3.3.100 \(\int \frac {x^{3/2}}{\sqrt {a x^2+b x^5}} \, dx\) [300]

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]

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

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Rubi [A]
time = 0.03, antiderivative size = 36, normalized size of antiderivative = 1.00, 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 = {2054, 212} \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x^{5/2}}{\sqrt {a x^2+b x^5}}\right )}{3 \sqrt {b}} \end {gather*}

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 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 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 {x^{3/2}}{\sqrt {a x^2+b x^5}} \, dx &=\frac {2}{3} \text {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*}

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Mathematica [A]
time = 0.01, size = 59, normalized size = 1.64 \begin {gather*} \frac {2 x \sqrt {a+b x^3} \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {b} x^{3/2}}\right )}{3 \sqrt {b} \sqrt {x^2 \left (a+b x^3\right )}} \end {gather*}

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

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.52, size = 480, normalized size = 13.33

method	result	size
default	\(-\frac {4 x^{\frac {3}{2}} \left (b \,x^{3}+a \right ) \left (-1+i \sqrt {3}\right ) \sqrt {-\frac {\left (i \sqrt {3}-3\right ) x b}{\left (-1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )^{2} \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+2 b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \sqrt {\frac {i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-2 b x -\left (-a \,b^{2}\right )^{\frac {1}{3}}}{\left (-1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}\, \left (\EllipticF \left (\sqrt {-\frac {\left (i \sqrt {3}-3\right ) x b}{\left (-1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}, \sqrt {\frac {\left (i \sqrt {3}+3\right ) \left (-1+i \sqrt {3}\right )}{\left (1+i \sqrt {3}\right ) \left (i \sqrt {3}-3\right )}}\right )-\EllipticPi \left (\sqrt {-\frac {\left (i \sqrt {3}-3\right ) x b}{\left (-1+i \sqrt {3}\right ) \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}}, \frac {-1+i \sqrt {3}}{i \sqrt {3}-3}, \sqrt {\frac {\left (i \sqrt {3}+3\right ) \left (-1+i \sqrt {3}\right )}{\left (1+i \sqrt {3}\right ) \left (i \sqrt {3}-3\right )}}\right )\right )}{\sqrt {b \,x^{5}+a \,x^{2}}\, b^{2} \sqrt {x \left (b \,x^{3}+a \right )}\, \left (i \sqrt {3}-3\right ) \sqrt {\frac {x \left (-b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}+2 b x +\left (-a \,b^{2}\right )^{\frac {1}{3}}\right ) \left (i \sqrt {3}\, \left (-a \,b^{2}\right )^{\frac {1}{3}}-2 b x -\left (-a \,b^{2}\right )^{\frac {1}{3}}\right )}{b^{2}}}}\)	\(480\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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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(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 = 2.76, size = 101, normalized size = 2.81 \begin {gather*} \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 {gather*}

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

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 = 0.53, size = 40, normalized size = 1.11 \begin {gather*} \frac {\log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{3 \, \sqrt {b}} - \frac {2 \, \log \left ({\left | -\sqrt {b} x^{\frac {3}{2}} + \sqrt {b x^{3} + a} \right |}\right )}{3 \, \sqrt {b} \mathrm {sgn}\left (x\right )} \end {gather*}

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]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x^{3/2}}{\sqrt {b\,x^5+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^5)^(1/2),x)

[Out]

int(x^(3/2)/(a*x^2 + b*x^5)^(1/2), x)

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