\(\int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx\) [108]

Optimal result

Integrand size = 19, antiderivative size = 142 \[ \int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx=-\frac {4 \sqrt {b \sqrt {x}+a x}}{9 b x^{5/2}}+\frac {32 a \sqrt {b \sqrt {x}+a x}}{63 b^2 x^2}-\frac {64 a^2 \sqrt {b \sqrt {x}+a x}}{105 b^3 x^{3/2}}+\frac {256 a^3 \sqrt {b \sqrt {x}+a x}}{315 b^4 x}-\frac {512 a^4 \sqrt {b \sqrt {x}+a x}}{315 b^5 \sqrt {x}} \]

[Out]

-4/9*(b*x^(1/2)+a*x)^(1/2)/b/x^(5/2)+32/63*a*(b*x^(1/2)+a*x)^(1/2)/b^2/x^2-64/105*a^2*(b*x^(1/2)+a*x)^(1/2)/b^
3/x^(3/2)+256/315*a^3*(b*x^(1/2)+a*x)^(1/2)/b^4/x-512/315*a^4*(b*x^(1/2)+a*x)^(1/2)/b^5/x^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2041, 2039} \[ \int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx=-\frac {512 a^4 \sqrt {a x+b \sqrt {x}}}{315 b^5 \sqrt {x}}+\frac {256 a^3 \sqrt {a x+b \sqrt {x}}}{315 b^4 x}-\frac {64 a^2 \sqrt {a x+b \sqrt {x}}}{105 b^3 x^{3/2}}+\frac {32 a \sqrt {a x+b \sqrt {x}}}{63 b^2 x^2}-\frac {4 \sqrt {a x+b \sqrt {x}}}{9 b x^{5/2}} \]

[In]

Int[1/(x^3*Sqrt[b*Sqrt[x] + a*x]),x]

[Out]

(-4*Sqrt[b*Sqrt[x] + a*x])/(9*b*x^(5/2)) + (32*a*Sqrt[b*Sqrt[x] + a*x])/(63*b^2*x^2) - (64*a^2*Sqrt[b*Sqrt[x]
+ a*x])/(105*b^3*x^(3/2)) + (256*a^3*Sqrt[b*Sqrt[x] + a*x])/(315*b^4*x) - (512*a^4*Sqrt[b*Sqrt[x] + a*x])/(315
*b^5*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}}{9 b x^{5/2}}-\frac {(8 a) \int \frac {1}{x^{5/2} \sqrt {b \sqrt {x}+a x}} \, dx}{9 b} \\ & = -\frac {4 \sqrt {b \sqrt {x}+a x}}{9 b x^{5/2}}+\frac {32 a \sqrt {b \sqrt {x}+a x}}{63 b^2 x^2}+\frac {\left (16 a^2\right ) \int \frac {1}{x^2 \sqrt {b \sqrt {x}+a x}} \, dx}{21 b^2} \\ & = -\frac {4 \sqrt {b \sqrt {x}+a x}}{9 b x^{5/2}}+\frac {32 a \sqrt {b \sqrt {x}+a x}}{63 b^2 x^2}-\frac {64 a^2 \sqrt {b \sqrt {x}+a x}}{105 b^3 x^{3/2}}-\frac {\left (64 a^3\right ) \int \frac {1}{x^{3/2} \sqrt {b \sqrt {x}+a x}} \, dx}{105 b^3} \\ & = -\frac {4 \sqrt {b \sqrt {x}+a x}}{9 b x^{5/2}}+\frac {32 a \sqrt {b \sqrt {x}+a x}}{63 b^2 x^2}-\frac {64 a^2 \sqrt {b \sqrt {x}+a x}}{105 b^3 x^{3/2}}+\frac {256 a^3 \sqrt {b \sqrt {x}+a x}}{315 b^4 x}+\frac {\left (128 a^4\right ) \int \frac {1}{x \sqrt {b \sqrt {x}+a x}} \, dx}{315 b^4} \\ & = -\frac {4 \sqrt {b \sqrt {x}+a x}}{9 b x^{5/2}}+\frac {32 a \sqrt {b \sqrt {x}+a x}}{63 b^2 x^2}-\frac {64 a^2 \sqrt {b \sqrt {x}+a x}}{105 b^3 x^{3/2}}+\frac {256 a^3 \sqrt {b \sqrt {x}+a x}}{315 b^4 x}-\frac {512 a^4 \sqrt {b \sqrt {x}+a x}}{315 b^5 \sqrt {x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx=-\frac {4 \sqrt {b \sqrt {x}+a x} \left (35 b^4-40 a b^3 \sqrt {x}+48 a^2 b^2 x-64 a^3 b x^{3/2}+128 a^4 x^2\right )}{315 b^5 x^{5/2}} \]

[In]

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

[Out]

(-4*Sqrt[b*Sqrt[x] + a*x]*(35*b^4 - 40*a*b^3*Sqrt[x] + 48*a^2*b^2*x - 64*a^3*b*x^(3/2) + 128*a^4*x^2))/(315*b^
5*x^(5/2))

Maple [A] (verified)

Time = 2.22 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(-\frac {4 \sqrt {b \sqrt {x}+a x}}{9 b \,x^{\frac {5}{2}}}-\frac {16 a \left (-\frac {2 \sqrt {b \sqrt {x}+a x}}{7 b \,x^{2}}-\frac {6 a \left (-\frac {2 \sqrt {b \sqrt {x}+a x}}{5 b \,x^{\frac {3}{2}}}-\frac {4 a \left (-\frac {2 \sqrt {b \sqrt {x}+a x}}{3 b x}+\frac {4 a \sqrt {b \sqrt {x}+a x}}{3 b^{2} \sqrt {x}}\right )}{5 b}\right )}{7 b}\right )}{9 b}\)	\(119\)
default	\(-\frac {\sqrt {b \sqrt {x}+a x}\, \left (1260 \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} x^{\frac {9}{2}} a^{\frac {9}{2}}-630 \sqrt {b \sqrt {x}+a x}\, x^{\frac {11}{2}} a^{\frac {11}{2}}-315 x^{\frac {11}{2}} \ln \left (\frac {2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+2 a \sqrt {x}+b}{2 \sqrt {a}}\right ) a^{5} b -630 x^{\frac {11}{2}} a^{\frac {11}{2}} \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}+315 x^{\frac {11}{2}} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{5} b +492 \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} x^{\frac {7}{2}} a^{\frac {5}{2}} b^{2}+140 \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} x^{\frac {5}{2}} \sqrt {a}\, b^{4}-748 a^{\frac {7}{2}} \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} b \,x^{4}-300 \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{3} x^{3}\right )}{315 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, b^{6} x^{\frac {11}{2}} \sqrt {a}}\)	\(262\)

[In]

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

[Out]

-4/9*(b*x^(1/2)+a*x)^(1/2)/b/x^(5/2)-16/9*a/b*(-2/7*(b*x^(1/2)+a*x)^(1/2)/b/x^2-6/7*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 = 0.33 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx=\frac {4 \, {\left (64 \, a^{3} b x^{2} + 40 \, a b^{3} x - {\left (128 \, a^{4} x^{2} + 48 \, a^{2} b^{2} x + 35 \, b^{4}\right )} \sqrt {x}\right )} \sqrt {a x + b \sqrt {x}}}{315 \, b^{5} x^{3}} \]

[In]

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

[Out]

4/315*(64*a^3*b*x^2 + 40*a*b^3*x - (128*a^4*x^2 + 48*a^2*b^2*x + 35*b^4)*sqrt(x))*sqrt(a*x + b*sqrt(x))/(b^5*x
^3)

Sympy [F]

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

[In]

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

[Out]

Integral(1/(x**3*sqrt(a*x + b*sqrt(x))), x)

Maxima [F]

\[ \int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx=\int { \frac {1}{\sqrt {a x + b \sqrt {x}} x^{3}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx=\frac {4 \, {\left (1008 \, a^{2} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{4} + 1680 \, a^{\frac {3}{2}} b {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{3} + 1080 \, a b^{2} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{2} + 315 \, \sqrt {a} b^{3} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} + 35 \, b^{4}\right )}}{315 \, {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )}^{9}} \]

[In]

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

[Out]

4/315*(1008*a^2*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x)))^4 + 1680*a^(3/2)*b*(sqrt(a)*sqrt(x) - sqrt(a*x + b*s
qrt(x)))^3 + 1080*a*b^2*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x)))^2 + 315*sqrt(a)*b^3*(sqrt(a)*sqrt(x) - sqrt(
a*x + b*sqrt(x))) + 35*b^4)/(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x)))^9

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx=\int \frac {1}{x^3\,\sqrt {a\,x+b\,\sqrt {x}}} \,d x \]

[In]

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

[Out]

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