3.5.24 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{x^9} \, dx\)

Optimal. Leaf size=250 \[ \frac {b^5 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {5 a b^4 \log (x) \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2 \left (a+b x^2\right )}-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^4 \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.07, antiderivative size = 250, 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 {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^4 \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2 \left (a+b x^2\right )}+\frac {b^5 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {5 a b^4 \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^9,x]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*x^8*(a + b*x^2)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(6*x^6*
(a + b*x^2)) - (5*a^3*b^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*x^4*(a + b*x^2)) - (5*a^2*b^3*Sqrt[a^2 + 2*a*b*x
^2 + b^2*x^4])/(x^2*(a + b*x^2)) + (b^5*x^2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*(a + b*x^2)) + (5*a*b^4*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 )^{5/2}}{x^9} \, 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^9} \, 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^5} \, 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 (b^{10}+\frac {a^5 b^5}{x^5}+\frac {5 a^4 b^6}{x^4}+\frac {10 a^3 b^7}{x^3}+\frac {10 a^2 b^8}{x^2}+\frac {5 a b^9}{x}\right ) \, dx,x,x^2\right )}{2 b^4 \left (a b+b^2 x^2\right )}\\ &=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 x^8 \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{6 x^6 \left (a+b x^2\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^4 \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2 \left (a+b x^2\right )}+\frac {b^5 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}+\frac {5 a b^4 \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 = 85, normalized size = 0.34 \begin {gather*} -\frac {\sqrt {\left (a+b x^2\right )^2} \left (3 a^5+20 a^4 b x^2+60 a^3 b^2 x^4+120 a^2 b^3 x^6-120 a b^4 x^8 \log (x)-12 b^5 x^{10}\right )}{24 x^8 \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^9,x]

[Out]

-1/24*(Sqrt[(a + b*x^2)^2]*(3*a^5 + 20*a^4*b*x^2 + 60*a^3*b^2*x^4 + 120*a^2*b^3*x^6 - 12*b^5*x^10 - 120*a*b^4*
x^8*Log[x]))/(x^8*(a + b*x^2))

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IntegrateAlgebraic [B]  time = 2.75, size = 2027, normalized size = 8.11 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^9,x]

[Out]

(-(b^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(3*a^8*b + 29*a^7*b^2*x^2 + 129*a^6*b^3*x^4 + 363*a^5*b^4*x^6 + 554*a^4
*b^5*x^8 + 390*a^3*b^6*x^10 + 66*a^2*b^7*x^12 - 42*a*b^8*x^14 - 12*b^9*x^16)) - b^3*Sqrt[b^2]*(-3*a^9 - 32*a^8
*b*x^2 - 158*a^7*b^2*x^4 - 492*a^6*b^3*x^6 - 917*a^5*b^4*x^8 - 944*a^4*b^5*x^10 - 456*a^3*b^6*x^12 - 24*a^2*b^
7*x^14 + 54*a*b^8*x^16 + 12*b^9*x^18))/(3*Sqrt[b^2]*x^8*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-8*a^3*b^3 - 24*a^2*b
^4*x^2 - 24*a*b^5*x^4 - 8*b^6*x^6) + 3*x^8*(8*a^4*b^4 + 32*a^3*b^5*x^2 + 48*a^2*b^6*x^4 + 32*a*b^7*x^6 + 8*b^8
*x^8)) + (5*a*b^4*Log[-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/4 - (5*a*b^3*Sqrt[b^2]*Log[-a - S
qrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/4 - (5*a^9*b^4*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4]])/(4*(-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4*(a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*
x^2 + b^2*x^4])^4) - (5*a^9*b^3*Sqrt[b^2]*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/(4*(-a - S
qrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4*(a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4) + (
5*a^7*b^4*(-(Sqrt[b^2]*x^2) + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^2*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4]])/((-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4*(a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x
^2 + b^2*x^4])^4) + (5*a^7*b^3*Sqrt[b^2]*(-(Sqrt[b^2]*x^2) + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^2*Log[a - Sqrt[b
^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/((-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4*(a - Sqr
t[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4) - (15*a^5*b^4*(-(Sqrt[b^2]*x^2) + Sqrt[a^2 + 2*a*b*x^2 + b^2*
x^4])^4*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/(2*(-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^
2 + b^2*x^4])^4*(a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4) - (15*a^5*b^3*Sqrt[b^2]*(-(Sqrt[b^2]*
x^2) + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/(2*(-a - S
qrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4*(a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4) + (
5*a^3*b^4*(-(Sqrt[b^2]*x^2) + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^6*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4]])/((-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4*(a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x
^2 + b^2*x^4])^4) + (5*a^3*b^3*Sqrt[b^2]*(-(Sqrt[b^2]*x^2) + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^6*Log[a - Sqrt[b
^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/((-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4*(a - Sqr
t[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4) - (5*a*b^4*(-(Sqrt[b^2]*x^2) + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4
])^8*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/(4*(-a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4])^4*(a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4) - (5*a*b^3*Sqrt[b^2]*(-(Sqrt[b^2]*x^2) +
Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^8*Log[a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]])/(4*(-a - Sqrt[b^
2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4*(a - Sqrt[b^2]*x^2 + Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])^4)

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fricas [A]  time = 0.83, size = 61, normalized size = 0.24 \begin {gather*} \frac {12 \, b^{5} x^{10} + 120 \, a b^{4} x^{8} \log \relax (x) - 120 \, a^{2} b^{3} x^{6} - 60 \, a^{3} b^{2} x^{4} - 20 \, a^{4} b x^{2} - 3 \, a^{5}}{24 \, x^{8}} \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^9,x, algorithm="fricas")

[Out]

1/24*(12*b^5*x^10 + 120*a*b^4*x^8*log(x) - 120*a^2*b^3*x^6 - 60*a^3*b^2*x^4 - 20*a^4*b*x^2 - 3*a^5)/x^8

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giac [A]  time = 0.19, size = 126, normalized size = 0.50 \begin {gather*} \frac {1}{2} \, b^{5} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{2} \, a b^{4} \log \left (x^{2}\right ) \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {125 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 120 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 60 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 20 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 3 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{24 \, x^{8}} \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^9,x, algorithm="giac")

[Out]

1/2*b^5*x^2*sgn(b*x^2 + a) + 5/2*a*b^4*log(x^2)*sgn(b*x^2 + a) - 1/24*(125*a*b^4*x^8*sgn(b*x^2 + a) + 120*a^2*
b^3*x^6*sgn(b*x^2 + a) + 60*a^3*b^2*x^4*sgn(b*x^2 + a) + 20*a^4*b*x^2*sgn(b*x^2 + a) + 3*a^5*sgn(b*x^2 + a))/x
^8

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maple [A]  time = 0.01, size = 82, normalized size = 0.33 \begin {gather*} \frac {\left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} \left (12 b^{5} x^{10}+120 a \,b^{4} x^{8} \ln \relax (x )-120 a^{2} b^{3} x^{6}-60 a^{3} b^{2} x^{4}-20 a^{4} b \,x^{2}-3 a^{5}\right )}{24 \left (b \,x^{2}+a \right )^{5} x^{8}} \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^9,x)

[Out]

1/24*((b*x^2+a)^2)^(5/2)*(12*b^5*x^10+120*a*b^4*ln(x)*x^8-120*a^2*b^3*x^6-60*a^3*b^2*x^4-20*a^4*b*x^2-3*a^5)/(
b*x^2+a)^5/x^8

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maxima [A]  time = 1.36, size = 56, normalized size = 0.22 \begin {gather*} \frac {1}{2} \, b^{5} x^{2} + 5 \, a b^{4} \log \relax (x) - \frac {5 \, a^{2} b^{3}}{x^{2}} - \frac {5 \, a^{3} b^{2}}{2 \, x^{4}} - \frac {5 \, a^{4} b}{6 \, x^{6}} - \frac {a^{5}}{8 \, x^{8}} \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^9,x, algorithm="maxima")

[Out]

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

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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^9} \,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^9,x)

[Out]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^9, 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^{9}}\, 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**9,x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/x**9, x)

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