3.205 \(\int \frac{1}{x (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

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

[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])

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Rubi [A]  time = 0.0863049, antiderivative size = 194, normalized size of antiderivative = 1., 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, 44} \[ \frac{1}{2 a^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{1}{3 a^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{1}{4 a (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{1}{a^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\log (x) (a+b x)}{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(a+b x) \log (a+b x)}{a^5 \sqrt{a^2+2 a b x+b^2 x^2}} \]

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 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]

Rule 44

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] && L
tQ[m + n + 2, 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*}

Mathematica [A]  time = 0.0346088, size = 84, normalized size = 0.43 \[ \frac{a \left (52 a^2 b x+25 a^3+42 a b^2 x^2+12 b^3 x^3\right )+12 \log (x) (a+b x)^4-12 (a+b x)^4 \log (a+b x)}{12 a^5 (a+b x)^3 \sqrt{(a+b x)^2}} \]

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 = 0.238, size = 173, normalized size = 0.9 \begin{align*}{\frac{ \left ( 12\,\ln \left ( x \right ){x}^{4}{b}^{4}-12\,\ln \left ( bx+a \right ){x}^{4}{b}^{4}+48\,\ln \left ( x \right ){x}^{3}a{b}^{3}-48\,\ln \left ( bx+a \right ){x}^{3}a{b}^{3}+72\,\ln \left ( x \right ){x}^{2}{a}^{2}{b}^{2}-72\,\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{2}+12\,a{b}^{3}{x}^{3}+48\,\ln \left ( x \right ) x{a}^{3}b-48\,\ln \left ( bx+a \right ) x{a}^{3}b+42\,{x}^{2}{a}^{2}{b}^{2}+12\,\ln \left ( x \right ){a}^{4}-12\,{a}^{4}\ln \left ( bx+a \right ) +52\,x{a}^{3}b+25\,{a}^{4} \right ) \left ( bx+a \right ) }{12\,{a}^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

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)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.7908, size = 365, normalized size = 1.88 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (\left (a + b x\right )^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

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]

sage0*x