3.2687 \(\int \frac {x^{-1+\frac {7 n}{2}}}{\sqrt {a+b x^n}} \, dx\)

Optimal. Leaf size=129 \[ -\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a+b x^n}}\right )}{8 b^{7/2} n}+\frac {5 a^2 x^{n/2} \sqrt {a+b x^n}}{8 b^3 n}-\frac {5 a x^{3 n/2} \sqrt {a+b x^n}}{12 b^2 n}+\frac {x^{5 n/2} \sqrt {a+b x^n}}{3 b n} \]

[Out]

-5/8*a^3*arctanh(x^(1/2*n)*b^(1/2)/(a+b*x^n)^(1/2))/b^(7/2)/n+5/8*a^2*x^(1/2*n)*(a+b*x^n)^(1/2)/b^3/n-5/12*a*x
^(3/2*n)*(a+b*x^n)^(1/2)/b^2/n+1/3*x^(5/2*n)*(a+b*x^n)^(1/2)/b/n

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Rubi [A]  time = 0.06, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {355, 288, 206} \[ \frac {5 a^2 x^{n/2} \sqrt {a+b x^n}}{8 b^3 n}-\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a+b x^n}}\right )}{8 b^{7/2} n}-\frac {5 a x^{3 n/2} \sqrt {a+b x^n}}{12 b^2 n}+\frac {x^{5 n/2} \sqrt {a+b x^n}}{3 b n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + (7*n)/2)/Sqrt[a + b*x^n],x]

[Out]

(5*a^2*x^(n/2)*Sqrt[a + b*x^n])/(8*b^3*n) - (5*a*x^((3*n)/2)*Sqrt[a + b*x^n])/(12*b^2*n) + (x^((5*n)/2)*Sqrt[a
+ b*x^n])/(3*b*n) - (5*a^3*ArcTanh[(Sqrt[b]*x^(n/2))/Sqrt[a + b*x^n]])/(8*b^(7/2)*n)

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 288

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 355

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]

Rubi steps

\begin {align*} \int \frac {x^{-1+\frac {7 n}{2}}}{\sqrt {a+b x^n}} \, dx &=\frac {\left (2 a^3\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (1-b x^2\right )^4} \, dx,x,\frac {x^{n/2}}{\sqrt {a+b x^n}}\right )}{n}\\ &=\frac {x^{5 n/2} \sqrt {a+b x^n}}{3 b n}-\frac {\left (5 a^3\right ) \operatorname {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 )}{3 b n}\\ &=-\frac {5 a x^{3 n/2} \sqrt {a+b x^n}}{12 b^2 n}+\frac {x^{5 n/2} \sqrt {a+b x^n}}{3 b n}+\frac {\left (5 a^3\right ) \operatorname {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 )}{4 b^2 n}\\ &=\frac {5 a^2 x^{n/2} \sqrt {a+b x^n}}{8 b^3 n}-\frac {5 a x^{3 n/2} \sqrt {a+b x^n}}{12 b^2 n}+\frac {x^{5 n/2} \sqrt {a+b x^n}}{3 b n}-\frac {\left (5 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^{n/2}}{\sqrt {a+b x^n}}\right )}{8 b^3 n}\\ &=\frac {5 a^2 x^{n/2} \sqrt {a+b x^n}}{8 b^3 n}-\frac {5 a x^{3 n/2} \sqrt {a+b x^n}}{12 b^2 n}+\frac {x^{5 n/2} \sqrt {a+b x^n}}{3 b n}-\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a+b x^n}}\right )}{8 b^{7/2} n}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 100, normalized size = 0.78 \[ \frac {\sqrt {a+b x^n} \left (\sqrt {b} x^{n/2} \left (15 a^2-10 a b x^n+8 b^2 x^{2 n}\right )-\frac {15 a^{5/2} \sinh ^{-1}\left (\frac {\sqrt {b} x^{n/2}}{\sqrt {a}}\right )}{\sqrt {\frac {b x^n}{a}+1}}\right )}{24 b^{7/2} n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + (7*n)/2)/Sqrt[a + b*x^n],x]

[Out]

(Sqrt[a + b*x^n]*(Sqrt[b]*x^(n/2)*(15*a^2 - 10*a*b*x^n + 8*b^2*x^(2*n)) - (15*a^(5/2)*ArcSinh[(Sqrt[b]*x^(n/2)
)/Sqrt[a]])/Sqrt[1 + (b*x^n)/a]))/(24*b^(7/2)*n)

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fricas [A]  time = 0.63, size = 175, normalized size = 1.36 \[ \left [\frac {15 \, a^{3} \sqrt {b} \log \left (2 \, \sqrt {b x^{n} + a} \sqrt {b} x^{\frac {1}{2} \, n} - 2 \, b x^{n} - a\right ) + 2 \, {\left (8 \, b^{3} x^{\frac {5}{2} \, n} - 10 \, a b^{2} x^{\frac {3}{2} \, n} + 15 \, a^{2} b x^{\frac {1}{2} \, n}\right )} \sqrt {b x^{n} + a}}{48 \, b^{4} n}, \frac {15 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x^{\frac {1}{2} \, n}}{\sqrt {b x^{n} + a}}\right ) + {\left (8 \, b^{3} x^{\frac {5}{2} \, n} - 10 \, a b^{2} x^{\frac {3}{2} \, n} + 15 \, a^{2} b x^{\frac {1}{2} \, n}\right )} \sqrt {b x^{n} + a}}{24 \, b^{4} n}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+7/2*n)/(a+b*x^n)^(1/2),x, algorithm="fricas")

[Out]

[1/48*(15*a^3*sqrt(b)*log(2*sqrt(b*x^n + a)*sqrt(b)*x^(1/2*n) - 2*b*x^n - a) + 2*(8*b^3*x^(5/2*n) - 10*a*b^2*x
^(3/2*n) + 15*a^2*b*x^(1/2*n))*sqrt(b*x^n + a))/(b^4*n), 1/24*(15*a^3*sqrt(-b)*arctan(sqrt(-b)*x^(1/2*n)/sqrt(
b*x^n + a)) + (8*b^3*x^(5/2*n) - 10*a*b^2*x^(3/2*n) + 15*a^2*b*x^(1/2*n))*sqrt(b*x^n + a))/(b^4*n)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {7}{2} \, n - 1}}{\sqrt {b x^{n} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+7/2*n)/(a+b*x^n)^(1/2),x, algorithm="giac")

[Out]

integrate(x^(7/2*n - 1)/sqrt(b*x^n + a), x)

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maple [A]  time = 0.11, size = 98, normalized size = 0.76 \[ -\frac {5 a^{3} \ln \left (\sqrt {b}\, {\mathrm e}^{\frac {n \ln \relax (x )}{2}}+\sqrt {b \,{\mathrm e}^{n \ln \relax (x )}+a}\right )}{8 b^{\frac {7}{2}} n}+\frac {\left (-10 a b \,{\mathrm e}^{n \ln \relax (x )}+8 b^{2} {\mathrm e}^{2 n \ln \relax (x )}+15 a^{2}\right ) \sqrt {b \,{\mathrm e}^{n \ln \relax (x )}+a}\, {\mathrm e}^{\frac {n \ln \relax (x )}{2}}}{24 b^{3} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1+7/2*n)/(b*x^n+a)^(1/2),x)

[Out]

1/24*exp(1/2*n*ln(x))*(8*exp(1/2*n*ln(x))^4*b^2-10*a*exp(1/2*n*ln(x))^2*b+15*a^2)*(b*exp(1/2*n*ln(x))^2+a)^(1/
2)/b^3/n-5/8*a^3/b^(7/2)/n*ln(b^(1/2)*exp(1/2*n*ln(x))+(b*exp(1/2*n*ln(x))^2+a)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {7}{2} \, n - 1}}{\sqrt {b x^{n} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+7/2*n)/(a+b*x^n)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^(7/2*n - 1)/sqrt(b*x^n + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{\frac {7\,n}{2}-1}}{\sqrt {a+b\,x^n}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^((7*n)/2 - 1)/(a + b*x^n)^(1/2),x)

[Out]

int(x^((7*n)/2 - 1)/(a + b*x^n)^(1/2), x)

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sympy [A]  time = 16.51, size = 148, normalized size = 1.15 \[ \frac {5 a^{\frac {5}{2}} x^{\frac {n}{2}}}{8 b^{3} n \sqrt {1 + \frac {b x^{n}}{a}}} + \frac {5 a^{\frac {3}{2}} x^{\frac {3 n}{2}}}{24 b^{2} n \sqrt {1 + \frac {b x^{n}}{a}}} - \frac {\sqrt {a} x^{\frac {5 n}{2}}}{12 b n \sqrt {1 + \frac {b x^{n}}{a}}} - \frac {5 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {n}{2}}}{\sqrt {a}} \right )}}{8 b^{\frac {7}{2}} n} + \frac {x^{\frac {7 n}{2}}}{3 \sqrt {a} n \sqrt {1 + \frac {b x^{n}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+7/2*n)/(a+b*x**n)**(1/2),x)

[Out]

5*a**(5/2)*x**(n/2)/(8*b**3*n*sqrt(1 + b*x**n/a)) + 5*a**(3/2)*x**(3*n/2)/(24*b**2*n*sqrt(1 + b*x**n/a)) - sqr
t(a)*x**(5*n/2)/(12*b*n*sqrt(1 + b*x**n/a)) - 5*a**3*asinh(sqrt(b)*x**(n/2)/sqrt(a))/(8*b**(7/2)*n) + x**(7*n/
2)/(3*sqrt(a)*n*sqrt(1 + b*x**n/a))

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