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

Optimal. Leaf size=164 \[ \frac {3 a b^2 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {3 a^2 b \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {b^3 x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 164, 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} \[ \frac {b^3 x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {3 a b^2 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {3 a^2 b \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*x^2*(a + b*x^2)) + (3*a*b^2*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*
(a + b*x^2)) + (b^3*x^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*(a + b*x^2)) + (3*a^2*b*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^3} \, 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^3} \, 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^2} \, 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 b^5+\frac {a^3 b^3}{x^2}+\frac {3 a^2 b^4}{x}+b^6 x\right ) \, dx,x,x^2\right )}{2 b^2 \left (a b+b^2 x^2\right )}\\ &=-\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^2 \left (a+b x^2\right )}+\frac {3 a b^2 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^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 \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 = 62, normalized size = 0.38 \[ \frac {\sqrt {\left (a+b x^2\right )^2} \left (-2 a^3+12 a^2 b x^2 \log (x)+6 a b^2 x^4+b^3 x^6\right )}{4 x^2 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x^2)^2]*(-2*a^3 + 6*a*b^2*x^4 + b^3*x^6 + 12*a^2*b*x^2*Log[x]))/(4*x^2*(a + b*x^2))

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fricas [A]  time = 0.80, size = 38, normalized size = 0.23 \[ \frac {b^{3} x^{6} + 6 \, a b^{2} x^{4} + 12 \, a^{2} b x^{2} \log \relax (x) - 2 \, a^{3}}{4 \, x^{2}} \]

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^3,x, algorithm="fricas")

[Out]

1/4*(b^3*x^6 + 6*a*b^2*x^4 + 12*a^2*b*x^2*log(x) - 2*a^3)/x^2

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giac [A]  time = 0.16, size = 87, normalized size = 0.53 \[ \frac {1}{4} \, b^{3} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {3}{2} \, a b^{2} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {3}{2} \, a^{2} b \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {3 \, a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{2 \, x^{2}} \]

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^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 59, normalized size = 0.36 \[ \frac {\left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} \left (b^{3} x^{6}+6 a \,b^{2} x^{4}+12 a^{2} b \,x^{2} \ln \relax (x )-2 a^{3}\right )}{4 \left (b \,x^{2}+a \right )^{3} x^{2}} \]

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^3,x)

[Out]

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

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maxima [A]  time = 1.29, size = 34, normalized size = 0.21 \[ \frac {1}{4} \, b^{3} x^{4} + \frac {3}{2} \, a b^{2} x^{2} + 3 \, a^{2} b \log \relax (x) - \frac {a^{3}}{2 \, x^{2}} \]

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^3,x, algorithm="maxima")

[Out]

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

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

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^3,x)

[Out]

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

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

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**3,x)

[Out]

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

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