3.698 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^{5/2}}{x^9} \, dx\)

Optimal. Leaf size=130 \[ -\frac{b \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-4 a B)}{168 a^3 x^6}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-4 a B)}{28 a^2 x^7}-\frac{A \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5}{8 a x^8} \]

[Out]

-(A*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*a*x^8) + ((A*b - 4*a*B)*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2
*x^2])/(28*a^2*x^7) - (b*(A*b - 4*a*B)*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(168*a^3*x^6)

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Rubi [A]  time = 0.0729824, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {770, 78, 45, 37} \[ -\frac{b \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-4 a B)}{168 a^3 x^6}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-4 a B)}{28 a^2 x^7}-\frac{A \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5}{8 a x^8} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^9,x]

[Out]

-(A*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*a*x^8) + ((A*b - 4*a*B)*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2
*x^2])/(28*a^2*x^7) - (b*(A*b - 4*a*B)*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(168*a^3*x^6)

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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{(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{x^9} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (a b+b^2 x\right )^5 (A+B x)}{x^9} \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=-\frac{A (a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{8 a x^8}+\frac{\left (\left (-2 A b^2+8 a b B\right ) \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \frac{\left (a b+b^2 x\right )^5}{x^8} \, dx}{8 a b^5 \left (a b+b^2 x\right )}\\ &=-\frac{A (a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{8 a x^8}+\frac{(A b-4 a B) (a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{28 a^2 x^7}-\frac{\left (\left (-2 A b^2+8 a b B\right ) \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \frac{\left (a b+b^2 x\right )^5}{x^7} \, dx}{56 a^2 b^4 \left (a b+b^2 x\right )}\\ &=-\frac{A (a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{8 a x^8}+\frac{(A b-4 a B) (a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{28 a^2 x^7}-\frac{b (A b-4 a B) (a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{168 a^3 x^6}\\ \end{align*}

Mathematica [A]  time = 0.0397723, size = 125, normalized size = 0.96 \[ -\frac{\sqrt{(a+b x)^2} \left (56 a^3 b^2 x^2 (5 A+6 B x)+84 a^2 b^3 x^3 (4 A+5 B x)+20 a^4 b x (6 A+7 B x)+3 a^5 (7 A+8 B x)+70 a b^4 x^4 (3 A+4 B x)+28 b^5 x^5 (2 A+3 B x)\right )}{168 x^8 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^9,x]

[Out]

-(Sqrt[(a + b*x)^2]*(28*b^5*x^5*(2*A + 3*B*x) + 70*a*b^4*x^4*(3*A + 4*B*x) + 84*a^2*b^3*x^3*(4*A + 5*B*x) + 56
*a^3*b^2*x^2*(5*A + 6*B*x) + 20*a^4*b*x*(6*A + 7*B*x) + 3*a^5*(7*A + 8*B*x)))/(168*x^8*(a + b*x))

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Maple [A]  time = 0.007, size = 140, normalized size = 1.1 \begin{align*} -{\frac{84\,B{b}^{5}{x}^{6}+56\,A{x}^{5}{b}^{5}+280\,B{x}^{5}a{b}^{4}+210\,A{x}^{4}a{b}^{4}+420\,B{x}^{4}{a}^{2}{b}^{3}+336\,A{x}^{3}{a}^{2}{b}^{3}+336\,B{x}^{3}{a}^{3}{b}^{2}+280\,A{x}^{2}{a}^{3}{b}^{2}+140\,B{x}^{2}{a}^{4}b+120\,A{a}^{4}bx+24\,B{a}^{5}x+21\,A{a}^{5}}{168\,{x}^{8} \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*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^9,x)

[Out]

-1/168*(84*B*b^5*x^6+56*A*b^5*x^5+280*B*a*b^4*x^5+210*A*a*b^4*x^4+420*B*a^2*b^3*x^4+336*A*a^2*b^3*x^3+336*B*a^
3*b^2*x^3+280*A*a^3*b^2*x^2+140*B*a^4*b*x^2+120*A*a^4*b*x+24*B*a^5*x+21*A*a^5)*((b*x+a)^2)^(5/2)/x^8/(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*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^9,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54642, size = 265, normalized size = 2.04 \begin{align*} -\frac{84 \, B b^{5} x^{6} + 21 \, A a^{5} + 56 \,{\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 210 \,{\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 336 \,{\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 140 \,{\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 24 \,{\left (B a^{5} + 5 \, A a^{4} b\right )} x}{168 \, x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^9,x, algorithm="fricas")

[Out]

-1/168*(84*B*b^5*x^6 + 21*A*a^5 + 56*(5*B*a*b^4 + A*b^5)*x^5 + 210*(2*B*a^2*b^3 + A*a*b^4)*x^4 + 336*(B*a^3*b^
2 + A*a^2*b^3)*x^3 + 140*(B*a^4*b + 2*A*a^3*b^2)*x^2 + 24*(B*a^5 + 5*A*a^4*b)*x)/x^8

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac{5}{2}}}{x^{9}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(5/2)/x**9,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(5/2)/x**9, x)

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Giac [B]  time = 1.22791, size = 298, normalized size = 2.29 \begin{align*} \frac{{\left (4 \, B a b^{7} - A b^{8}\right )} \mathrm{sgn}\left (b x + a\right )}{168 \, a^{3}} - \frac{84 \, B b^{5} x^{6} \mathrm{sgn}\left (b x + a\right ) + 280 \, B a b^{4} x^{5} \mathrm{sgn}\left (b x + a\right ) + 56 \, A b^{5} x^{5} \mathrm{sgn}\left (b x + a\right ) + 420 \, B a^{2} b^{3} x^{4} \mathrm{sgn}\left (b x + a\right ) + 210 \, A a b^{4} x^{4} \mathrm{sgn}\left (b x + a\right ) + 336 \, B a^{3} b^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + 336 \, A a^{2} b^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + 140 \, B a^{4} b x^{2} \mathrm{sgn}\left (b x + a\right ) + 280 \, A a^{3} b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 24 \, B a^{5} x \mathrm{sgn}\left (b x + a\right ) + 120 \, A a^{4} b x \mathrm{sgn}\left (b x + a\right ) + 21 \, A a^{5} \mathrm{sgn}\left (b x + a\right )}{168 \, x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^9,x, algorithm="giac")

[Out]

1/168*(4*B*a*b^7 - A*b^8)*sgn(b*x + a)/a^3 - 1/168*(84*B*b^5*x^6*sgn(b*x + a) + 280*B*a*b^4*x^5*sgn(b*x + a) +
56*A*b^5*x^5*sgn(b*x + a) + 420*B*a^2*b^3*x^4*sgn(b*x + a) + 210*A*a*b^4*x^4*sgn(b*x + a) + 336*B*a^3*b^2*x^3
*sgn(b*x + a) + 336*A*a^2*b^3*x^3*sgn(b*x + a) + 140*B*a^4*b*x^2*sgn(b*x + a) + 280*A*a^3*b^2*x^2*sgn(b*x + a)
+ 24*B*a^5*x*sgn(b*x + a) + 120*A*a^4*b*x*sgn(b*x + a) + 21*A*a^5*sgn(b*x + a))/x^8