3.5.20 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x} \, dx\)

Optimal. Leaf size=251 \[ \frac {b^5 x^{10} \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 \left (a+b x^2\right )}+\frac {5 a b^4 x^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac {5 a^2 b^3 x^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {a^5 \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {5 a^4 b x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {5 a^3 b^2 x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.07, antiderivative size = 251, 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} \begin {gather*} \frac {b^5 x^{10} \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 \left (a+b x^2\right )}+\frac {5 a b^4 x^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac {5 a^2 b^3 x^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {5 a^3 b^2 x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {5 a^4 b x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {a^5 \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x,x]

[Out]

(5*a^4*b*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*(a + b*x^2)) + (5*a^3*b^2*x^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]
)/(2*(a + b*x^2)) + (5*a^2*b^3*x^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*(a + b*x^2)) + (5*a*b^4*x^8*Sqrt[a^2 +
2*a*b*x^2 + b^2*x^4])/(8*(a + b*x^2)) + (b^5*x^10*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(10*(a + b*x^2)) + (a^5*Sqr
t[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 )^{5/2}}{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 )^5}{x} \, dx}{b^4 \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 )^5}{x} \, dx,x,x^2\right )}{2 b^4 \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 (5 a^4 b^6+\frac {a^5 b^5}{x}+10 a^3 b^7 x+10 a^2 b^8 x^2+5 a b^9 x^3+b^{10} x^4\right ) \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )}\\ &=\frac {5 a^4 b x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {5 a^3 b^2 x^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {5 a^2 b^3 x^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {5 a b^4 x^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac {b^5 x^{10} \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 \left (a+b x^2\right )}+\frac {a^5 \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 = 82, normalized size = 0.33 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (120 a^5 \log (x)+b x^2 \left (300 a^4+300 a^3 b x^2+200 a^2 b^2 x^4+75 a b^3 x^6+12 b^4 x^8\right )\right )}{120 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x,x]

[Out]

(Sqrt[(a + b*x^2)^2]*(b*x^2*(300*a^4 + 300*a^3*b*x^2 + 200*a^2*b^2*x^4 + 75*a*b^3*x^6 + 12*b^4*x^8) + 120*a^5*
Log[x]))/(120*(a + b*x^2))

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IntegrateAlgebraic [A]  time = 0.56, size = 314, normalized size = 1.25 \begin {gather*} \frac {1}{4} a^5 \log \left (\sqrt {a^2+2 a b x^2+b^2 x^4}-a-\sqrt {b^2} x^2\right )-\frac {a^5 \left (\sqrt {b^2}+b\right ) \log \left (\sqrt {a^2+2 a b x^2+b^2 x^4}+a-\sqrt {b^2} x^2\right )}{4 b}-\frac {a^5 \sqrt {b^2} \log \left (b \sqrt {a^2+2 a b x^2+b^2 x^4}-a b-b \sqrt {b^2} x^2\right )}{4 b}+\frac {1}{240} \sqrt {a^2+2 a b x^2+b^2 x^4} \left (137 a^4+163 a^3 b x^2+137 a^2 b^2 x^4+63 a b^3 x^6+12 b^4 x^8\right )+\frac {1}{240} \left (-300 a^4 \sqrt {b^2} x^2-300 a^3 b \sqrt {b^2} x^4-200 a^2 \left (b^2\right )^{3/2} x^6-75 a b^3 \sqrt {b^2} x^8-12 b^4 \sqrt {b^2} x^{10}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x,x]

[Out]

(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(137*a^4 + 163*a^3*b*x^2 + 137*a^2*b^2*x^4 + 63*a*b^3*x^6 + 12*b^4*x^8))/240
+ (-300*a^4*Sqrt[b^2]*x^2 - 300*a^3*b*Sqrt[b^2]*x^4 - 200*a^2*(b^2)^(3/2)*x^6 - 75*a*b^3*Sqrt[b^2]*x^8 - 12*b^
4*Sqrt[b^2]*x^10)/240 + (a^5*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/4 - (a^5*(b + Sqrt[b^2
])*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/(4*b) - (a^5*Sqrt[b^2]*Log[-(a*b) - b*Sqrt[b^2]*x
^2 + b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/(4*b)

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fricas [A]  time = 0.86, size = 55, normalized size = 0.22 \begin {gather*} \frac {1}{10} \, b^{5} x^{10} + \frac {5}{8} \, a b^{4} x^{8} + \frac {5}{3} \, a^{2} b^{3} x^{6} + \frac {5}{2} \, a^{3} b^{2} x^{4} + \frac {5}{2} \, a^{4} b x^{2} + a^{5} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x,x, algorithm="fricas")

[Out]

1/10*b^5*x^10 + 5/8*a*b^4*x^8 + 5/3*a^2*b^3*x^6 + 5/2*a^3*b^2*x^4 + 5/2*a^4*b*x^2 + a^5*log(x)

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giac [A]  time = 0.19, size = 106, normalized size = 0.42 \begin {gather*} \frac {1}{10} \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{8} \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{3} \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{2} \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{2} \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{2} \, a^{5} \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x,x, algorithm="giac")

[Out]

1/10*b^5*x^10*sgn(b*x^2 + a) + 5/8*a*b^4*x^8*sgn(b*x^2 + a) + 5/3*a^2*b^3*x^6*sgn(b*x^2 + a) + 5/2*a^3*b^2*x^4
*sgn(b*x^2 + a) + 5/2*a^4*b*x^2*sgn(b*x^2 + a) + 1/2*a^5*log(x^2)*sgn(b*x^2 + a)

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maple [A]  time = 0.01, size = 79, normalized size = 0.31 \begin {gather*} \frac {\left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} \left (12 b^{5} x^{10}+75 a \,b^{4} x^{8}+200 a^{2} b^{3} x^{6}+300 a^{3} b^{2} x^{4}+300 a^{4} b \,x^{2}+120 a^{5} \ln \relax (x )\right )}{120 \left (b \,x^{2}+a \right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x,x)

[Out]

1/120*((b*x^2+a)^2)^(5/2)*(12*b^5*x^10+75*a*b^4*x^8+200*a^2*b^3*x^6+300*a^3*b^2*x^4+300*a^4*b*x^2+120*a^5*ln(x
))/(b*x^2+a)^5

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maxima [A]  time = 1.36, size = 55, normalized size = 0.22 \begin {gather*} \frac {1}{10} \, b^{5} x^{10} + \frac {5}{8} \, a b^{4} x^{8} + \frac {5}{3} \, a^{2} b^{3} x^{6} + \frac {5}{2} \, a^{3} b^{2} x^{4} + \frac {5}{2} \, a^{4} b x^{2} + a^{5} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x,x, algorithm="maxima")

[Out]

1/10*b^5*x^10 + 5/8*a*b^4*x^8 + 5/3*a^2*b^3*x^6 + 5/2*a^3*b^2*x^4 + 5/2*a^4*b*x^2 + a^5*log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x,x)

[Out]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/x,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x, x)

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