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

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

[Out]

5/32*b^4*arctanh((a+b*x^(1/2))^(1/2)/a^(1/2))/a^(7/2)-1/2*(a+b*x^(1/2))^(1/2)/x^2-1/12*b*(a+b*x^(1/2))^(1/2)/a
/x^(3/2)+5/48*b^2*(a+b*x^(1/2))^(1/2)/a^2/x-5/32*b^3*(a+b*x^(1/2))^(1/2)/a^3/x^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 133, normalized size of antiderivative = 1.00, 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 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]

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]]

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*}

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Mathematica [C]  time = 0.01, 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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fricas [A]  time = 1.10, size = 184, normalized size = 1.38 \[ \left [\frac {15 \, \sqrt {a} b^{4} x^{2} \log \left (\frac {b x + 2 \, \sqrt {b \sqrt {x} + a} \sqrt {a} \sqrt {x} + 2 \, a \sqrt {x}}{x}\right ) + 2 \, {\left (10 \, a^{2} b^{2} x - 48 \, a^{4} - {\left (15 \, a b^{3} x + 8 \, a^{3} b\right )} \sqrt {x}\right )} \sqrt {b \sqrt {x} + a}}{192 \, a^{4} x^{2}}, -\frac {15 \, \sqrt {-a} b^{4} x^{2} \arctan \left (\frac {\sqrt {b \sqrt {x} + a} \sqrt {-a}}{a}\right ) - {\left (10 \, a^{2} b^{2} x - 48 \, a^{4} - {\left (15 \, a b^{3} x + 8 \, a^{3} b\right )} \sqrt {x}\right )} \sqrt {b \sqrt {x} + a}}{96 \, a^{4} x^{2}}\right ] \]

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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giac [A]  time = 0.17, size = 109, normalized size = 0.82 \[ -\frac {\frac {15 \, b^{5} \arctan \left (\frac {\sqrt {b \sqrt {x} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {15 \, {\left (b \sqrt {x} + a\right )}^{\frac {7}{2}} b^{5} - 55 \, {\left (b \sqrt {x} + a\right )}^{\frac {5}{2}} a b^{5} + 73 \, {\left (b \sqrt {x} + a\right )}^{\frac {3}{2}} a^{2} b^{5} + 15 \, \sqrt {b \sqrt {x} + a} a^{3} b^{5}}{a^{3} b^{4} x^{2}}}{96 \, b} \]

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

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maple [A]  time = 0.01, size = 87, normalized size = 0.65 \[ 4 \left (\frac {5 \arctanh \left (\frac {\sqrt {b \sqrt {x}+a}}{\sqrt {a}}\right )}{128 a^{\frac {7}{2}}}+\frac {-\frac {73 \left (b \sqrt {x}+a \right )^{\frac {3}{2}}}{384 a}+\frac {55 \left (b \sqrt {x}+a \right )^{\frac {5}{2}}}{384 a^{2}}-\frac {5 \left (b \sqrt {x}+a \right )^{\frac {7}{2}}}{128 a^{3}}-\frac {5 \sqrt {b \sqrt {x}+a}}{128}}{b^{4} x^{2}}\right ) b^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.90, size = 166, normalized size = 1.25 \[ -\frac {5 \, b^{4} \log \left (\frac {\sqrt {b \sqrt {x} + a} - \sqrt {a}}{\sqrt {b \sqrt {x} + a} + \sqrt {a}}\right )}{64 \, a^{\frac {7}{2}}} - \frac {15 \, {\left (b \sqrt {x} + a\right )}^{\frac {7}{2}} b^{4} - 55 \, {\left (b \sqrt {x} + a\right )}^{\frac {5}{2}} a b^{4} + 73 \, {\left (b \sqrt {x} + a\right )}^{\frac {3}{2}} a^{2} b^{4} + 15 \, \sqrt {b \sqrt {x} + a} a^{3} b^{4}}{96 \, {\left ({\left (b \sqrt {x} + a\right )}^{4} a^{3} - 4 \, {\left (b \sqrt {x} + a\right )}^{3} a^{4} + 6 \, {\left (b \sqrt {x} + a\right )}^{2} a^{5} - 4 \, {\left (b \sqrt {x} + a\right )} a^{6} + a^{7}\right )}} \]

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]

-5/64*b^4*log((sqrt(b*sqrt(x) + a) - sqrt(a))/(sqrt(b*sqrt(x) + a) + sqrt(a)))/a^(7/2) - 1/96*(15*(b*sqrt(x) +
a)^(7/2)*b^4 - 55*(b*sqrt(x) + a)^(5/2)*a*b^4 + 73*(b*sqrt(x) + a)^(3/2)*a^2*b^4 + 15*sqrt(b*sqrt(x) + a)*a^3
*b^4)/((b*sqrt(x) + a)^4*a^3 - 4*(b*sqrt(x) + a)^3*a^4 + 6*(b*sqrt(x) + a)^2*a^5 - 4*(b*sqrt(x) + a)*a^6 + a^7
)

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mupad [B]  time = 1.42, size = 91, normalized size = 0.68 \[ \frac {55\,{\left (a+b\,\sqrt {x}\right )}^{5/2}}{96\,a^2\,x^2}-\frac {73\,{\left (a+b\,\sqrt {x}\right )}^{3/2}}{96\,a\,x^2}-\frac {5\,\sqrt {a+b\,\sqrt {x}}}{32\,x^2}-\frac {5\,{\left (a+b\,\sqrt {x}\right )}^{7/2}}{32\,a^3\,x^2}-\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {a+b\,\sqrt {x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{32\,a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^(1/2))^(1/2)/x^3,x)

[Out]

(55*(a + b*x^(1/2))^(5/2))/(96*a^2*x^2) - (b^4*atan(((a + b*x^(1/2))^(1/2)*1i)/a^(1/2))*5i)/(32*a^(7/2)) - (73
*(a + b*x^(1/2))^(3/2))/(96*a*x^2) - (5*(a + b*x^(1/2))^(1/2))/(32*x^2) - (5*(a + b*x^(1/2))^(7/2))/(32*a^3*x^
2)

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sympy [A]  time = 10.26, size = 170, normalized size = 1.28 \[ - \frac {a}{2 \sqrt {b} x^{\frac {9}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} - \frac {7 \sqrt {b}}{12 x^{\frac {7}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {b^{\frac {3}{2}}}{48 a x^{\frac {5}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} - \frac {5 b^{\frac {5}{2}}}{96 a^{2} x^{\frac {3}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} - \frac {5 b^{\frac {7}{2}}}{32 a^{3} \sqrt [4]{x} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {5 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt [4]{x}} \right )}}{32 a^{\frac {7}{2}}} \]

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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