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

Optimal. Leaf size=222 \[ \frac {b^5 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {5 a b^4 x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {10 a^2 b^3 \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}-\frac {a^5 \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)} \]

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Rubi [A]  time = 0.05, antiderivative size = 222, 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}}{3 x^3 (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {5 a b^4 x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {b^5 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {10 a^2 b^3 \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^4,x]

[Out]

-(a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*x^3*(a + b*x)) - (5*a^4*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*x^2*(a + b
*x)) - (10*a^3*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x*(a + b*x)) + (5*a*b^4*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a
+ b*x) + (b^5*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*(a + b*x)) + (10*a^2*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Lo
g[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^4} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5}{x^4} \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (5 a b^9+\frac {a^5 b^5}{x^4}+\frac {5 a^4 b^6}{x^3}+\frac {10 a^3 b^7}{x^2}+\frac {10 a^2 b^8}{x}+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}}{3 x^3 (a+b x)}-\frac {5 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {5 a b^4 x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {b^5 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 (a+b x)}+\frac {10 a^2 b^3 \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 (-2 a^5-15 a^4 b x-60 a^3 b^2 x^2+60 a^2 b^3 x^3 \log (x)+30 a b^4 x^4+3 b^5 x^5\right )}{6 x^3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(-2*a^5 - 15*a^4*b*x - 60*a^3*b^2*x^2 + 30*a*b^4*x^4 + 3*b^5*x^5 + 60*a^2*b^3*x^3*Log[x]))/
(6*x^3*(a + b*x))

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IntegrateAlgebraic [A]  time = 1.82, size = 338, normalized size = 1.52 \begin {gather*} -5 a^2 \left (b^2\right )^{3/2} \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )-5 a^2 \left (b^2\right )^{3/2} \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )+10 a^2 b^3 \tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 a b x+b^2 x^2}}{a}\right )+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-8 a^5 b-60 a^4 b^2 x-240 a^3 b^3 x^2-391 a^2 b^4 x^3+120 a b^5 x^4+12 b^6 x^5\right )+\sqrt {b^2} \left (8 a^6+68 a^5 b x+300 a^4 b^2 x^2+631 a^3 b^3 x^3+271 a^2 b^4 x^4-132 a b^5 x^5-12 b^6 x^6\right )}{24 x^3 \left (a b+b^2 x\right )-24 \sqrt {b^2} x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-8*a^5*b - 60*a^4*b^2*x - 240*a^3*b^3*x^2 - 391*a^2*b^4*x^3 + 120*a*b^5*x^4 +
12*b^6*x^5) + Sqrt[b^2]*(8*a^6 + 68*a^5*b*x + 300*a^4*b^2*x^2 + 631*a^3*b^3*x^3 + 271*a^2*b^4*x^4 - 132*a*b^5*
x^5 - 12*b^6*x^6))/(24*x^3*(a*b + b^2*x) - 24*Sqrt[b^2]*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + 10*a^2*b^3*ArcTan
h[(Sqrt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a] - 5*a^2*(b^2)^(3/2)*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a
*b*x + b^2*x^2]] - 5*a^2*(b^2)^(3/2)*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]]

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fricas [A]  time = 0.40, size = 59, normalized size = 0.27 \begin {gather*} \frac {3 \, b^{5} x^{5} + 30 \, a b^{4} x^{4} + 60 \, a^{2} b^{3} x^{3} \log \relax (x) - 60 \, a^{3} b^{2} x^{2} - 15 \, a^{4} b x - 2 \, a^{5}}{6 \, x^{3}} \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^4,x, algorithm="fricas")

[Out]

1/6*(3*b^5*x^5 + 30*a*b^4*x^4 + 60*a^2*b^3*x^3*log(x) - 60*a^3*b^2*x^2 - 15*a^4*b*x - 2*a^5)/x^3

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giac [A]  time = 0.18, size = 92, normalized size = 0.41 \begin {gather*} \frac {1}{2} \, b^{5} x^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, a b^{4} x \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) - \frac {60 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 15 \, a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{5} \mathrm {sgn}\left (b x + a\right )}{6 \, x^{3}} \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^4,x, algorithm="giac")

[Out]

1/2*b^5*x^2*sgn(b*x + a) + 5*a*b^4*x*sgn(b*x + a) + 10*a^2*b^3*log(abs(x))*sgn(b*x + a) - 1/6*(60*a^3*b^2*x^2*
sgn(b*x + a) + 15*a^4*b*x*sgn(b*x + a) + 2*a^5*sgn(b*x + a))/x^3

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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 (3 b^{5} x^{5}+60 a^{2} b^{3} x^{3} \ln \relax (x )+30 a \,b^{4} x^{4}-60 a^{3} b^{2} x^{2}-15 a^{4} b x -2 a^{5}\right )}{6 \left (b x +a \right )^{5} x^{3}} \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^4,x)

[Out]

1/6*((b*x+a)^2)^(5/2)*(3*b^5*x^5+60*a^2*b^3*ln(x)*x^3+30*a*b^4*x^4-60*a^3*b^2*x^2-15*a^4*b*x-2*a^5)/(b*x+a)^5/
x^3

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maxima [A]  time = 1.55, size = 284, normalized size = 1.28 \begin {gather*} 10 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} a^{2} b^{3} \log \left (2 \, b^{2} x + 2 \, a b\right ) - 10 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} a^{2} b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + 5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4} x + 15 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a b^{3} + \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4} x}{2 \, a^{2}} + \frac {35 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{3}}{6 \, a} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{3}}{6 \, a^{3}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{2}}{6 \, a^{2} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} b}{6 \, a^{3} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}}}{3 \, a^{2} x^{3}} \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^4,x, algorithm="maxima")

[Out]

10*(-1)^(2*b^2*x + 2*a*b)*a^2*b^3*log(2*b^2*x + 2*a*b) - 10*(-1)^(2*a*b*x + 2*a^2)*a^2*b^3*log(2*a*b*x/abs(x)
+ 2*a^2/abs(x)) + 5*sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4*x + 15*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*b^3 + 5/2*(b^2*x^
2 + 2*a*b*x + a^2)^(3/2)*b^4*x/a^2 + 35/6*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^3/a + 1/6*(b^2*x^2 + 2*a*b*x + a^2
)^(5/2)*b^3/a^3 - 11/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^2/(a^2*x) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*b/(a^
3*x^2) - 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)/(a^2*x^3)

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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^4} \,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^4,x)

[Out]

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

[Out]

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

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