3.270 \(\int \frac{x^{5/2}}{\sqrt{a x^2+b x^3}} \, dx\)

Optimal. Leaf size=95 \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a x^2+b x^3}}\right )}{4 b^{5/2}}-\frac{3 a \sqrt{a x^2+b x^3}}{4 b^2 \sqrt{x}}+\frac{\sqrt{x} \sqrt{a x^2+b x^3}}{2 b} \]

[Out]

(-3*a*Sqrt[a*x^2 + b*x^3])/(4*b^2*Sqrt[x]) + (Sqrt[x]*Sqrt[a*x^2 + b*x^3])/(2*b) + (3*a^2*ArcTanh[(Sqrt[b]*x^(
3/2))/Sqrt[a*x^2 + b*x^3]])/(4*b^(5/2))

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Rubi [A]  time = 0.125441, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2024, 2029, 206} \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a x^2+b x^3}}\right )}{4 b^{5/2}}-\frac{3 a \sqrt{a x^2+b x^3}}{4 b^2 \sqrt{x}}+\frac{\sqrt{x} \sqrt{a x^2+b x^3}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/Sqrt[a*x^2 + b*x^3],x]

[Out]

(-3*a*Sqrt[a*x^2 + b*x^3])/(4*b^2*Sqrt[x]) + (Sqrt[x]*Sqrt[a*x^2 + b*x^3])/(2*b) + (3*a^2*ArcTanh[(Sqrt[b]*x^(
3/2))/Sqrt[a*x^2 + b*x^3]])/(4*b^(5/2))

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]

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])

Rubi steps

\begin{align*} \int \frac{x^{5/2}}{\sqrt{a x^2+b x^3}} \, dx &=\frac{\sqrt{x} \sqrt{a x^2+b x^3}}{2 b}-\frac{(3 a) \int \frac{x^{3/2}}{\sqrt{a x^2+b x^3}} \, dx}{4 b}\\ &=-\frac{3 a \sqrt{a x^2+b x^3}}{4 b^2 \sqrt{x}}+\frac{\sqrt{x} \sqrt{a x^2+b x^3}}{2 b}+\frac{\left (3 a^2\right ) \int \frac{\sqrt{x}}{\sqrt{a x^2+b x^3}} \, dx}{8 b^2}\\ &=-\frac{3 a \sqrt{a x^2+b x^3}}{4 b^2 \sqrt{x}}+\frac{\sqrt{x} \sqrt{a x^2+b x^3}}{2 b}+\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^{3/2}}{\sqrt{a x^2+b x^3}}\right )}{4 b^2}\\ &=-\frac{3 a \sqrt{a x^2+b x^3}}{4 b^2 \sqrt{x}}+\frac{\sqrt{x} \sqrt{a x^2+b x^3}}{2 b}+\frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a x^2+b x^3}}\right )}{4 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0519435, size = 90, normalized size = 0.95 \[ \frac{\sqrt{b} x^{3/2} \left (-3 a^2-a b x+2 b^2 x^2\right )+3 a^{5/2} x \sqrt{\frac{b x}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{5/2} \sqrt{x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/Sqrt[a*x^2 + b*x^3],x]

[Out]

(Sqrt[b]*x^(3/2)*(-3*a^2 - a*b*x + 2*b^2*x^2) + 3*a^(5/2)*x*Sqrt[1 + (b*x)/a]*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a
]])/(4*b^(5/2)*Sqrt[x^2*(a + b*x)])

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Maple [A]  time = 0.005, size = 92, normalized size = 1. \begin{align*}{\frac{1}{8}\sqrt{x} \left ( 4\,{b}^{7/2}{x}^{3}-2\,{b}^{5/2}{x}^{2}a-6\,{b}^{3/2}x{a}^{2}+3\,\sqrt{x \left ( bx+a \right ) }\ln \left ( 1/2\,{\frac{2\,\sqrt{b{x}^{2}+ax}\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{2}b \right ){\frac{1}{\sqrt{b{x}^{3}+a{x}^{2}}}}{b}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(b*x^3+a*x^2)^(1/2),x)

[Out]

1/8*x^(1/2)*(4*b^(7/2)*x^3-2*b^(5/2)*x^2*a-6*b^(3/2)*x*a^2+3*(x*(b*x+a))^(1/2)*ln(1/2*(2*(b*x^2+a*x)^(1/2)*b^(
1/2)+2*b*x+a)/b^(1/2))*a^2*b)/(b*x^3+a*x^2)^(1/2)/b^(7/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{5}{2}}}{\sqrt{b x^{3} + a x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^(5/2)/sqrt(b*x^3 + a*x^2), x)

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Fricas [A]  time = 0.830067, size = 375, normalized size = 3.95 \begin{align*} \left [\frac{3 \, a^{2} \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}}{\left (2 \, b^{2} x - 3 \, a b\right )} \sqrt{x}}{8 \, b^{3} x}, -\frac{3 \, a^{2} \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}}{\left (2 \, b^{2} x - 3 \, a b\right )} \sqrt{x}}{4 \, b^{3} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(3*a^2*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)*(
2*b^2*x - 3*a*b)*sqrt(x))/(b^3*x), -1/4*(3*a^2*sqrt(-b)*x*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-b)/(b*x^(3/2))) - s
qrt(b*x^3 + a*x^2)*(2*b^2*x - 3*a*b)*sqrt(x))/(b^3*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{5}{2}}}{\sqrt{x^{2} \left (a + b x\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)/(b*x**3+a*x**2)**(1/2),x)

[Out]

Integral(x**(5/2)/sqrt(x**2*(a + b*x)), x)

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Giac [A]  time = 1.29259, size = 70, normalized size = 0.74 \begin{align*} \frac{1}{4} \, \sqrt{b x + a} \sqrt{x}{\left (\frac{2 \, x}{b} - \frac{3 \, a}{b^{2}}\right )} - \frac{3 \, a^{2} \log \left ({\left | -\sqrt{b} \sqrt{x} + \sqrt{b x + a} \right |}\right )}{4 \, b^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(b*x^3+a*x^2)^(1/2),x, algorithm="giac")

[Out]

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