\(\int \frac {1}{\sqrt {b x^{2/3}+a x}} \, dx\) [189]

Optimal result

Integrand size = 15, antiderivative size = 47 \[ \int \frac {1}{\sqrt {b x^{2/3}+a x}} \, dx=\frac {2 \sqrt {b x^{2/3}+a x}}{a}-\frac {4 b \sqrt {b x^{2/3}+a x}}{a^2 \sqrt [3]{x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2027, 2039} \[ \int \frac {1}{\sqrt {b x^{2/3}+a x}} \, dx=\frac {2 \sqrt {a x+b x^{2/3}}}{a}-\frac {4 b \sqrt {a x+b x^{2/3}}}{a^2 \sqrt [3]{x}} \]

[In]

Int[1/Sqrt[b*x^(2/3) + a*x],x]

[Out]

(2*Sqrt[b*x^(2/3) + a*x])/a - (4*b*Sqrt[b*x^(2/3) + a*x])/(a^2*x^(1/3))

Rule 2027

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

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])

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sqrt {b x^{2/3}+a x}}{a}-\frac {(2 b) \int \frac {1}{\sqrt [3]{x} \sqrt {b x^{2/3}+a x}} \, dx}{3 a} \\ & = \frac {2 \sqrt {b x^{2/3}+a x}}{a}-\frac {4 b \sqrt {b x^{2/3}+a x}}{a^2 \sqrt [3]{x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77 \[ \int \frac {1}{\sqrt {b x^{2/3}+a x}} \, dx=\frac {2 \left (-2 b+a \sqrt [3]{x}\right ) \sqrt {b x^{2/3}+a x}}{a^2 \sqrt [3]{x}} \]

[In]

Integrate[1/Sqrt[b*x^(2/3) + a*x],x]

[Out]

(2*(-2*b + a*x^(1/3))*Sqrt[b*x^(2/3) + a*x])/(a^2*x^(1/3))

Maple [A] (verified)

Time = 1.79 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {2 x^{\frac {1}{3}} \left (b +a \,x^{\frac {1}{3}}\right ) \left (a \,x^{\frac {1}{3}}-2 b \right )}{\sqrt {b \,x^{\frac {2}{3}}+a x}\, a^{2}}\)	\(36\)
default	\(\frac {2 x^{\frac {1}{3}} \left (b +a \,x^{\frac {1}{3}}\right ) \left (a \,x^{\frac {1}{3}}-2 b \right )}{\sqrt {b \,x^{\frac {2}{3}}+a x}\, a^{2}}\)	\(36\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (37) = 74\).

Time = 138.46 (sec) , antiderivative size = 238, normalized size of antiderivative = 5.06 \[ \int \frac {1}{\sqrt {b x^{2/3}+a x}} \, dx=\frac {{\left (50331648 \, b^{7} + 10485760 \, b^{6} + 49152 \, {\left (512 \, a^{3} - 3\right )} b^{4} - 983040 \, b^{5} + 256 \, {\left (24576 \, a^{3} + 53\right )} b^{3} + 11648 \, a^{3} - 96 \, {\left (2048 \, a^{3} + 1\right )} b^{2} - 3 \, {\left (155648 \, a^{3} + 3\right )} b\right )} x + 2 \, {\left ({\left (16777216 \, a b^{6} + 6291456 \, a b^{5} + 196608 \, a b^{4} - 262144 \, a^{4} - 114688 \, a b^{3} - 2304 \, a b^{2} + 864 \, a b - 27 \, a\right )} x - 2 \, {\left (16777216 \, b^{7} + 6291456 \, b^{6} + 196608 \, b^{5} - 114688 \, b^{4} - 2304 \, b^{3} - {\left (262144 \, a^{3} + 27\right )} b + 864 \, b^{2}\right )} x^{\frac {2}{3}}\right )} \sqrt {a x + b x^{\frac {2}{3}}}}{{\left (16777216 \, a^{2} b^{6} + 6291456 \, a^{2} b^{5} + 196608 \, a^{2} b^{4} - 262144 \, a^{5} - 114688 \, a^{2} b^{3} - 2304 \, a^{2} b^{2} + 864 \, a^{2} b - 27 \, a^{2}\right )} x} \]

[In]

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

[Out]

((50331648*b^7 + 10485760*b^6 + 49152*(512*a^3 - 3)*b^4 - 983040*b^5 + 256*(24576*a^3 + 53)*b^3 + 11648*a^3 -
96*(2048*a^3 + 1)*b^2 - 3*(155648*a^3 + 3)*b)*x + 2*((16777216*a*b^6 + 6291456*a*b^5 + 196608*a*b^4 - 262144*a
^4 - 114688*a*b^3 - 2304*a*b^2 + 864*a*b - 27*a)*x - 2*(16777216*b^7 + 6291456*b^6 + 196608*b^5 - 114688*b^4 -
2304*b^3 - (262144*a^3 + 27)*b + 864*b^2)*x^(2/3))*sqrt(a*x + b*x^(2/3)))/((16777216*a^2*b^6 + 6291456*a^2*b^
5 + 196608*a^2*b^4 - 262144*a^5 - 114688*a^2*b^3 - 2304*a^2*b^2 + 864*a^2*b - 27*a^2)*x)

Sympy [F]

\[ \int \frac {1}{\sqrt {b x^{2/3}+a x}} \, dx=\int \frac {1}{\sqrt {a x + b x^{\frac {2}{3}}}}\, dx \]

[In]

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

[Out]

Integral(1/sqrt(a*x + b*x**(2/3)), x)

Maxima [F]

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

[In]

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

[Out]

integrate(1/sqrt(a*x + b*x^(2/3)), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77 \[ \int \frac {1}{\sqrt {b x^{2/3}+a x}} \, dx=\frac {4 \, b^{\frac {3}{2}}}{a^{2}} + \frac {2 \, {\left ({\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} - 3 \, \sqrt {a x^{\frac {1}{3}} + b} b\right )}}{a^{2}} \]

[In]

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

[Out]

4*b^(3/2)/a^2 + 2*((a*x^(1/3) + b)^(3/2) - 3*sqrt(a*x^(1/3) + b)*b)/a^2

Mupad [B] (verification not implemented)

Time = 9.36 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\sqrt {b x^{2/3}+a x}} \, dx=\frac {3\,x\,\sqrt {\frac {a\,x^{1/3}}{b}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},2;\ 3;\ -\frac {a\,x^{1/3}}{b}\right )}{2\,\sqrt {a\,x+b\,x^{2/3}}} \]

[In]

int(1/(a*x + b*x^(2/3))^(1/2),x)

[Out]

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