3.3.5 \(\int \frac {1}{x (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [205]

Optimal. Leaf size=194 \[ \frac {1}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {1}{4 a (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {1}{3 a^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {1}{2 a^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) \log (x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(a+b x) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 194, 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 = {660, 46} \begin {gather*} \frac {1}{3 a^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {1}{4 a (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\log (x) (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(a+b x) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {1}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {1}{2 a^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

1/(a^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + 1/(4*a*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + 1/(3*a^2*(a + b*x)
^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + 1/(2*a^3*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((a + b*x)*Log[x])/(a^
5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - ((a + b*x)*Log[a + b*x])/(a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 660

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 {1}{x \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {1}{x \left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac {1}{a^5 b^5 x}-\frac {1}{a b^4 (a+b x)^5}-\frac {1}{a^2 b^4 (a+b x)^4}-\frac {1}{a^3 b^4 (a+b x)^3}-\frac {1}{a^4 b^4 (a+b x)^2}-\frac {1}{a^5 b^4 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {1}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {1}{4 a (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {1}{3 a^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {1}{2 a^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) \log (x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(a+b x) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 84, normalized size = 0.43 \begin {gather*} \frac {a \left (25 a^3+52 a^2 b x+42 a b^2 x^2+12 b^3 x^3\right )+12 (a+b x)^4 \log (x)-12 (a+b x)^4 \log (a+b x)}{12 a^5 (a+b x)^3 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

(a*(25*a^3 + 52*a^2*b*x + 42*a*b^2*x^2 + 12*b^3*x^3) + 12*(a + b*x)^4*Log[x] - 12*(a + b*x)^4*Log[a + b*x])/(1
2*a^5*(a + b*x)^3*Sqrt[(a + b*x)^2])

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Maple [A]
time = 0.58, size = 173, normalized size = 0.89

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {b^{3} x^{3}}{a^{4}}+\frac {7 b^{2} x^{2}}{2 a^{3}}+\frac {13 b x}{3 a^{2}}+\frac {25}{12 a}\right )}{\left (b x +a \right )^{5}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \ln \left (b x +a \right )}{\left (b x +a \right ) a^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \ln \left (-x \right )}{\left (b x +a \right ) a^{5}}\)	\(104\)
default	\(-\frac {\left (12 \ln \left (b x +a \right ) b^{4} x^{4}-12 \ln \left (x \right ) b^{4} x^{4}+48 \ln \left (b x +a \right ) a \,b^{3} x^{3}-48 \ln \left (x \right ) a \,b^{3} x^{3}+72 \ln \left (b x +a \right ) a^{2} b^{2} x^{2}-72 \ln \left (x \right ) a^{2} b^{2} x^{2}-12 a \,b^{3} x^{3}+48 \ln \left (b x +a \right ) a^{3} b x -48 \ln \left (x \right ) a^{3} b x -42 a^{2} b^{2} x^{2}+12 a^{4} \ln \left (b x +a \right )-12 \ln \left (x \right ) a^{4}-52 a^{3} b x -25 a^{4}\right ) \left (b x +a \right )}{12 a^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(173\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/12*(12*ln(b*x+a)*b^4*x^4-12*ln(x)*b^4*x^4+48*ln(b*x+a)*a*b^3*x^3-48*ln(x)*a*b^3*x^3+72*ln(b*x+a)*a^2*b^2*x^
2-72*ln(x)*a^2*b^2*x^2-12*a*b^3*x^3+48*ln(b*x+a)*a^3*b*x-48*ln(x)*a^3*b*x-42*a^2*b^2*x^2+12*a^4*ln(b*x+a)-12*l
n(x)*a^4-52*a^3*b*x-25*a^4)*(b*x+a)/a^5/((b*x+a)^2)^(5/2)

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Maxima [A]
time = 0.27, size = 118, normalized size = 0.61 \begin {gather*} -\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{5}} + \frac {1}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2}} + \frac {1}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4}} + \frac {1}{2 \, a^{3} b^{2} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {1}{4 \, a b^{4} {\left (x + \frac {a}{b}\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.79, size = 168, normalized size = 0.87 \begin {gather*} \frac {12 \, a b^{3} x^{3} + 42 \, a^{2} b^{2} x^{2} + 52 \, a^{3} b x + 25 \, a^{4} - 12 \, {\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \log \left (b x + a\right ) + 12 \, {\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \log \left (x\right )}{12 \, {\left (a^{5} b^{4} x^{4} + 4 \, a^{6} b^{3} x^{3} + 6 \, a^{7} b^{2} x^{2} + 4 \, a^{8} b x + a^{9}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/12*(12*a*b^3*x^3 + 42*a^2*b^2*x^2 + 52*a^3*b*x + 25*a^4 - 12*(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*
b*x + a^4)*log(b*x + a) + 12*(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4)*log(x))/(a^5*b^4*x^4 +
4*a^6*b^3*x^3 + 6*a^7*b^2*x^2 + 4*a^8*b*x + a^9)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral(1/(x*((a + b*x)**2)**(5/2)), x)

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Giac [A]
time = 0.63, size = 89, normalized size = 0.46 \begin {gather*} -\frac {\log \left ({\left | b x + a \right |}\right )}{a^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {\log \left ({\left | x \right |}\right )}{a^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {12 \, a b^{3} x^{3} + 42 \, a^{2} b^{2} x^{2} + 52 \, a^{3} b x + 25 \, a^{4}}{12 \, {\left (b x + a\right )}^{4} a^{5} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

-log(abs(b*x + a))/(a^5*sgn(b*x + a)) + log(abs(x))/(a^5*sgn(b*x + a)) + 1/12*(12*a*b^3*x^3 + 42*a^2*b^2*x^2 +
52*a^3*b*x + 25*a^4)/((b*x + a)^4*a^5*sgn(b*x + a))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)),x)

[Out]

int(1/(x*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)), x)

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