### 3.170 $$\int \frac{(a^2+2 a b x+b^2 x^2)^{5/2}}{x^2} \, dx$$

Optimal. Leaf size=220 $-\frac{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{10 a^3 b^2 x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{5 a^2 b^3 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{5 a b^4 x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{b^5 x^4 \sqrt{a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac{5 a^4 b \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}$

[Out]

-((a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x*(a + b*x))) + (10*a^3*b^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x)
+ (5*a^2*b^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (5*a*b^4*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a
+ b*x)) + (b^5*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*(a + b*x)) + (5*a^4*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[x
])/(a + b*x)

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Rubi [A]  time = 0.0504543, antiderivative size = 220, 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, 43} $-\frac{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac{10 a^3 b^2 x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{5 a^2 b^3 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{5 a b^4 x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{b^5 x^4 \sqrt{a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac{5 a^4 b \log (x) \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x*(a + b*x))) + (10*a^3*b^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x)
+ (5*a^2*b^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (5*a*b^4*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*(a
+ b*x)) + (b^5*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*(a + b*x)) + (5*a^4*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[x
])/(a + b*x)

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 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])

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^2} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (a b+b^2 x\right )^5}{x^2} \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (10 a^3 b^7+\frac{a^5 b^5}{x^2}+\frac{5 a^4 b^6}{x}+10 a^2 b^8 x+5 a b^9 x^2+b^{10} x^3\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}}{x (a+b x)}+\frac{10 a^3 b^2 x \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{5 a^2 b^3 x^2 \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{5 a b^4 x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{b^5 x^4 \sqrt{a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac{5 a^4 b \sqrt{a^2+2 a b x+b^2 x^2} \log (x)}{a+b x}\\ \end{align*}

Mathematica [A]  time = 0.0254742, size = 79, normalized size = 0.36 $\frac{\sqrt{(a+b x)^2} \left (120 a^3 b^2 x^2+60 a^2 b^3 x^3+60 a^4 b x \log (x)-12 a^5+20 a b^4 x^4+3 b^5 x^5\right )}{12 x (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(-12*a^5 + 120*a^3*b^2*x^2 + 60*a^2*b^3*x^3 + 20*a*b^4*x^4 + 3*b^5*x^5 + 60*a^4*b*x*Log[x])
)/(12*x*(a + b*x))

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Maple [A]  time = 0.224, size = 76, normalized size = 0.4 \begin{align*}{\frac{3\,{b}^{5}{x}^{5}+20\,a{b}^{4}{x}^{4}+60\,{a}^{2}{b}^{3}{x}^{3}+60\,{a}^{4}b\ln \left ( x \right ) x+120\,{a}^{3}{b}^{2}{x}^{2}-12\,{a}^{5}}{12\, \left ( bx+a \right ) ^{5}x} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

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^2,x)

[Out]

1/12*((b*x+a)^2)^(5/2)*(3*b^5*x^5+20*a*b^4*x^4+60*a^2*b^3*x^3+60*a^4*b*ln(x)*x+120*a^3*b^2*x^2-12*a^5)/(b*x+a)
^5/x

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.66037, size = 134, normalized size = 0.61 \begin{align*} \frac{3 \, b^{5} x^{5} + 20 \, a b^{4} x^{4} + 60 \, a^{2} b^{3} x^{3} + 120 \, a^{3} b^{2} x^{2} + 60 \, a^{4} b x \log \left (x\right ) - 12 \, a^{5}}{12 \, x} \end{align*}

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^2,x, algorithm="fricas")

[Out]

1/12*(3*b^5*x^5 + 20*a*b^4*x^4 + 60*a^2*b^3*x^3 + 120*a^3*b^2*x^2 + 60*a^4*b*x*log(x) - 12*a^5)/x

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

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**2,x)

[Out]

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

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Giac [A]  time = 1.18102, size = 123, normalized size = 0.56 \begin{align*} \frac{1}{4} \, b^{5} x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{3} \, a b^{4} x^{3} \mathrm{sgn}\left (b x + a\right ) + 5 \, a^{2} b^{3} x^{2} \mathrm{sgn}\left (b x + a\right ) + 10 \, a^{3} b^{2} x \mathrm{sgn}\left (b x + a\right ) + 5 \, a^{4} b \log \left ({\left | x \right |}\right ) \mathrm{sgn}\left (b x + a\right ) - \frac{a^{5} \mathrm{sgn}\left (b x + a\right )}{x} \end{align*}

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^2,x, algorithm="giac")

[Out]

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