3.2.81 \(\int \frac {(b x^{2/3}+a x)^{3/2}}{x^3} \, dx\) [181]

Optimal. Leaf size=113 \[ -\frac {3 a \sqrt {b x^{2/3}+a x}}{4 x}-\frac {3 a^2 \sqrt {b x^{2/3}+a x}}{8 b x^{2/3}}-\frac {\left (b x^{2/3}+a x\right )^{3/2}}{x^2}+\frac {3 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {b x^{2/3}+a x}}\right )}{8 b^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2045, 2050, 2054, 212} \begin {gather*} \frac {3 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {a x+b x^{2/3}}}\right )}{8 b^{3/2}}-\frac {3 a^2 \sqrt {a x+b x^{2/3}}}{8 b x^{2/3}}-\frac {3 a \sqrt {a x+b x^{2/3}}}{4 x}-\frac {\left (a x+b x^{2/3}\right )^{3/2}}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^(2/3) + a*x)^(3/2)/x^3,x]

[Out]

(-3*a*Sqrt[b*x^(2/3) + a*x])/(4*x) - (3*a^2*Sqrt[b*x^(2/3) + a*x])/(8*b*x^(2/3)) - (b*x^(2/3) + a*x)^(3/2)/x^2
+ (3*a^3*ArcTanh[(Sqrt[b]*x^(1/3))/Sqrt[b*x^(2/3) + a*x]])/(8*b^(3/2))

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 2045

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

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*} \int \frac {\left (b x^{2/3}+a x\right )^{3/2}}{x^3} \, dx &=-\frac {\left (b x^{2/3}+a x\right )^{3/2}}{x^2}+\frac {1}{2} a \int \frac {\sqrt {b x^{2/3}+a x}}{x^2} \, dx\\ &=-\frac {3 a \sqrt {b x^{2/3}+a x}}{4 x}-\frac {\left (b x^{2/3}+a x\right )^{3/2}}{x^2}+\frac {1}{8} a^2 \int \frac {1}{x \sqrt {b x^{2/3}+a x}} \, dx\\ &=-\frac {3 a \sqrt {b x^{2/3}+a x}}{4 x}-\frac {3 a^2 \sqrt {b x^{2/3}+a x}}{8 b x^{2/3}}-\frac {\left (b x^{2/3}+a x\right )^{3/2}}{x^2}-\frac {a^3 \int \frac {1}{x^{2/3} \sqrt {b x^{2/3}+a x}} \, dx}{16 b}\\ &=-\frac {3 a \sqrt {b x^{2/3}+a x}}{4 x}-\frac {3 a^2 \sqrt {b x^{2/3}+a x}}{8 b x^{2/3}}-\frac {\left (b x^{2/3}+a x\right )^{3/2}}{x^2}+\frac {\left (3 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}\\ &=-\frac {3 a \sqrt {b x^{2/3}+a x}}{4 x}-\frac {3 a^2 \sqrt {b x^{2/3}+a x}}{8 b x^{2/3}}-\frac {\left (b x^{2/3}+a x\right )^{3/2}}{x^2}+\frac {3 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [3]{x}}{\sqrt {b x^{2/3}+a x}}\right )}{8 b^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 10.06, size = 61, normalized size = 0.54 \begin {gather*} \frac {6 a^3 \left (b+a \sqrt [3]{x}\right )^2 \sqrt {b x^{2/3}+a x} \, _2F_1\left (\frac {5}{2},4;\frac {7}{2};1+\frac {a \sqrt [3]{x}}{b}\right )}{5 b^4 \sqrt [3]{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a^3*(b + a*x^(1/3))^2*Sqrt[b*x^(2/3) + a*x]*Hypergeometric2F1[5/2, 4, 7/2, 1 + (a*x^(1/3))/b])/(5*b^4*x^(1/
3))

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Maple [A]
time = 0.39, size = 93, normalized size = 0.82

method	result	size
derivativedivides	\(-\frac {\left (b \,x^{\frac {2}{3}}+a x \right )^{\frac {3}{2}} \left (3 \left (b +a \,x^{\frac {1}{3}}\right )^{\frac {5}{2}} b^{\frac {3}{2}}-3 \arctanh \left (\frac {\sqrt {b +a \,x^{\frac {1}{3}}}}{\sqrt {b}}\right ) b \,a^{3} x +8 \left (b +a \,x^{\frac {1}{3}}\right )^{\frac {3}{2}} b^{\frac {5}{2}}-3 \sqrt {b +a \,x^{\frac {1}{3}}}\, b^{\frac {7}{2}}\right )}{8 x^{2} \left (b +a \,x^{\frac {1}{3}}\right )^{\frac {3}{2}} b^{\frac {5}{2}}}\)	\(93\)
default	\(\frac {\left (b \,x^{\frac {2}{3}}+a x \right )^{\frac {3}{2}} \left (-3 \left (b +a \,x^{\frac {1}{3}}\right )^{\frac {5}{2}} b^{\frac {3}{2}}-8 \left (b +a \,x^{\frac {1}{3}}\right )^{\frac {3}{2}} b^{\frac {5}{2}}+3 \arctanh \left (\frac {\sqrt {b +a \,x^{\frac {1}{3}}}}{\sqrt {b}}\right ) b \,a^{3} x +3 \sqrt {b +a \,x^{\frac {1}{3}}}\, b^{\frac {7}{2}}\right )}{8 x^{2} \left (b +a \,x^{\frac {1}{3}}\right )^{\frac {3}{2}} b^{\frac {5}{2}}}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a x + b x^{\frac {2}{3}}\right )^{\frac {3}{2}}}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.58, size = 92, normalized size = 0.81 \begin {gather*} -\frac {\frac {3 \, a^{4} \arctan \left (\frac {\sqrt {a x^{\frac {1}{3}} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} + \frac {3 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} a^{4} + 8 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} a^{4} b - 3 \, \sqrt {a x^{\frac {1}{3}} + b} a^{4} b^{2}}{a^{3} b x}}{8 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*a^4*arctan(sqrt(a*x^(1/3) + b)/sqrt(-b))/(sqrt(-b)*b) + (3*(a*x^(1/3) + b)^(5/2)*a^4 + 8*(a*x^(1/3) +
b)^(3/2)*a^4*b - 3*sqrt(a*x^(1/3) + b)*a^4*b^2)/(a^3*b*x))/a

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a\,x+b\,x^{2/3}\right )}^{3/2}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^(2/3))^(3/2)/x^3,x)

[Out]

int((a*x + b*x^(2/3))^(3/2)/x^3, x)

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