3.1308 \(\int \frac{x^{23/2}}{\sqrt{a+b x^5}} \, dx\)

Optimal. Leaf size=83 \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} x^{5/2}}{\sqrt{a+b x^5}}\right )}{20 b^{5/2}}-\frac{3 a x^{5/2} \sqrt{a+b x^5}}{20 b^2}+\frac{x^{15/2} \sqrt{a+b x^5}}{10 b} \]

[Out]

(-3*a*x^(5/2)*Sqrt[a + b*x^5])/(20*b^2) + (x^(15/2)*Sqrt[a + b*x^5])/(10*b) + (3*a^2*ArcTanh[(Sqrt[b]*x^(5/2))
/Sqrt[a + b*x^5]])/(20*b^(5/2))

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Rubi [A]  time = 0.0455901, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {321, 329, 275, 217, 206} \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} x^{5/2}}{\sqrt{a+b x^5}}\right )}{20 b^{5/2}}-\frac{3 a x^{5/2} \sqrt{a+b x^5}}{20 b^2}+\frac{x^{15/2} \sqrt{a+b x^5}}{10 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(23/2)/Sqrt[a + b*x^5],x]

[Out]

(-3*a*x^(5/2)*Sqrt[a + b*x^5])/(20*b^2) + (x^(15/2)*Sqrt[a + b*x^5])/(10*b) + (3*a^2*ArcTanh[(Sqrt[b]*x^(5/2))
/Sqrt[a + b*x^5]])/(20*b^(5/2))

Rule 321

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

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]

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 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 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^{23/2}}{\sqrt{a+b x^5}} \, dx &=\frac{x^{15/2} \sqrt{a+b x^5}}{10 b}-\frac{(3 a) \int \frac{x^{13/2}}{\sqrt{a+b x^5}} \, dx}{4 b}\\ &=-\frac{3 a x^{5/2} \sqrt{a+b x^5}}{20 b^2}+\frac{x^{15/2} \sqrt{a+b x^5}}{10 b}+\frac{\left (3 a^2\right ) \int \frac{x^{3/2}}{\sqrt{a+b x^5}} \, dx}{8 b^2}\\ &=-\frac{3 a x^{5/2} \sqrt{a+b x^5}}{20 b^2}+\frac{x^{15/2} \sqrt{a+b x^5}}{10 b}+\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{a+b x^{10}}} \, dx,x,\sqrt{x}\right )}{4 b^2}\\ &=-\frac{3 a x^{5/2} \sqrt{a+b x^5}}{20 b^2}+\frac{x^{15/2} \sqrt{a+b x^5}}{10 b}+\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,x^{5/2}\right )}{20 b^2}\\ &=-\frac{3 a x^{5/2} \sqrt{a+b x^5}}{20 b^2}+\frac{x^{15/2} \sqrt{a+b x^5}}{10 b}+\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^{5/2}}{\sqrt{a+b x^5}}\right )}{20 b^2}\\ &=-\frac{3 a x^{5/2} \sqrt{a+b x^5}}{20 b^2}+\frac{x^{15/2} \sqrt{a+b x^5}}{10 b}+\frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} x^{5/2}}{\sqrt{a+b x^5}}\right )}{20 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0299345, size = 70, normalized size = 0.84 \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} x^{5/2}}{\sqrt{a+b x^5}}\right )+\sqrt{b} x^{5/2} \sqrt{a+b x^5} \left (2 b x^5-3 a\right )}{20 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(23/2)/Sqrt[a + b*x^5],x]

[Out]

(Sqrt[b]*x^(5/2)*Sqrt[a + b*x^5]*(-3*a + 2*b*x^5) + 3*a^2*ArcTanh[(Sqrt[b]*x^(5/2))/Sqrt[a + b*x^5]])/(20*b^(5
/2))

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Maple [F]  time = 0.057, size = 0, normalized size = 0. \begin{align*} \int{{x}^{{\frac{23}{2}}}{\frac{1}{\sqrt{b{x}^{5}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(23/2)/(b*x^5+a)^(1/2),x)

[Out]

int(x^(23/2)/(b*x^5+a)^(1/2),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(23/2)/(b*x^5+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 6.18219, size = 416, normalized size = 5.01 \begin{align*} \left [\frac{3 \, a^{2} \sqrt{b} \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 ) + 4 \,{\left (2 \, b^{2} x^{7} - 3 \, a b x^{2}\right )} \sqrt{b x^{5} + a} \sqrt{x}}{80 \, b^{3}}, -\frac{3 \, a^{2} \sqrt{-b} \arctan \left (\frac{2 \, \sqrt{b x^{5} + a} \sqrt{-b} x^{\frac{5}{2}}}{2 \, b x^{5} + a}\right ) - 2 \,{\left (2 \, b^{2} x^{7} - 3 \, a b x^{2}\right )} \sqrt{b x^{5} + a} \sqrt{x}}{40 \, b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(23/2)/(b*x^5+a)^(1/2),x, algorithm="fricas")

[Out]

[1/80*(3*a^2*sqrt(b)*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)
+ 4*(2*b^2*x^7 - 3*a*b*x^2)*sqrt(b*x^5 + a)*sqrt(x))/b^3, -1/40*(3*a^2*sqrt(-b)*arctan(2*sqrt(b*x^5 + a)*sqrt(
-b)*x^(5/2)/(2*b*x^5 + a)) - 2*(2*b^2*x^7 - 3*a*b*x^2)*sqrt(b*x^5 + a)*sqrt(x))/b^3]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(23/2)/(b*x**5+a)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.45586, size = 78, normalized size = 0.94 \begin{align*} \frac{1}{20} \, \sqrt{b x^{5} + a}{\left (\frac{2 \, x^{5}}{b} - \frac{3 \, a}{b^{2}}\right )} x^{\frac{5}{2}} - \frac{3 \, a^{2} \log \left ({\left | -\sqrt{b} x^{\frac{5}{2}} + \sqrt{b x^{5} + a} \right |}\right )}{20 \, b^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(23/2)/(b*x^5+a)^(1/2),x, algorithm="giac")

[Out]

1/20*sqrt(b*x^5 + a)*(2*x^5/b - 3*a/b^2)*x^(5/2) - 3/20*a^2*log(abs(-sqrt(b)*x^(5/2) + sqrt(b*x^5 + a)))/b^(5/
2)