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

Optimal. Leaf size=219 \[ \frac {b^5 x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {5 a b^4 \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}-\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{x^2 (a+b x)} \]

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Rubi [A]  time = 0.05, antiderivative size = 219, 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}}{4 x^4 (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{x^2 (a+b x)}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^5 x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {5 a b^4 \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^5,x]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*x^4*(a + b*x)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*x^3*(a + b
*x)) - (5*a^3*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x^2*(a + b*x)) - (10*a^2*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/
(x*(a + b*x)) + (b^5*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (5*a*b^4*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^5} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5}{x^5} \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \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}{b^4 \left (a b+b^2 x\right )}\\ &=-\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{x^2 (a+b x)}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^5 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} \log (x)}{a+b x}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 79, normalized size = 0.36 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (3 a^5+20 a^4 b x+60 a^3 b^2 x^2+120 a^2 b^3 x^3-60 a b^4 x^4 \log (x)-12 b^5 x^5\right )}{12 x^4 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(Sqrt[(a + b*x)^2]*(3*a^5 + 20*a^4*b*x + 60*a^3*b^2*x^2 + 120*a^2*b^3*x^3 - 12*b^5*x^5 - 60*a*b^4*x^4*Lo
g[x]))/(x^4*(a + b*x))

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IntegrateAlgebraic [B]  time = 2.66, size = 1851, normalized size = 8.45

result too large to display

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(3*a^8*b + 29*a^7*b^2*x + 129*a^6*b^3*x^2 + 363*a^5*b^4*x^3 + 554*a^4*b^
5*x^4 + 390*a^3*b^6*x^5 + 66*a^2*b^7*x^6 - 42*a*b^8*x^7 - 12*b^9*x^8) - 2*b^3*Sqrt[b^2]*(-3*a^9 - 32*a^8*b*x -
158*a^7*b^2*x^2 - 492*a^6*b^3*x^3 - 917*a^5*b^4*x^4 - 944*a^4*b^5*x^5 - 456*a^3*b^6*x^6 - 24*a^2*b^7*x^7 + 54
*a*b^8*x^8 + 12*b^9*x^9))/(3*Sqrt[b^2]*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-8*a^3*b^3 - 24*a^2*b^4*x - 24*a*b^5
*x^2 - 8*b^6*x^3) + 3*x^4*(8*a^4*b^4 + 32*a^3*b^5*x + 48*a^2*b^6*x^2 + 32*a*b^7*x^3 + 8*b^8*x^4)) + (5*a*b^4*L
og[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/2 - (5*a*b^3*Sqrt[b^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 +
2*a*b*x + b^2*x^2]])/2 - (5*a^9*b^4*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])^4*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^4) - (5*a^9*b^3*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])
^4*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^4) + (10*a^7*b^4*(-(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]])/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2
*x^2])^4*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^4) + (10*a^7*b^3*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]])/((-a - Sqrt[b^2]*x + Sqrt[a^2
+ 2*a*b*x + b^2*x^2])^4*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^4) - (15*a^5*b^4*(-(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 + Sq
rt[a^2 + 2*a*b*x + b^2*x^2])^4*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^4) - (15*a^5*b^3*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])^4*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^4) + (10*a^3*
b^4*(-(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])^4*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^4) + (1
0*a^3*b^3*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])^4*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b
^2*x^2])^4) - (5*a*b^4*(-(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])^4*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*
x + b^2*x^2])^4) - (5*a*b^3*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])^4*(a - Sqrt[b^2]*x + Sq
rt[a^2 + 2*a*b*x + b^2*x^2])^4)

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fricas [A]  time = 0.40, size = 59, normalized size = 0.27 \begin {gather*} \frac {12 \, b^{5} x^{5} + 60 \, a b^{4} x^{4} \log \relax (x) - 120 \, a^{2} b^{3} x^{3} - 60 \, a^{3} b^{2} x^{2} - 20 \, a^{4} b x - 3 \, a^{5}}{12 \, x^{4}} \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^5,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 91, normalized size = 0.42 \begin {gather*} b^{5} x \mathrm {sgn}\left (b x + a\right ) + 5 \, a b^{4} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) - \frac {120 \, a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 60 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 20 \, a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{5} \mathrm {sgn}\left (b x + a\right )}{12 \, x^{4}} \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^5,x, algorithm="giac")

[Out]

b^5*x*sgn(b*x + a) + 5*a*b^4*log(abs(x))*sgn(b*x + a) - 1/12*(120*a^2*b^3*x^3*sgn(b*x + a) + 60*a^3*b^2*x^2*sg
n(b*x + a) + 20*a^4*b*x*sgn(b*x + a) + 3*a^5*sgn(b*x + a))/x^4

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maple [A]  time = 0.06, size = 76, normalized size = 0.35 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (60 a \,b^{4} x^{4} \ln \relax (x )+12 b^{5} x^{5}-120 a^{2} b^{3} x^{3}-60 a^{3} b^{2} x^{2}-20 a^{4} b x -3 a^{5}\right )}{12 \left (b x +a \right )^{5} x^{4}} \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^5,x)

[Out]

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

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maxima [B]  time = 1.56, size = 311, normalized size = 1.42 \begin {gather*} 5 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} a b^{4} \log \left (2 \, b^{2} x + 2 \, a b\right ) - 5 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} a b^{4} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5} x}{2 \, a} + \frac {15}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4} + \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{5} x}{4 \, a^{3}} + \frac {35 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}}{12 \, a^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{4}}{3 \, a^{4}} - \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{3}}{3 \, a^{3} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b^{2}}{3 \, a^{4} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b}{12 \, a^{3} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}}}{4 \, a^{2} x^{4}} \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^5,x, algorithm="maxima")

[Out]

5*(-1)^(2*b^2*x + 2*a*b)*a*b^4*log(2*b^2*x + 2*a*b) - 5*(-1)^(2*a*b*x + 2*a^2)*a*b^4*log(2*a*b*x/abs(x) + 2*a^
2/abs(x)) + 5/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^5*x/a + 15/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4 + 5/4*(b^2*x^2
+ 2*a*b*x + a^2)^(3/2)*b^5*x/a^3 + 35/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4/a^2 + 1/3*(b^2*x^2 + 2*a*b*x + a^
2)^(5/2)*b^4/a^4 - 2/3*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^3/(a^3*x) - 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b^2/(
a^4*x^2) + 1/12*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b/(a^3*x^3) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)/(a^2*x^4)

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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^5} \,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^5,x)

[Out]

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

[Out]

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

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