3.1309 \(\int \frac {x^{13/2}}{\sqrt {a+b x^5}} \, dx\)

Optimal. Leaf size=57 \[ \frac {x^{5/2} \sqrt {a+b x^5}}{5 b}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^{5/2}}{\sqrt {a+b x^5}}\right )}{5 b^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ \frac {x^{5/2} \sqrt {a+b x^5}}{5 b}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^{5/2}}{\sqrt {a+b x^5}}\right )}{5 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(13/2)/Sqrt[a + b*x^5],x]

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 57, normalized size = 1.00 \[ \frac {x^{5/2} \sqrt {a+b x^5}}{5 b}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^{5/2}}{\sqrt {a+b x^5}}\right )}{5 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(13/2)/Sqrt[a + b*x^5],x]

[Out]

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

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fricas [A]  time = 1.97, size = 136, normalized size = 2.39 \[ \left [\frac {4 \, \sqrt {b x^{5} + a} b x^{\frac {5}{2}} + a \sqrt {b} \log \left (-8 \, b^{2} x^{10} - 8 \, a b x^{5} + 4 \, {\left (2 \, b x^{7} + a x^{2}\right )} \sqrt {b x^{5} + a} \sqrt {b} \sqrt {x} - a^{2}\right )}{20 \, b^{2}}, \frac {2 \, \sqrt {b x^{5} + a} b x^{\frac {5}{2}} + a \sqrt {-b} \arctan \left (\frac {2 \, \sqrt {b x^{5} + a} \sqrt {-b} x^{\frac {5}{2}}}{2 \, b x^{5} + a}\right )}{10 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)/(b*x^5+a)^(1/2),x, algorithm="fricas")

[Out]

[1/20*(4*sqrt(b*x^5 + a)*b*x^(5/2) + a*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))/b^2, 1/10*(2*sqrt(b*x^5 + a)*b*x^(5/2) + a*sqrt(-b)*arctan(2*sqrt(b*x^5 + a)*sqrt(-
b)*x^(5/2)/(2*b*x^5 + a)))/b^2]

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giac [A]  time = 0.29, size = 44, normalized size = 0.77 \[ \frac {\sqrt {b x^{5} + a} x^{\frac {5}{2}}}{5 \, b} + \frac {a \log \left ({\left | -\sqrt {b} x^{\frac {5}{2}} + \sqrt {b x^{5} + a} \right |}\right )}{5 \, b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)/(b*x^5+a)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {13}{2}}}{\sqrt {b \,x^{5}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(13/2)/(b*x^5+a)^(1/2),x)

[Out]

int(x^(13/2)/(b*x^5+a)^(1/2),x)

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maxima [A]  time = 2.35, size = 81, normalized size = 1.42 \[ \frac {a \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x^{5} + a}}{x^{\frac {5}{2}}}}{\sqrt {b} + \frac {\sqrt {b x^{5} + a}}{x^{\frac {5}{2}}}}\right )}{10 \, b^{\frac {3}{2}}} - \frac {\sqrt {b x^{5} + a} a}{5 \, {\left (b^{2} - \frac {{\left (b x^{5} + a\right )} b}{x^{5}}\right )} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)/(b*x^5+a)^(1/2),x, algorithm="maxima")

[Out]

1/10*a*log(-(sqrt(b) - sqrt(b*x^5 + a)/x^(5/2))/(sqrt(b) + sqrt(b*x^5 + a)/x^(5/2)))/b^(3/2) - 1/5*sqrt(b*x^5
+ a)*a/((b^2 - (b*x^5 + a)*b/x^5)*x^(5/2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^{13/2}}{\sqrt {b\,x^5+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(13/2)/(a + b*x^5)^(1/2),x)

[Out]

int(x^(13/2)/(a + b*x^5)^(1/2), x)

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sympy [A]  time = 165.09, size = 49, normalized size = 0.86 \[ \frac {\sqrt {a} x^{\frac {5}{2}} \sqrt {1 + \frac {b x^{5}}{a}}}{5 b} - \frac {a \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {5}{2}}}{\sqrt {a}} \right )}}{5 b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(13/2)/(b*x**5+a)**(1/2),x)

[Out]

sqrt(a)*x**(5/2)*sqrt(1 + b*x**5/a)/(5*b) - a*asinh(sqrt(b)*x**(5/2)/sqrt(a))/(5*b**(3/2))

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