\(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^6} \, dx\) [189]

Optimal result

Integrand size = 26, antiderivative size = 133 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^6} \, dx=\frac {(2 b B+3 A c) \sqrt {b x^2+c x^4}}{2 x}+\frac {(2 b B+3 A c) \left (b x^2+c x^4\right )^{3/2}}{6 b x^3}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{2 b x^7}-\frac {1}{2} \sqrt {b} (2 b B+3 A c) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right ) \] Output:

1/2*(3*A*c+2*B*b)*(c*x^4+b*x^2)^(1/2)/x+1/6*(3*A*c+2*B*b)*(c*x^4+b*x^2)^(3 
/2)/b/x^3-1/2*A*(c*x^4+b*x^2)^(5/2)/b/x^7-1/2*b^(1/2)*(3*A*c+2*B*b)*arctan 
h(b^(1/2)*x/(c*x^4+b*x^2)^(1/2))
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.82 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^6} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {b+c x^2} \left (-3 A b+8 b B x^2+6 A c x^2+2 B c x^4\right )-3 \sqrt {b} (2 b B+3 A c) x^2 \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{6 x^3 \sqrt {b+c x^2}} \] Input:

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^6,x]
 

Output:

(Sqrt[x^2*(b + c*x^2)]*(Sqrt[b + c*x^2]*(-3*A*b + 8*b*B*x^2 + 6*A*c*x^2 + 
2*B*c*x^4) - 3*Sqrt[b]*(2*b*B + 3*A*c)*x^2*ArcTanh[Sqrt[b + c*x^2]/Sqrt[b] 
]))/(6*x^3*Sqrt[b + c*x^2])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 118, 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, 1426, 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.

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

\(\Big \downarrow \) 1944

\(\displaystyle \frac {(3 A c+2 b B) \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^4}dx}{2 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{2 b x^7}\)

\(\Big \downarrow \) 1426

\(\displaystyle \frac {(3 A c+2 b B) \left (b \int \frac {\sqrt {c x^4+b x^2}}{x^2}dx+\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^3}\right )}{2 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{2 b x^7}\)

\(\Big \downarrow \) 1426

\(\displaystyle \frac {(3 A c+2 b B) \left (b \left (b \int \frac {1}{\sqrt {c x^4+b x^2}}dx+\frac {\sqrt {b x^2+c x^4}}{x}\right )+\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^3}\right )}{2 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{2 b x^7}\)

\(\Big \downarrow \) 1400

\(\displaystyle \frac {(3 A c+2 b B) \left (b \left (\frac {\sqrt {b x^2+c x^4}}{x}-b \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}\right )+\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^3}\right )}{2 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{2 b x^7}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(3 A c+2 b B) \left (b \left (\frac {\sqrt {b x^2+c x^4}}{x}-\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )\right )+\frac {\left (b x^2+c x^4\right )^{3/2}}{3 x^3}\right )}{2 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{2 b x^7}\)

Input:

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^6,x]
 

Output:

-1/2*(A*(b*x^2 + c*x^4)^(5/2))/(b*x^7) + ((2*b*B + 3*A*c)*((b*x^2 + c*x^4) 
^(3/2)/(3*x^3) + b*(Sqrt[b*x^2 + c*x^4]/x - Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sq 
rt[b*x^2 + c*x^4]])))/(2*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 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 = 0.89 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.12

method	result	size
risch	\(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, A b}{2 x^{3}}+\frac {\left (-\frac {\sqrt {b}\, \left (3 A c +2 B b \right ) \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right )}{2}+A \sqrt {c \,x^{2}+b}\, c +B \,c^{2} \left (\frac {x^{2} \sqrt {c \,x^{2}+b}}{3 c}-\frac {2 b \sqrt {c \,x^{2}+b}}{3 c^{2}}\right )+2 B b \sqrt {c \,x^{2}+b}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{x \sqrt {c \,x^{2}+b}}\)	\(149\)
default	\(-\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (-3 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c \,x^{2}+9 A \,b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c \,x^{2}-2 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,x^{2}+6 B \,b^{\frac {5}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) x^{2}+3 A \left (c \,x^{2}+b \right )^{\frac {5}{2}}-9 A \sqrt {c \,x^{2}+b}\, b c \,x^{2}-6 B \sqrt {c \,x^{2}+b}\, b^{2} x^{2}\right )}{6 x^{5} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b}\)	\(172\)

Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^6,x,method=_RETURNVERBOSE)
 

Output:

-1/2/x^3*(x^2*(c*x^2+b))^(1/2)*A*b+(-1/2*b^(1/2)*(3*A*c+2*B*b)*ln((2*b+2*b 
^(1/2)*(c*x^2+b)^(1/2))/x)+A*(c*x^2+b)^(1/2)*c+B*c^2*(1/3*x^2/c*(c*x^2+b)^ 
(1/2)-2/3*b/c^2*(c*x^2+b)^(1/2))+2*B*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 = 190, normalized size of antiderivative = 1.43 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^6} \, dx=\left [\frac {3 \, {\left (2 \, B b + 3 \, A c\right )} \sqrt {b} x^{3} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, {\left (2 \, B c x^{4} + 2 \, {\left (4 \, B b + 3 \, A c\right )} x^{2} - 3 \, A b\right )} \sqrt {c x^{4} + b x^{2}}}{12 \, x^{3}}, \frac {3 \, {\left (2 \, B b + 3 \, A c\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) + {\left (2 \, B c x^{4} + 2 \, {\left (4 \, B b + 3 \, A c\right )} x^{2} - 3 \, A b\right )} \sqrt {c x^{4} + b x^{2}}}{6 \, x^{3}}\right ] \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^6,x, algorithm="fricas")
 

Output:

[1/12*(3*(2*B*b + 3*A*c)*sqrt(b)*x^3*log(-(c*x^3 + 2*b*x - 2*sqrt(c*x^4 + 
b*x^2)*sqrt(b))/x^3) + 2*(2*B*c*x^4 + 2*(4*B*b + 3*A*c)*x^2 - 3*A*b)*sqrt( 
c*x^4 + b*x^2))/x^3, 1/6*(3*(2*B*b + 3*A*c)*sqrt(-b)*x^3*arctan(sqrt(c*x^4 
 + b*x^2)*sqrt(-b)/(b*x)) + (2*B*c*x^4 + 2*(4*B*b + 3*A*c)*x^2 - 3*A*b)*sq 
rt(c*x^4 + b*x^2))/x^3]
 

Sympy [F]

\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^6} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{6}}\, dx \] Input:

integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**6,x)
 

Output:

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**6, x)
 

Maxima [F]

\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^6} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{6}} \,d x } \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^6,x, algorithm="maxima")
 

Output:

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^6, x)
 

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.93 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^6} \, dx=\frac {1}{6} \, c {\left (\frac {3 \, {\left (2 \, B b^{2} \mathrm {sgn}\left (x\right ) + 3 \, A b c \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} c} - \frac {3 \, \sqrt {c x^{2} + b} A b \mathrm {sgn}\left (x\right )}{c x^{2}} + \frac {2 \, {\left ({\left (c x^{2} + b\right )}^{\frac {3}{2}} B c^{2} \mathrm {sgn}\left (x\right ) + 3 \, \sqrt {c x^{2} + b} B b c^{2} \mathrm {sgn}\left (x\right ) + 3 \, \sqrt {c x^{2} + b} A c^{3} \mathrm {sgn}\left (x\right )\right )}}{c^{3}}\right )} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^6,x, algorithm="giac")
 

Output:

1/6*c*(3*(2*B*b^2*sgn(x) + 3*A*b*c*sgn(x))*arctan(sqrt(c*x^2 + b)/sqrt(-b) 
)/(sqrt(-b)*c) - 3*sqrt(c*x^2 + b)*A*b*sgn(x)/(c*x^2) + 2*((c*x^2 + b)^(3/ 
2)*B*c^2*sgn(x) + 3*sqrt(c*x^2 + b)*B*b*c^2*sgn(x) + 3*sqrt(c*x^2 + b)*A*c 
^3*sgn(x))/c^3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^6} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^6} \,d x \] Input:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^6,x)
 

Output:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^6, x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.43 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^6} \, dx=\frac {-3 \sqrt {c \,x^{2}+b}\, a b +6 \sqrt {c \,x^{2}+b}\, a c \,x^{2}+8 \sqrt {c \,x^{2}+b}\, b^{2} x^{2}+2 \sqrt {c \,x^{2}+b}\, b c \,x^{4}+9 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a c \,x^{2}+6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} x^{2}-9 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a c \,x^{2}-6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} x^{2}}{6 x^{2}} \] Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^6,x)
 

Output:

( - 3*sqrt(b + c*x**2)*a*b + 6*sqrt(b + c*x**2)*a*c*x**2 + 8*sqrt(b + c*x* 
*2)*b**2*x**2 + 2*sqrt(b + c*x**2)*b*c*x**4 + 9*sqrt(b)*log((sqrt(b + c*x* 
*2) - sqrt(b) + sqrt(c)*x)/sqrt(b))*a*c*x**2 + 6*sqrt(b)*log((sqrt(b + c*x 
**2) - sqrt(b) + sqrt(c)*x)/sqrt(b))*b**2*x**2 - 9*sqrt(b)*log((sqrt(b + c 
*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*a*c*x**2 - 6*sqrt(b)*log((sqrt(b + 
c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*b**2*x**2)/(6*x**2)