3.6.26 \(\int \frac {(a^2+2 a b x^n+b^2 x^{2 n})^{3/2}}{x} \, dx\) [526]

Optimal. Leaf size=196 \[ \frac {3 a^2 b^2 x^n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{n \left (a b+b^2 x^n\right )}+\frac {3 a b^3 x^{2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{2 n \left (a b+b^2 x^n\right )}+\frac {b^4 x^{3 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{3 n \left (a b+b^2 x^n\right )}+\frac {a^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \log (x)}{a+b x^n} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {1369, 272, 45} \begin {gather*} \frac {3 a^2 b^2 x^n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{n \left (a b+b^2 x^n\right )}+\frac {b^4 x^{3 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{3 n \left (a b+b^2 x^n\right )}+\frac {3 a b^3 x^{2 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{2 n \left (a b+b^2 x^n\right )}+\frac {a^3 \log (x) \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{a+b x^n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

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Mathematica [A]
time = 0.04, size = 67, normalized size = 0.34 \begin {gather*} \frac {\sqrt {\left (a+b x^n\right )^2} \left (b x^n \left (18 a^2+9 a b x^n+2 b^2 x^{2 n}\right )+6 a^3 \log \left (x^n\right )\right )}{6 n \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,x]

[Out]

(Sqrt[(a + b*x^n)^2]*(b*x^n*(18*a^2 + 9*a*b*x^n + 2*b^2*x^(2*n)) + 6*a^3*Log[x^n]))/(6*n*(a + b*x^n))

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Maple [A]
time = 0.02, size = 127, normalized size = 0.65

method	result	size
risch	\(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{3} \ln \left (x \right )}{a +b \,x^{n}}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{3} x^{3 n}}{3 \left (a +b \,x^{n}\right ) n}+\frac {3 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, a \,b^{2} x^{2 n}}{2 \left (a +b \,x^{n}\right ) n}+\frac {3 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{2} b \,x^{n}}{\left (a +b \,x^{n}\right ) n}\)	\(127\)

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,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 43, normalized size = 0.22 \begin {gather*} a^{3} \log \left (x\right ) + \frac {2 \, b^{3} x^{3 \, n} + 9 \, a b^{2} x^{2 \, n} + 18 \, a^{2} b x^{n}}{6 \, n} \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,x, algorithm="maxima")

[Out]

a^3*log(x) + 1/6*(2*b^3*x^(3*n) + 9*a*b^2*x^(2*n) + 18*a^2*b*x^n)/n

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Fricas [A]
time = 0.37, size = 44, normalized size = 0.22 \begin {gather*} \frac {6 \, a^{3} n \log \left (x\right ) + 2 \, b^{3} x^{3 \, n} + 9 \, a b^{2} x^{2 \, n} + 18 \, a^{2} b x^{n}}{6 \, n} \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,x, algorithm="fricas")

[Out]

1/6*(6*a^3*n*log(x) + 2*b^3*x^(3*n) + 9*a*b^2*x^(2*n) + 18*a^2*b*x^n)/n

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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}\, 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,x)

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{3/2}}{x} \,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,x)

[Out]

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

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