3.30.46 \(\int \frac {\sqrt {a+b (c x^2)^{3/2}}}{x^4} \, dx\) [2946]

3.30.46.1 Optimal result

Integrand size = 21, antiderivative size = 71 \[ \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^4} \, dx=-\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{3 x^3}-\frac {b \left (c x^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{\sqrt {a}}\right )}{3 \sqrt {a} x^3} \]

output

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

3.30.46.2 Mathematica [A] (verified)

Time = 1.57 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.31 \[ \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^4} \, dx=\frac {-a-b \left (c x^2\right )^{3/2}-b \left (c x^2\right )^{3/2} \sqrt {1+\frac {b \left (c x^2\right )^{3/2}}{a}} \text {arctanh}\left (\sqrt {1+\frac {b \left (c x^2\right )^{3/2}}{a}}\right )}{3 x^3 \sqrt {a+b \left (c x^2\right )^{3/2}}} \]

input

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

output

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

3.30.46.3 Rubi [A] (warning: unable to verify)

Time = 0.19 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {892, 798, 51, 73, 221}

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.

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

\(\Big \downarrow \) 892

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

\(\Big \downarrow \) 798

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

input

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

output

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

3.30.46.3.1 Defintions of rubi rules used

rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 798

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n   Subst 
[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, 
b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 892

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbo 
l] :> Simp[(d*x)^(m + 1)/(d*((c*x^q)^(1/q))^(m + 1))   Subst[Int[x^m*(a + b 
*x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x 
] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]
 

3.30.46.4 Maple [F]

\[\int \frac {\sqrt {a +b \left (c \,x^{2}\right )^{\frac {3}{2}}}}{x^{4}}d x\]

input

int((a+b*(c*x^2)^(3/2))^(1/2)/x^4,x)
 

output

int((a+b*(c*x^2)^(3/2))^(1/2)/x^4,x)
 

3.30.46.5 Fricas [A] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (53) = 106\).

Time = 0.28 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.90 \[ \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^4} \, dx=\left [\frac {b c x^{3} \sqrt {\frac {c}{a}} \log \left (\frac {b c^{2} x^{4} - 2 \, \sqrt {\sqrt {c x^{2}} b c x^{2} + a} a x \sqrt {\frac {c}{a}} + 2 \, \sqrt {c x^{2}} a}{x^{4}}\right ) - 2 \, \sqrt {\sqrt {c x^{2}} b c x^{2} + a}}{6 \, x^{3}}, -\frac {b c x^{3} \sqrt {-\frac {c}{a}} \arctan \left (-\frac {{\left (a b c^{2} x^{4} \sqrt {-\frac {c}{a}} - \sqrt {c x^{2}} a^{2} \sqrt {-\frac {c}{a}}\right )} \sqrt {\sqrt {c x^{2}} b c x^{2} + a}}{b^{2} c^{4} x^{7} - a^{2} c x}\right ) + \sqrt {\sqrt {c x^{2}} b c x^{2} + a}}{3 \, x^{3}}\right ] \]

input

integrate((a+b*(c*x^2)^(3/2))^(1/2)/x^4,x, algorithm="fricas")
 

output

[1/6*(b*c*x^3*sqrt(c/a)*log((b*c^2*x^4 - 2*sqrt(sqrt(c*x^2)*b*c*x^2 + a)*a 
*x*sqrt(c/a) + 2*sqrt(c*x^2)*a)/x^4) - 2*sqrt(sqrt(c*x^2)*b*c*x^2 + a))/x^ 
3, -1/3*(b*c*x^3*sqrt(-c/a)*arctan(-(a*b*c^2*x^4*sqrt(-c/a) - sqrt(c*x^2)* 
a^2*sqrt(-c/a))*sqrt(sqrt(c*x^2)*b*c*x^2 + a)/(b^2*c^4*x^7 - a^2*c*x)) + s 
qrt(sqrt(c*x^2)*b*c*x^2 + a))/x^3]
 

3.30.46.6 Sympy [F]

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

input

integrate((a+b*(c*x**2)**(3/2))**(1/2)/x**4,x)
 

output

Integral(sqrt(a + b*(c*x**2)**(3/2))/x**4, x)
 

3.30.46.7 Maxima [F]

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

input

integrate((a+b*(c*x^2)^(3/2))^(1/2)/x^4,x, algorithm="maxima")
 

output

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

3.30.46.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^4} \, dx=\frac {\frac {b^{2} c^{3} \arctan \left (\frac {\sqrt {b c^{\frac {3}{2}} x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {\sqrt {b c^{\frac {3}{2}} x^{3} + a} b c^{\frac {3}{2}}}{x^{3}}}{3 \, b c^{\frac {3}{2}}} \]

input

integrate((a+b*(c*x^2)^(3/2))^(1/2)/x^4,x, algorithm="giac")
 

output

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

3.30.46.9 Mupad [F(-1)]

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

input

int((a + b*(c*x^2)^(3/2))^(1/2)/x^4,x)
 

output

int((a + b*(c*x^2)^(3/2))^(1/2)/x^4, x)