3.161 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{x^8} \, dx\)

Optimal. Leaf size=151 \[ -\frac {a^2 b \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^6 (a+b x)}-\frac {3 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {a^3 \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)} \]

[Out]

-1/7*a^3*((b*x+a)^2)^(1/2)/x^7/(b*x+a)-1/2*a^2*b*((b*x+a)^2)^(1/2)/x^6/(b*x+a)-3/5*a*b^2*((b*x+a)^2)^(1/2)/x^5
/(b*x+a)-1/4*b^3*((b*x+a)^2)^(1/2)/x^4/(b*x+a)

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Rubi [A]  time = 0.04, 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} \[ -\frac {a^3 \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {a^2 b \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^6 (a+b x)}-\frac {3 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {b^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/x^8,x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 55, normalized size = 0.36 \[ -\frac {\sqrt {(a+b x)^2} \left (20 a^3+70 a^2 b x+84 a b^2 x^2+35 b^3 x^3\right )}{140 x^7 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/x^8,x]

[Out]

-1/140*(Sqrt[(a + b*x)^2]*(20*a^3 + 70*a^2*b*x + 84*a*b^2*x^2 + 35*b^3*x^3))/(x^7*(a + b*x))

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fricas [A]  time = 0.91, size = 35, normalized size = 0.23 \[ -\frac {35 \, b^{3} x^{3} + 84 \, a b^{2} x^{2} + 70 \, a^{2} b x + 20 \, a^{3}}{140 \, x^{7}} \]

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^8,x, algorithm="fricas")

[Out]

-1/140*(35*b^3*x^3 + 84*a*b^2*x^2 + 70*a^2*b*x + 20*a^3)/x^7

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giac [A]  time = 0.19, size = 74, normalized size = 0.49 \[ \frac {b^{7} \mathrm {sgn}\left (b x + a\right )}{140 \, a^{4}} - \frac {35 \, b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 84 \, a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 70 \, a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 20 \, a^{3} \mathrm {sgn}\left (b x + a\right )}{140 \, x^{7}} \]

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^8,x, algorithm="giac")

[Out]

1/140*b^7*sgn(b*x + a)/a^4 - 1/140*(35*b^3*x^3*sgn(b*x + a) + 84*a*b^2*x^2*sgn(b*x + a) + 70*a^2*b*x*sgn(b*x +
a) + 20*a^3*sgn(b*x + a))/x^7

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maple [A]  time = 0.05, size = 52, normalized size = 0.34 \[ -\frac {\left (35 b^{3} x^{3}+84 a \,b^{2} x^{2}+70 a^{2} b x +20 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{140 \left (b x +a \right )^{3} x^{7}} \]

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^8,x)

[Out]

-1/140*(35*b^3*x^3+84*a*b^2*x^2+70*a^2*b*x+20*a^3)*((b*x+a)^2)^(3/2)/x^7/(b*x+a)^3

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maxima [B]  time = 1.48, size = 225, normalized size = 1.49 \[ -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{7}}{4 \, a^{7}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{6}}{4 \, a^{6} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{5}}{4 \, a^{7} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{4}}{4 \, a^{6} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{3}}{4 \, a^{5} x^{4}} - \frac {17 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b^{2}}{70 \, a^{4} x^{5}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} b}{14 \, a^{3} x^{6}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}{7 \, a^{2} x^{7}} \]

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^8,x, algorithm="maxima")

[Out]

-1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^7/a^7 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^6/(a^6*x) + 1/4*(b^2*x^2
+ 2*a*b*x + a^2)^(5/2)*b^5/(a^7*x^2) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^4/(a^6*x^3) + 1/4*(b^2*x^2 + 2*a*
b*x + a^2)^(5/2)*b^3/(a^5*x^4) - 17/70*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*b^2/(a^4*x^5) + 3/14*(b^2*x^2 + 2*a*b*x
+ a^2)^(5/2)*b/(a^3*x^6) - 1/7*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)/(a^2*x^7)

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mupad [B]  time = 0.20, size = 135, normalized size = 0.89 \[ -\frac {a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,x^4\,\left (a+b\,x\right )}-\frac {3\,a\,b^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{5\,x^5\,\left (a+b\,x\right )}-\frac {a^2\,b\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,x^6\,\left (a+b\,x\right )} \]

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^8,x)

[Out]

- (a^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(7*x^7*(a + b*x)) - (b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*x^4*(a +
b*x)) - (3*a*b^2*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(5*x^5*(a + b*x)) - (a^2*b*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/
(2*x^6*(a + b*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{8}}\, dx \]

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**8,x)

[Out]

Integral(((a + b*x)**2)**(3/2)/x**8, x)

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