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

Optimal. Leaf size=221 \[ \frac {b^5 x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {5 a b^4 x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {10 a^2 b^3 x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {a^5 \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {5 a^4 b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {5 a^3 b^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x} \]

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Rubi [A]  time = 0.05, antiderivative size = 221, 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 {5 a^4 b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {5 a^3 b^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {10 a^2 b^3 x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {5 a b^4 x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {b^5 x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {a^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,x]

[Out]

(5*a^4*b*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (5*a^3*b^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x)
+ (10*a^2*b^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (5*a*b^4*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(
4*(a + b*x)) + (b^5*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(a + b*x)) + (a^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} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5}{x} \, 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^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}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {5 a^4 b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {5 a^3 b^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {10 a^2 b^3 x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {5 a b^4 x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {b^5 x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {a^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 = 74, normalized size = 0.33 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (60 a^5 \log (x)+b x \left (300 a^4+300 a^3 b x+200 a^2 b^2 x^2+75 a b^3 x^3+12 b^4 x^4\right )\right )}{60 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.53, size = 296, normalized size = 1.34 \begin {gather*} \frac {1}{2} a^5 \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )-\frac {a^5 \left (\sqrt {b^2}+b\right ) \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )}{2 b}-\frac {a^5 \sqrt {b^2} \log \left (b \sqrt {a^2+2 a b x+b^2 x^2}-a b-\sqrt {b^2} b x\right )}{2 b}+\frac {1}{120} \sqrt {a^2+2 a b x+b^2 x^2} \left (137 a^4+163 a^3 b x+137 a^2 b^2 x^2+63 a b^3 x^3+12 b^4 x^4\right )+\frac {1}{120} \left (-300 a^4 \sqrt {b^2} x-300 a^3 b \sqrt {b^2} x^2-200 a^2 \left (b^2\right )^{3/2} x^3-75 a b^3 \sqrt {b^2} x^4-12 b^4 \sqrt {b^2} x^5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(137*a^4 + 163*a^3*b*x + 137*a^2*b^2*x^2 + 63*a*b^3*x^3 + 12*b^4*x^4))/120 + (-
300*a^4*Sqrt[b^2]*x - 300*a^3*b*Sqrt[b^2]*x^2 - 200*a^2*(b^2)^(3/2)*x^3 - 75*a*b^3*Sqrt[b^2]*x^4 - 12*b^4*Sqrt
[b^2]*x^5)/120 + (a^5*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/2 - (a^5*(b + Sqrt[b^2])*Log[a -
Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*b) - (a^5*Sqrt[b^2]*Log[-(a*b) - b*Sqrt[b^2]*x + b*Sqrt[a^2 +
2*a*b*x + b^2*x^2]])/(2*b)

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fricas [A]  time = 0.39, size = 53, normalized size = 0.24 \begin {gather*} \frac {1}{5} \, b^{5} x^{5} + \frac {5}{4} \, a b^{4} x^{4} + \frac {10}{3} \, a^{2} b^{3} x^{3} + 5 \, a^{3} b^{2} x^{2} + 5 \, a^{4} b x + a^{5} \log \relax (x) \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,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 90, normalized size = 0.41 \begin {gather*} \frac {1}{5} \, b^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {10}{3} \, a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b x \mathrm {sgn}\left (b x + a\right ) + a^{5} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) \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,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 73, normalized size = 0.33 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (12 b^{5} x^{5}+75 a \,b^{4} x^{4}+200 a^{2} b^{3} x^{3}+300 a^{3} b^{2} x^{2}+60 a^{5} \ln \relax (x )+300 a^{4} b x \right )}{60 \left (b x +a \right )^{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,x)

[Out]

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

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maxima [A]  time = 1.39, size = 182, normalized size = 0.82 \begin {gather*} \left (-1\right )^{2 \, b^{2} x + 2 \, a b} a^{5} \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} a^{5} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} b x + \frac {3}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4} + \frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a b x + \frac {7}{12} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} + \frac {1}{5} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} \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,x, algorithm="maxima")

[Out]

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

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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} \,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,x)

[Out]

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

[Out]

Integral(((a + b*x)**2)**(5/2)/x, x)

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