\(\int \frac {(A+B x^2) \sqrt {b x^2+c x^4}}{x^6} \, dx\) [170]

Optimal result

Integrand size = 26, antiderivative size = 103 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^6} \, dx=-\frac {(4 b B-A c) \sqrt {b x^2+c x^4}}{8 b x^3}-\frac {A \left (b x^2+c x^4\right )^{3/2}}{4 b x^7}-\frac {c (4 b B-A c) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 b^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.03 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^6} \, dx=-\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {b} \sqrt {b+c x^2} \left (2 A b+4 b B x^2+A c x^2\right )+c (4 b B-A c) x^4 \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{8 b^{3/2} x^5 \sqrt {b+c x^2}} \] Input:

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

Output:

-1/8*(Sqrt[x^2*(b + c*x^2)]*(Sqrt[b]*Sqrt[b + c*x^2]*(2*A*b + 4*b*B*x^2 + 
A*c*x^2) + c*(4*b*B - A*c)*x^4*ArcTanh[Sqrt[b + c*x^2]/Sqrt[b]]))/(b^(3/2) 
*x^5*Sqrt[b + c*x^2])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1944, 1425, 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.

\(\displaystyle \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^6} \, dx\)

\(\Big \downarrow \) 1944

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

\(\Big \downarrow \) 1425

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

\(\Big \downarrow \) 1400

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-1/4*(A*(b*x^2 + c*x^4)^(3/2))/(b*x^7) + ((4*b*B - A*c)*(-1/2*Sqrt[b*x^2 + 
 c*x^4]/x^3 - (c*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(2*Sqrt[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 1425

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 2*p + 1))), x] - Simp[2*c*(p/(d^4 
*(m + 2*p + 1)))   Int[(d*x)^(m + 4)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre 
eQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && GtQ[p, 0] && 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 = 0.87 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00

method	result	size
risch	\(-\frac {\left (A c \,x^{2}+4 B b \,x^{2}+2 A b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 x^{5} b}+\frac {\left (A c -4 B b \right ) c \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 b^{\frac {3}{2}} x \sqrt {c \,x^{2}+b}}\)	\(103\)
default	\(-\frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (-A \sqrt {b}\, \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c^{2} x^{4}+4 B \,b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c \,x^{4}+A \sqrt {c \,x^{2}+b}\, c^{2} x^{4}-4 B \sqrt {c \,x^{2}+b}\, b c \,x^{4}-A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c \,x^{2}+4 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,x^{2}+2 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \right )}{8 x^{5} \sqrt {c \,x^{2}+b}\, b^{2}}\)	\(175\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.87 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^6} \, dx=\left [-\frac {{\left (4 \, B b c - A c^{2}\right )} \sqrt {b} x^{5} \log \left (-\frac {c x^{3} + 2 \, b x + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (2 \, A b^{2} + {\left (4 \, B b^{2} + A b c\right )} x^{2}\right )}}{16 \, b^{2} x^{5}}, \frac {{\left (4 \, B b c - A c^{2}\right )} \sqrt {-b} x^{5} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) - \sqrt {c x^{4} + b x^{2}} {\left (2 \, A b^{2} + {\left (4 \, B b^{2} + A b c\right )} x^{2}\right )}}{8 \, b^{2} x^{5}}\right ] \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^6} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{6}}\, dx \] Input:

integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x**6,x)
 

Output:

Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**6, x)
 

Maxima [F]

\[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^6} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}} {\left (B x^{2} + A\right )}}{x^{6}} \,d x } \] Input:

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

Output:

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^6, x)
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.28 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^6} \, dx=\frac {\frac {{\left (4 \, B b c^{2} \mathrm {sgn}\left (x\right ) - A c^{3} \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} - \frac {4 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b c^{2} \mathrm {sgn}\left (x\right ) - 4 \, \sqrt {c x^{2} + b} B b^{2} c^{2} \mathrm {sgn}\left (x\right ) + {\left (c x^{2} + b\right )}^{\frac {3}{2}} A c^{3} \mathrm {sgn}\left (x\right ) + \sqrt {c x^{2} + b} A b c^{3} \mathrm {sgn}\left (x\right )}{b c^{2} x^{4}}}{8 \, c} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.81 \[ \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^6} \, dx=\frac {-2 \sqrt {c \,x^{2}+b}\, a \,b^{2}-\sqrt {c \,x^{2}+b}\, a b c \,x^{2}-4 \sqrt {c \,x^{2}+b}\, b^{3} x^{2}-\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,c^{2} x^{4}+4 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c \,x^{4}+\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,c^{2} x^{4}-4 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c \,x^{4}}{8 b^{2} x^{4}} \] Input:

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

Output:

( - 2*sqrt(b + c*x**2)*a*b**2 - sqrt(b + c*x**2)*a*b*c*x**2 - 4*sqrt(b + c 
*x**2)*b**3*x**2 - sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) + sqrt(c)*x)/sq 
rt(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 + 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) + sqr 
t(c)*x)/sqrt(b))*b**2*c*x**4)/(8*b**2*x**4)