\(\int \frac {A+B x^2+C x^4+D x^6}{\sqrt {a+b x^2}} \, dx\) [135]

Optimal result

Integrand size = 29, antiderivative size = 146 \[ \int \frac {A+B x^2+C x^4+D x^6}{\sqrt {a+b x^2}} \, dx=\frac {\left (8 b^2 B-6 a b C+5 a^2 D\right ) x \sqrt {a+b x^2}}{16 b^3}+\frac {(6 b C-5 a D) x^3 \sqrt {a+b x^2}}{24 b^2}+\frac {D x^5 \sqrt {a+b x^2}}{6 b}+\frac {\left (16 A b^3-a \left (8 b^2 B-6 a b C+5 a^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{7/2}} \] Output:

1/16*(8*B*b^2-6*C*a*b+5*D*a^2)*x*(b*x^2+a)^(1/2)/b^3+1/24*(6*C*b-5*D*a)*x^ 
3*(b*x^2+a)^(1/2)/b^2+1/6*D*x^5*(b*x^2+a)^(1/2)/b+1/16*(16*A*b^3-a*(8*B*b^ 
2-6*C*a*b+5*D*a^2))*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x^2+C x^4+D x^6}{\sqrt {a+b x^2}} \, dx=\frac {x \sqrt {a+b x^2} \left (24 b^2 B-18 a b C+15 a^2 D+12 b^2 C x^2-10 a b D x^2+8 b^2 D x^4\right )}{48 b^3}+\frac {\left (16 A b^3-8 a b^2 B+6 a^2 b C-5 a^3 D\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{8 b^{7/2}} \] Input:

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

Output:

(x*Sqrt[a + b*x^2]*(24*b^2*B - 18*a*b*C + 15*a^2*D + 12*b^2*C*x^2 - 10*a*b 
*D*x^2 + 8*b^2*D*x^4))/(48*b^3) + ((16*A*b^3 - 8*a*b^2*B + 6*a^2*b*C - 5*a 
^3*D)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/(8*b^(7/2))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2346, 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.

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

\(\Big \downarrow \) 2346

\(\displaystyle \frac {\int \frac {(6 b C-5 a D) x^4+6 b B x^2+6 A b}{\sqrt {b x^2+a}}dx}{6 b}+\frac {D x^5 \sqrt {a+b x^2}}{6 b}\)

\(\Big \downarrow \) 1473

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {3 \left (\frac {\left (16 A b^3-a \left (5 a^2 D-6 a b C+8 b^2 B\right )\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}+\frac {x \sqrt {a+b x^2} \left (5 a^2 D-6 a b C+8 b^2 B\right )}{2 b}\right )}{4 b}+\frac {x^3 \sqrt {a+b x^2} (6 b C-5 a D)}{4 b}}{6 b}+\frac {D x^5 \sqrt {a+b x^2}}{6 b}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {3 \left (\frac {\left (16 A b^3-a \left (5 a^2 D-6 a b C+8 b^2 B\right )\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}+\frac {x \sqrt {a+b x^2} \left (5 a^2 D-6 a b C+8 b^2 B\right )}{2 b}\right )}{4 b}+\frac {x^3 \sqrt {a+b x^2} (6 b C-5 a D)}{4 b}}{6 b}+\frac {D x^5 \sqrt {a+b x^2}}{6 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {3 \left (\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (16 A b^3-a \left (5 a^2 D-6 a b C+8 b^2 B\right )\right )}{2 b^{3/2}}+\frac {x \sqrt {a+b x^2} \left (5 a^2 D-6 a b C+8 b^2 B\right )}{2 b}\right )}{4 b}+\frac {x^3 \sqrt {a+b x^2} (6 b C-5 a D)}{4 b}}{6 b}+\frac {D x^5 \sqrt {a+b x^2}}{6 b}\)

Input:

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

Output:

(D*x^5*Sqrt[a + b*x^2])/(6*b) + (((6*b*C - 5*a*D)*x^3*Sqrt[a + b*x^2])/(4* 
b) + (3*(((8*b^2*B - 6*a*b*C + 5*a^2*D)*x*Sqrt[a + b*x^2])/(2*b) + ((16*A* 
b^3 - a*(8*b^2*B - 6*a*b*C + 5*a^2*D))*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2] 
])/(2*b^(3/2))))/(4*b))/(6*b)
 

Defintions of rubi rules used

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 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.73

method	result	size
pseudoelliptic	\(\frac {\left (b^{3} A -\frac {1}{2} a \,b^{2} B +\frac {3}{8} a^{2} b C -\frac {5}{16} a^{3} D\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+\frac {5 \sqrt {b \,x^{2}+a}\, \left (\frac {4 \left (\frac {2}{3} D x^{4}+C \,x^{2}+2 B \right ) b^{\frac {5}{2}}}{5}+\left (2 \left (-\frac {D x^{2}}{3}-\frac {3 C}{5}\right ) b^{\frac {3}{2}}+D a \sqrt {b}\right ) a \right ) x}{16}}{b^{\frac {7}{2}}}\)	\(107\)
default	\(\frac {A \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+C \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+D \left (\frac {x^{5} \sqrt {b \,x^{2}+a}}{6 b}-\frac {5 a \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )+B \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )\)	\(215\)

Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/b^(7/2)*((b^3*A-1/2*a*b^2*B+3/8*a^2*b*C-5/16*a^3*D)*arctanh((b*x^2+a)^(1 
/2)/x/b^(1/2))+5/16*(b*x^2+a)^(1/2)*(4/5*(2/3*D*x^4+C*x^2+2*B)*b^(5/2)+(2* 
(-1/3*D*x^2-3/5*C)*b^(3/2)+D*a*b^(1/2))*a)*x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.71 \[ \int \frac {A+B x^2+C x^4+D x^6}{\sqrt {a+b x^2}} \, dx=\left [-\frac {3 \, {\left (5 \, D a^{3} - 6 \, C a^{2} b + 8 \, B a b^{2} - 16 \, A b^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (8 \, D b^{3} x^{5} - 2 \, {\left (5 \, D a b^{2} - 6 \, C b^{3}\right )} x^{3} + 3 \, {\left (5 \, D a^{2} b - 6 \, C a b^{2} + 8 \, B b^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{4}}, \frac {3 \, {\left (5 \, D a^{3} - 6 \, C a^{2} b + 8 \, B a b^{2} - 16 \, A b^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, D b^{3} x^{5} - 2 \, {\left (5 \, D a b^{2} - 6 \, C b^{3}\right )} x^{3} + 3 \, {\left (5 \, D a^{2} b - 6 \, C a b^{2} + 8 \, B b^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{4}}\right ] \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

[-1/96*(3*(5*D*a^3 - 6*C*a^2*b + 8*B*a*b^2 - 16*A*b^3)*sqrt(b)*log(-2*b*x^ 
2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(8*D*b^3*x^5 - 2*(5*D*a*b^2 - 6*C 
*b^3)*x^3 + 3*(5*D*a^2*b - 6*C*a*b^2 + 8*B*b^3)*x)*sqrt(b*x^2 + a))/b^4, 1 
/48*(3*(5*D*a^3 - 6*C*a^2*b + 8*B*a*b^2 - 16*A*b^3)*sqrt(-b)*arctan(sqrt(- 
b)*x/sqrt(b*x^2 + a)) + (8*D*b^3*x^5 - 2*(5*D*a*b^2 - 6*C*b^3)*x^3 + 3*(5* 
D*a^2*b - 6*C*a*b^2 + 8*B*b^3)*x)*sqrt(b*x^2 + a))/b^4]
 

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.06 \[ \int \frac {A+B x^2+C x^4+D x^6}{\sqrt {a+b x^2}} \, dx=\begin {cases} \left (A - \frac {a \left (B - \frac {3 a \left (C - \frac {5 D a}{6 b}\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) + \sqrt {a + b x^{2}} \left (\frac {D x^{5}}{6 b} + \frac {x^{3} \left (C - \frac {5 D a}{6 b}\right )}{4 b} + \frac {x \left (B - \frac {3 a \left (C - \frac {5 D a}{6 b}\right )}{4 b}\right )}{2 b}\right ) & \text {for}\: b \neq 0 \\\frac {A x + \frac {B x^{3}}{3} + \frac {C x^{5}}{5} + \frac {D x^{7}}{7}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:

integrate((D*x**6+C*x**4+B*x**2+A)/(b*x**2+a)**(1/2),x)
 

Output:

Piecewise(((A - a*(B - 3*a*(C - 5*D*a/(6*b))/(4*b))/(2*b))*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), True)) + sqrt(a + b*x**2)*(D*x**5/(6*b) + x**3*(C - 5*D*a/(6*b))/(4* 
b) + x*(B - 3*a*(C - 5*D*a/(6*b))/(4*b))/(2*b)), Ne(b, 0)), ((A*x + B*x**3 
/3 + C*x**5/5 + D*x**7/7)/sqrt(a), True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.19 \[ \int \frac {A+B x^2+C x^4+D x^6}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} D x^{5}}{6 \, b} - \frac {5 \, \sqrt {b x^{2} + a} D a x^{3}}{24 \, b^{2}} + \frac {\sqrt {b x^{2} + a} C x^{3}}{4 \, b} + \frac {5 \, \sqrt {b x^{2} + a} D a^{2} x}{16 \, b^{3}} - \frac {3 \, \sqrt {b x^{2} + a} C a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} B x}{2 \, b} - \frac {5 \, D a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {7}{2}}} + \frac {3 \, C a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {B a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} + \frac {A \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="maxima")
 

Output:

1/6*sqrt(b*x^2 + a)*D*x^5/b - 5/24*sqrt(b*x^2 + a)*D*a*x^3/b^2 + 1/4*sqrt( 
b*x^2 + a)*C*x^3/b + 5/16*sqrt(b*x^2 + a)*D*a^2*x/b^3 - 3/8*sqrt(b*x^2 + a 
)*C*a*x/b^2 + 1/2*sqrt(b*x^2 + a)*B*x/b - 5/16*D*a^3*arcsinh(b*x/sqrt(a*b) 
)/b^(7/2) + 3/8*C*a^2*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 1/2*B*a*arcsinh(b*x 
/sqrt(a*b))/b^(3/2) + A*arcsinh(b*x/sqrt(a*b))/sqrt(b)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x^2+C x^4+D x^6}{\sqrt {a+b x^2}} \, dx=\frac {1}{48} \, {\left (2 \, {\left (\frac {4 \, D x^{2}}{b} - \frac {5 \, D a b^{3} - 6 \, C b^{4}}{b^{5}}\right )} x^{2} + \frac {3 \, {\left (5 \, D a^{2} b^{2} - 6 \, C a b^{3} + 8 \, B b^{4}\right )}}{b^{5}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (5 \, D a^{3} - 6 \, C a^{2} b + 8 \, B a b^{2} - 16 \, A b^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {7}{2}}} \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="giac")
 

Output:

1/48*(2*(4*D*x^2/b - (5*D*a*b^3 - 6*C*b^4)/b^5)*x^2 + 3*(5*D*a^2*b^2 - 6*C 
*a*b^3 + 8*B*b^4)/b^5)*sqrt(b*x^2 + a)*x + 1/16*(5*D*a^3 - 6*C*a^2*b + 8*B 
*a*b^2 - 16*A*b^3)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

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

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(1/2),x)
 

Output:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.27 \[ \int \frac {A+B x^2+C x^4+D x^6}{\sqrt {a+b x^2}} \, dx=\frac {15 \sqrt {b \,x^{2}+a}\, a^{2} b d x -18 \sqrt {b \,x^{2}+a}\, a \,b^{2} c x -10 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{3}+24 \sqrt {b \,x^{2}+a}\, b^{4} x +12 \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{3}+8 \sqrt {b \,x^{2}+a}\, b^{3} d \,x^{5}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} d +18 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b c +24 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{3}}{48 b^{4}} \] Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x)
 

Output:

(15*sqrt(a + b*x**2)*a**2*b*d*x - 18*sqrt(a + b*x**2)*a*b**2*c*x - 10*sqrt 
(a + b*x**2)*a*b**2*d*x**3 + 24*sqrt(a + b*x**2)*b**4*x + 12*sqrt(a + b*x* 
*2)*b**3*c*x**3 + 8*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 + 18*sqrt(b)*log((sqrt(a + b*x**2) 
+ sqrt(b)*x)/sqrt(a))*a**2*b*c + 24*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b 
)*x)/sqrt(a))*a*b**3)/(48*b**4)