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

Optimal. Leaf size=36 \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} x^{5/2}}{\sqrt{a x^2+b x^5}}\right )}{3 \sqrt{b}} \]

[Out]

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

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Rubi [A]  time = 0.0475052, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2029, 206} \[ \frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} x^{5/2}}{\sqrt{a x^2+b x^5}}\right )}{3 \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/Sqrt[a*x^2 + b*x^5],x]

[Out]

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

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

Mathematica [A]  time = 0.0120326, size = 59, normalized size = 1.64 \[ \frac{2 x \sqrt{a+b x^3} \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a+b x^3}}\right )}{3 \sqrt{b} \sqrt{x^2 \left (a+b x^3\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.118, size = 480, normalized size = 13.3 \begin{align*} -4\,{\frac{{x}^{3/2} \left ( b{x}^{3}+a \right ) \left ( -1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{-{b}^{2}a} \right ) ^{2}}{\sqrt{b{x}^{5}+a{x}^{2}}{b}^{2}\sqrt{ \left ( b{x}^{3}+a \right ) x} \left ( i\sqrt{3}-3 \right ) }\sqrt{-{\frac{ \left ( i\sqrt{3}-3 \right ) xb}{ \left ( -1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{-{b}^{2}a} \right ) }}}\sqrt{{\frac{i\sqrt{3}\sqrt [3]{-{b}^{2}a}+2\,bx+\sqrt [3]{-{b}^{2}a}}{ \left ( 1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{-{b}^{2}a} \right ) }}}\sqrt{{\frac{i\sqrt{3}\sqrt [3]{-{b}^{2}a}-2\,bx-\sqrt [3]{-{b}^{2}a}}{ \left ( -1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{-{b}^{2}a} \right ) }}} \left ({\it EllipticF} \left ( \sqrt{-{\frac{ \left ( i\sqrt{3}-3 \right ) xb}{ \left ( -1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{-{b}^{2}a} \right ) }}},\sqrt{{\frac{ \left ( i\sqrt{3}+3 \right ) \left ( -1+i\sqrt{3} \right ) }{ \left ( i\sqrt{3}-3 \right ) \left ( 1+i\sqrt{3} \right ) }}} \right ) -{\it EllipticPi} \left ( \sqrt{-{\frac{ \left ( i\sqrt{3}-3 \right ) xb}{ \left ( -1+i\sqrt{3} \right ) \left ( -bx+\sqrt [3]{-{b}^{2}a} \right ) }}},{\frac{-1+i\sqrt{3}}{i\sqrt{3}-3}},\sqrt{{\frac{ \left ( i\sqrt{3}+3 \right ) \left ( -1+i\sqrt{3} \right ) }{ \left ( i\sqrt{3}-3 \right ) \left ( 1+i\sqrt{3} \right ) }}} \right ) \right ){\frac{1}{\sqrt{{\frac{x \left ( -bx+\sqrt [3]{-{b}^{2}a} \right ) \left ( i\sqrt{3}\sqrt [3]{-{b}^{2}a}+2\,bx+\sqrt [3]{-{b}^{2}a} \right ) \left ( i\sqrt{3}\sqrt [3]{-{b}^{2}a}-2\,bx-\sqrt [3]{-{b}^{2}a} \right ) }{{b}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*x^(3/2)*(b*x^3+a)*(-1+I*3^(1/2))*(-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*(-b*x+(-b^
2*a)^(1/3))^2*((I*3^(1/2)*(-b^2*a)^(1/3)+2*b*x+(-b^2*a)^(1/3))/(1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*((I*
3^(1/2)*(-b^2*a)^(1/3)-2*b*x-(-b^2*a)^(1/3))/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2)*(EllipticF((-(I*3^(1/
2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3
))^(1/2))-EllipticPi((-(I*3^(1/2)-3)*x*b/(-1+I*3^(1/2))/(-b*x+(-b^2*a)^(1/3)))^(1/2),(-1+I*3^(1/2))/(I*3^(1/2)
-3),((I*3^(1/2)+3)*(-1+I*3^(1/2))/(1+I*3^(1/2))/(I*3^(1/2)-3))^(1/2)))/(b*x^5+a*x^2)^(1/2)/b^2/((b*x^3+a)*x)^(
1/2)/(I*3^(1/2)-3)/(1/b^2*x*(-b*x+(-b^2*a)^(1/3))*(I*3^(1/2)*(-b^2*a)^(1/3)+2*b*x+(-b^2*a)^(1/3))*(I*3^(1/2)*(
-b^2*a)^(1/3)-2*b*x-(-b^2*a)^(1/3)))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.33068, size = 247, normalized size = 6.86 \begin{align*} \left [\frac{\log \left (-8 \, b^{2} x^{6} - 8 \, a b x^{3} - 4 \, \sqrt{b x^{5} + a x^{2}}{\left (2 \, b x^{3} + a\right )} \sqrt{b} \sqrt{x} - a^{2}\right )}{6 \, \sqrt{b}}, -\frac{\sqrt{-b} \arctan \left (\frac{2 \, \sqrt{b x^{5} + a x^{2}} \sqrt{-b} \sqrt{x}}{2 \, b x^{3} + a}\right )}{3 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*log(-8*b^2*x^6 - 8*a*b*x^3 - 4*sqrt(b*x^5 + a*x^2)*(2*b*x^3 + a)*sqrt(b)*sqrt(x) - a^2)/sqrt(b), -1/3*sqr
t(-b)*arctan(2*sqrt(b*x^5 + a*x^2)*sqrt(-b)*sqrt(x)/(2*b*x^3 + a))/b]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28855, size = 55, normalized size = 1.53 \begin{align*} -\frac{2 \, \arctan \left (\frac{\sqrt{b + \frac{a}{x^{3}}}}{\sqrt{-b}}\right )}{3 \, \sqrt{-b}} + \frac{2 \, \arctan \left (\frac{\sqrt{b}}{\sqrt{-b}}\right )}{3 \, \sqrt{-b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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