\(\int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx\) [2244]

Optimal result

Integrand size = 17, antiderivative size = 136 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx=-\frac {\sqrt {a+b \sqrt {x}}}{2 a x^2}+\frac {7 b \sqrt {a+b \sqrt {x}}}{12 a^2 x^{3/2}}-\frac {35 b^2 \sqrt {a+b \sqrt {x}}}{48 a^3 x}+\frac {35 b^3 \sqrt {a+b \sqrt {x}}}{32 a^4 \sqrt {x}}-\frac {35 b^4 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{32 a^{9/2}} \]

[Out]

-35/32*b^4*arctanh((a+b*x^(1/2))^(1/2)/a^(1/2))/a^(9/2)-1/2*(a+b*x^(1/2))^(1/2)/a/x^2+7/12*b*(a+b*x^(1/2))^(1/
2)/a^2/x^(3/2)-35/48*b^2*(a+b*x^(1/2))^(1/2)/a^3/x+35/32*b^3*(a+b*x^(1/2))^(1/2)/a^4/x^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {272, 44, 65, 214} \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx=-\frac {35 b^4 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{32 a^{9/2}}+\frac {35 b^3 \sqrt {a+b \sqrt {x}}}{32 a^4 \sqrt {x}}-\frac {35 b^2 \sqrt {a+b \sqrt {x}}}{48 a^3 x}+\frac {7 b \sqrt {a+b \sqrt {x}}}{12 a^2 x^{3/2}}-\frac {\sqrt {a+b \sqrt {x}}}{2 a x^2} \]

[In]

Int[1/(Sqrt[a + b*Sqrt[x]]*x^3),x]

[Out]

-1/2*Sqrt[a + b*Sqrt[x]]/(a*x^2) + (7*b*Sqrt[a + b*Sqrt[x]])/(12*a^2*x^(3/2)) - (35*b^2*Sqrt[a + b*Sqrt[x]])/(
48*a^3*x) + (35*b^3*Sqrt[a + b*Sqrt[x]])/(32*a^4*Sqrt[x]) - (35*b^4*ArcTanh[Sqrt[a + b*Sqrt[x]]/Sqrt[a]])/(32*
a^(9/2))

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 272

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*} \text {integral}& = 2 \text {Subst}\left (\int \frac {1}{x^5 \sqrt {a+b x}} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {\sqrt {a+b \sqrt {x}}}{2 a x^2}-\frac {(7 b) \text {Subst}\left (\int \frac {1}{x^4 \sqrt {a+b x}} \, dx,x,\sqrt {x}\right )}{4 a} \\ & = -\frac {\sqrt {a+b \sqrt {x}}}{2 a x^2}+\frac {7 b \sqrt {a+b \sqrt {x}}}{12 a^2 x^{3/2}}+\frac {\left (35 b^2\right ) \text {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x}} \, dx,x,\sqrt {x}\right )}{24 a^2} \\ & = -\frac {\sqrt {a+b \sqrt {x}}}{2 a x^2}+\frac {7 b \sqrt {a+b \sqrt {x}}}{12 a^2 x^{3/2}}-\frac {35 b^2 \sqrt {a+b \sqrt {x}}}{48 a^3 x}-\frac {\left (35 b^3\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\sqrt {x}\right )}{32 a^3} \\ & = -\frac {\sqrt {a+b \sqrt {x}}}{2 a x^2}+\frac {7 b \sqrt {a+b \sqrt {x}}}{12 a^2 x^{3/2}}-\frac {35 b^2 \sqrt {a+b \sqrt {x}}}{48 a^3 x}+\frac {35 b^3 \sqrt {a+b \sqrt {x}}}{32 a^4 \sqrt {x}}+\frac {\left (35 b^4\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sqrt {x}\right )}{64 a^4} \\ & = -\frac {\sqrt {a+b \sqrt {x}}}{2 a x^2}+\frac {7 b \sqrt {a+b \sqrt {x}}}{12 a^2 x^{3/2}}-\frac {35 b^2 \sqrt {a+b \sqrt {x}}}{48 a^3 x}+\frac {35 b^3 \sqrt {a+b \sqrt {x}}}{32 a^4 \sqrt {x}}+\frac {\left (35 b^3\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sqrt {x}}\right )}{32 a^4} \\ & = -\frac {\sqrt {a+b \sqrt {x}}}{2 a x^2}+\frac {7 b \sqrt {a+b \sqrt {x}}}{12 a^2 x^{3/2}}-\frac {35 b^2 \sqrt {a+b \sqrt {x}}}{48 a^3 x}+\frac {35 b^3 \sqrt {a+b \sqrt {x}}}{32 a^4 \sqrt {x}}-\frac {35 b^4 \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{32 a^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx=\frac {\sqrt {a+b \sqrt {x}} \left (-48 a^3+56 a^2 b \sqrt {x}-70 a b^2 x+105 b^3 x^{3/2}\right )}{96 a^4 x^2}-\frac {35 b^4 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{32 a^{9/2}} \]

[In]

Integrate[1/(Sqrt[a + b*Sqrt[x]]*x^3),x]

[Out]

(Sqrt[a + b*Sqrt[x]]*(-48*a^3 + 56*a^2*b*Sqrt[x] - 70*a*b^2*x + 105*b^3*x^(3/2)))/(96*a^4*x^2) - (35*b^4*ArcTa
nh[Sqrt[a + b*Sqrt[x]]/Sqrt[a]])/(32*a^(9/2))

Maple [A] (verified)

Time = 3.62 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(4 b^{4} \left (-\frac {\sqrt {a +b \sqrt {x}}}{8 a \,b^{4} x^{2}}-\frac {7 \left (-\frac {\sqrt {a +b \sqrt {x}}}{6 a \,b^{3} x^{\frac {3}{2}}}+\frac {\frac {5 \sqrt {a +b \sqrt {x}}}{24 a \,b^{2} x}+\frac {5 \left (-\frac {3 \sqrt {a +b \sqrt {x}}}{8 a b \sqrt {x}}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \sqrt {x}}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )}{6 a}}{a}\right )}{8 a}\right )\)	\(124\)
default	\(4 b^{4} \left (-\frac {\sqrt {a +b \sqrt {x}}}{8 a \,b^{4} x^{2}}-\frac {7 \left (-\frac {\sqrt {a +b \sqrt {x}}}{6 a \,b^{3} x^{\frac {3}{2}}}+\frac {\frac {5 \sqrt {a +b \sqrt {x}}}{24 a \,b^{2} x}+\frac {5 \left (-\frac {3 \sqrt {a +b \sqrt {x}}}{8 a b \sqrt {x}}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \sqrt {x}}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )}{6 a}}{a}\right )}{8 a}\right )\)	\(124\)

[In]

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

[Out]

4*b^4*(-1/8*(a+b*x^(1/2))^(1/2)/a/b^4/x^2-7/8/a*(-1/6*(a+b*x^(1/2))^(1/2)/a/b^3/x^(3/2)+5/6/a*(1/4*(a+b*x^(1/2
))^(1/2)/a/b^2/x+3/4/a*(-1/2*(a+b*x^(1/2))^(1/2)/a/b/x^(1/2)+1/2/a^(3/2)*arctanh((a+b*x^(1/2))^(1/2)/a^(1/2)))
)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.35 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx=\left [\frac {105 \, \sqrt {a} b^{4} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b \sqrt {x} + a} \sqrt {a} \sqrt {x} + 2 \, a \sqrt {x}}{x}\right ) - 2 \, {\left (70 \, a^{2} b^{2} x + 48 \, a^{4} - 7 \, {\left (15 \, a b^{3} x + 8 \, a^{3} b\right )} \sqrt {x}\right )} \sqrt {b \sqrt {x} + a}}{192 \, a^{5} x^{2}}, \frac {105 \, \sqrt {-a} b^{4} x^{2} \arctan \left (\frac {\sqrt {b \sqrt {x} + a} \sqrt {-a}}{a}\right ) - {\left (70 \, a^{2} b^{2} x + 48 \, a^{4} - 7 \, {\left (15 \, a b^{3} x + 8 \, a^{3} b\right )} \sqrt {x}\right )} \sqrt {b \sqrt {x} + a}}{96 \, a^{5} x^{2}}\right ] \]

[In]

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

[Out]

[1/192*(105*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*(70*a^2*b^2
*x + 48*a^4 - 7*(15*a*b^3*x + 8*a^3*b)*sqrt(x))*sqrt(b*sqrt(x) + a))/(a^5*x^2), 1/96*(105*sqrt(-a)*b^4*x^2*arc
tan(sqrt(b*sqrt(x) + a)*sqrt(-a)/a) - (70*a^2*b^2*x + 48*a^4 - 7*(15*a*b^3*x + 8*a^3*b)*sqrt(x))*sqrt(b*sqrt(x
) + a))/(a^5*x^2)]

Sympy [A] (verification not implemented)

Time = 34.21 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.27 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx=- \frac {1}{2 \sqrt {b} x^{\frac {9}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {\sqrt {b}}{12 a x^{\frac {7}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} - \frac {7 b^{\frac {3}{2}}}{48 a^{2} x^{\frac {5}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {35 b^{\frac {5}{2}}}{96 a^{3} x^{\frac {3}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {35 b^{\frac {7}{2}}}{32 a^{4} \sqrt [4]{x} \sqrt {\frac {a}{b \sqrt {x}} + 1}} - \frac {35 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt [4]{x}} \right )}}{32 a^{\frac {9}{2}}} \]

[In]

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

[Out]

-1/(2*sqrt(b)*x**(9/4)*sqrt(a/(b*sqrt(x)) + 1)) + sqrt(b)/(12*a*x**(7/4)*sqrt(a/(b*sqrt(x)) + 1)) - 7*b**(3/2)
/(48*a**2*x**(5/4)*sqrt(a/(b*sqrt(x)) + 1)) + 35*b**(5/2)/(96*a**3*x**(3/4)*sqrt(a/(b*sqrt(x)) + 1)) + 35*b**(
7/2)/(32*a**4*x**(1/4)*sqrt(a/(b*sqrt(x)) + 1)) - 35*b**4*asinh(sqrt(a)/(sqrt(b)*x**(1/4)))/(32*a**(9/2))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.22 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx=\frac {35 \, b^{4} \log \left (\frac {\sqrt {b \sqrt {x} + a} - \sqrt {a}}{\sqrt {b \sqrt {x} + a} + \sqrt {a}}\right )}{64 \, a^{\frac {9}{2}}} + \frac {105 \, {\left (b \sqrt {x} + a\right )}^{\frac {7}{2}} b^{4} - 385 \, {\left (b \sqrt {x} + a\right )}^{\frac {5}{2}} a b^{4} + 511 \, {\left (b \sqrt {x} + a\right )}^{\frac {3}{2}} a^{2} b^{4} - 279 \, \sqrt {b \sqrt {x} + a} a^{3} b^{4}}{96 \, {\left ({\left (b \sqrt {x} + a\right )}^{4} a^{4} - 4 \, {\left (b \sqrt {x} + a\right )}^{3} a^{5} + 6 \, {\left (b \sqrt {x} + a\right )}^{2} a^{6} - 4 \, {\left (b \sqrt {x} + a\right )} a^{7} + a^{8}\right )}} \]

[In]

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

[Out]

35/64*b^4*log((sqrt(b*sqrt(x) + a) - sqrt(a))/(sqrt(b*sqrt(x) + a) + sqrt(a)))/a^(9/2) + 1/96*(105*(b*sqrt(x)
+ a)^(7/2)*b^4 - 385*(b*sqrt(x) + a)^(5/2)*a*b^4 + 511*(b*sqrt(x) + a)^(3/2)*a^2*b^4 - 279*sqrt(b*sqrt(x) + a)
*a^3*b^4)/((b*sqrt(x) + a)^4*a^4 - 4*(b*sqrt(x) + a)^3*a^5 + 6*(b*sqrt(x) + a)^2*a^6 - 4*(b*sqrt(x) + a)*a^7 +
a^8)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx=\frac {\frac {105 \, b^{5} \arctan \left (\frac {\sqrt {b \sqrt {x} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{4}} + \frac {105 \, {\left (b \sqrt {x} + a\right )}^{\frac {7}{2}} b^{5} - 385 \, {\left (b \sqrt {x} + a\right )}^{\frac {5}{2}} a b^{5} + 511 \, {\left (b \sqrt {x} + a\right )}^{\frac {3}{2}} a^{2} b^{5} - 279 \, \sqrt {b \sqrt {x} + a} a^{3} b^{5}}{a^{4} b^{4} x^{2}}}{96 \, b} \]

[In]

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

[Out]

1/96*(105*b^5*arctan(sqrt(b*sqrt(x) + a)/sqrt(-a))/(sqrt(-a)*a^4) + (105*(b*sqrt(x) + a)^(7/2)*b^5 - 385*(b*sq
rt(x) + a)^(5/2)*a*b^5 + 511*(b*sqrt(x) + a)^(3/2)*a^2*b^5 - 279*sqrt(b*sqrt(x) + a)*a^3*b^5)/(a^4*b^4*x^2))/b

Mupad [B] (verification not implemented)

Time = 5.86 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\sqrt {a+b \sqrt {x}} x^3} \, dx=\frac {511\,{\left (a+b\,\sqrt {x}\right )}^{3/2}}{96\,a^2\,x^2}-\frac {93\,\sqrt {a+b\,\sqrt {x}}}{32\,a\,x^2}-\frac {385\,{\left (a+b\,\sqrt {x}\right )}^{5/2}}{96\,a^3\,x^2}+\frac {35\,{\left (a+b\,\sqrt {x}\right )}^{7/2}}{32\,a^4\,x^2}+\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {a+b\,\sqrt {x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,35{}\mathrm {i}}{32\,a^{9/2}} \]

[In]

int(1/(x^3*(a + b*x^(1/2))^(1/2)),x)

[Out]

(b^4*atan(((a + b*x^(1/2))^(1/2)*1i)/a^(1/2))*35i)/(32*a^(9/2)) - (93*(a + b*x^(1/2))^(1/2))/(32*a*x^2) + (511
*(a + b*x^(1/2))^(3/2))/(96*a^2*x^2) - (385*(a + b*x^(1/2))^(5/2))/(96*a^3*x^2) + (35*(a + b*x^(1/2))^(7/2))/(
32*a^4*x^2)