\(\int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx\) [2782]

Optimal result

Integrand size = 23, antiderivative size = 137 \[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=-\frac {3 a x^{-2 n} (c x)^{5 n/2} \sqrt {a+b x^n}}{4 b^2 c n}+\frac {x^{-n} (c x)^{5 n/2} \sqrt {a+b x^n}}{2 b c n}+\frac {3 a^2 x^{-5 n/2} (c x)^{5 n/2} \text {arctanh}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a+b x^n}}\right )}{4 b^{5/2} c n} \]

[Out]

3/4*a^2*(c*x)^(5/2*n)*arctanh(x^(1/2*n)*b^(1/2)/(a+b*x^n)^(1/2))/b^(5/2)/c/n/(x^(5/2*n))-3/4*a*(c*x)^(5/2*n)*(
a+b*x^n)^(1/2)/b^2/c/n/(x^(2*n))+1/2*(c*x)^(5/2*n)*(a+b*x^n)^(1/2)/b/c/n/(x^n)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {364, 362, 294, 212} \[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=\frac {3 a^2 x^{-5 n/2} (c x)^{5 n/2} \text {arctanh}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a+b x^n}}\right )}{4 b^{5/2} c n}-\frac {3 a x^{-2 n} (c x)^{5 n/2} \sqrt {a+b x^n}}{4 b^2 c n}+\frac {x^{-n} (c x)^{5 n/2} \sqrt {a+b x^n}}{2 b c n} \]

[In]

Int[(c*x)^(-1 + (5*n)/2)/Sqrt[a + b*x^n],x]

[Out]

(-3*a*(c*x)^((5*n)/2)*Sqrt[a + b*x^n])/(4*b^2*c*n*x^(2*n)) + ((c*x)^((5*n)/2)*Sqrt[a + b*x^n])/(2*b*c*n*x^n) +
(3*a^2*(c*x)^((5*n)/2)*ArcTanh[(Sqrt[b]*x^(n/2))/Sqrt[a + b*x^n]])/(4*b^(5/2)*c*n*x^((5*n)/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 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 362

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

Rule 364

Int[((c_)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[c^IntPart[m]*((c*x)^FracPart[m]/x^FracPa
rt[m]), Int[x^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && IntegerQ[p + Simplify[(m + 1)/n]] && LtQ
[-1, p, 0]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.88 \[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=\frac {a x^{-5 n/2} (c x)^{5 n/2} \sqrt {1+\frac {b x^n}{a}} \left (\sqrt {b} x^{n/2} \left (-3 a+2 b x^n\right ) \sqrt {1+\frac {b x^n}{a}}+3 a^{3/2} \text {arcsinh}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a}}\right )\right )}{4 b^{5/2} c n \sqrt {a+b x^n}} \]

[In]

Integrate[(c*x)^(-1 + (5*n)/2)/Sqrt[a + b*x^n],x]

[Out]

(a*(c*x)^((5*n)/2)*Sqrt[1 + (b*x^n)/a]*(Sqrt[b]*x^(n/2)*(-3*a + 2*b*x^n)*Sqrt[1 + (b*x^n)/a] + 3*a^(3/2)*ArcSi
nh[(Sqrt[b]*x^(n/2))/Sqrt[a]]))/(4*b^(5/2)*c*n*x^((5*n)/2)*Sqrt[a + b*x^n])

Maple [F]

\[\int \frac {\left (c x \right )^{-1+\frac {5 n}{2}}}{\sqrt {a +b \,x^{n}}}d x\]

[In]

int((c*x)^(-1+5/2*n)/(a+b*x^n)^(1/2),x)

[Out]

int((c*x)^(-1+5/2*n)/(a+b*x^n)^(1/2),x)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.40 \[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=\left [\frac {3 \, a^{2} \sqrt {b} c^{\frac {5}{2} \, n - 1} \log \left (-2 \, \sqrt {b x^{n} + a} \sqrt {b} x^{\frac {1}{2} \, n} - 2 \, b x^{n} - a\right ) + 2 \, {\left (2 \, b^{2} c^{\frac {5}{2} \, n - 1} x^{\frac {3}{2} \, n} - 3 \, a b c^{\frac {5}{2} \, n - 1} x^{\frac {1}{2} \, n}\right )} \sqrt {b x^{n} + a}}{8 \, b^{3} n}, -\frac {3 \, a^{2} \sqrt {-b} c^{\frac {5}{2} \, n - 1} \arctan \left (\frac {\sqrt {-b} x^{\frac {1}{2} \, n}}{\sqrt {b x^{n} + a}}\right ) - {\left (2 \, b^{2} c^{\frac {5}{2} \, n - 1} x^{\frac {3}{2} \, n} - 3 \, a b c^{\frac {5}{2} \, n - 1} x^{\frac {1}{2} \, n}\right )} \sqrt {b x^{n} + a}}{4 \, b^{3} n}\right ] \]

[In]

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

[Out]

[1/8*(3*a^2*sqrt(b)*c^(5/2*n - 1)*log(-2*sqrt(b*x^n + a)*sqrt(b)*x^(1/2*n) - 2*b*x^n - a) + 2*(2*b^2*c^(5/2*n
- 1)*x^(3/2*n) - 3*a*b*c^(5/2*n - 1)*x^(1/2*n))*sqrt(b*x^n + a))/(b^3*n), -1/4*(3*a^2*sqrt(-b)*c^(5/2*n - 1)*a
rctan(sqrt(-b)*x^(1/2*n)/sqrt(b*x^n + a)) - (2*b^2*c^(5/2*n - 1)*x^(3/2*n) - 3*a*b*c^(5/2*n - 1)*x^(1/2*n))*sq
rt(b*x^n + a))/(b^3*n)]

Sympy [A] (verification not implemented)

Time = 2.78 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.09 \[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=- \frac {3 a^{\frac {3}{2}} c^{\frac {5 n}{2} - 1} x^{\frac {n}{2}}}{4 b^{2} n \sqrt {1 + \frac {b x^{n}}{a}}} - \frac {\sqrt {a} c^{\frac {5 n}{2} - 1} x^{\frac {3 n}{2}}}{4 b n \sqrt {1 + \frac {b x^{n}}{a}}} + \frac {3 a^{2} c^{\frac {5 n}{2} - 1} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {n}{2}}}{\sqrt {a}} \right )}}{4 b^{\frac {5}{2}} n} + \frac {c^{\frac {5 n}{2} - 1} x^{\frac {5 n}{2}}}{2 \sqrt {a} n \sqrt {1 + \frac {b x^{n}}{a}}} \]

[In]

integrate((c*x)**(-1+5/2*n)/(a+b*x**n)**(1/2),x)

[Out]

-3*a**(3/2)*c**(5*n/2 - 1)*x**(n/2)/(4*b**2*n*sqrt(1 + b*x**n/a)) - sqrt(a)*c**(5*n/2 - 1)*x**(3*n/2)/(4*b*n*s
qrt(1 + b*x**n/a)) + 3*a**2*c**(5*n/2 - 1)*asinh(sqrt(b)*x**(n/2)/sqrt(a))/(4*b**(5/2)*n) + c**(5*n/2 - 1)*x**
(5*n/2)/(2*sqrt(a)*n*sqrt(1 + b*x**n/a))

Maxima [F]

\[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=\int { \frac {\left (c x\right )^{\frac {5}{2} \, n - 1}}{\sqrt {b x^{n} + a}} \,d x } \]

[In]

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

[Out]

integrate((c*x)^(5/2*n - 1)/sqrt(b*x^n + a), x)

Giac [F]

\[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=\int { \frac {\left (c x\right )^{\frac {5}{2} \, n - 1}}{\sqrt {b x^{n} + a}} \,d x } \]

[In]

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

[Out]

integrate((c*x)^(5/2*n - 1)/sqrt(b*x^n + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^{-1+\frac {5 n}{2}}}{\sqrt {a+b x^n}} \, dx=\int \frac {{\left (c\,x\right )}^{\frac {5\,n}{2}-1}}{\sqrt {a+b\,x^n}} \,d x \]

[In]

int((c*x)^((5*n)/2 - 1)/(a + b*x^n)^(1/2),x)

[Out]

int((c*x)^((5*n)/2 - 1)/(a + b*x^n)^(1/2), x)