### 3.176 $$\int \frac{(a^2+2 a b x+b^2 x^2)^{5/2}}{x^8} \, dx$$

Optimal. Leaf size=76 $\frac{b (a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{42 a^2 x^6}-\frac{(a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{7 a x^7}$

[Out]

-((a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*a*x^7) + (b*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(42*a^2
*x^6)

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Rubi [A]  time = 0.0230251, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {646, 45, 37} $\frac{b (a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{42 a^2 x^6}-\frac{(a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{7 a x^7}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^8,x]

[Out]

-((a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*a*x^7) + (b*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(42*a^2
*x^6)

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 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^8} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (a b+b^2 x\right )^5}{x^8} \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=-\frac{(a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{7 a x^7}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (a b+b^2 x\right )^5}{x^7} \, dx}{7 a b^3 \left (a b+b^2 x\right )}\\ &=-\frac{(a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{7 a x^7}+\frac{b (a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{42 a^2 x^6}\\ \end{align*}

Mathematica [A]  time = 0.0157314, size = 77, normalized size = 1.01 $-\frac{\sqrt{(a+b x)^2} \left (84 a^3 b^2 x^2+105 a^2 b^3 x^3+35 a^4 b x+6 a^5+70 a b^4 x^4+21 b^5 x^5\right )}{42 x^7 (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/x^8,x]

[Out]

-(Sqrt[(a + b*x)^2]*(6*a^5 + 35*a^4*b*x + 84*a^3*b^2*x^2 + 105*a^2*b^3*x^3 + 70*a*b^4*x^4 + 21*b^5*x^5))/(42*x
^7*(a + b*x))

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Maple [A]  time = 0.203, size = 74, normalized size = 1. \begin{align*} -{\frac{21\,{b}^{5}{x}^{5}+70\,a{b}^{4}{x}^{4}+105\,{a}^{2}{b}^{3}{x}^{3}+84\,{a}^{3}{b}^{2}{x}^{2}+35\,{a}^{4}bx+6\,{a}^{5}}{42\,{x}^{7} \left ( bx+a \right ) ^{5}} \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^8,x)

[Out]

-1/42*(21*b^5*x^5+70*a*b^4*x^4+105*a^2*b^3*x^3+84*a^3*b^2*x^2+35*a^4*b*x+6*a^5)*((b*x+a)^2)^(5/2)/x^7/(b*x+a)^
5

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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^8,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.66511, size = 128, normalized size = 1.68 \begin{align*} -\frac{21 \, b^{5} x^{5} + 70 \, a b^{4} x^{4} + 105 \, a^{2} b^{3} x^{3} + 84 \, a^{3} b^{2} x^{2} + 35 \, a^{4} b x + 6 \, a^{5}}{42 \, x^{7}} \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^8,x, algorithm="fricas")

[Out]

-1/42*(21*b^5*x^5 + 70*a*b^4*x^4 + 105*a^2*b^3*x^3 + 84*a^3*b^2*x^2 + 35*a^4*b*x + 6*a^5)/x^7

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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^{8}}\, 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**8,x)

[Out]

Integral(((a + b*x)**2)**(5/2)/x**8, x)

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Giac [B]  time = 1.38857, size = 146, normalized size = 1.92 \begin{align*} \frac{b^{7} \mathrm{sgn}\left (b x + a\right )}{42 \, a^{2}} - \frac{21 \, b^{5} x^{5} \mathrm{sgn}\left (b x + a\right ) + 70 \, a b^{4} x^{4} \mathrm{sgn}\left (b x + a\right ) + 105 \, a^{2} b^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + 84 \, a^{3} b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 35 \, a^{4} b x \mathrm{sgn}\left (b x + a\right ) + 6 \, a^{5} \mathrm{sgn}\left (b x + a\right )}{42 \, x^{7}} \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^8,x, algorithm="giac")

[Out]

1/42*b^7*sgn(b*x + a)/a^2 - 1/42*(21*b^5*x^5*sgn(b*x + a) + 70*a*b^4*x^4*sgn(b*x + a) + 105*a^2*b^3*x^3*sgn(b*
x + a) + 84*a^3*b^2*x^2*sgn(b*x + a) + 35*a^4*b*x*sgn(b*x + a) + 6*a^5*sgn(b*x + a))/x^7