\(\int \frac {A+B x^2}{x^2 (b x^2+c x^4)^{3/2}} \, dx\) [223]

Optimal result

Integrand size = 26, antiderivative size = 137 \[ \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {A}{4 b x^3 \sqrt {b x^2+c x^4}}+\frac {4 b B-5 A c}{4 b^2 x \sqrt {b x^2+c x^4}}-\frac {3 (4 b B-5 A c) \sqrt {b x^2+c x^4}}{8 b^3 x^3}+\frac {3 c (4 b B-5 A c) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{7/2}} \] Output:

-1/4*A/b/x^3/(c*x^4+b*x^2)^(1/2)+1/4*(-5*A*c+4*B*b)/b^2/x/(c*x^4+b*x^2)^(1 
/2)-3/8*(-5*A*c+4*B*b)*(c*x^4+b*x^2)^(1/2)/b^3/x^3+3/8*c*(-5*A*c+4*B*b)*ar 
ctanh(b^(1/2)*x/(c*x^4+b*x^2)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {\sqrt {b} \left (-4 b B x^2 \left (b+3 c x^2\right )+A \left (-2 b^2+5 b c x^2+15 c^2 x^4\right )\right )+3 c (4 b B-5 A c) x^4 \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )}{8 b^{7/2} x^3 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

Integrate[(A + B*x^2)/(x^2*(b*x^2 + c*x^4)^(3/2)),x]
 

Output:

(Sqrt[b]*(-4*b*B*x^2*(b + 3*c*x^2) + A*(-2*b^2 + 5*b*c*x^2 + 15*c^2*x^4)) 
+ 3*c*(4*b*B - 5*A*c)*x^4*Sqrt[b + c*x^2]*ArcTanh[Sqrt[b + c*x^2]/Sqrt[b]] 
)/(8*b^(7/2)*x^3*Sqrt[x^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1944, 1401, 1430, 1400, 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.

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

\(\Big \downarrow \) 1944

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

\(\Big \downarrow \) 1401

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

\(\Big \downarrow \) 1430

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

\(\Big \downarrow \) 1400

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

\(\Big \downarrow \) 219

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

Input:

Int[(A + B*x^2)/(x^2*(b*x^2 + c*x^4)^(3/2)),x]
 

Output:

-1/4*A/(b*x^3*Sqrt[b*x^2 + c*x^4]) + ((4*b*B - 5*A*c)*(1/(b*x*Sqrt[b*x^2 + 
 c*x^4]) + (3*(-1/2*Sqrt[b*x^2 + c*x^4]/(b*x^3) + (c*ArcTanh[(Sqrt[b]*x)/S 
qrt[b*x^2 + c*x^4]])/(2*b^(3/2))))/b))/(4*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 1400

Int[1/Sqrt[(b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> -Subst[Int[1/(1 - b*x 
^2), x], x, x/Sqrt[b*x^2 + c*x^4]] /; FreeQ[{b, c}, x]
 

rule 1401

Int[((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[-(b*x^2 + c*x^4)^ 
(p + 1)/(2*b*(p + 1)*x), x] + Simp[(4*p + 3)/(2*b*(p + 1))   Int[(b*x^2 + c 
*x^4)^(p + 1)/x^2, x], x] /; FreeQ[{b, c}, x] &&  !IntegerQ[p] && LtQ[p, -1 
]
 

rule 1430

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[d*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[c*( 
(m + 4*p + 3)/(b*d^2*(m + 2*p + 1)))   Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^p, 
 x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && LtQ[m + 2*p + 1, 0 
]
 

rule 1944

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j 
+ b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b 
*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1))   Int[(e*x)^(m + n)*(a*x^ 
j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j 
+ n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 
] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( 
GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 
, 0]
 

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.12

method	result	size
risch	\(-\frac {\left (c \,x^{2}+b \right ) \left (-7 A c \,x^{2}+4 B b \,x^{2}+2 A b \right )}{8 b^{3} x^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {c \left (-\frac {7 A c -4 B b}{\sqrt {c \,x^{2}+b}}+3 b \left (5 A c -4 B b \right ) \left (\frac {1}{b \sqrt {c \,x^{2}+b}}-\frac {\ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right )}{b^{\frac {3}{2}}}\right )\right ) x \sqrt {c \,x^{2}+b}}{8 b^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(153\)
default	\(-\frac {\left (c \,x^{2}+b \right ) \left (15 A \sqrt {c \,x^{2}+b}\, \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) b \,c^{2} x^{4}-15 A \,b^{\frac {3}{2}} c^{2} x^{4}-12 B \sqrt {c \,x^{2}+b}\, \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) b^{2} c \,x^{4}+12 B \,b^{\frac {5}{2}} c \,x^{4}-5 A \,b^{\frac {5}{2}} c \,x^{2}+4 B \,b^{\frac {7}{2}} x^{2}+2 A \,b^{\frac {7}{2}}\right )}{8 x \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{\frac {9}{2}}}\)	\(157\)

Input:

int((B*x^2+A)/x^2/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/8*(c*x^2+b)*(-7*A*c*x^2+4*B*b*x^2+2*A*b)/b^3/x^3/(x^2*(c*x^2+b))^(1/2)+ 
1/8*c/b^3*(-(7*A*c-4*B*b)/(c*x^2+b)^(1/2)+3*b*(5*A*c-4*B*b)*(1/b/(c*x^2+b) 
^(1/2)-1/b^(3/2)*ln((2*b+2*b^(1/2)*(c*x^2+b)^(1/2))/x)))*x/(x^2*(c*x^2+b)) 
^(1/2)*(c*x^2+b)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.26 \[ \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left ({\left (4 \, B b c^{2} - 5 \, A c^{3}\right )} x^{7} + {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{5}\right )} \sqrt {b} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, {\left (3 \, {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{4} + 2 \, A b^{3} + {\left (4 \, B b^{3} - 5 \, A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{16 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}, -\frac {3 \, {\left ({\left (4 \, B b c^{2} - 5 \, A c^{3}\right )} x^{7} + {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{5}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) + {\left (3 \, {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{4} + 2 \, A b^{3} + {\left (4 \, B b^{3} - 5 \, A b^{2} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{8 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}\right ] \] Input:

integrate((B*x^2+A)/x^2/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
 

Output:

[-1/16*(3*((4*B*b*c^2 - 5*A*c^3)*x^7 + (4*B*b^2*c - 5*A*b*c^2)*x^5)*sqrt(b 
)*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) + 2*(3*(4*B*b^ 
2*c - 5*A*b*c^2)*x^4 + 2*A*b^3 + (4*B*b^3 - 5*A*b^2*c)*x^2)*sqrt(c*x^4 + b 
*x^2))/(b^4*c*x^7 + b^5*x^5), -1/8*(3*((4*B*b*c^2 - 5*A*c^3)*x^7 + (4*B*b^ 
2*c - 5*A*b*c^2)*x^5)*sqrt(-b)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(b*x)) 
+ (3*(4*B*b^2*c - 5*A*b*c^2)*x^4 + 2*A*b^3 + (4*B*b^3 - 5*A*b^2*c)*x^2)*sq 
rt(c*x^4 + b*x^2))/(b^4*c*x^7 + b^5*x^5)]
 

Sympy [F]

\[ \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {A + B x^{2}}{x^{2} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((B*x**2+A)/x**2/(c*x**4+b*x**2)**(3/2),x)
 

Output:

Integral((A + B*x**2)/(x**2*(x**2*(b + c*x**2))**(3/2)), x)
 

Maxima [F]

\[ \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {B x^{2} + A}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2}} \,d x } \] Input:

integrate((B*x^2+A)/x^2/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((B*x^2 + A)/((c*x^4 + b*x^2)^(3/2)*x^2), x)
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.09 \[ \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {3 \, {\left (4 \, B b c - 5 \, A c^{2}\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{8 \, \sqrt {-b} b^{3} \mathrm {sgn}\left (x\right )} - \frac {B b c - A c^{2}}{\sqrt {c x^{2} + b} b^{3} \mathrm {sgn}\left (x\right )} - \frac {4 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b c - 4 \, \sqrt {c x^{2} + b} B b^{2} c - 7 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A c^{2} + 9 \, \sqrt {c x^{2} + b} A b c^{2}}{8 \, b^{3} c^{2} x^{4} \mathrm {sgn}\left (x\right )} \] Input:

integrate((B*x^2+A)/x^2/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
 

Output:

-3/8*(4*B*b*c - 5*A*c^2)*arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b^3*sg 
n(x)) - (B*b*c - A*c^2)/(sqrt(c*x^2 + b)*b^3*sgn(x)) - 1/8*(4*(c*x^2 + b)^ 
(3/2)*B*b*c - 4*sqrt(c*x^2 + b)*B*b^2*c - 7*(c*x^2 + b)^(3/2)*A*c^2 + 9*sq 
rt(c*x^2 + b)*A*b*c^2)/(b^3*c^2*x^4*sgn(x))
 

Mupad [B] (verification not implemented)

Time = 9.79 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.65 \[ \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {A\,{\left (\frac {b}{c\,x^2}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {7}{2};\ \frac {9}{2};\ -\frac {b}{c\,x^2}\right )}{7\,x\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}-\frac {B\,x\,{\left (\frac {b}{c\,x^2}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {5}{2};\ \frac {7}{2};\ -\frac {b}{c\,x^2}\right )}{5\,{\left (c\,x^4+b\,x^2\right )}^{3/2}} \] Input:

int((A + B*x^2)/(x^2*(b*x^2 + c*x^4)^(3/2)),x)
 

Output:

- (A*(b/(c*x^2) + 1)^(3/2)*hypergeom([3/2, 7/2], 9/2, -b/(c*x^2)))/(7*x*(b 
*x^2 + c*x^4)^(3/2)) - (B*x*(b/(c*x^2) + 1)^(3/2)*hypergeom([3/2, 5/2], 7/ 
2, -b/(c*x^2)))/(5*(b*x^2 + c*x^4)^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.71 \[ \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-2 \sqrt {c \,x^{2}+b}\, a \,b^{3}+5 \sqrt {c \,x^{2}+b}\, a \,b^{2} c \,x^{2}+15 \sqrt {c \,x^{2}+b}\, a b \,c^{2} x^{4}-4 \sqrt {c \,x^{2}+b}\, b^{4} x^{2}-12 \sqrt {c \,x^{2}+b}\, b^{3} c \,x^{4}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a b \,c^{2} x^{4}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,c^{3} x^{6}-12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{3} c \,x^{4}-12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c^{2} x^{6}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a b \,c^{2} x^{4}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,c^{3} x^{6}+12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{3} c \,x^{4}+12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c^{2} x^{6}}{8 b^{4} x^{4} \left (c \,x^{2}+b \right )} \] Input:

int((B*x^2+A)/x^2/(c*x^4+b*x^2)^(3/2),x)
 

Output:

( - 2*sqrt(b + c*x**2)*a*b**3 + 5*sqrt(b + c*x**2)*a*b**2*c*x**2 + 15*sqrt 
(b + c*x**2)*a*b*c**2*x**4 - 4*sqrt(b + c*x**2)*b**4*x**2 - 12*sqrt(b + c* 
x**2)*b**3*c*x**4 + 15*sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) + sqrt(c)*x 
)/sqrt(b))*a*b*c**2*x**4 + 15*sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) + sq 
rt(c)*x)/sqrt(b))*a*c**3*x**6 - 12*sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) 
 + sqrt(c)*x)/sqrt(b))*b**3*c*x**4 - 12*sqrt(b)*log((sqrt(b + c*x**2) - sq 
rt(b) + sqrt(c)*x)/sqrt(b))*b**2*c**2*x**6 - 15*sqrt(b)*log((sqrt(b + c*x* 
*2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*a*b*c**2*x**4 - 15*sqrt(b)*log((sqrt(b 
 + c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*a*c**3*x**6 + 12*sqrt(b)*log((s 
qrt(b + c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*b**3*c*x**4 + 12*sqrt(b)*l 
og((sqrt(b + c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*b**2*c**2*x**6)/(8*b* 
*4*x**4*(b + c*x**2))