3.7.69 \(\int \frac {1}{x (-a+b (c x)^n)^{5/2}} \, dx\) [669]

Optimal. Leaf size=81 \[ -\frac {2}{3 a n \left (-a+b (c x)^n\right )^{3/2}}+\frac {2}{a^2 n \sqrt {-a+b (c x)^n}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt {-a+b (c x)^n}}{\sqrt {a}}\right )}{a^{5/2} n} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {374, 12, 272, 53, 65, 211} \begin {gather*} \frac {2 \text {ArcTan}\left (\frac {\sqrt {b (c x)^n-a}}{\sqrt {a}}\right )}{a^{5/2} n}+\frac {2}{a^2 n \sqrt {b (c x)^n-a}}-\frac {2}{3 a n \left (b (c x)^n-a\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(-a + b*(c*x)^n)^(5/2)),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 272

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

Rule 374

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_)*(x_))^(n_))^(p_.), x_Symbol] :> Dist[1/c, Subst[Int[(d*(x/c))^m*(a
+ b*x^n)^p, x], x, c*x], x] /; FreeQ[{a, b, c, d, m, n, p}, x]

Rubi steps

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

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Mathematica [A]
time = 0.19, size = 70, normalized size = 0.86 \begin {gather*} \frac {2 \left (\frac {\sqrt {a} \left (-4 a+3 b (c x)^n\right )}{\left (-a+b (c x)^n\right )^{3/2}}+3 \tan ^{-1}\left (\frac {\sqrt {-a+b (c x)^n}}{\sqrt {a}}\right )\right )}{3 a^{5/2} n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.43, size = 65, normalized size = 0.80

method	result	size
derivativedivides	\(\frac {\frac {2 \arctan \left (\frac {\sqrt {-a +b \left (c x \right )^{n}}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}}-\frac {2}{3 a \left (-a +b \left (c x \right )^{n}\right )^{\frac {3}{2}}}+\frac {2}{a^{2} \sqrt {-a +b \left (c x \right )^{n}}}}{n}\)	\(65\)
default	\(\frac {\frac {2 \arctan \left (\frac {\sqrt {-a +b \left (c x \right )^{n}}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}}-\frac {2}{3 a \left (-a +b \left (c x \right )^{n}\right )^{\frac {3}{2}}}+\frac {2}{a^{2} \sqrt {-a +b \left (c x \right )^{n}}}}{n}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 277, normalized size = 3.42 \begin {gather*} \left [-\frac {3 \, {\left (2 \, \left (c x\right )^{n} \sqrt {-a} a b - \left (c x\right )^{2 \, n} \sqrt {-a} b^{2} - \sqrt {-a} a^{2}\right )} \log \left (\frac {\left (c x\right )^{n} b - 2 \, \sqrt {\left (c x\right )^{n} b - a} \sqrt {-a} - 2 \, a}{\left (c x\right )^{n}}\right ) + 2 \, {\left (3 \, \left (c x\right )^{n} a b - 4 \, a^{2}\right )} \sqrt {\left (c x\right )^{n} b - a}}{3 \, {\left (2 \, \left (c x\right )^{n} a^{4} b n - \left (c x\right )^{2 \, n} a^{3} b^{2} n - a^{5} n\right )}}, \frac {2 \, {\left (3 \, {\left (2 \, \left (c x\right )^{n} a^{\frac {3}{2}} b - \left (c x\right )^{2 \, n} \sqrt {a} b^{2} - a^{\frac {5}{2}}\right )} \arctan \left (\frac {\sqrt {\left (c x\right )^{n} b - a}}{\sqrt {a}}\right ) - {\left (3 \, \left (c x\right )^{n} a b - 4 \, a^{2}\right )} \sqrt {\left (c x\right )^{n} b - a}\right )}}{3 \, {\left (2 \, \left (c x\right )^{n} a^{4} b n - \left (c x\right )^{2 \, n} a^{3} b^{2} n - a^{5} n\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 7.00, size = 63, normalized size = 0.78 \begin {gather*} - \frac {2}{3 a n \left (- a + b \left (c x\right )^{n}\right )^{\frac {3}{2}}} + \frac {2}{a^{2} n \sqrt {- a + b \left (c x\right )^{n}}} + \frac {2 \operatorname {atan}{\left (\frac {\sqrt {- a + b \left (c x\right )^{n}}}{\sqrt {a}} \right )}}{a^{\frac {5}{2}} n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(3*a*n*(-a + b*(c*x)**n)**(3/2)) + 2/(a**2*n*sqrt(-a + b*(c*x)**n)) + 2*atan(sqrt(-a + b*(c*x)**n)/sqrt(a))
/(a**(5/2)*n)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x\,{\left (b\,{\left (c\,x\right )}^n-a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(b*(c*x)^n - a)^(5/2)),x)

[Out]

int(1/(x*(b*(c*x)^n - a)^(5/2)), x)

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