b C-5 a D)}{2 \sqrt {b}}+\frac {1}{2} D x \sqrt {a+b x^2}\right )}{b^3}}{3 a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{3 a \left (a+b x^2\right )^{3/2}}\)

Input:

Int[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^(5/2),x]
 

Output:

((A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*x)/(3*a*(a + b*x^2)^(3/2)) + ((((2* 
A)/a + (b^2*B - 4*a*b*C + 7*a^2*D)/b^3)*x)/Sqrt[a + b*x^2] + (3*a*((D*x*Sq 
rt[a + b*x^2])/2 + ((2*b*C - 5*a*D)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/ 
(2*Sqrt[b])))/b^3)/(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 1471

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 
, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x 
, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q 
 + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), 
x], x], x]] /; FreeQ[{a, b, c, d, e}, 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 = 0.65 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.86

method	result	size
pseudoelliptic	\(\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2} \left (C b -\frac {5 D a}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+\frac {2 \left (\frac {15 D a^{4} \sqrt {b}}{4}+b^{\frac {3}{2}} \left (\left (5 D x^{2}-\frac {3 C}{2}\right ) a^{3}-2 \left (-\frac {3 D x^{2}}{8}+C \right ) x^{2} b \,a^{2}+\frac {3 a \left (\frac {x^{2} B}{3}+A \right ) b^{2}}{2}+A \,x^{2} b^{3}\right )\right ) x}{3}}{b^{\frac {7}{2}} \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2}}\)	\(128\)
default	\(A \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )+C \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )+D \left (\frac {x^{5}}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {5 a \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )}{2 b}\right )+B \left (-\frac {x}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {a \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )}{2 b}\right )\)	\(234\)

Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 415, normalized size of antiderivative = 2.79 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left (5 \, D a^{5} - 2 \, C a^{4} b + {\left (5 \, D a^{3} b^{2} - 2 \, C a^{2} b^{3}\right )} x^{4} + 2 \, {\left (5 \, D a^{4} b - 2 \, C a^{3} b^{2}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (3 \, D a^{2} b^{3} x^{5} + 2 \, {\left (10 \, D a^{3} b^{2} - 4 \, C a^{2} b^{3} + B a b^{4} + 2 \, A b^{5}\right )} x^{3} + 3 \, {\left (5 \, D a^{4} b - 2 \, C a^{3} b^{2} + 2 \, A a b^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{12 \, {\left (a^{2} b^{6} x^{4} + 2 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}}, \frac {3 \, {\left (5 \, D a^{5} - 2 \, C a^{4} b + {\left (5 \, D a^{3} b^{2} - 2 \, C a^{2} b^{3}\right )} x^{4} + 2 \, {\left (5 \, D a^{4} b - 2 \, C a^{3} b^{2}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, D a^{2} b^{3} x^{5} + 2 \, {\left (10 \, D a^{3} b^{2} - 4 \, C a^{2} b^{3} + B a b^{4} + 2 \, A b^{5}\right )} x^{3} + 3 \, {\left (5 \, D a^{4} b - 2 \, C a^{3} b^{2} + 2 \, A a b^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{2} b^{6} x^{4} + 2 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}}\right ] \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(5/2),x, algorithm="fricas")
 

Output:

[-1/12*(3*(5*D*a^5 - 2*C*a^4*b + (5*D*a^3*b^2 - 2*C*a^2*b^3)*x^4 + 2*(5*D* 
a^4*b - 2*C*a^3*b^2)*x^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b) 
*x - a) - 2*(3*D*a^2*b^3*x^5 + 2*(10*D*a^3*b^2 - 4*C*a^2*b^3 + B*a*b^4 + 2 
*A*b^5)*x^3 + 3*(5*D*a^4*b - 2*C*a^3*b^2 + 2*A*a*b^4)*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*(5*D*a^5 - 2*C*a^4*b + (5* 
D*a^3*b^2 - 2*C*a^2*b^3)*x^4 + 2*(5*D*a^4*b - 2*C*a^3*b^2)*x^2)*sqrt(-b)*a 
rctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (3*D*a^2*b^3*x^5 + 2*(10*D*a^3*b^2 - 4 
*C*a^2*b^3 + B*a*b^4 + 2*A*b^5)*x^3 + 3*(5*D*a^4*b - 2*C*a^3*b^2 + 2*A*a*b 
^4)*x)*sqrt(b*x^2 + a))/(a^2*b^6*x^4 + 2*a^3*b^5*x^2 + a^4*b^4)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 821 vs. \(2 (138) = 276\).

Time = 10.28 (sec) , antiderivative size = 821, normalized size of antiderivative = 5.51 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate((D*x**6+C*x**4+B*x**2+A)/(b*x**2+a)**(5/2),x)
 

Output:

A*(3*a*x/(3*a**(7/2)*sqrt(1 + b*x**2/a) + 3*a**(5/2)*b*x**2*sqrt(1 + b*x** 
2/a)) + 2*b*x**3/(3*a**(7/2)*sqrt(1 + b*x**2/a) + 3*a**(5/2)*b*x**2*sqrt(1 
 + b*x**2/a))) + B*x**3/(3*a**(5/2)*sqrt(1 + b*x**2/a) + 3*a**(3/2)*b*x**2 
*sqrt(1 + b*x**2/a)) + C*(3*a**(39/2)*b**11*sqrt(1 + b*x**2/a)*asinh(sqrt( 
b)*x/sqrt(a))/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**( 
29/2)*x**2*sqrt(1 + b*x**2/a)) + 3*a**(37/2)*b**12*x**2*sqrt(1 + b*x**2/a) 
*asinh(sqrt(b)*x/sqrt(a))/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a* 
*(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) - 3*a**19*b**(23/2)*x/(3*a**(39 
/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x 
**2/a)) - 4*a**18*b**(25/2)*x**3/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) 
 + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a))) + D*(-15*a**(81/2)*b**2 
2*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(6*a**(79/2)*b**(51/2)*sqrt( 
1 + b*x**2/a) + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)) - 15*a**(79 
/2)*b**23*x**2*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(6*a**(79/2)*b* 
*(51/2)*sqrt(1 + b*x**2/a) + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a) 
) + 15*a**40*b**(45/2)*x/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6*a** 
(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)) + 20*a**39*b**(47/2)*x**3/(6*a** 
(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + 
b*x**2/a)) + 3*a**38*b**(49/2)*x**5/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2 
/a) + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.52 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {D x^{5}}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {1}{3} \, C x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} + \frac {5 \, D a x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )}}{6 \, b} + \frac {2 \, A x}{3 \, \sqrt {b x^{2} + a} a^{2}} + \frac {A x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} + \frac {5 \, D a x}{6 \, \sqrt {b x^{2} + a} b^{3}} - \frac {C x}{3 \, \sqrt {b x^{2} + a} b^{2}} - \frac {B x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {B x}{3 \, \sqrt {b x^{2} + a} a b} - \frac {5 \, D a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {7}{2}}} + \frac {C \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(5/2),x, algorithm="maxima")
 

Output:

1/2*D*x^5/((b*x^2 + a)^(3/2)*b) - 1/3*C*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2 
*a/((b*x^2 + a)^(3/2)*b^2)) + 5/6*D*a*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a 
/((b*x^2 + a)^(3/2)*b^2))/b + 2/3*A*x/(sqrt(b*x^2 + a)*a^2) + 1/3*A*x/((b* 
x^2 + a)^(3/2)*a) + 5/6*D*a*x/(sqrt(b*x^2 + a)*b^3) - 1/3*C*x/(sqrt(b*x^2 
+ a)*b^2) - 1/3*B*x/((b*x^2 + a)^(3/2)*b) + 1/3*B*x/(sqrt(b*x^2 + a)*a*b) 
- 5/2*D*a*arcsinh(b*x/sqrt(a*b))/b^(7/2) + C*arcsinh(b*x/sqrt(a*b))/b^(5/2 
)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {{\left ({\left (\frac {3 \, D x^{2}}{b} + \frac {2 \, {\left (10 \, D a^{3} b^{3} - 4 \, C a^{2} b^{4} + B a b^{5} + 2 \, A b^{6}\right )}}{a^{2} b^{5}}\right )} x^{2} + \frac {3 \, {\left (5 \, D a^{4} b^{2} - 2 \, C a^{3} b^{3} + 2 \, A a b^{5}\right )}}{a^{2} b^{5}}\right )} x}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} + \frac {{\left (5 \, D a - 2 \, C b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {7}{2}}} \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(5/2),x, algorithm="giac")
 

Output:

1/6*((3*D*x^2/b + 2*(10*D*a^3*b^3 - 4*C*a^2*b^4 + B*a*b^5 + 2*A*b^6)/(a^2* 
b^5))*x^2 + 3*(5*D*a^4*b^2 - 2*C*a^3*b^3 + 2*A*a*b^5)/(a^2*b^5))*x/(b*x^2 
+ a)^(3/2) + 1/2*(5*D*a - 2*C*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^ 
(7/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/2}} \, dx=\int \frac {A+B\,x^2+C\,x^4+x^6\,D}{{\left (b\,x^2+a\right )}^{5/2}} \,d x \] Input:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(5/2),x)
 

Output:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 398, normalized size of antiderivative = 2.67 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {30 \sqrt {b \,x^{2}+a}\, a^{3} b d x -12 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} c x +40 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} d \,x^{3}+12 \sqrt {b \,x^{2}+a}\, a \,b^{4} x -16 \sqrt {b \,x^{2}+a}\, a \,b^{3} c \,x^{3}+6 \sqrt {b \,x^{2}+a}\, a \,b^{3} d \,x^{5}+12 \sqrt {b \,x^{2}+a}\, b^{5} x^{3}-30 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{4} d +12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b c -60 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b d \,x^{2}+24 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2} c \,x^{2}-30 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2} d \,x^{4}+12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{3} c \,x^{4}-5 \sqrt {b}\, a^{4} d -10 \sqrt {b}\, a^{3} b d \,x^{2}-4 \sqrt {b}\, a^{2} b^{3}-5 \sqrt {b}\, a^{2} b^{2} d \,x^{4}-8 \sqrt {b}\, a \,b^{4} x^{2}-4 \sqrt {b}\, b^{5} x^{4}}{12 a \,b^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(5/2),x)
 

Output:

(30*sqrt(a + b*x**2)*a**3*b*d*x - 12*sqrt(a + b*x**2)*a**2*b**2*c*x + 40*s 
qrt(a + b*x**2)*a**2*b**2*d*x**3 + 12*sqrt(a + b*x**2)*a*b**4*x - 16*sqrt( 
a + b*x**2)*a*b**3*c*x**3 + 6*sqrt(a + b*x**2)*a*b**3*d*x**5 + 12*sqrt(a + 
 b*x**2)*b**5*x**3 - 30*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a) 
)*a**4*d + 12*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*b*c 
 - 60*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*b*d*x**2 + 
24*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b**2*c*x**2 - 
30*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b**2*d*x**4 + 
12*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**3*c*x**4 - 5*s 
qrt(b)*a**4*d - 10*sqrt(b)*a**3*b*d*x**2 - 4*sqrt(b)*a**2*b**3 - 5*sqrt(b) 
*a**2*b**2*d*x**4 - 8*sqrt(b)*a*b**4*x**2 - 4*sqrt(b)*b**5*x**4)/(12*a*b** 
4*(a**2 + 2*a*b*x**2 + b**2*x**4))