3.1310 \(\int \frac {x^{3/2}}{\sqrt {a+b x^5}} \, dx\)

Optimal. Leaf size=32 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x^{5/2}}{\sqrt {a+b x^5}}\right )}{5 \sqrt {b}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {329, 275, 217, 206} \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x^{5/2}}{\sqrt {a+b x^5}}\right )}{5 \sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/Sqrt[a + b*x^5],x]

[Out]

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

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {x^{3/2}}{\sqrt {a+b x^5}} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {a+b x^{10}}} \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{5} \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^{5/2}\right )\\ &=\frac {2}{5} \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^{5/2}}{\sqrt {a+b x^5}}\right )\\ &=\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x^{5/2}}{\sqrt {a+b x^5}}\right )}{5 \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 32, normalized size = 1.00 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x^{5/2}}{\sqrt {a+b x^5}}\right )}{5 \sqrt {b}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/Sqrt[a + b*x^5],x]

[Out]

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

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fricas [A]  time = 1.89, size = 97, normalized size = 3.03 \[ \left [\frac {\log \left (-8 \, b^{2} x^{10} - 8 \, a b x^{5} - 4 \, {\left (2 \, b x^{7} + a x^{2}\right )} \sqrt {b x^{5} + a} \sqrt {b} \sqrt {x} - a^{2}\right )}{10 \, \sqrt {b}}, -\frac {\sqrt {-b} \arctan \left (\frac {2 \, \sqrt {b x^{5} + a} \sqrt {-b} x^{\frac {5}{2}}}{2 \, b x^{5} + a}\right )}{5 \, b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/10*log(-8*b^2*x^10 - 8*a*b*x^5 - 4*(2*b*x^7 + a*x^2)*sqrt(b*x^5 + a)*sqrt(b)*sqrt(x) - a^2)/sqrt(b), -1/5*s
qrt(-b)*arctan(2*sqrt(b*x^5 + a)*sqrt(-b)*x^(5/2)/(2*b*x^5 + a))/b]

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giac [A]  time = 0.18, size = 25, normalized size = 0.78 \[ -\frac {2 \, \log \left ({\left | -\sqrt {b} x^{\frac {5}{2}} + \sqrt {b x^{5} + a} \right |}\right )}{5 \, \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {3}{2}}}{\sqrt {b \,x^{5}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 2.27, size = 45, normalized size = 1.41 \[ -\frac {\log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x^{5} + a}}{x^{\frac {5}{2}}}}{\sqrt {b} + \frac {\sqrt {b x^{5} + a}}{x^{\frac {5}{2}}}}\right )}{5 \, \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {x^{3/2}}{\sqrt {b\,x^5+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(a + b*x^5)^(1/2),x)

[Out]

int(x^(3/2)/(a + b*x^5)^(1/2), x)

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sympy [A]  time = 2.32, size = 24, normalized size = 0.75 \[ \frac {2 \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {5}{2}}}{\sqrt {a}} \right )}}{5 \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*asinh(sqrt(b)*x**(5/2)/sqrt(a))/(5*sqrt(b))

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