3.2946 \(\int \frac {\sqrt {a+b (c x^2)^{3/2}}}{x^4} \, dx\)

Optimal. Leaf size=71 \[ -\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{3 x^3}-\frac {b \left (c x^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{\sqrt {a}}\right )}{3 \sqrt {a} x^3} \]

[Out]

-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

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {368, 266, 47, 63, 208} \[ -\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{3 x^3}-\frac {b \left (c x^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{\sqrt {a}}\right )}{3 \sqrt {a} x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*(c*x^2)^(3/2)]/x^4,x]

[Out]

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

Rule 47

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] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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 368

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> Dist[(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)]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^4} \, dx &=\frac {\left (c x^2\right )^{3/2} \operatorname {Subst}\left (\int \frac {\sqrt {a+b x^3}}{x^4} \, dx,x,\sqrt {c x^2}\right )}{x^3}\\ &=\frac {\left (c x^2\right )^{3/2} \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^2} \, dx,x,\left (c x^2\right )^{3/2}\right )}{3 x^3}\\ &=-\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{3 x^3}+\frac {\left (b \left (c x^2\right )^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\left (c x^2\right )^{3/2}\right )}{6 x^3}\\ &=-\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{3 x^3}+\frac {\left (c x^2\right )^{3/2} \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \left (c x^2\right )^{3/2}}\right )}{3 x^3}\\ &=-\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{3 x^3}-\frac {b \left (c x^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{\sqrt {a}}\right )}{3 \sqrt {a} x^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 93, normalized size = 1.31 \[ \frac {-b \left (c x^2\right )^{3/2} \sqrt {\frac {b \left (c x^2\right )^{3/2}}{a}+1} \tanh ^{-1}\left (\sqrt {\frac {b \left (c x^2\right )^{3/2}}{a}+1}\right )-a-b \left (c x^2\right )^{3/2}}{3 x^3 \sqrt {a+b \left (c x^2\right )^{3/2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 1.08, size = 206, normalized size = 2.90 \[ \left [\frac {b c x^{3} \sqrt {\frac {c}{a}} \log \left (\frac {b c^{2} x^{4} - 2 \, \sqrt {\sqrt {c x^{2}} b c x^{2} + a} a x \sqrt {\frac {c}{a}} + 2 \, \sqrt {c x^{2}} a}{x^{4}}\right ) - 2 \, \sqrt {\sqrt {c x^{2}} b c x^{2} + a}}{6 \, x^{3}}, -\frac {b c x^{3} \sqrt {-\frac {c}{a}} \arctan \left (-\frac {{\left (a b c^{2} x^{4} \sqrt {-\frac {c}{a}} - \sqrt {c x^{2}} a^{2} \sqrt {-\frac {c}{a}}\right )} \sqrt {\sqrt {c x^{2}} b c x^{2} + a}}{b^{2} c^{4} x^{7} - a^{2} c x}\right ) + \sqrt {\sqrt {c x^{2}} b c x^{2} + a}}{3 \, x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[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)) + sqrt(sqrt(c*x^2)*b*c*x^2 + a))/x^3]

________________________________________________________________________________________

giac [A]  time = 0.17, size = 61, normalized size = 0.86 \[ \frac {\frac {b^{2} c^{3} \arctan \left (\frac {\sqrt {b c^{\frac {3}{2}} x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {\sqrt {b c^{\frac {3}{2}} x^{3} + a} b c^{\frac {3}{2}}}{x^{3}}}{3 \, b c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

________________________________________________________________________________________

maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a +\left (c \,x^{2}\right )^{\frac {3}{2}} b}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\left (c x^{2}\right )^{\frac {3}{2}} b + a}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+b\,{\left (c\,x^2\right )}^{3/2}}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + b \left (c x^{2}\right )^{\frac {3}{2}}}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________