\(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^8} \, dx\) [190]

Optimal result

Integrand size = 26, antiderivative size = 135 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^8} \, dx=\frac {3 c (4 b B+A c) \sqrt {b x^2+c x^4}}{8 b x}-\frac {(4 b B+A c) \left (b x^2+c x^4\right )^{3/2}}{8 b x^5}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{4 b x^9}-\frac {3 c (4 b B+A c) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{8 \sqrt {b}} \] Output:

3/8*c*(A*c+4*B*b)*(c*x^4+b*x^2)^(1/2)/b/x-1/8*(A*c+4*B*b)*(c*x^4+b*x^2)^(3 
/2)/b/x^5-1/4*A*(c*x^4+b*x^2)^(5/2)/b/x^9-3/8*c*(A*c+4*B*b)*arctanh(b^(1/2 
)*x/(c*x^4+b*x^2)^(1/2))/b^(1/2)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.84 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^8} \, dx=-\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {b} \sqrt {b+c x^2} \left (2 A b+4 b B x^2+5 A c x^2-8 B c x^4\right )+3 c (4 b B+A c) x^4 \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{8 \sqrt {b} x^5 \sqrt {b+c x^2}} \] Input:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1944, 1425, 1426, 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.

\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^8} \, dx\)

\(\Big \downarrow \) 1944

\(\displaystyle \frac {(A c+4 b B) \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^6}dx}{4 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{4 b x^9}\)

\(\Big \downarrow \) 1425

\(\displaystyle \frac {(A c+4 b B) \left (\frac {3}{2} c \int \frac {\sqrt {c x^4+b x^2}}{x^2}dx-\frac {\left (b x^2+c x^4\right )^{3/2}}{2 x^5}\right )}{4 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{4 b x^9}\)

\(\Big \downarrow \) 1426

\(\displaystyle \frac {(A c+4 b B) \left (\frac {3}{2} c \left (b \int \frac {1}{\sqrt {c x^4+b x^2}}dx+\frac {\sqrt {b x^2+c x^4}}{x}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{2 x^5}\right )}{4 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{4 b x^9}\)

\(\Big \downarrow \) 1400

\(\displaystyle \frac {(A c+4 b B) \left (\frac {3}{2} c \left (\frac {\sqrt {b x^2+c x^4}}{x}-b \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{2 x^5}\right )}{4 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{4 b x^9}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(A c+4 b B) \left (\frac {3}{2} c \left (\frac {\sqrt {b x^2+c x^4}}{x}-\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{2 x^5}\right )}{4 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{4 b x^9}\)

Input:

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^8,x]
 

Output:

-1/4*(A*(b*x^2 + c*x^4)^(5/2))/(b*x^9) + ((4*b*B + A*c)*(-1/2*(b*x^2 + c*x 
^4)^(3/2)/x^5 + (3*c*(Sqrt[b*x^2 + c*x^4]/x - Sqrt[b]*ArcTanh[(Sqrt[b]*x)/ 
Sqrt[b*x^2 + c*x^4]]))/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 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 1426

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 + 4*p + 1))), x] + Simp[2*b*(p/(d^2 
*(m + 4*p + 1)))   Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre 
eQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && GtQ[p, 0] && NeQ[m + 4*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 = 1.03 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.87

method	result	size
risch	\(-\frac {\left (5 A c \,x^{2}+4 B b \,x^{2}+2 A b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 x^{5}}+\frac {c \left (-\frac {\left (3 A c +12 B b \right ) \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right )}{\sqrt {b}}+8 B \sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{8 x \sqrt {c \,x^{2}+b}}\)	\(117\)
default	\(-\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (-A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{2} x^{4}+3 A \,b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c^{2} x^{4}-4 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} b c \,x^{4}+12 B \,b^{\frac {5}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c \,x^{4}+A \left (c \,x^{2}+b \right )^{\frac {5}{2}} c \,x^{2}-3 A \sqrt {c \,x^{2}+b}\, b \,c^{2} x^{4}+4 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \,x^{2}-12 B \sqrt {c \,x^{2}+b}\, b^{2} c \,x^{4}+2 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \right )}{8 x^{7} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{2}}\)	\(213\)

Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^8,x,method=_RETURNVERBOSE)
 

Output:

-1/8*(5*A*c*x^2+4*B*b*x^2+2*A*b)/x^5*(x^2*(c*x^2+b))^(1/2)+1/8*c*(-(3*A*c+ 
12*B*b)/b^(1/2)*ln((2*b+2*b^(1/2)*(c*x^2+b)^(1/2))/x)+8*B*(c*x^2+b)^(1/2)) 
*(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 = 212, normalized size of antiderivative = 1.57 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^8} \, dx=\left [\frac {3 \, {\left (4 \, B b c + A c^{2}\right )} \sqrt {b} x^{5} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, {\left (8 \, B b c x^{4} - 2 \, A b^{2} - {\left (4 \, B b^{2} + 5 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{16 \, b x^{5}}, \frac {3 \, {\left (4 \, B b c + A c^{2}\right )} \sqrt {-b} x^{5} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) + {\left (8 \, B b c x^{4} - 2 \, A b^{2} - {\left (4 \, B b^{2} + 5 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{8 \, b x^{5}}\right ] \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^8,x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^8} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{8}}\, dx \] Input:

integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**8,x)
 

Output:

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**8, x)
 

Maxima [F]

\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^8} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{8}} \,d x } \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^8,x, algorithm="maxima")
 

Output:

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^8, x)
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.07 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^8} \, dx=\frac {8 \, \sqrt {c x^{2} + b} B c^{2} \mathrm {sgn}\left (x\right ) + \frac {3 \, {\left (4 \, B b c^{2} \mathrm {sgn}\left (x\right ) + A c^{3} \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {4 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b c^{2} \mathrm {sgn}\left (x\right ) - 4 \, \sqrt {c x^{2} + b} B b^{2} c^{2} \mathrm {sgn}\left (x\right ) + 5 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A c^{3} \mathrm {sgn}\left (x\right ) - 3 \, \sqrt {c x^{2} + b} A b c^{3} \mathrm {sgn}\left (x\right )}{c^{2} x^{4}}}{8 \, c} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^8,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^8} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^8} \,d x \] Input:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^8,x)
 

Output:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^8, x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.51 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^8} \, dx=\frac {-2 \sqrt {c \,x^{2}+b}\, a \,b^{2}-5 \sqrt {c \,x^{2}+b}\, a b c \,x^{2}-4 \sqrt {c \,x^{2}+b}\, b^{3} x^{2}+8 \sqrt {c \,x^{2}+b}\, b^{2} c \,x^{4}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,c^{2} x^{4}+12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c \,x^{4}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,c^{2} x^{4}-12 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c \,x^{4}}{8 b \,x^{4}} \] Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^8,x)
 

Output:

( - 2*sqrt(b + c*x**2)*a*b**2 - 5*sqrt(b + c*x**2)*a*b*c*x**2 - 4*sqrt(b + 
 c*x**2)*b**3*x**2 + 8*sqrt(b + c*x**2)*b**2*c*x**4 + 3*sqrt(b)*log((sqrt( 
b + c*x**2) - sqrt(b) + sqrt(c)*x)/sqrt(b))*a*c**2*x**4 + 12*sqrt(b)*log(( 
sqrt(b + c*x**2) - sqrt(b) + sqrt(c)*x)/sqrt(b))*b**2*c*x**4 - 3*sqrt(b)*l 
og((sqrt(b + c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*a*c**2*x**4 - 12*sqrt 
(b)*log((sqrt(b + c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*b**2*c*x**4)/(8* 
b*x**4)