\(\int \frac {A+B x^2}{x^4 \sqrt {b x^2+c x^4}} \, dx\) [209]

Optimal result

Integrand size = 26, antiderivative size = 103 \[ \int \frac {A+B x^2}{x^4 \sqrt {b x^2+c x^4}} \, dx=-\frac {A \sqrt {b x^2+c x^4}}{4 b x^5}-\frac {(4 b B-3 A c) \sqrt {b x^2+c x^4}}{8 b^2 x^3}+\frac {c (4 b B-3 A c) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int \frac {A+B x^2}{x^4 \sqrt {b x^2+c x^4}} \, dx=\frac {-\sqrt {b} \left (b+c x^2\right ) \left (2 A b+4 b B x^2-3 A c x^2\right )+c (4 b B-3 A c) x^4 \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )}{8 b^{5/2} x^3 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1944, 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.

\(\displaystyle \int \frac {A+B x^2}{x^4 \sqrt {b x^2+c x^4}} \, dx\)

\(\Big \downarrow \) 1944

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

\(\Big \downarrow \) 1430

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

\(\Big \downarrow \) 1400

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-1/4*(A*Sqrt[b*x^2 + c*x^4])/(b*x^5) + ((4*b*B - 3*A*c)*(-1/2*Sqrt[b*x^2 + 
 c*x^4]/(b*x^3) + (c*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(2*b^(3/2)) 
))/(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 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 = 0.76 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.07

method	result	size
risch	\(-\frac {\left (c \,x^{2}+b \right ) \left (-3 A c \,x^{2}+4 B b \,x^{2}+2 A b \right )}{8 b^{2} x^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {\left (3 A c -4 B b \right ) c \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) x \sqrt {c \,x^{2}+b}}{8 b^{\frac {5}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(110\)
default	\(-\frac {\sqrt {c \,x^{2}+b}\, \left (3 A \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) b \,c^{2} x^{4}-4 B \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) b^{2} c \,x^{4}-3 A \sqrt {c \,x^{2}+b}\, b^{\frac {3}{2}} c \,x^{2}+4 B \sqrt {c \,x^{2}+b}\, b^{\frac {5}{2}} x^{2}+2 A \sqrt {c \,x^{2}+b}\, b^{\frac {5}{2}}\right )}{8 x^{3} \sqrt {c \,x^{4}+b \,x^{2}}\, b^{\frac {7}{2}}}\)	\(146\)

Input:

int((B*x^2+A)/x^4/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/8*(c*x^2+b)*(-3*A*c*x^2+4*B*b*x^2+2*A*b)/b^2/x^3/(x^2*(c*x^2+b))^(1/2)- 
1/8*(3*A*c-4*B*b)*c/b^(5/2)*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 = 0.11 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.88 \[ \int \frac {A+B x^2}{x^4 \sqrt {b x^2+c x^4}} \, dx=\left [-\frac {{\left (4 \, B b c - 3 \, A c^{2}\right )} \sqrt {b} x^{5} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (2 \, A b^{2} + {\left (4 \, B b^{2} - 3 \, A b c\right )} x^{2}\right )}}{16 \, b^{3} x^{5}}, -\frac {{\left (4 \, B b c - 3 \, A c^{2}\right )} \sqrt {-b} x^{5} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (2 \, A b^{2} + {\left (4 \, B b^{2} - 3 \, A b c\right )} x^{2}\right )}}{8 \, b^{3} x^{5}}\right ] \] Input:

integrate((B*x^2+A)/x^4/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((B*x**2+A)/x**4/(c*x**4+b*x**2)**(1/2),x)
 

Output:

Integral((A + B*x**2)/(x**4*sqrt(x**2*(b + c*x**2))), x)
 

Maxima [F]

\[ \int \frac {A+B x^2}{x^4 \sqrt {b x^2+c x^4}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {c x^{4} + b x^{2}} x^{4}} \,d x } \] Input:

integrate((B*x^2+A)/x^4/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.21 \[ \int \frac {A+B x^2}{x^4 \sqrt {b x^2+c x^4}} \, dx=-\frac {\frac {{\left (4 \, B b c^{2} - 3 \, A c^{3}\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {4 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b c^{2} - 4 \, \sqrt {c x^{2} + b} B b^{2} c^{2} - 3 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A c^{3} + 5 \, \sqrt {c x^{2} + b} A b c^{3}}{b^{2} c^{2} x^{4}}}{8 \, c \mathrm {sgn}\left (x\right )} \] Input:

integrate((B*x^2+A)/x^4/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
 

Output:

-1/8*((4*B*b*c^2 - 3*A*c^3)*arctan(sqrt(c*x^2 + b)/sqrt(-b))/(sqrt(-b)*b^2 
) + (4*(c*x^2 + b)^(3/2)*B*b*c^2 - 4*sqrt(c*x^2 + b)*B*b^2*c^2 - 3*(c*x^2 
+ b)^(3/2)*A*c^3 + 5*sqrt(c*x^2 + b)*A*b*c^3)/(b^2*c^2*x^4))/(c*sgn(x))
 

Mupad [F(-1)]

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

int((A + B*x^2)/(x^4*(b*x^2 + c*x^4)^(1/2)),x)
 

Output:

int((A + B*x^2)/(x^4*(b*x^2 + c*x^4)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.82 \[ \int \frac {A+B x^2}{x^4 \sqrt {b x^2+c x^4}} \, dx=\frac {-2 \sqrt {c \,x^{2}+b}\, a \,b^{2}+3 \sqrt {c \,x^{2}+b}\, a b c \,x^{2}-4 \sqrt {c \,x^{2}+b}\, b^{3} x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,c^{2} x^{4}-4 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c \,x^{4}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,c^{2} x^{4}+4 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c \,x^{4}}{8 b^{3} x^{4}} \] Input:

int((B*x^2+A)/x^4/(c*x^4+b*x^2)^(1/2),x)
 

Output:

( - 2*sqrt(b + c*x**2)*a*b**2 + 3*sqrt(b + c*x**2)*a*b*c*x**2 - 4*sqrt(b + 
 c*x**2)*b**3*x**2 + 3*sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) + sqrt(c)*x 
)/sqrt(b))*a*c**2*x**4 - 4*sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) + sqrt( 
c)*x)/sqrt(b))*b**2*c*x**4 - 3*sqrt(b)*log((sqrt(b + c*x**2) + sqrt(b) + s 
qrt(c)*x)/sqrt(b))*a*c**2*x**4 + 4*sqrt(b)*log((sqrt(b + c*x**2) + sqrt(b) 
 + sqrt(c)*x)/sqrt(b))*b**2*c*x**4)/(8*b**3*x**4)