3.162 \(\int \frac{(a^2+2 a b x+b^2 x^2)^{3/2}}{x^9} \, dx\)

Optimal. Leaf size=151 \[ -\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac{3 a^2 b \sqrt{a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac{a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^6 (a+b x)}-\frac{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)} \]

[Out]

-(a^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*x^8*(a + b*x)) - (3*a^2*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*x^7*(a + b
*x)) - (a*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*x^6*(a + b*x)) - (b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*x^5*(a
+ b*x))

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Rubi [A]  time = 0.0348462, antiderivative size = 151, 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} \[ -\frac{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac{3 a^2 b \sqrt{a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac{a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^6 (a+b x)}-\frac{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/x^9,x]

[Out]

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

Mathematica [A]  time = 0.0169757, size = 55, normalized size = 0.36 \[ -\frac{\sqrt{(a+b x)^2} \left (120 a^2 b x+35 a^3+140 a b^2 x^2+56 b^3 x^3\right )}{280 x^8 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/x^9,x]

[Out]

-(Sqrt[(a + b*x)^2]*(35*a^3 + 120*a^2*b*x + 140*a*b^2*x^2 + 56*b^3*x^3))/(280*x^8*(a + b*x))

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Maple [A]  time = 0.181, size = 52, normalized size = 0.3 \begin{align*} -{\frac{56\,{b}^{3}{x}^{3}+140\,a{b}^{2}{x}^{2}+120\,b{a}^{2}x+35\,{a}^{3}}{280\,{x}^{8} \left ( bx+a \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{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)^(3/2)/x^9,x)

[Out]

-1/280*(56*b^3*x^3+140*a*b^2*x^2+120*a^2*b*x+35*a^3)*((b*x+a)^2)^(3/2)/x^8/(b*x+a)^3

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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)^(3/2)/x^9,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.57423, size = 86, normalized size = 0.57 \begin{align*} -\frac{56 \, b^{3} x^{3} + 140 \, a b^{2} x^{2} + 120 \, a^{2} b x + 35 \, a^{3}}{280 \, x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/x^9,x, algorithm="fricas")

[Out]

-1/280*(56*b^3*x^3 + 140*a*b^2*x^2 + 120*a^2*b*x + 35*a^3)/x^8

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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{3}{2}}}{x^{9}}\, 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)**(3/2)/x**9,x)

[Out]

Integral(((a + b*x)**2)**(3/2)/x**9, x)

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Giac [A]  time = 1.39493, size = 100, normalized size = 0.66 \begin{align*} -\frac{b^{8} \mathrm{sgn}\left (b x + a\right )}{280 \, a^{5}} - \frac{56 \, b^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + 140 \, a b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 120 \, a^{2} b x \mathrm{sgn}\left (b x + a\right ) + 35 \, a^{3} \mathrm{sgn}\left (b x + a\right )}{280 \, x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/x^9,x, algorithm="giac")

[Out]

-1/280*b^8*sgn(b*x + a)/a^5 - 1/280*(56*b^3*x^3*sgn(b*x + a) + 140*a*b^2*x^2*sgn(b*x + a) + 120*a^2*b*x*sgn(b*
x + a) + 35*a^3*sgn(b*x + a))/x^8