\(\int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^4} \, dx\) [2940]

Optimal result

Integrand size = 21, antiderivative size = 144 \[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^4} \, dx=-\frac {\sqrt {a+b \sqrt {c x^2}}}{3 x^3}+\frac {b^2 c \sqrt {a+b \sqrt {c x^2}}}{8 a^2 x}-\frac {b \left (c x^2\right )^{3/2} \sqrt {a+b \sqrt {c x^2}}}{12 a c x^5}-\frac {b^3 \left (c x^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {c x^2}}}{\sqrt {a}}\right )}{8 a^{5/2} x^3} \]

[Out]

-1/8*b^3*(c*x^2)^(3/2)*arctanh((a+b*(c*x^2)^(1/2))^(1/2)/a^(1/2))/a^(5/2)/x^3-1/3*(a+b*(c*x^2)^(1/2))^(1/2)/x^
3+1/8*b^2*c*(a+b*(c*x^2)^(1/2))^(1/2)/a^2/x-1/12*b*(c*x^2)^(3/2)*(a+b*(c*x^2)^(1/2))^(1/2)/a/c/x^5

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {375, 43, 44, 65, 214} \[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^4} \, dx=-\frac {b^3 \left (c x^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {c x^2}}}{\sqrt {a}}\right )}{8 a^{5/2} x^3}+\frac {b^2 c \sqrt {a+b \sqrt {c x^2}}}{8 a^2 x}-\frac {b \left (c x^2\right )^{3/2} \sqrt {a+b \sqrt {c x^2}}}{12 a c x^5}-\frac {\sqrt {a+b \sqrt {c x^2}}}{3 x^3} \]

[In]

Int[Sqrt[a + b*Sqrt[c*x^2]]/x^4,x]

[Out]

-1/3*Sqrt[a + b*Sqrt[c*x^2]]/x^3 + (b^2*c*Sqrt[a + b*Sqrt[c*x^2]])/(8*a^2*x) - (b*(c*x^2)^(3/2)*Sqrt[a + b*Sqr
t[c*x^2]])/(12*a*c*x^5) - (b^3*(c*x^2)^(3/2)*ArcTanh[Sqrt[a + b*Sqrt[c*x^2]]/Sqrt[a]])/(8*a^(5/2)*x^3)

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 214

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 375

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*} \text {integral}& = \frac {\left (c x^2\right )^{3/2} \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x^4} \, dx,x,\sqrt {c x^2}\right )}{x^3} \\ & = -\frac {\sqrt {a+b \sqrt {c x^2}}}{3 x^3}+\frac {\left (b \left (c x^2\right )^{3/2}\right ) \text {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x}} \, dx,x,\sqrt {c x^2}\right )}{6 x^3} \\ & = -\frac {\sqrt {a+b \sqrt {c x^2}}}{3 x^3}-\frac {b \left (c x^2\right )^{3/2} \sqrt {a+b \sqrt {c x^2}}}{12 a c x^5}-\frac {\left (b^2 \left (c x^2\right )^{3/2}\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\sqrt {c x^2}\right )}{8 a x^3} \\ & = -\frac {\sqrt {a+b \sqrt {c x^2}}}{3 x^3}+\frac {b^2 c \sqrt {a+b \sqrt {c x^2}}}{8 a^2 x}-\frac {b \left (c x^2\right )^{3/2} \sqrt {a+b \sqrt {c x^2}}}{12 a c x^5}+\frac {\left (b^3 \left (c x^2\right )^{3/2}\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sqrt {c x^2}\right )}{16 a^2 x^3} \\ & = -\frac {\sqrt {a+b \sqrt {c x^2}}}{3 x^3}+\frac {b^2 c \sqrt {a+b \sqrt {c x^2}}}{8 a^2 x}-\frac {b \left (c x^2\right )^{3/2} \sqrt {a+b \sqrt {c x^2}}}{12 a c x^5}+\frac {\left (b^2 \left (c x^2\right )^{3/2}\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sqrt {c x^2}}\right )}{8 a^2 x^3} \\ & = -\frac {\sqrt {a+b \sqrt {c x^2}}}{3 x^3}+\frac {b^2 c \sqrt {a+b \sqrt {c x^2}}}{8 a^2 x}-\frac {b \left (c x^2\right )^{3/2} \sqrt {a+b \sqrt {c x^2}}}{12 a c x^5}-\frac {b^3 \left (c x^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c x^2}}}{\sqrt {a}}\right )}{8 a^{5/2} x^3} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.44 \[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^4} \, dx=\frac {2 b^3 \left (c x^2\right )^{3/2} \left (a+b \sqrt {c x^2}\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},4,\frac {5}{2},1+\frac {b \sqrt {c x^2}}{a}\right )}{3 a^4 x^3} \]

[In]

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

[Out]

(2*b^3*(c*x^2)^(3/2)*(a + b*Sqrt[c*x^2])^(3/2)*Hypergeometric2F1[3/2, 4, 5/2, 1 + (b*Sqrt[c*x^2])/a])/(3*a^4*x
^3)

Maple [A] (verified)

Time = 3.93 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.67

method	result	size
default	\(-\frac {3 a^{\frac {9}{2}} \sqrt {a +b \sqrt {c \,x^{2}}}+8 a^{\frac {7}{2}} \left (a +b \sqrt {c \,x^{2}}\right )^{\frac {3}{2}}-3 a^{\frac {5}{2}} \left (a +b \sqrt {c \,x^{2}}\right )^{\frac {5}{2}}+3 \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \sqrt {c \,x^{2}}}}{\sqrt {a}}\right ) a^{2} b^{3} \left (c \,x^{2}\right )^{\frac {3}{2}}}{24 x^{3} a^{\frac {9}{2}}}\)	\(97\)

[In]

int((a+b*(c*x^2)^(1/2))^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/24*(3*a^(9/2)*(a+b*(c*x^2)^(1/2))^(1/2)+8*a^(7/2)*(a+b*(c*x^2)^(1/2))^(3/2)-3*a^(5/2)*(a+b*(c*x^2)^(1/2))^(
5/2)+3*arctanh((a+b*(c*x^2)^(1/2))^(1/2)/a^(1/2))*a^2*b^3*(c*x^2)^(3/2))/x^3/a^(9/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.75 \[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^4} \, dx=\left [\frac {3 \, b^{3} c x^{3} \sqrt {\frac {c}{a}} \log \left (\frac {b c x^{2} - 2 \, \sqrt {\sqrt {c x^{2}} b + a} a x \sqrt {\frac {c}{a}} + 2 \, \sqrt {c x^{2}} a}{x^{2}}\right ) + 2 \, {\left (3 \, b^{2} c x^{2} - 2 \, \sqrt {c x^{2}} a b - 8 \, a^{2}\right )} \sqrt {\sqrt {c x^{2}} b + a}}{48 \, a^{2} x^{3}}, -\frac {3 \, b^{3} c x^{3} \sqrt {-\frac {c}{a}} \arctan \left (-\frac {{\left (a b c x^{2} \sqrt {-\frac {c}{a}} - \sqrt {c x^{2}} a^{2} \sqrt {-\frac {c}{a}}\right )} \sqrt {\sqrt {c x^{2}} b + a}}{b^{2} c^{2} x^{3} - a^{2} c x}\right ) - {\left (3 \, b^{2} c x^{2} - 2 \, \sqrt {c x^{2}} a b - 8 \, a^{2}\right )} \sqrt {\sqrt {c x^{2}} b + a}}{24 \, a^{2} x^{3}}\right ] \]

[In]

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

[Out]

[1/48*(3*b^3*c*x^3*sqrt(c/a)*log((b*c*x^2 - 2*sqrt(sqrt(c*x^2)*b + a)*a*x*sqrt(c/a) + 2*sqrt(c*x^2)*a)/x^2) +
2*(3*b^2*c*x^2 - 2*sqrt(c*x^2)*a*b - 8*a^2)*sqrt(sqrt(c*x^2)*b + a))/(a^2*x^3), -1/24*(3*b^3*c*x^3*sqrt(-c/a)*
arctan(-(a*b*c*x^2*sqrt(-c/a) - sqrt(c*x^2)*a^2*sqrt(-c/a))*sqrt(sqrt(c*x^2)*b + a)/(b^2*c^2*x^3 - a^2*c*x)) -
(3*b^2*c*x^2 - 2*sqrt(c*x^2)*a*b - 8*a^2)*sqrt(sqrt(c*x^2)*b + a))/(a^2*x^3)]

Sympy [F]

\[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^4} \, dx=\int \frac {\sqrt {a + b \sqrt {c x^{2}}}}{x^{4}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^4} \, dx=\int { \frac {\sqrt {\sqrt {c x^{2}} b + a}}{x^{4}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^4} \, dx=\frac {\frac {3 \, b^{4} c^{2} \arctan \left (\frac {\sqrt {b \sqrt {c} x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b \sqrt {c} x + a\right )}^{\frac {5}{2}} b^{4} c^{2} - 8 \, {\left (b \sqrt {c} x + a\right )}^{\frac {3}{2}} a b^{4} c^{2} - 3 \, \sqrt {b \sqrt {c} x + a} a^{2} b^{4} c^{2}}{a^{2} b^{3} c^{\frac {3}{2}} x^{3}}}{24 \, b \sqrt {c}} \]

[In]

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

[Out]

1/24*(3*b^4*c^2*arctan(sqrt(b*sqrt(c)*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*sqrt(c)*x + a)^(5/2)*b^4*c^2 - 8
*(b*sqrt(c)*x + a)^(3/2)*a*b^4*c^2 - 3*sqrt(b*sqrt(c)*x + a)*a^2*b^4*c^2)/(a^2*b^3*c^(3/2)*x^3))/(b*sqrt(c))

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^4} \, dx=\int \frac {\sqrt {a+b\,\sqrt {c\,x^2}}}{x^4} \,d x \]

[In]

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

[Out]

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