3.563 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{3/2}}{x} \, dx\)

Optimal. Leaf size=163 \[ \frac {3 a b^2 x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {3 a^2 b x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {b^3 x^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )}+\frac {a^3 \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2} \]

[Out]

3/2*a^2*b*x^2*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+3/4*a*b^2*x^4*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+1/6*b^3*x^6*((b*x^2+a)
^2)^(1/2)/(b*x^2+a)+a^3*ln(x)*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

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Rubi [A]  time = 0.05, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1112, 266, 43} \[ \frac {b^3 x^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 \left (a+b x^2\right )}+\frac {3 a b^2 x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {3 a^2 b x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {a^3 \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x,x]

[Out]

(3*a^2*b*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*(a + b*x^2)) + (3*a*b^2*x^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/
(4*(a + b*x^2)) + (b^3*x^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(6*(a + b*x^2)) + (a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*
x^4]*Log[x])/(a + b*x^2)

Rule 43

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 266

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 1112

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 60, normalized size = 0.37 \[ \frac {\sqrt {\left (a+b x^2\right )^2} \left (12 a^3 \log (x)+b x^2 \left (18 a^2+9 a b x^2+2 b^2 x^4\right )\right )}{12 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/x,x]

[Out]

(Sqrt[(a + b*x^2)^2]*(b*x^2*(18*a^2 + 9*a*b*x^2 + 2*b^2*x^4) + 12*a^3*Log[x]))/(12*(a + b*x^2))

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fricas [A]  time = 0.84, size = 33, normalized size = 0.20 \[ \frac {1}{6} \, b^{3} x^{6} + \frac {3}{4} \, a b^{2} x^{4} + \frac {3}{2} \, a^{2} b x^{2} + a^{3} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x,x, algorithm="fricas")

[Out]

1/6*b^3*x^6 + 3/4*a*b^2*x^4 + 3/2*a^2*b*x^2 + a^3*log(x)

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giac [A]  time = 0.19, size = 68, normalized size = 0.42 \[ \frac {1}{6} \, b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {3}{4} \, a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {3}{2} \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{2} \, a^{3} \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x,x, algorithm="giac")

[Out]

1/6*b^3*x^6*sgn(b*x^2 + a) + 3/4*a*b^2*x^4*sgn(b*x^2 + a) + 3/2*a^2*b*x^2*sgn(b*x^2 + a) + 1/2*a^3*log(x^2)*sg
n(b*x^2 + a)

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maple [A]  time = 0.01, size = 57, normalized size = 0.35 \[ \frac {\left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} \left (2 b^{3} x^{6}+9 a \,b^{2} x^{4}+18 a^{2} b \,x^{2}+12 a^{3} \ln \relax (x )\right )}{12 \left (b \,x^{2}+a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x,x)

[Out]

1/12*((b*x^2+a)^2)^(3/2)*(2*b^3*x^6+9*a*b^2*x^4+18*a^2*b*x^2+12*a^3*ln(x))/(b*x^2+a)^3

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maxima [A]  time = 1.33, size = 33, normalized size = 0.20 \[ \frac {1}{6} \, b^{3} x^{6} + \frac {3}{4} \, a b^{2} x^{4} + \frac {3}{2} \, a^{2} b x^{2} + a^{3} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/x,x, algorithm="maxima")

[Out]

1/6*b^3*x^6 + 3/4*a*b^2*x^4 + 3/2*a^2*b*x^2 + a^3*log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x,x)

[Out]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/x,x)

[Out]

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

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