\(\int \frac {1}{x (b x^{2/3}+a x)^{3/2}} \, dx\) [199]

Optimal result

Integrand size = 19, antiderivative size = 146 \[ \int \frac {1}{x \left (b x^{2/3}+a x\right )^{3/2}} \, dx=\frac {6}{b x^{2/3} \sqrt {b x^{2/3}+a x}}-\frac {7 \sqrt {b x^{2/3}+a x}}{b^2 x^{4/3}}+\frac {35 a \sqrt {b x^{2/3}+a x}}{4 b^3 x}-\frac {105 a^2 \sqrt {b x^{2/3}+a x}}{8 b^4 x^{2/3}}+\frac {105 a^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {b x^{2/3}+a x}}\right )}{8 b^{9/2}} \]

[Out]

105/8*a^3*arctanh(x^(1/3)*b^(1/2)/(b*x^(2/3)+a*x)^(1/2))/b^(9/2)+6/b/x^(2/3)/(b*x^(2/3)+a*x)^(1/2)-7*(b*x^(2/3
)+a*x)^(1/2)/b^2/x^(4/3)+35/4*a*(b*x^(2/3)+a*x)^(1/2)/b^3/x-105/8*a^2*(b*x^(2/3)+a*x)^(1/2)/b^4/x^(2/3)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2048, 2050, 2054, 212} \[ \int \frac {1}{x \left (b x^{2/3}+a x\right )^{3/2}} \, dx=\frac {105 a^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {a x+b x^{2/3}}}\right )}{8 b^{9/2}}-\frac {105 a^2 \sqrt {a x+b x^{2/3}}}{8 b^4 x^{2/3}}+\frac {35 a \sqrt {a x+b x^{2/3}}}{4 b^3 x}-\frac {7 \sqrt {a x+b x^{2/3}}}{b^2 x^{4/3}}+\frac {6}{b x^{2/3} \sqrt {a x+b x^{2/3}}} \]

[In]

Int[1/(x*(b*x^(2/3) + a*x)^(3/2)),x]

[Out]

6/(b*x^(2/3)*Sqrt[b*x^(2/3) + a*x]) - (7*Sqrt[b*x^(2/3) + a*x])/(b^2*x^(4/3)) + (35*a*Sqrt[b*x^(2/3) + a*x])/(
4*b^3*x) - (105*a^2*Sqrt[b*x^(2/3) + a*x])/(8*b^4*x^(2/3)) + (105*a^3*ArcTanh[(Sqrt[b]*x^(1/3))/Sqrt[b*x^(2/3)
+ a*x]])/(8*b^(9/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2048

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] + Dist[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1)))
, Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n]
&& (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1]

Rule 2050

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, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2054

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rubi steps \begin{align*} \text {integral}& = \frac {6}{b x^{2/3} \sqrt {b x^{2/3}+a x}}+\frac {7 \int \frac {1}{x^{5/3} \sqrt {b x^{2/3}+a x}} \, dx}{b} \\ & = \frac {6}{b x^{2/3} \sqrt {b x^{2/3}+a x}}-\frac {7 \sqrt {b x^{2/3}+a x}}{b^2 x^{4/3}}-\frac {(35 a) \int \frac {1}{x^{4/3} \sqrt {b x^{2/3}+a x}} \, dx}{6 b^2} \\ & = \frac {6}{b x^{2/3} \sqrt {b x^{2/3}+a x}}-\frac {7 \sqrt {b x^{2/3}+a x}}{b^2 x^{4/3}}+\frac {35 a \sqrt {b x^{2/3}+a x}}{4 b^3 x}+\frac {\left (35 a^2\right ) \int \frac {1}{x \sqrt {b x^{2/3}+a x}} \, dx}{8 b^3} \\ & = \frac {6}{b x^{2/3} \sqrt {b x^{2/3}+a x}}-\frac {7 \sqrt {b x^{2/3}+a x}}{b^2 x^{4/3}}+\frac {35 a \sqrt {b x^{2/3}+a x}}{4 b^3 x}-\frac {105 a^2 \sqrt {b x^{2/3}+a x}}{8 b^4 x^{2/3}}-\frac {\left (35 a^3\right ) \int \frac {1}{x^{2/3} \sqrt {b x^{2/3}+a x}} \, dx}{16 b^4} \\ & = \frac {6}{b x^{2/3} \sqrt {b x^{2/3}+a x}}-\frac {7 \sqrt {b x^{2/3}+a x}}{b^2 x^{4/3}}+\frac {35 a \sqrt {b x^{2/3}+a x}}{4 b^3 x}-\frac {105 a^2 \sqrt {b x^{2/3}+a x}}{8 b^4 x^{2/3}}+\frac {\left (105 a^3\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt [3]{x}}{\sqrt {b x^{2/3}+a x}}\right )}{8 b^4} \\ & = \frac {6}{b x^{2/3} \sqrt {b x^{2/3}+a x}}-\frac {7 \sqrt {b x^{2/3}+a x}}{b^2 x^{4/3}}+\frac {35 a \sqrt {b x^{2/3}+a x}}{4 b^3 x}-\frac {105 a^2 \sqrt {b x^{2/3}+a x}}{8 b^4 x^{2/3}}+\frac {105 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {b x^{2/3}+a x}}\right )}{8 b^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.70 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x \left (b x^{2/3}+a x\right )^{3/2}} \, dx=\frac {-\sqrt {b} \left (8 b^3-14 a b^2 \sqrt [3]{x}+35 a^2 b x^{2/3}+105 a^3 x\right )+105 a^3 \sqrt {b+a \sqrt [3]{x}} x \text {arctanh}\left (\frac {\sqrt {b+a \sqrt [3]{x}}}{\sqrt {b}}\right )}{8 b^{9/2} x^{2/3} \sqrt {b x^{2/3}+a x}} \]

[In]

Integrate[1/(x*(b*x^(2/3) + a*x)^(3/2)),x]

[Out]

(-(Sqrt[b]*(8*b^3 - 14*a*b^2*x^(1/3) + 35*a^2*b*x^(2/3) + 105*a^3*x)) + 105*a^3*Sqrt[b + a*x^(1/3)]*x*ArcTanh[
Sqrt[b + a*x^(1/3)]/Sqrt[b]])/(8*b^(9/2)*x^(2/3)*Sqrt[b*x^(2/3) + a*x])

Maple [A] (verified)

Time = 1.79 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.60

method	result	size
derivativedivides	\(\frac {\left (b +a \,x^{\frac {1}{3}}\right ) \left (105 \,\operatorname {arctanh}\left (\frac {\sqrt {b +a \,x^{\frac {1}{3}}}}{\sqrt {b}}\right ) \sqrt {b +a \,x^{\frac {1}{3}}}\, a^{3} x +14 b^{\frac {5}{2}} a \,x^{\frac {1}{3}}-35 b^{\frac {3}{2}} a^{2} x^{\frac {2}{3}}-105 \sqrt {b}\, a^{3} x -8 b^{\frac {7}{2}}\right )}{8 \left (b \,x^{\frac {2}{3}}+a x \right )^{\frac {3}{2}} b^{\frac {9}{2}}}\)	\(88\)
default	\(-\frac {\left (b +a \,x^{\frac {1}{3}}\right ) \left (105 \sqrt {b}\, a^{3} x +35 b^{\frac {3}{2}} a^{2} x^{\frac {2}{3}}-14 b^{\frac {5}{2}} a \,x^{\frac {1}{3}}-105 \,\operatorname {arctanh}\left (\frac {\sqrt {b +a \,x^{\frac {1}{3}}}}{\sqrt {b}}\right ) \sqrt {b +a \,x^{\frac {1}{3}}}\, a^{3} x +8 b^{\frac {7}{2}}\right )}{8 \left (b \,x^{\frac {2}{3}}+a x \right )^{\frac {3}{2}} b^{\frac {9}{2}}}\)	\(88\)

[In]

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

[Out]

1/8*(b+a*x^(1/3))*(105*arctanh((b+a*x^(1/3))^(1/2)/b^(1/2))*(b+a*x^(1/3))^(1/2)*a^3*x+14*b^(5/2)*a*x^(1/3)-35*
b^(3/2)*a^2*x^(2/3)-105*b^(1/2)*a^3*x-8*b^(7/2))/(b*x^(2/3)+a*x)^(3/2)/b^(9/2)

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{x \left (b x^{2/3}+a x\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{x \left (b x^{2/3}+a x\right )^{3/2}} \, dx=\int \frac {1}{x \left (a x + b x^{\frac {2}{3}}\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{x \left (b x^{2/3}+a x\right )^{3/2}} \, dx=\int { \frac {1}{{\left (a x + b x^{\frac {2}{3}}\right )}^{\frac {3}{2}} x} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.72 \[ \int \frac {1}{x \left (b x^{2/3}+a x\right )^{3/2}} \, dx=-\frac {105 \, a^{3} \arctan \left (\frac {\sqrt {a x^{\frac {1}{3}} + b}}{\sqrt {-b}}\right )}{8 \, \sqrt {-b} b^{4}} - \frac {6 \, a^{3}}{\sqrt {a x^{\frac {1}{3}} + b} b^{4}} - \frac {57 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} a^{3} - 136 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} a^{3} b + 87 \, \sqrt {a x^{\frac {1}{3}} + b} a^{3} b^{2}}{8 \, a^{3} b^{4} x} \]

[In]

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

[Out]

-105/8*a^3*arctan(sqrt(a*x^(1/3) + b)/sqrt(-b))/(sqrt(-b)*b^4) - 6*a^3/(sqrt(a*x^(1/3) + b)*b^4) - 1/8*(57*(a*
x^(1/3) + b)^(5/2)*a^3 - 136*(a*x^(1/3) + b)^(3/2)*a^3*b + 87*sqrt(a*x^(1/3) + b)*a^3*b^2)/(a^3*b^4*x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x \left (b x^{2/3}+a x\right )^{3/2}} \, dx=\int \frac {1}{x\,{\left (a\,x+b\,x^{2/3}\right )}^{3/2}} \,d x \]

[In]

int(1/(x*(a*x + b*x^(2/3))^(3/2)),x)

[Out]

int(1/(x*(a*x + b*x^(2/3))^(3/2)), x)