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

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

[Out]

-1/8*a^3*((b*x+a)^2)^(1/2)/x^8/(b*x+a)-3/7*a^2*b*((b*x+a)^2)^(1/2)/x^7/(b*x+a)-1/2*a*b^2*((b*x+a)^2)^(1/2)/x^6
/(b*x+a)-1/5*b^3*((b*x+a)^2)^(1/2)/x^5/(b*x+a)

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Rubi [A]  time = 0.03, antiderivative size = 151, 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 = {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 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])

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]

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*}

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Mathematica [A]  time = 0.02, size = 55, normalized size = 0.36 \[ -\frac {\sqrt {(a+b x)^2} \left (35 a^3+120 a^2 b x+140 a b^2 x^2+56 b^3 x^3\right )}{280 x^8 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.93, size = 35, normalized size = 0.23 \[ -\frac {56 \, b^{3} x^{3} + 140 \, a b^{2} x^{2} + 120 \, a^{2} b x + 35 \, a^{3}}{280 \, x^{8}} \]

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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giac [A]  time = 0.18, size = 74, normalized size = 0.49 \[ -\frac {b^{8} \mathrm {sgn}\left (b x + a\right )}{280 \, a^{5}} - \frac {56 \, b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 140 \, a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 120 \, a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 35 \, a^{3} \mathrm {sgn}\left (b x + a\right )}{280 \, x^{8}} \]

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

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maple [A]  time = 0.05, size = 52, normalized size = 0.34 \[ -\frac {\left (56 b^{3} x^{3}+140 a \,b^{2} x^{2}+120 a^{2} b x +35 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{280 \left (b x +a \right )^{3} x^{8}} \]

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 [B]  time = 1.54, size = 254, normalized size = 1.68 \[ \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{8}}{4 \, a^{8}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{7}}{4 \, a^{7} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{6}}{4 \, a^{8} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{5}}{4 \, a^{7} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{4}}{4 \, a^{6} x^{4}} + \frac {69 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{3}}{280 \, a^{5} x^{5}} - \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{2}}{56 \, a^{4} x^{6}} + \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b}{56 \, a^{3} x^{7}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}{8 \, a^{2} x^{8}} \]

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]

1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^8/a^8 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^7/(a^7*x) - 1/4*(b^2*x^2 +
2*a*b*x + a^2)^(5/2)*b^6/(a^8*x^2) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^5/(a^7*x^3) - 1/4*(b^2*x^2 + 2*a*b
*x + a^2)^(5/2)*b^4/(a^6*x^4) + 69/280*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^3/(a^5*x^5) - 13/56*(b^2*x^2 + 2*a*b*
x + a^2)^(5/2)*b^2/(a^4*x^6) + 11/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b/(a^3*x^7) - 1/8*(b^2*x^2 + 2*a*b*x + a^
2)^(5/2)/(a^2*x^8)

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mupad [B]  time = 0.20, size = 135, normalized size = 0.89 \[ -\frac {a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{8\,x^8\,\left (a+b\,x\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{5\,x^5\,\left (a+b\,x\right )}-\frac {a\,b^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,x^6\,\left (a+b\,x\right )}-\frac {3\,a^2\,b\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (a^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(8*x^8*(a + b*x)) - (b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(5*x^5*(a +
b*x)) - (a*b^2*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(2*x^6*(a + b*x)) - (3*a^2*b*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/
(7*x^7*(a + b*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{9}}\, dx \]

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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