\(\int \frac {(b x^2+c x^4)^{3/2}}{x^6} \, dx\) [187]

Optimal result

Integrand size = 19, antiderivative size = 79 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^6} \, dx=\frac {3 c \sqrt {b x^2+c x^4}}{2 x}-\frac {\left (b x^2+c x^4\right )^{3/2}}{2 x^5}-\frac {3}{2} \sqrt {b} c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right ) \] Output:

3/2*c*(c*x^4+b*x^2)^(1/2)/x-1/2*(c*x^4+b*x^2)^(3/2)/x^5-3/2*b^(1/2)*c*arct 
anh(b^(1/2)*x/(c*x^4+b*x^2)^(1/2))
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.05 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^6} \, dx=-\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\left (b-2 c x^2\right ) \sqrt {b+c x^2}+3 \sqrt {b} c x^2 \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{2 x^3 \sqrt {b+c x^2}} \] Input:

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

Output:

-1/2*(Sqrt[x^2*(b + c*x^2)]*((b - 2*c*x^2)*Sqrt[b + c*x^2] + 3*Sqrt[b]*c*x 
^2*ArcTanh[Sqrt[b + c*x^2]/Sqrt[b]]))/(x^3*Sqrt[b + c*x^2])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {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.

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

\(\Big \downarrow \) 1425

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

\(\Big \downarrow \) 1426

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

\(\Big \downarrow \) 1400

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-1/2*(b*x^2 + c*x^4)^(3/2)/x^5 + (3*c*(Sqrt[b*x^2 + c*x^4]/x - Sqrt[b]*Arc 
Tanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]]))/2
 

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]
 

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.11

method	result	size
risch	\(-\frac {b \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{2 x^{3}}+\frac {\left (-\frac {3 \sqrt {b}\, \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c}{2}+\sqrt {c \,x^{2}+b}\, c \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{x \sqrt {c \,x^{2}+b}}\)	\(88\)
default	\(-\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3 b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c \,x^{2}-\left (c \,x^{2}+b \right )^{\frac {3}{2}} c \,x^{2}+\left (c \,x^{2}+b \right )^{\frac {5}{2}}-3 \sqrt {c \,x^{2}+b}\, b c \,x^{2}\right )}{2 x^{5} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b}\)	\(102\)

Input:

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

Output:

-1/2*b/x^3*(x^2*(c*x^2+b))^(1/2)+(-3/2*b^(1/2)*ln((2*b+2*b^(1/2)*(c*x^2+b) 
^(1/2))/x)*c+(c*x^2+b)^(1/2)*c)*(x^2*(c*x^2+b))^(1/2)/x/(c*x^2+b)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.80 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^6} \, dx=\left [\frac {3 \, \sqrt {b} c x^{3} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (2 \, c x^{2} - b\right )}}{4 \, x^{3}}, \frac {3 \, \sqrt {-b} c x^{3} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (2 \, c x^{2} - b\right )}}{2 \, x^{3}}\right ] \] Input:

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

Output:

[1/4*(3*sqrt(b)*c*x^3*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*c*x^2 - b))/x^3, 1/2*(3*sqrt(-b)*c*x^3*ar 
ctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(b*x)) + sqrt(c*x^4 + b*x^2)*(2*c*x^2 - 
b))/x^3]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.80 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^6} \, dx=\frac {1}{2} \, {\left (\frac {3 \, b \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b}} + 2 \, \sqrt {c x^{2} + b} \mathrm {sgn}\left (x\right ) - \frac {\sqrt {c x^{2} + b} b \mathrm {sgn}\left (x\right )}{c x^{2}}\right )} c \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.15 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^6} \, dx=\frac {-\sqrt {c \,x^{2}+b}\, b +2 \sqrt {c \,x^{2}+b}\, c \,x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) c \,x^{2}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) c \,x^{2}}{2 x^{2}} \] Input:

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

Output:

( - sqrt(b + c*x**2)*b + 2*sqrt(b + c*x**2)*c*x**2 + 3*sqrt(b)*log((sqrt(b 
 + c*x**2) - sqrt(b) + sqrt(c)*x)/sqrt(b))*c*x**2 - 3*sqrt(b)*log((sqrt(b 
+ c*x**2) + sqrt(b) + sqrt(c)*x)/sqrt(b))*c*x**2)/(2*x**2)