\(\int \frac {1}{x^{5/2} \sqrt {b \sqrt {x}+a x}} \, dx\) [122]

Optimal result

Integrand size = 21, antiderivative size = 112 \[ \int \frac {1}{x^{5/2} \sqrt {b \sqrt {x}+a x}} \, dx=-\frac {4 \sqrt {b \sqrt {x}+a x}}{7 b x^2}+\frac {24 a \sqrt {b \sqrt {x}+a x}}{35 b^2 x^{3/2}}-\frac {32 a^2 \sqrt {b \sqrt {x}+a x}}{35 b^3 x}+\frac {64 a^3 \sqrt {b \sqrt {x}+a x}}{35 b^4 \sqrt {x}} \]

[Out]

-4/7*(b*x^(1/2)+a*x)^(1/2)/b/x^2+24/35*a*(b*x^(1/2)+a*x)^(1/2)/b^2/x^(3/2)-32/35*a^2*(b*x^(1/2)+a*x)^(1/2)/b^3
/x+64/35*a^3*(b*x^(1/2)+a*x)^(1/2)/b^4/x^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2041, 2039} \[ \int \frac {1}{x^{5/2} \sqrt {b \sqrt {x}+a x}} \, dx=\frac {64 a^3 \sqrt {a x+b \sqrt {x}}}{35 b^4 \sqrt {x}}-\frac {32 a^2 \sqrt {a x+b \sqrt {x}}}{35 b^3 x}+\frac {24 a \sqrt {a x+b \sqrt {x}}}{35 b^2 x^{3/2}}-\frac {4 \sqrt {a x+b \sqrt {x}}}{7 b x^2} \]

[In]

Int[1/(x^(5/2)*Sqrt[b*Sqrt[x] + a*x]),x]

[Out]

(-4*Sqrt[b*Sqrt[x] + a*x])/(7*b*x^2) + (24*a*Sqrt[b*Sqrt[x] + a*x])/(35*b^2*x^(3/2)) - (32*a^2*Sqrt[b*Sqrt[x]
+ a*x])/(35*b^3*x) + (64*a^3*Sqrt[b*Sqrt[x] + a*x])/(35*b^4*Sqrt[x])

Rule 2039

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-c^(j - 1))*(c*x)^(m - j
+ 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2041

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps \begin{align*} \text {integral}& = -\frac {4 \sqrt {b \sqrt {x}+a x}}{7 b x^2}-\frac {(6 a) \int \frac {1}{x^2 \sqrt {b \sqrt {x}+a x}} \, dx}{7 b} \\ & = -\frac {4 \sqrt {b \sqrt {x}+a x}}{7 b x^2}+\frac {24 a \sqrt {b \sqrt {x}+a x}}{35 b^2 x^{3/2}}+\frac {\left (24 a^2\right ) \int \frac {1}{x^{3/2} \sqrt {b \sqrt {x}+a x}} \, dx}{35 b^2} \\ & = -\frac {4 \sqrt {b \sqrt {x}+a x}}{7 b x^2}+\frac {24 a \sqrt {b \sqrt {x}+a x}}{35 b^2 x^{3/2}}-\frac {32 a^2 \sqrt {b \sqrt {x}+a x}}{35 b^3 x}-\frac {\left (16 a^3\right ) \int \frac {1}{x \sqrt {b \sqrt {x}+a x}} \, dx}{35 b^3} \\ & = -\frac {4 \sqrt {b \sqrt {x}+a x}}{7 b x^2}+\frac {24 a \sqrt {b \sqrt {x}+a x}}{35 b^2 x^{3/2}}-\frac {32 a^2 \sqrt {b \sqrt {x}+a x}}{35 b^3 x}+\frac {64 a^3 \sqrt {b \sqrt {x}+a x}}{35 b^4 \sqrt {x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x^{5/2} \sqrt {b \sqrt {x}+a x}} \, dx=-\frac {4 \sqrt {b \sqrt {x}+a x} \left (5 b^3-6 a b^2 \sqrt {x}+8 a^2 b x-16 a^3 x^{3/2}\right )}{35 b^4 x^2} \]

[In]

Integrate[1/(x^(5/2)*Sqrt[b*Sqrt[x] + a*x]),x]

[Out]

(-4*Sqrt[b*Sqrt[x] + a*x]*(5*b^3 - 6*a*b^2*Sqrt[x] + 8*a^2*b*x - 16*a^3*x^(3/2)))/(35*b^4*x^2)

Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(-\frac {4 \sqrt {b \sqrt {x}+a x}}{7 b \,x^{2}}-\frac {12 a \left (-\frac {2 \sqrt {b \sqrt {x}+a x}}{5 b \,x^{\frac {3}{2}}}-\frac {4 a \left (-\frac {2 \sqrt {b \sqrt {x}+a x}}{3 b x}+\frac {4 a \sqrt {b \sqrt {x}+a x}}{3 b^{2} \sqrt {x}}\right )}{5 b}\right )}{7 b}\)	\(93\)
default	\(-\frac {\sqrt {b \sqrt {x}+a x}\, \left (70 x^{\frac {9}{2}} \sqrt {b \sqrt {x}+a x}\, a^{\frac {9}{2}}+70 x^{\frac {9}{2}} a^{\frac {9}{2}} \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}-140 x^{\frac {7}{2}} \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {7}{2}}+35 x^{\frac {9}{2}} \ln \left (\frac {2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+2 a \sqrt {x}+b}{2 \sqrt {a}}\right ) a^{4} b -35 x^{\frac {9}{2}} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{4} b -44 x^{\frac {5}{2}} \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{2}+76 a^{\frac {5}{2}} \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} b \,x^{3}+20 \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} \sqrt {a}\, b^{3} x^{2}\right )}{35 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, b^{5} x^{\frac {9}{2}} \sqrt {a}}\)	\(240\)

[In]

int(1/x^(5/2)/(b*x^(1/2)+a*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x^{5/2} \sqrt {b \sqrt {x}+a x}} \, dx=-\frac {4 \, {\left (8 \, a^{2} b x + 5 \, b^{3} - 2 \, {\left (8 \, a^{3} x + 3 \, a b^{2}\right )} \sqrt {x}\right )} \sqrt {a x + b \sqrt {x}}}{35 \, b^{4} x^{2}} \]

[In]

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

[Out]

-4/35*(8*a^2*b*x + 5*b^3 - 2*(8*a^3*x + 3*a*b^2)*sqrt(x))*sqrt(a*x + b*sqrt(x))/(b^4*x^2)

Sympy [F]

\[ \int \frac {1}{x^{5/2} \sqrt {b \sqrt {x}+a x}} \, dx=\int \frac {1}{x^{\frac {5}{2}} \sqrt {a x + b \sqrt {x}}}\, dx \]

[In]

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

[Out]

Integral(1/(x**(5/2)*sqrt(a*x + b*sqrt(x))), x)

Maxima [F]

\[ \int \frac {1}{x^{5/2} \sqrt {b \sqrt {x}+a x}} \, dx=\int { \frac {1}{\sqrt {a x + b \sqrt {x}} x^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^{5/2} \sqrt {b \sqrt {x}+a x}} \, dx=\frac {4 \, {\left (70 \, a^{\frac {3}{2}} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{3} + 84 \, a b {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{2} + 35 \, \sqrt {a} b^{2} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} + 5 \, b^{3}\right )}}{35 \, {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{7}} \]

[In]

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

[Out]

4/35*(70*a^(3/2)*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x)))^3 + 84*a*b*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x))
)^2 + 35*sqrt(a)*b^2*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x))) + 5*b^3)/(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x
)))^7

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^{5/2} \sqrt {b \sqrt {x}+a x}} \, dx=\int \frac {1}{x^{5/2}\,\sqrt {a\,x+b\,\sqrt {x}}} \,d x \]

[In]

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

[Out]

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