3.6.27 \(\int \frac {(a^2+2 a b x^n+b^2 x^{2 n})^{3/2}}{x^2} \, dx\) [527]

Optimal. Leaf size=212 \[ -\frac {a^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{x \left (a+b x^n\right )}-\frac {3 a^2 b^2 x^{-1+n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1-n) \left (a b+b^2 x^n\right )}-\frac {3 a b^3 x^{-1+2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1-2 n) \left (a b+b^2 x^n\right )}-\frac {b^4 x^{-1+3 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1-3 n) \left (a b+b^2 x^n\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1369, 276} \begin {gather*} -\frac {3 a^2 b^2 x^{n-1} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1-n) \left (a b+b^2 x^n\right )}-\frac {b^4 x^{3 n-1} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1-3 n) \left (a b+b^2 x^n\right )}-\frac {3 a b^3 x^{2 n-1} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(1-2 n) \left (a b+b^2 x^n\right )}-\frac {a^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{x \left (a+b x^n\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2)/x^2,x]

[Out]

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

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 1369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.08, size = 124, normalized size = 0.58 \begin {gather*} \frac {\sqrt {\left (a+b x^n\right )^2} \left (a^3 \left (1-6 n+11 n^2-6 n^3\right )+3 a^2 b \left (1-5 n+6 n^2\right ) x^n+3 a b^2 \left (1-4 n+3 n^2\right ) x^{2 n}+b^3 \left (1-3 n+2 n^2\right ) x^{3 n}\right )}{(-1+n) (-1+2 n) (-1+3 n) x \left (a+b x^n\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(3/2)/x^2,x]

[Out]

(Sqrt[(a + b*x^n)^2]*(a^3*(1 - 6*n + 11*n^2 - 6*n^3) + 3*a^2*b*(1 - 5*n + 6*n^2)*x^n + 3*a*b^2*(1 - 4*n + 3*n^
2)*x^(2*n) + b^3*(1 - 3*n + 2*n^2)*x^(3*n)))/((-1 + n)*(-1 + 2*n)*(-1 + 3*n)*x*(a + b*x^n))

________________________________________________________________________________________

Maple [A]
time = 0.02, size = 147, normalized size = 0.69

method	result	size
risch	\(-\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{3}}{\left (a +b \,x^{n}\right ) x}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{3} x^{3 n}}{\left (a +b \,x^{n}\right ) \left (-1+3 n \right ) x}+\frac {3 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, a \,b^{2} x^{2 n}}{\left (a +b \,x^{n}\right ) \left (-1+2 n \right ) x}+\frac {3 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{2} b \,x^{n}}{\left (a +b \,x^{n}\right ) \left (-1+n \right ) x}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.32, size = 101, normalized size = 0.48 \begin {gather*} \frac {{\left (2 \, n^{2} - 3 \, n + 1\right )} b^{3} x^{3 \, n} + 3 \, {\left (3 \, n^{2} - 4 \, n + 1\right )} a b^{2} x^{2 \, n} + 3 \, {\left (6 \, n^{2} - 5 \, n + 1\right )} a^{2} b x^{n} - {\left (6 \, n^{3} - 11 \, n^{2} + 6 \, n - 1\right )} a^{3}}{{\left (6 \, n^{3} - 11 \, n^{2} + 6 \, n - 1\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2)/x^2,x, algorithm="maxima")

[Out]

((2*n^2 - 3*n + 1)*b^3*x^(3*n) + 3*(3*n^2 - 4*n + 1)*a*b^2*x^(2*n) + 3*(6*n^2 - 5*n + 1)*a^2*b*x^n - (6*n^3 -
11*n^2 + 6*n - 1)*a^3)/((6*n^3 - 11*n^2 + 6*n - 1)*x)

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 131, normalized size = 0.62 \begin {gather*} -\frac {6 \, a^{3} n^{3} - 11 \, a^{3} n^{2} + 6 \, a^{3} n - a^{3} - {\left (2 \, b^{3} n^{2} - 3 \, b^{3} n + b^{3}\right )} x^{3 \, n} - 3 \, {\left (3 \, a b^{2} n^{2} - 4 \, a b^{2} n + a b^{2}\right )} x^{2 \, n} - 3 \, {\left (6 \, a^{2} b n^{2} - 5 \, a^{2} b n + a^{2} b\right )} x^{n}}{{\left (6 \, n^{3} - 11 \, n^{2} + 6 \, n - 1\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2)/x^2,x, algorithm="fricas")

[Out]

-(6*a^3*n^3 - 11*a^3*n^2 + 6*a^3*n - a^3 - (2*b^3*n^2 - 3*b^3*n + b^3)*x^(3*n) - 3*(3*a*b^2*n^2 - 4*a*b^2*n +
a*b^2)*x^(2*n) - 3*(6*a^2*b*n^2 - 5*a^2*b*n + a^2*b)*x^n)/((6*n^3 - 11*n^2 + 6*n - 1)*x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (a + b x^{n}\right )^{2}\right )^{\frac {3}{2}}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2+2*a*b*x**n+b**2*x**(2*n))**(3/2)/x**2,x)

[Out]

Integral(((a + b*x**n)**2)**(3/2)/x**2, x)

________________________________________________________________________________________

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((a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2)/x^2,x, algorithm="giac")

[Out]

integrate((b^2*x^(2*n) + 2*a*b*x^n + a^2)^(3/2)/x^2, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{3/2}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(3/2)/x^2,x)

[Out]

int((a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(3/2)/x^2, x)

________________________________________________________________________________________