3.2.92 \(\int \frac {x^{3/2}}{\sqrt {a x^2+b x^3}} \, dx\)

Optimal. Leaf size=60 \[ \frac {\sqrt {a x^2+b x^3}}{b \sqrt {x}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{b^{3/2}} \]

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Rubi [A]  time = 0.08, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2024, 2029, 206} \begin {gather*} \frac {\sqrt {a x^2+b x^3}}{b \sqrt {x}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a x^2+b x^3}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/Sqrt[a*x^2 + b*x^3],x]

[Out]

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

Rule 206

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

Rule 2024

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

Rule 2029

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

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Mathematica [A]  time = 0.04, size = 73, normalized size = 1.22 \begin {gather*} \frac {\sqrt {b} x^{3/2} (a+b x)-a^{3/2} x \sqrt {\frac {b x}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2} \sqrt {x^2 (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.14, size = 75, normalized size = 1.25 \begin {gather*} \frac {a \log \left (\sqrt {a x^2+b x^3}-\sqrt {b} x^{3/2}\right )}{b^{3/2}}-\frac {2 a \log \left (\sqrt {x}\right )}{b^{3/2}}+\frac {\sqrt {a x^2+b x^3}}{b \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 131, normalized size = 2.18 \begin {gather*} \left [\frac {a \sqrt {b} x \log \left (\frac {2 \, b x^{2} + a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {b} \sqrt {x}}{x}\right ) + 2 \, \sqrt {b x^{3} + a x^{2}} b \sqrt {x}}{2 \, b^{2} x}, \frac {a \sqrt {-b} x \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-b}}{b x^{\frac {3}{2}}}\right ) + \sqrt {b x^{3} + a x^{2}} b \sqrt {x}}{b^{2} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(a*sqrt(b)*x*log((2*b*x^2 + a*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(b)*sqrt(x))/x) + 2*sqrt(b*x^3 + a*x^2)*b*sqr
t(x))/(b^2*x), (a*sqrt(-b)*x*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-b)/(b*x^(3/2))) + sqrt(b*x^3 + a*x^2)*b*sqrt(x))
/(b^2*x)]

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giac [A]  time = 0.18, size = 38, normalized size = 0.63 \begin {gather*} \frac {a \log \left ({\left | -\sqrt {b} \sqrt {x} + \sqrt {b x + a} \right |}\right )}{b^{\frac {3}{2}}} + \frac {\sqrt {b x + a} \sqrt {x}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log(abs(-sqrt(b)*sqrt(x) + sqrt(b*x + a)))/b^(3/2) + sqrt(b*x + a)*sqrt(x)/b

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maple [A]  time = 0.05, size = 78, normalized size = 1.30 \begin {gather*} -\frac {\left (-2 b^{\frac {5}{2}} x^{2}-2 a \,b^{\frac {3}{2}} x +\sqrt {\left (b x +a \right ) x}\, a b \ln \left (\frac {2 b x +a +2 \sqrt {b \,x^{2}+a x}\, \sqrt {b}}{2 \sqrt {b}}\right )\right ) \sqrt {x}}{2 \sqrt {b \,x^{3}+a \,x^{2}}\, b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^(1/2)*(-2*b^(5/2)*x^2-2*a*b^(3/2)*x+a*((b*x+a)*x)^(1/2)*ln(1/2*(2*b*x+a+2*(b*x^2+a*x)^(1/2)*b^(1/2))/b^
(1/2))*b)/(b*x^3+a*x^2)^(1/2)/b^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {3}{2}}}{\sqrt {b x^{3} + a x^{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^{3/2}}{\sqrt {b\,x^3+a\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(a*x^2 + b*x^3)^(1/2),x)

[Out]

int(x^(3/2)/(a*x^2 + b*x^3)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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