\(\int \frac {A+B x^2+C x^4+D x^6}{(a+b x^2)^{7/2}} \, dx\) [138]

Optimal result

Integrand size = 29, antiderivative size = 167 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {\left (\frac {A}{a}-\frac {b^2 B-a b C+a^2 D}{b^3}\right ) x}{5 \left (a+b x^2\right )^{5/2}}+\frac {\left (4 A b^3+a \left (b^2 B-6 a b C+11 a^2 D\right )\right ) x}{15 a^2 b^3 \left (a+b x^2\right )^{3/2}}+\frac {\left (8 A b^3+a \left (2 b^2 B+3 a b C-23 a^2 D\right )\right ) x}{15 a^3 b^3 \sqrt {a+b x^2}}+\frac {D \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{7/2}} \] Output:

1/5*(A/a-(B*b^2-C*a*b+D*a^2)/b^3)*x/(b*x^2+a)^(5/2)+1/15*(4*A*b^3+a*(B*b^2 
-6*C*a*b+11*D*a^2))*x/a^2/b^3/(b*x^2+a)^(3/2)+1/15*(8*A*b^3+a*(2*B*b^2+3*C 
*a*b-23*D*a^2))*x/a^3/b^3/(b*x^2+a)^(1/2)+D*arctanh(b^(1/2)*x/(b*x^2+a)^(1 
/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {-15 a^5 D x-35 a^4 b D x^3+8 A b^5 x^5-23 a^3 b^2 D x^5+2 a b^4 x^3 \left (10 A+B x^2\right )+a^2 b^3 x \left (15 A+5 B x^2+3 C x^4\right )}{15 a^3 b^3 \left (a+b x^2\right )^{5/2}}-\frac {D \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{b^{7/2}} \] Input:

Integrate[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^(7/2),x]
 

Output:

(-15*a^5*D*x - 35*a^4*b*D*x^3 + 8*A*b^5*x^5 - 23*a^3*b^2*D*x^5 + 2*a*b^4*x 
^3*(10*A + B*x^2) + a^2*b^3*x*(15*A + 5*B*x^2 + 3*C*x^4))/(15*a^3*b^3*(a + 
 b*x^2)^(5/2)) - (D*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/b^(7/2)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2345, 25, 1471, 25, 27, 298, 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.

\(\displaystyle \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{7/2}} \, dx\)

\(\Big \downarrow \) 2345

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1471

\(\displaystyle \frac {\frac {x \left (\frac {11 a^2 D-6 a b C+b^2 B}{b^3}+\frac {4 A}{a}\right )}{3 \left (a+b x^2\right )^{3/2}}-\frac {\int -\frac {-8 D a^3+15 b D x^2 a^2+3 b C a^2+2 b^2 B a+8 A b^3}{b^3 \left (b x^2+a\right )^{3/2}}dx}{3 a}}{5 a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{5 a \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {8 A b^3+15 a^2 D x^2 b+a \left (-8 D a^2+3 b C a+2 b^2 B\right )}{b^3 \left (b x^2+a\right )^{3/2}}dx}{3 a}+\frac {x \left (\frac {11 a^2 D-6 a b C+b^2 B}{b^3}+\frac {4 A}{a}\right )}{3 \left (a+b x^2\right )^{3/2}}}{5 a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{5 a \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {8 A b^3+15 a^2 D x^2 b+a \left (-8 D a^2+3 b C a+2 b^2 B\right )}{\left (b x^2+a\right )^{3/2}}dx}{3 a b^3}+\frac {x \left (\frac {11 a^2 D-6 a b C+b^2 B}{b^3}+\frac {4 A}{a}\right )}{3 \left (a+b x^2\right )^{3/2}}}{5 a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{5 a \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 298

\(\displaystyle \frac {\frac {15 a^2 D \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {x \left (a \left (-23 a^2 D+3 a b C+2 b^2 B\right )+8 A b^3\right )}{a \sqrt {a+b x^2}}}{3 a b^3}+\frac {x \left (\frac {11 a^2 D-6 a b C+b^2 B}{b^3}+\frac {4 A}{a}\right )}{3 \left (a+b x^2\right )^{3/2}}}{5 a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{5 a \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {15 a^2 D \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {x \left (a \left (-23 a^2 D+3 a b C+2 b^2 B\right )+8 A b^3\right )}{a \sqrt {a+b x^2}}}{3 a b^3}+\frac {x \left (\frac {11 a^2 D-6 a b C+b^2 B}{b^3}+\frac {4 A}{a}\right )}{3 \left (a+b x^2\right )^{3/2}}}{5 a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{5 a \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {x \left (a \left (-23 a^2 D+3 a b C+2 b^2 B\right )+8 A b^3\right )}{a \sqrt {a+b x^2}}+\frac {15 a^2 D \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}}{3 a b^3}+\frac {x \left (\frac {11 a^2 D-6 a b C+b^2 B}{b^3}+\frac {4 A}{a}\right )}{3 \left (a+b x^2\right )^{3/2}}}{5 a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{5 a \left (a+b x^2\right )^{5/2}}\)

Input:

Int[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^(7/2),x]
 

Output:

((A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*x)/(5*a*(a + b*x^2)^(5/2)) + ((((4* 
A)/a + (b^2*B - 6*a*b*C + 11*a^2*D)/b^3)*x)/(3*(a + b*x^2)^(3/2)) + (((8*A 
*b^3 + a*(2*b^2*B + 3*a*b*C - 23*a^2*D))*x)/(a*Sqrt[a + b*x^2]) + (15*a^2* 
D*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b])/(3*a*b^3))/(5*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 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 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 = 0.75 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.80

method	result	size
pseudoelliptic	\(\frac {a^{3} D \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right ) b^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}+\frac {8 b^{\frac {7}{2}} \left (A \,b^{5} x^{4}+\frac {5 \left (\frac {x^{2} B}{10}+A \right ) x^{2} a \,b^{4}}{2}+\frac {15 \left (\frac {1}{5} C \,x^{4}+\frac {1}{3} x^{2} B +A \right ) a^{2} b^{3}}{8}-\frac {23 D a^{3} b^{2} x^{4}}{8}-\frac {35 D a^{4} b \,x^{2}}{8}-\frac {15 D a^{5}}{8}\right ) x}{15}}{\left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{\frac {13}{2}} a^{3}}\)	\(134\)
default	\(A \left (\frac {x}{5 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}}{a}\right )+C \left (-\frac {x^{3}}{2 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {3 a \left (-\frac {x}{4 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {a \left (\frac {x}{5 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}}{a}\right )}{4 b}\right )}{2 b}\right )+D \left (-\frac {x^{5}}{5 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}}{b}\right )+B \left (-\frac {x}{4 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {a \left (\frac {x}{5 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}}{a}\right )}{4 b}\right )\)	\(314\)

Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

1/(b*x^2+a)^(5/2)*(a^3*D*arctanh((b*x^2+a)^(1/2)/x/b^(1/2))*b^3*(b*x^2+a)^ 
(5/2)+8/15*b^(7/2)*(A*b^5*x^4+5/2*(1/10*x^2*B+A)*x^2*a*b^4+15/8*(1/5*C*x^4 
+1/3*x^2*B+A)*a^2*b^3-23/8*D*a^3*b^2*x^4-35/8*D*a^4*b*x^2-15/8*D*a^5)*x)/b 
^(13/2)/a^3
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 419, normalized size of antiderivative = 2.51 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{7/2}} \, dx=\left [\frac {15 \, {\left (D a^{3} b^{3} x^{6} + 3 \, D a^{4} b^{2} x^{4} + 3 \, D a^{5} b x^{2} + D a^{6}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left ({\left (23 \, D a^{3} b^{3} - 3 \, C a^{2} b^{4} - 2 \, B a b^{5} - 8 \, A b^{6}\right )} x^{5} + 5 \, {\left (7 \, D a^{4} b^{2} - B a^{2} b^{4} - 4 \, A a b^{5}\right )} x^{3} + 15 \, {\left (D a^{5} b - A a^{2} b^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{30 \, {\left (a^{3} b^{7} x^{6} + 3 \, a^{4} b^{6} x^{4} + 3 \, a^{5} b^{5} x^{2} + a^{6} b^{4}\right )}}, -\frac {15 \, {\left (D a^{3} b^{3} x^{6} + 3 \, D a^{4} b^{2} x^{4} + 3 \, D a^{5} b x^{2} + D a^{6}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left ({\left (23 \, D a^{3} b^{3} - 3 \, C a^{2} b^{4} - 2 \, B a b^{5} - 8 \, A b^{6}\right )} x^{5} + 5 \, {\left (7 \, D a^{4} b^{2} - B a^{2} b^{4} - 4 \, A a b^{5}\right )} x^{3} + 15 \, {\left (D a^{5} b - A a^{2} b^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{15 \, {\left (a^{3} b^{7} x^{6} + 3 \, a^{4} b^{6} x^{4} + 3 \, a^{5} b^{5} x^{2} + a^{6} b^{4}\right )}}\right ] \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x, algorithm="fricas")
 

Output:

[1/30*(15*(D*a^3*b^3*x^6 + 3*D*a^4*b^2*x^4 + 3*D*a^5*b*x^2 + D*a^6)*sqrt(b 
)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*((23*D*a^3*b^3 - 3*C 
*a^2*b^4 - 2*B*a*b^5 - 8*A*b^6)*x^5 + 5*(7*D*a^4*b^2 - B*a^2*b^4 - 4*A*a*b 
^5)*x^3 + 15*(D*a^5*b - A*a^2*b^4)*x)*sqrt(b*x^2 + a))/(a^3*b^7*x^6 + 3*a^ 
4*b^6*x^4 + 3*a^5*b^5*x^2 + a^6*b^4), -1/15*(15*(D*a^3*b^3*x^6 + 3*D*a^4*b 
^2*x^4 + 3*D*a^5*b*x^2 + D*a^6)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a) 
) + ((23*D*a^3*b^3 - 3*C*a^2*b^4 - 2*B*a*b^5 - 8*A*b^6)*x^5 + 5*(7*D*a^4*b 
^2 - B*a^2*b^4 - 4*A*a*b^5)*x^3 + 15*(D*a^5*b - A*a^2*b^4)*x)*sqrt(b*x^2 + 
 a))/(a^3*b^7*x^6 + 3*a^4*b^6*x^4 + 3*a^5*b^5*x^2 + a^6*b^4)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1690 vs. \(2 (162) = 324\).

Time = 21.65 (sec) , antiderivative size = 1690, normalized size of antiderivative = 10.12 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{7/2}} \, dx=\text {Too large to display} \] Input:

integrate((D*x**6+C*x**4+B*x**2+A)/(b*x**2+a)**(7/2),x)
 

Output:

A*(15*a**5*x/(15*a**(17/2)*sqrt(1 + b*x**2/a) + 45*a**(15/2)*b*x**2*sqrt(1 
 + b*x**2/a) + 45*a**(13/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 15*a**(11/2)*b* 
*3*x**6*sqrt(1 + b*x**2/a)) + 35*a**4*b*x**3/(15*a**(17/2)*sqrt(1 + b*x**2 
/a) + 45*a**(15/2)*b*x**2*sqrt(1 + b*x**2/a) + 45*a**(13/2)*b**2*x**4*sqrt 
(1 + b*x**2/a) + 15*a**(11/2)*b**3*x**6*sqrt(1 + b*x**2/a)) + 28*a**3*b**2 
*x**5/(15*a**(17/2)*sqrt(1 + b*x**2/a) + 45*a**(15/2)*b*x**2*sqrt(1 + b*x* 
*2/a) + 45*a**(13/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 15*a**(11/2)*b**3*x**6 
*sqrt(1 + b*x**2/a)) + 8*a**2*b**3*x**7/(15*a**(17/2)*sqrt(1 + b*x**2/a) + 
 45*a**(15/2)*b*x**2*sqrt(1 + b*x**2/a) + 45*a**(13/2)*b**2*x**4*sqrt(1 + 
b*x**2/a) + 15*a**(11/2)*b**3*x**6*sqrt(1 + b*x**2/a))) + B*(5*a*x**3/(15* 
a**(9/2)*sqrt(1 + b*x**2/a) + 30*a**(7/2)*b*x**2*sqrt(1 + b*x**2/a) + 15*a 
**(5/2)*b**2*x**4*sqrt(1 + b*x**2/a)) + 2*b*x**5/(15*a**(9/2)*sqrt(1 + b*x 
**2/a) + 30*a**(7/2)*b*x**2*sqrt(1 + b*x**2/a) + 15*a**(5/2)*b**2*x**4*sqr 
t(1 + b*x**2/a))) + C*x**5/(5*a**(7/2)*sqrt(1 + b*x**2/a) + 10*a**(5/2)*b* 
x**2*sqrt(1 + b*x**2/a) + 5*a**(3/2)*b**2*x**4*sqrt(1 + b*x**2/a)) + D*(15 
*a**(99/2)*b**25*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(15*a**(99/2) 
*b**(57/2)*sqrt(1 + b*x**2/a) + 45*a**(97/2)*b**(59/2)*x**2*sqrt(1 + b*x** 
2/a) + 45*a**(95/2)*b**(61/2)*x**4*sqrt(1 + b*x**2/a) + 15*a**(93/2)*b**(6 
3/2)*x**6*sqrt(1 + b*x**2/a)) + 45*a**(97/2)*b**26*x**2*sqrt(1 + b*x**2/a) 
*asinh(sqrt(b)*x/sqrt(a))/(15*a**(99/2)*b**(57/2)*sqrt(1 + b*x**2/a) + ...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 317 vs. \(2 (150) = 300\).

Time = 0.05 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.90 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{7/2}} \, dx=-\frac {1}{15} \, D x {\left (\frac {15 \, x^{4}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} b} + \frac {20 \, a x^{2}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3}}\right )} - \frac {D x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )}}{3 \, b} - \frac {C x^{3}}{2 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b} + \frac {8 \, A x}{15 \, \sqrt {b x^{2} + a} a^{3}} + \frac {4 \, A x}{15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} + \frac {A x}{5 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a} + \frac {7 \, D x}{15 \, \sqrt {b x^{2} + a} b^{3}} - \frac {4 \, D a x}{15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}} + \frac {C x}{10 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} + \frac {C x}{5 \, \sqrt {b x^{2} + a} a b^{2}} - \frac {3 \, C a x}{10 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}} - \frac {B x}{5 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b} + \frac {2 \, B x}{15 \, \sqrt {b x^{2} + a} a^{2} b} + \frac {B x}{15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b} + \frac {D \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {7}{2}}} \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x, algorithm="maxima")
 

Output:

-1/15*D*x*(15*x^4/((b*x^2 + a)^(5/2)*b) + 20*a*x^2/((b*x^2 + a)^(5/2)*b^2) 
 + 8*a^2/((b*x^2 + a)^(5/2)*b^3)) - 1/3*D*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 
 2*a/((b*x^2 + a)^(3/2)*b^2))/b - 1/2*C*x^3/((b*x^2 + a)^(5/2)*b) + 8/15*A 
*x/(sqrt(b*x^2 + a)*a^3) + 4/15*A*x/((b*x^2 + a)^(3/2)*a^2) + 1/5*A*x/((b* 
x^2 + a)^(5/2)*a) + 7/15*D*x/(sqrt(b*x^2 + a)*b^3) - 4/15*D*a*x/((b*x^2 + 
a)^(3/2)*b^3) + 1/10*C*x/((b*x^2 + a)^(3/2)*b^2) + 1/5*C*x/(sqrt(b*x^2 + a 
)*a*b^2) - 3/10*C*a*x/((b*x^2 + a)^(5/2)*b^2) - 1/5*B*x/((b*x^2 + a)^(5/2) 
*b) + 2/15*B*x/(sqrt(b*x^2 + a)*a^2*b) + 1/15*B*x/((b*x^2 + a)^(3/2)*a*b) 
+ D*arcsinh(b*x/sqrt(a*b))/b^(7/2)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.87 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{7/2}} \, dx=-\frac {{\left (x^{2} {\left (\frac {{\left (23 \, D a^{3} b^{4} - 3 \, C a^{2} b^{5} - 2 \, B a b^{6} - 8 \, A b^{7}\right )} x^{2}}{a^{3} b^{5}} + \frac {5 \, {\left (7 \, D a^{4} b^{3} - B a^{2} b^{5} - 4 \, A a b^{6}\right )}}{a^{3} b^{5}}\right )} + \frac {15 \, {\left (D a^{5} b^{2} - A a^{2} b^{5}\right )}}{a^{3} b^{5}}\right )} x}{15 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}}} - \frac {D \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {7}{2}}} \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x, algorithm="giac")
 

Output:

-1/15*(x^2*((23*D*a^3*b^4 - 3*C*a^2*b^5 - 2*B*a*b^6 - 8*A*b^7)*x^2/(a^3*b^ 
5) + 5*(7*D*a^4*b^3 - B*a^2*b^5 - 4*A*a*b^6)/(a^3*b^5)) + 15*(D*a^5*b^2 - 
A*a^2*b^5)/(a^3*b^5))*x/(b*x^2 + a)^(5/2) - D*log(abs(-sqrt(b)*x + sqrt(b* 
x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{7/2}} \, dx=\int \frac {A+B\,x^2+C\,x^4+x^6\,D}{{\left (b\,x^2+a\right )}^{7/2}} \,d x \] Input:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(7/2),x)
 

Output:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(7/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.57 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {-15 \sqrt {b \,x^{2}+a}\, a^{4} b d x -35 \sqrt {b \,x^{2}+a}\, a^{3} b^{2} d \,x^{3}+15 \sqrt {b \,x^{2}+a}\, a^{2} b^{4} x -23 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} d \,x^{5}+25 \sqrt {b \,x^{2}+a}\, a \,b^{5} x^{3}+3 \sqrt {b \,x^{2}+a}\, a \,b^{4} c \,x^{5}+10 \sqrt {b \,x^{2}+a}\, b^{6} x^{5}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{5} d +45 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{4} b d \,x^{2}+45 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b^{2} d \,x^{4}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{3} d \,x^{6}+5 \sqrt {b}\, a^{5} d +3 \sqrt {b}\, a^{4} b c +15 \sqrt {b}\, a^{4} b d \,x^{2}-10 \sqrt {b}\, a^{3} b^{3}+9 \sqrt {b}\, a^{3} b^{2} c \,x^{2}+15 \sqrt {b}\, a^{3} b^{2} d \,x^{4}-30 \sqrt {b}\, a^{2} b^{4} x^{2}+9 \sqrt {b}\, a^{2} b^{3} c \,x^{4}+5 \sqrt {b}\, a^{2} b^{3} d \,x^{6}-30 \sqrt {b}\, a \,b^{5} x^{4}+3 \sqrt {b}\, a \,b^{4} c \,x^{6}-10 \sqrt {b}\, b^{6} x^{6}}{15 a^{2} b^{4} \left (b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}\right )} \] Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x)
 

Output:

( - 15*sqrt(a + b*x**2)*a**4*b*d*x - 35*sqrt(a + b*x**2)*a**3*b**2*d*x**3 
+ 15*sqrt(a + b*x**2)*a**2*b**4*x - 23*sqrt(a + b*x**2)*a**2*b**3*d*x**5 + 
 25*sqrt(a + b*x**2)*a*b**5*x**3 + 3*sqrt(a + b*x**2)*a*b**4*c*x**5 + 10*s 
qrt(a + b*x**2)*b**6*x**5 + 15*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/ 
sqrt(a))*a**5*d + 45*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a 
**4*b*d*x**2 + 45*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3 
*b**2*d*x**4 + 15*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2 
*b**3*d*x**6 + 5*sqrt(b)*a**5*d + 3*sqrt(b)*a**4*b*c + 15*sqrt(b)*a**4*b*d 
*x**2 - 10*sqrt(b)*a**3*b**3 + 9*sqrt(b)*a**3*b**2*c*x**2 + 15*sqrt(b)*a** 
3*b**2*d*x**4 - 30*sqrt(b)*a**2*b**4*x**2 + 9*sqrt(b)*a**2*b**3*c*x**4 + 5 
*sqrt(b)*a**2*b**3*d*x**6 - 30*sqrt(b)*a*b**5*x**4 + 3*sqrt(b)*a*b**4*c*x* 
*6 - 10*sqrt(b)*b**6*x**6)/(15*a**2*b**4*(a**3 + 3*a**2*b*x**2 + 3*a*b**2* 
x**4 + b**3*x**6))