3.2935 \(\int \frac{\sqrt{a+b \sqrt{c x^2}}}{x^5} \, dx\)

Optimal. Leaf size=171 \[ -\frac{5 b^3 c^2 \sqrt{a+b \sqrt{c x^2}}}{64 a^3 \sqrt{c x^2}}+\frac{5 b^4 c^2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c x^2}}}{\sqrt{a}}\right )}{64 a^{7/2}}+\frac{5 b^2 c \sqrt{a+b \sqrt{c x^2}}}{96 a^2 x^2}-\frac{b c^2 \sqrt{a+b \sqrt{c x^2}}}{24 a \left (c x^2\right )^{3/2}}-\frac{\sqrt{a+b \sqrt{c x^2}}}{4 x^4} \]

[Out]

-Sqrt[a + b*Sqrt[c*x^2]]/(4*x^4) + (5*b^2*c*Sqrt[a + b*Sqrt[c*x^2]])/(96*a^2*x^2) - (b*c^2*Sqrt[a + b*Sqrt[c*x
^2]])/(24*a*(c*x^2)^(3/2)) - (5*b^3*c^2*Sqrt[a + b*Sqrt[c*x^2]])/(64*a^3*Sqrt[c*x^2]) + (5*b^4*c^2*ArcTanh[Sqr
t[a + b*Sqrt[c*x^2]]/Sqrt[a]])/(64*a^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0781574, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {368, 47, 51, 63, 208} \[ -\frac{5 b^3 c^2 \sqrt{a+b \sqrt{c x^2}}}{64 a^3 \sqrt{c x^2}}+\frac{5 b^4 c^2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c x^2}}}{\sqrt{a}}\right )}{64 a^{7/2}}+\frac{5 b^2 c \sqrt{a+b \sqrt{c x^2}}}{96 a^2 x^2}-\frac{b c^2 \sqrt{a+b \sqrt{c x^2}}}{24 a \left (c x^2\right )^{3/2}}-\frac{\sqrt{a+b \sqrt{c x^2}}}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[c*x^2]]/x^5,x]

[Out]

-Sqrt[a + b*Sqrt[c*x^2]]/(4*x^4) + (5*b^2*c*Sqrt[a + b*Sqrt[c*x^2]])/(96*a^2*x^2) - (b*c^2*Sqrt[a + b*Sqrt[c*x
^2]])/(24*a*(c*x^2)^(3/2)) - (5*b^3*c^2*Sqrt[a + b*Sqrt[c*x^2]])/(64*a^3*Sqrt[c*x^2]) + (5*b^4*c^2*ArcTanh[Sqr
t[a + b*Sqrt[c*x^2]]/Sqrt[a]])/(64*a^(7/2))

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)]

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 51

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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]

Rubi steps

\begin{align*} \int \frac{\sqrt{a+b \sqrt{c x^2}}}{x^5} \, dx &=c^2 \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x^5} \, dx,x,\sqrt{c x^2}\right )\\ &=-\frac{\sqrt{a+b \sqrt{c x^2}}}{4 x^4}+\frac{1}{8} \left (b c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^4 \sqrt{a+b x}} \, dx,x,\sqrt{c x^2}\right )\\ &=-\frac{\sqrt{a+b \sqrt{c x^2}}}{4 x^4}-\frac{b c^2 \sqrt{a+b \sqrt{c x^2}}}{24 a \left (c x^2\right )^{3/2}}-\frac{\left (5 b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{a+b x}} \, dx,x,\sqrt{c x^2}\right )}{48 a}\\ &=-\frac{\sqrt{a+b \sqrt{c x^2}}}{4 x^4}+\frac{5 b^2 c \sqrt{a+b \sqrt{c x^2}}}{96 a^2 x^2}-\frac{b c^2 \sqrt{a+b \sqrt{c x^2}}}{24 a \left (c x^2\right )^{3/2}}+\frac{\left (5 b^3 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,\sqrt{c x^2}\right )}{64 a^2}\\ &=-\frac{\sqrt{a+b \sqrt{c x^2}}}{4 x^4}+\frac{5 b^2 c \sqrt{a+b \sqrt{c x^2}}}{96 a^2 x^2}-\frac{b c^2 \sqrt{a+b \sqrt{c x^2}}}{24 a \left (c x^2\right )^{3/2}}-\frac{5 b^3 c^2 \sqrt{a+b \sqrt{c x^2}}}{64 a^3 \sqrt{c x^2}}-\frac{\left (5 b^4 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sqrt{c x^2}\right )}{128 a^3}\\ &=-\frac{\sqrt{a+b \sqrt{c x^2}}}{4 x^4}+\frac{5 b^2 c \sqrt{a+b \sqrt{c x^2}}}{96 a^2 x^2}-\frac{b c^2 \sqrt{a+b \sqrt{c x^2}}}{24 a \left (c x^2\right )^{3/2}}-\frac{5 b^3 c^2 \sqrt{a+b \sqrt{c x^2}}}{64 a^3 \sqrt{c x^2}}-\frac{\left (5 b^3 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sqrt{c x^2}}\right )}{64 a^3}\\ &=-\frac{\sqrt{a+b \sqrt{c x^2}}}{4 x^4}+\frac{5 b^2 c \sqrt{a+b \sqrt{c x^2}}}{96 a^2 x^2}-\frac{b c^2 \sqrt{a+b \sqrt{c x^2}}}{24 a \left (c x^2\right )^{3/2}}-\frac{5 b^3 c^2 \sqrt{a+b \sqrt{c x^2}}}{64 a^3 \sqrt{c x^2}}+\frac{5 b^4 c^2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c x^2}}}{\sqrt{a}}\right )}{64 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0106537, size = 54, normalized size = 0.32 \[ -\frac{2 b^4 c^2 \left (a+b \sqrt{c x^2}\right )^{3/2} \, _2F_1\left (\frac{3}{2},5;\frac{5}{2};\frac{\sqrt{c x^2} b}{a}+1\right )}{3 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sqrt[c*x^2]]/x^5,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.013, size = 114, normalized size = 0.7 \begin{align*} -{\frac{1}{192\,{x}^{4}} \left ( 15\, \left ( a+b\sqrt{c{x}^{2}} \right ) ^{7/2}{a}^{7/2}-15\,{\it Artanh} \left ({\frac{\sqrt{a+b\sqrt{c{x}^{2}}}}{\sqrt{a}}} \right ){a}^{3}{b}^{4}{c}^{2}{x}^{4}-55\, \left ( a+b\sqrt{c{x}^{2}} \right ) ^{5/2}{a}^{9/2}+73\, \left ( a+b\sqrt{c{x}^{2}} \right ) ^{3/2}{a}^{11/2}+15\,\sqrt{a+b\sqrt{c{x}^{2}}}{a}^{13/2} \right ){a}^{-{\frac{13}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*(c*x^2)^(1/2))^(1/2)/x^5,x)

[Out]

-1/192*(15*(a+b*(c*x^2)^(1/2))^(7/2)*a^(7/2)-15*arctanh((a+b*(c*x^2)^(1/2))^(1/2)/a^(1/2))*a^3*b^4*c^2*x^4-55*
(a+b*(c*x^2)^(1/2))^(5/2)*a^(9/2)+73*(a+b*(c*x^2)^(1/2))^(3/2)*a^(11/2)+15*(a+b*(c*x^2)^(1/2))^(1/2)*a^(13/2))
/a^(13/2)/x^4

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x^2)^(1/2))^(1/2)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.36573, size = 554, normalized size = 3.24 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{4} c^{2} x^{4} \log \left (\frac{b c x^{2} + 2 \, \sqrt{c x^{2}} \sqrt{\sqrt{c x^{2}} b + a} \sqrt{a} + 2 \, \sqrt{c x^{2}} a}{x^{2}}\right ) + 2 \,{\left (10 \, a^{2} b^{2} c x^{2} - 48 \, a^{4} -{\left (15 \, a b^{3} c x^{2} + 8 \, a^{3} b\right )} \sqrt{c x^{2}}\right )} \sqrt{\sqrt{c x^{2}} b + a}}{384 \, a^{4} x^{4}}, -\frac{15 \, \sqrt{-a} b^{4} c^{2} x^{4} \arctan \left (\frac{\sqrt{\sqrt{c x^{2}} b + a} \sqrt{-a}}{a}\right ) -{\left (10 \, a^{2} b^{2} c x^{2} - 48 \, a^{4} -{\left (15 \, a b^{3} c x^{2} + 8 \, a^{3} b\right )} \sqrt{c x^{2}}\right )} \sqrt{\sqrt{c x^{2}} b + a}}{192 \, a^{4} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x^2)^(1/2))^(1/2)/x^5,x, algorithm="fricas")

[Out]

[1/384*(15*sqrt(a)*b^4*c^2*x^4*log((b*c*x^2 + 2*sqrt(c*x^2)*sqrt(sqrt(c*x^2)*b + a)*sqrt(a) + 2*sqrt(c*x^2)*a)
/x^2) + 2*(10*a^2*b^2*c*x^2 - 48*a^4 - (15*a*b^3*c*x^2 + 8*a^3*b)*sqrt(c*x^2))*sqrt(sqrt(c*x^2)*b + a))/(a^4*x
^4), -1/192*(15*sqrt(-a)*b^4*c^2*x^4*arctan(sqrt(sqrt(c*x^2)*b + a)*sqrt(-a)/a) - (10*a^2*b^2*c*x^2 - 48*a^4 -
(15*a*b^3*c*x^2 + 8*a^3*b)*sqrt(c*x^2))*sqrt(sqrt(c*x^2)*b + a))/(a^4*x^4)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + b \sqrt{c x^{2}}}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x**2)**(1/2))**(1/2)/x**5,x)

[Out]

Integral(sqrt(a + b*sqrt(c*x**2))/x**5, x)

________________________________________________________________________________________

Giac [A]  time = 1.19353, size = 182, normalized size = 1.06 \begin{align*} -\frac{\frac{15 \, b^{5} c^{\frac{5}{2}} \arctan \left (\frac{\sqrt{b \sqrt{c} x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} + \frac{15 \,{\left (b \sqrt{c} x + a\right )}^{\frac{7}{2}} b^{5} c^{\frac{5}{2}} - 55 \,{\left (b \sqrt{c} x + a\right )}^{\frac{5}{2}} a b^{5} c^{\frac{5}{2}} + 73 \,{\left (b \sqrt{c} x + a\right )}^{\frac{3}{2}} a^{2} b^{5} c^{\frac{5}{2}} + 15 \, \sqrt{b \sqrt{c} x + a} a^{3} b^{5} c^{\frac{5}{2}}}{a^{3} b^{4} c^{2} x^{4}}}{192 \, b \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(c*x^2)^(1/2))^(1/2)/x^5,x, algorithm="giac")

[Out]

-1/192*(15*b^5*c^(5/2)*arctan(sqrt(b*sqrt(c)*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + (15*(b*sqrt(c)*x + a)^(7/2)*b^5
*c^(5/2) - 55*(b*sqrt(c)*x + a)^(5/2)*a*b^5*c^(5/2) + 73*(b*sqrt(c)*x + a)^(3/2)*a^2*b^5*c^(5/2) + 15*sqrt(b*s
qrt(c)*x + a)*a^3*b^5*c^(5/2))/(a^3*b^4*c^2*x^4))/(b*sqrt(c))