### 3.206 $$\int \frac{1}{x^2 (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx$$

Optimal. Leaf size=235 $-\frac{3 b}{2 a^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 b}{3 a^3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{b}{4 a^2 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{4 b}{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a+b x}{a^5 x \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 b \log (x) (a+b x)}{a^6 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 b (a+b x) \log (a+b x)}{a^6 \sqrt{a^2+2 a b x+b^2 x^2}}$

[Out]

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

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Rubi [A]  time = 0.102017, antiderivative size = 235, 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} $-\frac{3 b}{2 a^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 b}{3 a^3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{b}{4 a^2 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{4 b}{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a+b x}{a^5 x \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 b \log (x) (a+b x)}{a^6 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 b (a+b x) \log (a+b x)}{a^6 \sqrt{a^2+2 a b x+b^2 x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-4*b)/(a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - b/(4*a^2*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (2*b)/(3*a^
3*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*b)/(2*a^4*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (a + b*
x)/(a^5*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*b*(a + b*x)*Log[x])/(a^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (5*b*(
a + b*x)*Log[a + b*x])/(a^6*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^2 \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^2 \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^2}-\frac{5}{a^6 b^4 x}+\frac{1}{a^2 b^3 (a+b x)^5}+\frac{2}{a^3 b^3 (a+b x)^4}+\frac{3}{a^4 b^3 (a+b x)^3}+\frac{4}{a^5 b^3 (a+b x)^2}+\frac{5}{a^6 b^3 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{4 b}{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{b}{4 a^2 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 b}{3 a^3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 b}{2 a^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{a+b x}{a^5 x \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 b (a+b x) \log (x)}{a^6 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 b (a+b x) \log (a+b x)}{a^6 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0402118, size = 103, normalized size = 0.44 $\frac{-a \left (260 a^2 b^2 x^2+125 a^3 b x+12 a^4+210 a b^3 x^3+60 b^4 x^4\right )-60 b x \log (x) (a+b x)^4+60 b x (a+b x)^4 \log (a+b x)}{12 a^6 x (a+b x)^3 \sqrt{(a+b x)^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-(a*(12*a^4 + 125*a^3*b*x + 260*a^2*b^2*x^2 + 210*a*b^3*x^3 + 60*b^4*x^4)) - 60*b*x*(a + b*x)^4*Log[x] + 60*b
*x*(a + b*x)^4*Log[a + b*x])/(12*a^6*x*(a + b*x)^3*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.225, size = 199, normalized size = 0.9 \begin{align*} -{\frac{ \left ( 60\,{b}^{5}\ln \left ( x \right ){x}^{5}-60\,\ln \left ( bx+a \right ){x}^{5}{b}^{5}+240\,a{b}^{4}\ln \left ( x \right ){x}^{4}-240\,\ln \left ( bx+a \right ){x}^{4}a{b}^{4}+360\,{a}^{2}{b}^{3}\ln \left ( x \right ){x}^{3}-360\,\ln \left ( bx+a \right ){x}^{3}{a}^{2}{b}^{3}+60\,a{b}^{4}{x}^{4}+240\,{a}^{3}{b}^{2}\ln \left ( x \right ){x}^{2}-240\,\ln \left ( bx+a \right ){x}^{2}{a}^{3}{b}^{2}+210\,{a}^{2}{b}^{3}{x}^{3}+60\,{a}^{4}b\ln \left ( x \right ) x-60\,\ln \left ( bx+a \right ) x{a}^{4}b+260\,{a}^{3}{b}^{2}{x}^{2}+125\,{a}^{4}bx+12\,{a}^{5} \right ) \left ( bx+a \right ) }{12\,x{a}^{6}} \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^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

-1/12*(60*b^5*ln(x)*x^5-60*ln(b*x+a)*x^5*b^5+240*a*b^4*ln(x)*x^4-240*ln(b*x+a)*x^4*a*b^4+360*a^2*b^3*ln(x)*x^3
-360*ln(b*x+a)*x^3*a^2*b^3+60*a*b^4*x^4+240*a^3*b^2*ln(x)*x^2-240*ln(b*x+a)*x^2*a^3*b^2+210*a^2*b^3*x^3+60*a^4
*b*ln(x)*x-60*ln(b*x+a)*x*a^4*b+260*a^3*b^2*x^2+125*a^4*b*x+12*a^5)*(b*x+a)/x/a^6/((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^2/(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 = 1.7869, size = 421, normalized size = 1.79 \begin{align*} -\frac{60 \, a b^{4} x^{4} + 210 \, a^{2} b^{3} x^{3} + 260 \, a^{3} b^{2} x^{2} + 125 \, a^{4} b x + 12 \, a^{5} - 60 \,{\left (b^{5} x^{5} + 4 \, a b^{4} x^{4} + 6 \, a^{2} b^{3} x^{3} + 4 \, a^{3} b^{2} x^{2} + a^{4} b x\right )} \log \left (b x + a\right ) + 60 \,{\left (b^{5} x^{5} + 4 \, a b^{4} x^{4} + 6 \, a^{2} b^{3} x^{3} + 4 \, a^{3} b^{2} x^{2} + a^{4} b x\right )} \log \left (x\right )}{12 \,{\left (a^{6} b^{4} x^{5} + 4 \, a^{7} b^{3} x^{4} + 6 \, a^{8} b^{2} x^{3} + 4 \, a^{9} b x^{2} + a^{10} x\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(60*a*b^4*x^4 + 210*a^2*b^3*x^3 + 260*a^3*b^2*x^2 + 125*a^4*b*x + 12*a^5 - 60*(b^5*x^5 + 4*a*b^4*x^4 + 6
*a^2*b^3*x^3 + 4*a^3*b^2*x^2 + a^4*b*x)*log(b*x + a) + 60*(b^5*x^5 + 4*a*b^4*x^4 + 6*a^2*b^3*x^3 + 4*a^3*b^2*x
^2 + a^4*b*x)*log(x))/(a^6*b^4*x^5 + 4*a^7*b^3*x^4 + 6*a^8*b^2*x^3 + 4*a^9*b*x^2 + a^10*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \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**2/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

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

[Out]

sage0*x