\(\int \frac {x^{23/2}}{\sqrt {a+b x^5}} \, dx\) [1308]

Optimal result

Integrand size = 17, antiderivative size = 83 \[ \int \frac {x^{23/2}}{\sqrt {a+b x^5}} \, dx=-\frac {3 a x^{5/2} \sqrt {a+b x^5}}{20 b^2}+\frac {x^{15/2} \sqrt {a+b x^5}}{10 b}+\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {b} x^{5/2}}{\sqrt {a+b x^5}}\right )}{20 b^{5/2}} \]

[Out]

3/20*a^2*arctanh(x^(5/2)*b^(1/2)/(b*x^5+a)^(1/2))/b^(5/2)-3/20*a*x^(5/2)*(b*x^5+a)^(1/2)/b^2+1/10*x^(15/2)*(b*
x^5+a)^(1/2)/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {327, 335, 281, 223, 212} \[ \int \frac {x^{23/2}}{\sqrt {a+b x^5}} \, dx=\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {b} x^{5/2}}{\sqrt {a+b x^5}}\right )}{20 b^{5/2}}-\frac {3 a x^{5/2} \sqrt {a+b x^5}}{20 b^2}+\frac {x^{15/2} \sqrt {a+b x^5}}{10 b} \]

[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 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 223

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 281

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 327

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 335

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*} \text {integral}& = \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 ) \text {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 ) \text {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 ) \text {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] (verified)

Time = 1.82 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.87 \[ \int \frac {x^{23/2}}{\sqrt {a+b x^5}} \, dx=\frac {\sqrt {a+b x^5} \left (-3 a x^{5/2}+2 b x^{15/2}\right )}{20 b^2}+\frac {3 a^2 \log \left (\sqrt {b} x^{5/2}+\sqrt {a+b x^5}\right )}{20 b^{5/2}} \]

[In]

Integrate[x^(23/2)/Sqrt[a + b*x^5],x]

[Out]

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

Maple [F]

\[\int \frac {x^{\frac {23}{2}}}{\sqrt {b \,x^{5}+a}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.65 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.07 \[ \int \frac {x^{23/2}}{\sqrt {a+b x^5}} \, dx=\left [\frac {3 \, a^{2} \sqrt {b} \log \left (-8 \, b^{2} x^{10} - 8 \, a b x^{5} - 4 \, {\left (2 \, b x^{7} + a x^{2}\right )} \sqrt {b x^{5} + a} \sqrt {b} \sqrt {x} - a^{2}\right ) + 4 \, {\left (2 \, b^{2} x^{7} - 3 \, a b x^{2}\right )} \sqrt {b x^{5} + a} \sqrt {x}}{80 \, b^{3}}, -\frac {3 \, a^{2} \sqrt {-b} \arctan \left (\frac {2 \, \sqrt {b x^{5} + a} \sqrt {-b} x^{\frac {5}{2}}}{2 \, b x^{5} + a}\right ) - 2 \, {\left (2 \, b^{2} x^{7} - 3 \, a b x^{2}\right )} \sqrt {b x^{5} + a} \sqrt {x}}{40 \, b^{3}}\right ] \]

[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]

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{23/2}}{\sqrt {a+b x^5}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (61) = 122\).

Time = 0.30 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.49 \[ \int \frac {x^{23/2}}{\sqrt {a+b x^5}} \, dx=-\frac {3 \, a^{2} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x^{5} + a}}{x^{\frac {5}{2}}}}{\sqrt {b} + \frac {\sqrt {b x^{5} + a}}{x^{\frac {5}{2}}}}\right )}{40 \, b^{\frac {5}{2}}} + \frac {\frac {5 \, \sqrt {b x^{5} + a} a^{2} b}{x^{\frac {5}{2}}} - \frac {3 \, {\left (b x^{5} + a\right )}^{\frac {3}{2}} a^{2}}{x^{\frac {15}{2}}}}{20 \, {\left (b^{4} - \frac {2 \, {\left (b x^{5} + a\right )} b^{3}}{x^{5}} + \frac {{\left (b x^{5} + a\right )}^{2} b^{2}}{x^{10}}\right )}} \]

[In]

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

[Out]

-3/40*a^2*log(-(sqrt(b) - sqrt(b*x^5 + a)/x^(5/2))/(sqrt(b) + sqrt(b*x^5 + a)/x^(5/2)))/b^(5/2) + 1/20*(5*sqrt
(b*x^5 + a)*a^2*b/x^(5/2) - 3*(b*x^5 + a)^(3/2)*a^2/x^(15/2))/(b^4 - 2*(b*x^5 + a)*b^3/x^5 + (b*x^5 + a)^2*b^2
/x^10)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.70 \[ \int \frac {x^{23/2}}{\sqrt {a+b x^5}} \, dx=\frac {1}{20} \, \sqrt {b x^{5} + a} {\left (\frac {2 \, x^{5}}{b} - \frac {3 \, a}{b^{2}}\right )} x^{\frac {5}{2}} - \frac {3 \, a^{2} \log \left ({\left | -\sqrt {b} x^{\frac {5}{2}} + \sqrt {b x^{5} + a} \right |}\right )}{20 \, b^{\frac {5}{2}}} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{23/2}}{\sqrt {a+b x^5}} \, dx=\int \frac {x^{23/2}}{\sqrt {b\,x^5+a}} \,d x \]

[In]

int(x^(23/2)/(a + b*x^5)^(1/2),x)

[Out]

int(x^(23/2)/(a + b*x^5)^(1/2), x)