3.2238 \(\int \frac{\sqrt{a+b \sqrt{x}}}{x^3} \, dx\)

Optimal. Leaf size=133 \[ -\frac{5 b^3 \sqrt{a+b \sqrt{x}}}{32 a^3 \sqrt{x}}+\frac{5 b^2 \sqrt{a+b \sqrt{x}}}{48 a^2 x}+\frac{5 b^4 \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{x}}}{\sqrt{a}}\right )}{32 a^{7/2}}-\frac{b \sqrt{a+b \sqrt{x}}}{12 a x^{3/2}}-\frac{\sqrt{a+b \sqrt{x}}}{2 x^2} \]

[Out]

-Sqrt[a + b*Sqrt[x]]/(2*x^2) - (b*Sqrt[a + b*Sqrt[x]])/(12*a*x^(3/2)) + (5*b^2*Sqrt[a + b*Sqrt[x]])/(48*a^2*x)
- (5*b^3*Sqrt[a + b*Sqrt[x]])/(32*a^3*Sqrt[x]) + (5*b^4*ArcTanh[Sqrt[a + b*Sqrt[x]]/Sqrt[a]])/(32*a^(7/2))

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Rubi [A]  time = 0.0608528, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {266, 47, 51, 63, 208} \[ -\frac{5 b^3 \sqrt{a+b \sqrt{x}}}{32 a^3 \sqrt{x}}+\frac{5 b^2 \sqrt{a+b \sqrt{x}}}{48 a^2 x}+\frac{5 b^4 \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{x}}}{\sqrt{a}}\right )}{32 a^{7/2}}-\frac{b \sqrt{a+b \sqrt{x}}}{12 a x^{3/2}}-\frac{\sqrt{a+b \sqrt{x}}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[x]]/x^3,x]

[Out]

-Sqrt[a + b*Sqrt[x]]/(2*x^2) - (b*Sqrt[a + b*Sqrt[x]])/(12*a*x^(3/2)) + (5*b^2*Sqrt[a + b*Sqrt[x]])/(48*a^2*x)
- (5*b^3*Sqrt[a + b*Sqrt[x]])/(32*a^3*Sqrt[x]) + (5*b^4*ArcTanh[Sqrt[a + b*Sqrt[x]]/Sqrt[a]])/(32*a^(7/2))

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 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{x}}}{x^3} \, dx &=2 \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x^5} \, dx,x,\sqrt{x}\right )\\ &=-\frac{\sqrt{a+b \sqrt{x}}}{2 x^2}+\frac{1}{4} b \operatorname{Subst}\left (\int \frac{1}{x^4 \sqrt{a+b x}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{\sqrt{a+b \sqrt{x}}}{2 x^2}-\frac{b \sqrt{a+b \sqrt{x}}}{12 a x^{3/2}}-\frac{\left (5 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{a+b x}} \, dx,x,\sqrt{x}\right )}{24 a}\\ &=-\frac{\sqrt{a+b \sqrt{x}}}{2 x^2}-\frac{b \sqrt{a+b \sqrt{x}}}{12 a x^{3/2}}+\frac{5 b^2 \sqrt{a+b \sqrt{x}}}{48 a^2 x}+\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,\sqrt{x}\right )}{32 a^2}\\ &=-\frac{\sqrt{a+b \sqrt{x}}}{2 x^2}-\frac{b \sqrt{a+b \sqrt{x}}}{12 a x^{3/2}}+\frac{5 b^2 \sqrt{a+b \sqrt{x}}}{48 a^2 x}-\frac{5 b^3 \sqrt{a+b \sqrt{x}}}{32 a^3 \sqrt{x}}-\frac{\left (5 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sqrt{x}\right )}{64 a^3}\\ &=-\frac{\sqrt{a+b \sqrt{x}}}{2 x^2}-\frac{b \sqrt{a+b \sqrt{x}}}{12 a x^{3/2}}+\frac{5 b^2 \sqrt{a+b \sqrt{x}}}{48 a^2 x}-\frac{5 b^3 \sqrt{a+b \sqrt{x}}}{32 a^3 \sqrt{x}}-\frac{\left (5 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sqrt{x}}\right )}{32 a^3}\\ &=-\frac{\sqrt{a+b \sqrt{x}}}{2 x^2}-\frac{b \sqrt{a+b \sqrt{x}}}{12 a x^{3/2}}+\frac{5 b^2 \sqrt{a+b \sqrt{x}}}{48 a^2 x}-\frac{5 b^3 \sqrt{a+b \sqrt{x}}}{32 a^3 \sqrt{x}}+\frac{5 b^4 \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{x}}}{\sqrt{a}}\right )}{32 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0081391, size = 43, normalized size = 0.32 \[ -\frac{4 b^4 \left (a+b \sqrt{x}\right )^{3/2} \, _2F_1\left (\frac{3}{2},5;\frac{5}{2};\frac{\sqrt{x} b}{a}+1\right )}{3 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sqrt[x]]/x^3,x]

[Out]

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

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Maple [A]  time = 0.009, size = 87, normalized size = 0.7 \begin{align*} 4\,{b}^{4} \left ({\frac{1}{{b}^{4}{x}^{2}} \left ( -{\frac{5\, \left ( a+b\sqrt{x} \right ) ^{7/2}}{128\,{a}^{3}}}+{\frac{55\, \left ( a+b\sqrt{x} \right ) ^{5/2}}{384\,{a}^{2}}}-{\frac{73\, \left ( a+b\sqrt{x} \right ) ^{3/2}}{384\,a}}-{\frac{5\,\sqrt{a+b\sqrt{x}}}{128}} \right ) }+{\frac{5}{128\,{a}^{7/2}}{\it Artanh} \left ({\frac{\sqrt{a+b\sqrt{x}}}{\sqrt{a}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*b^4*((-5/128/a^3*(a+b*x^(1/2))^(7/2)+55/384/a^2*(a+b*x^(1/2))^(5/2)-73/384/a*(a+b*x^(1/2))^(3/2)-5/128*(a+b*
x^(1/2))^(1/2))/b^4/x^2+5/128/a^(7/2)*arctanh((a+b*x^(1/2))^(1/2)/a^(1/2)))

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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*x^(1/2))^(1/2)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.34446, size = 468, normalized size = 3.52 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{4} x^{2} \log \left (\frac{b x + 2 \, \sqrt{b \sqrt{x} + a} \sqrt{a} \sqrt{x} + 2 \, a \sqrt{x}}{x}\right ) + 2 \,{\left (10 \, a^{2} b^{2} x - 48 \, a^{4} -{\left (15 \, a b^{3} x + 8 \, a^{3} b\right )} \sqrt{x}\right )} \sqrt{b \sqrt{x} + a}}{192 \, a^{4} x^{2}}, -\frac{15 \, \sqrt{-a} b^{4} x^{2} \arctan \left (\frac{\sqrt{b \sqrt{x} + a} \sqrt{-a}}{a}\right ) -{\left (10 \, a^{2} b^{2} x - 48 \, a^{4} -{\left (15 \, a b^{3} x + 8 \, a^{3} b\right )} \sqrt{x}\right )} \sqrt{b \sqrt{x} + a}}{96 \, a^{4} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/192*(15*sqrt(a)*b^4*x^2*log((b*x + 2*sqrt(b*sqrt(x) + a)*sqrt(a)*sqrt(x) + 2*a*sqrt(x))/x) + 2*(10*a^2*b^2*
x - 48*a^4 - (15*a*b^3*x + 8*a^3*b)*sqrt(x))*sqrt(b*sqrt(x) + a))/(a^4*x^2), -1/96*(15*sqrt(-a)*b^4*x^2*arctan
(sqrt(b*sqrt(x) + a)*sqrt(-a)/a) - (10*a^2*b^2*x - 48*a^4 - (15*a*b^3*x + 8*a^3*b)*sqrt(x))*sqrt(b*sqrt(x) + a
))/(a^4*x^2)]

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Sympy [A]  time = 12.7781, size = 170, normalized size = 1.28 \begin{align*} - \frac{a}{2 \sqrt{b} x^{\frac{9}{4}} \sqrt{\frac{a}{b \sqrt{x}} + 1}} - \frac{7 \sqrt{b}}{12 x^{\frac{7}{4}} \sqrt{\frac{a}{b \sqrt{x}} + 1}} + \frac{b^{\frac{3}{2}}}{48 a x^{\frac{5}{4}} \sqrt{\frac{a}{b \sqrt{x}} + 1}} - \frac{5 b^{\frac{5}{2}}}{96 a^{2} x^{\frac{3}{4}} \sqrt{\frac{a}{b \sqrt{x}} + 1}} - \frac{5 b^{\frac{7}{2}}}{32 a^{3} \sqrt [4]{x} \sqrt{\frac{a}{b \sqrt{x}} + 1}} + \frac{5 b^{4} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt [4]{x}} \right )}}{32 a^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/(2*sqrt(b)*x**(9/4)*sqrt(a/(b*sqrt(x)) + 1)) - 7*sqrt(b)/(12*x**(7/4)*sqrt(a/(b*sqrt(x)) + 1)) + b**(3/2)/(
48*a*x**(5/4)*sqrt(a/(b*sqrt(x)) + 1)) - 5*b**(5/2)/(96*a**2*x**(3/4)*sqrt(a/(b*sqrt(x)) + 1)) - 5*b**(7/2)/(3
2*a**3*x**(1/4)*sqrt(a/(b*sqrt(x)) + 1)) + 5*b**4*asinh(sqrt(a)/(sqrt(b)*x**(1/4)))/(32*a**(7/2))

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Giac [A]  time = 1.27545, size = 127, normalized size = 0.95 \begin{align*} -\frac{1}{96} \, b^{4}{\left (\frac{15 \, \arctan \left (\frac{\sqrt{b \sqrt{x} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} + \frac{15 \,{\left (b \sqrt{x} + a\right )}^{\frac{7}{2}} - 55 \,{\left (b \sqrt{x} + a\right )}^{\frac{5}{2}} a + 73 \,{\left (b \sqrt{x} + a\right )}^{\frac{3}{2}} a^{2} + 15 \, \sqrt{b \sqrt{x} + a} a^{3}}{a^{3} b^{4} x^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*b^4*(15*arctan(sqrt(b*sqrt(x) + a)/sqrt(-a))/(sqrt(-a)*a^3) + (15*(b*sqrt(x) + a)^(7/2) - 55*(b*sqrt(x)
+ a)^(5/2)*a + 73*(b*sqrt(x) + a)^(3/2)*a^2 + 15*sqrt(b*sqrt(x) + a)*a^3)/(a^3*b^4*x^2))