3.2.52 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{5/2}}{x^6} \, dx\)

Optimal. Leaf size=223 \[ \frac {b^5 \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}-\frac {5 a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{x^2 (a+b x)}-\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

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Rubi [A]  time = 0.05, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {646, 43} \begin {gather*} -\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{x^2 (a+b x)}-\frac {5 a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^5 \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^6,x]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*x^5*(a + b*x)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*x^4*(a + b
*x)) - (10*a^3*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*x^3*(a + b*x)) - (5*a^2*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)/(x^2*(a + b*x)) - (5*a*b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x*(a + b*x)) + (b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
*Log[x])/(a + b*x)

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 646

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

Rubi steps

\begin {align*} \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^6} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5}{x^6} \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {a^5 b^5}{x^6}+\frac {5 a^4 b^6}{x^5}+\frac {10 a^3 b^7}{x^4}+\frac {10 a^2 b^8}{x^3}+\frac {5 a b^9}{x^2}+\frac {b^{10}}{x}\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=-\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{x^2 (a+b x)}-\frac {5 a b^4 \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^5 \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 79, normalized size = 0.35 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (a \left (12 a^4+75 a^3 b x+200 a^2 b^2 x^2+300 a b^3 x^3+300 b^4 x^4\right )-60 b^5 x^5 \log (x)\right )}{60 x^5 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^6,x]

[Out]

-1/60*(Sqrt[(a + b*x)^2]*(a*(12*a^4 + 75*a^3*b*x + 200*a^2*b^2*x^2 + 300*a*b^3*x^3 + 300*b^4*x^4) - 60*b^5*x^5
*Log[x]))/(x^5*(a + b*x))

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IntegrateAlgebraic [B]  time = 3.01, size = 2174, normalized size = 9.75 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^6,x]

[Out]

(4*a*b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-12*a^8*b - 123*a^7*b^2*x - 572*a^6*b^3*x^2 - 1598*a^5*b^4*x^3 - 3012*
a^4*b^5*x^4 - 3875*a^3*b^6*x^5 - 3200*a^2*b^7*x^6 - 1500*a*b^8*x^7 - 300*b^9*x^8) + 4*a*b^4*Sqrt[b^2]*(12*a^9
+ 135*a^8*b*x + 695*a^7*b^2*x^2 + 2170*a^6*b^3*x^3 + 4610*a^5*b^4*x^4 + 6887*a^4*b^5*x^5 + 7075*a^3*b^6*x^6 +
4700*a^2*b^7*x^7 + 1800*a*b^8*x^8 + 300*b^9*x^9))/(15*Sqrt[b^2]*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-16*a^4*b^4
- 64*a^3*b^5*x - 96*a^2*b^6*x^2 - 64*a*b^7*x^3 - 16*b^8*x^4) + 15*x^5*(16*a^5*b^5 + 80*a^4*b^6*x + 160*a^3*b^
7*x^2 + 160*a^2*b^8*x^3 + 80*a*b^9*x^4 + 16*b^10*x^5)) + (b^5*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*
x^2]])/2 - (b^4*Sqrt[b^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/2 + (a^10*b^5*Log[a - Sqrt[b^
2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*(-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5*(a - Sqrt[b^2]*
x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5) + (a^10*b^4*Sqrt[b^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2
]])/(2*(-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^
5) - (5*a^8*b^5*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x +
b^2*x^2]])/(2*(-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2
*x^2])^5) - (5*a^8*b^4*Sqrt[b^2]*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2*Log[a - Sqrt[b^2]*x + Sqrt
[a^2 + 2*a*b*x + b^2*x^2]])/(2*(-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5*(a - Sqrt[b^2]*x + Sqrt[a^
2 + 2*a*b*x + b^2*x^2])^5) + (5*a^6*b^5*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^4*Log[a - Sqrt[b^2]*x
+ Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5*(a - Sqrt[b^2]*x + Sq
rt[a^2 + 2*a*b*x + b^2*x^2])^5) + (5*a^6*b^4*Sqrt[b^2]*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^4*Log[
a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5*(a - S
qrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5) - (5*a^4*b^5*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6
*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5*(
a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5) - (5*a^4*b^4*Sqrt[b^2]*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*
x + b^2*x^2])^6*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x
+ b^2*x^2])^5*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5) + (5*a^2*b^5*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2
*a*b*x + b^2*x^2])^8*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*(-a - Sqrt[b^2]*x + Sqrt[a^2 + 2
*a*b*x + b^2*x^2])^5*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5) + (5*a^2*b^4*Sqrt[b^2]*(-(Sqrt[b^2]*
x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^8*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*(-a - Sqrt[b^2]
*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5) - (b^5*(-(Sqrt[b^2]
*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^10*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*(-a - Sqrt[b^
2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5) - (b^4*Sqrt[b^2]*
(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^10*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*(
-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^5)

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fricas [A]  time = 0.40, size = 59, normalized size = 0.26 \begin {gather*} \frac {60 \, b^{5} x^{5} \log \relax (x) - 300 \, a b^{4} x^{4} - 300 \, a^{2} b^{3} x^{3} - 200 \, a^{3} b^{2} x^{2} - 75 \, a^{4} b x - 12 \, a^{5}}{60 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^6,x, algorithm="fricas")

[Out]

1/60*(60*b^5*x^5*log(x) - 300*a*b^4*x^4 - 300*a^2*b^3*x^3 - 200*a^3*b^2*x^2 - 75*a^4*b*x - 12*a^5)/x^5

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giac [A]  time = 0.17, size = 93, normalized size = 0.42 \begin {gather*} b^{5} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) - \frac {300 \, a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 300 \, a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 200 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 75 \, a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 12 \, a^{5} \mathrm {sgn}\left (b x + a\right )}{60 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^6,x, algorithm="giac")

[Out]

b^5*log(abs(x))*sgn(b*x + a) - 1/60*(300*a*b^4*x^4*sgn(b*x + a) + 300*a^2*b^3*x^3*sgn(b*x + a) + 200*a^3*b^2*x
^2*sgn(b*x + a) + 75*a^4*b*x*sgn(b*x + a) + 12*a^5*sgn(b*x + a))/x^5

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maple [A]  time = 0.06, size = 76, normalized size = 0.34 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (60 b^{5} x^{5} \ln \relax (x )-300 a \,b^{4} x^{4}-300 a^{2} b^{3} x^{3}-200 a^{3} b^{2} x^{2}-75 a^{4} b x -12 a^{5}\right )}{60 \left (b x +a \right )^{5} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^6,x)

[Out]

1/60*((b*x+a)^2)^(5/2)*(60*b^5*ln(x)*x^5-300*a*b^4*x^4-300*a^2*b^3*x^3-200*a^3*b^2*x^2-75*a^4*b*x-12*a^5)/(b*x
+a)^5/x^5

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maxima [B]  time = 1.58, size = 340, normalized size = 1.52 \begin {gather*} \left (-1\right )^{2 \, b^{2} x + 2 \, a b} b^{5} \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} b^{5} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{6} x}{2 \, a^{2}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}}{2 \, a} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{6} x}{4 \, a^{4}} + \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{5}}{12 \, a^{3}} - \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{5}}{15 \, a^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{4}}{3 \, a^{4} x} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{3}}{15 \, a^{5} x^{2}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{2}}{60 \, a^{4} x^{3}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b}{20 \, a^{3} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}}}{5 \, a^{2} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/x^6,x, algorithm="maxima")

[Out]

(-1)^(2*b^2*x + 2*a*b)*b^5*log(2*b^2*x + 2*a*b) - (-1)^(2*a*b*x + 2*a^2)*b^5*log(2*a*b*x/abs(x) + 2*a^2/abs(x)
) + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^6*x/a^2 + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^5/a + 1/4*(b^2*x^2 + 2*a
*b*x + a^2)^(3/2)*b^6*x/a^4 + 7/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^5/a^3 - 2/15*(b^2*x^2 + 2*a*b*x + a^2)^(5
/2)*b^5/a^5 - 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^4/(a^4*x) + 2/15*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^3/(a^5*
x^2) - 11/60*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^2/(a^4*x^3) + 3/20*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b/(a^3*x^4)
- 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)/(a^2*x^5)

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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+b^2\,x^2\right )}^{5/2}}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/x^6,x)

[Out]

int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/x^6, 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\right )^{2}\right )^{\frac {5}{2}}}{x^{6}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(5/2)/x**6,x)

[Out]

Integral(((a + b*x)**2)**(5/2)/x**6, x)

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