3.697 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^{5/2}}{x^8} \, dx\)

Optimal. Leaf size=77 \[ \frac{(a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2} (A b-a B)}{6 a^2 x^6}-\frac{A \left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 a^2 x^7} \]

[Out]

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

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Rubi [A]  time = 0.0466944, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {769, 646, 37} \[ \frac{(a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2} (A b-a B)}{6 a^2 x^6}-\frac{A \left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 a^2 x^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 769

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(-2*c*(e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)^2), x] + Dist[(2*c*f -
b*g)/(2*c*d - b*e), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x]
&& EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && NeQ[2*c*f - b*g, 0] && NeQ[2*c*d - b*e, 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]

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

Mathematica [A]  time = 0.0386734, size = 122, normalized size = 1.58 \[ -\frac{\sqrt{(a+b x)^2} \left (21 a^3 b^2 x^2 (4 A+5 B x)+35 a^2 b^3 x^3 (3 A+4 B x)+7 a^4 b x (5 A+6 B x)+a^5 (6 A+7 B x)+35 a b^4 x^4 (2 A+3 B x)+21 b^5 x^5 (A+2 B x)\right )}{42 x^7 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.007, size = 140, normalized size = 1.8 \begin{align*} -{\frac{42\,B{b}^{5}{x}^{6}+21\,A{x}^{5}{b}^{5}+105\,B{x}^{5}a{b}^{4}+70\,A{x}^{4}a{b}^{4}+140\,B{x}^{4}{a}^{2}{b}^{3}+105\,A{x}^{3}{a}^{2}{b}^{3}+105\,B{x}^{3}{a}^{3}{b}^{2}+84\,A{x}^{2}{a}^{3}{b}^{2}+42\,B{x}^{2}{a}^{4}b+35\,A{a}^{4}bx+7\,B{a}^{5}x+6\,A{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*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^8,x)

[Out]

-1/42*(42*B*b^5*x^6+21*A*b^5*x^5+105*B*a*b^4*x^5+70*A*a*b^4*x^4+140*B*a^2*b^3*x^4+105*A*a^2*b^3*x^3+105*B*a^3*
b^2*x^3+84*A*a^3*b^2*x^2+42*B*a^4*b*x^2+35*A*a^4*b*x+7*B*a^5*x+6*A*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*x+A)*(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 = 1.57644, size = 258, normalized size = 3.35 \begin{align*} -\frac{42 \, B b^{5} x^{6} + 6 \, A a^{5} + 21 \,{\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 70 \,{\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 105 \,{\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 42 \,{\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 7 \,{\left (B a^{5} + 5 \, A a^{4} b\right )} x}{42 \, x^{7}} \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^8,x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [B]  time = 1.26353, size = 298, normalized size = 3.87 \begin{align*} -\frac{{\left (7 \, B a b^{6} - A b^{7}\right )} \mathrm{sgn}\left (b x + a\right )}{42 \, a^{2}} - \frac{42 \, B b^{5} x^{6} \mathrm{sgn}\left (b x + a\right ) + 105 \, B a b^{4} x^{5} \mathrm{sgn}\left (b x + a\right ) + 21 \, A b^{5} x^{5} \mathrm{sgn}\left (b x + a\right ) + 140 \, B a^{2} b^{3} x^{4} \mathrm{sgn}\left (b x + a\right ) + 70 \, A a b^{4} x^{4} \mathrm{sgn}\left (b x + a\right ) + 105 \, B a^{3} b^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + 105 \, A a^{2} b^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + 42 \, B a^{4} b x^{2} \mathrm{sgn}\left (b x + a\right ) + 84 \, A a^{3} b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 7 \, B a^{5} x \mathrm{sgn}\left (b x + a\right ) + 35 \, A a^{4} b x \mathrm{sgn}\left (b x + a\right ) + 6 \, A 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*x+A)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/x^8,x, algorithm="giac")

[Out]

-1/42*(7*B*a*b^6 - A*b^7)*sgn(b*x + a)/a^2 - 1/42*(42*B*b^5*x^6*sgn(b*x + a) + 105*B*a*b^4*x^5*sgn(b*x + a) +
21*A*b^5*x^5*sgn(b*x + a) + 140*B*a^2*b^3*x^4*sgn(b*x + a) + 70*A*a*b^4*x^4*sgn(b*x + a) + 105*B*a^3*b^2*x^3*s
gn(b*x + a) + 105*A*a^2*b^3*x^3*sgn(b*x + a) + 42*B*a^4*b*x^2*sgn(b*x + a) + 84*A*a^3*b^2*x^2*sgn(b*x + a) + 7
*B*a^5*x*sgn(b*x + a) + 35*A*a^4*b*x*sgn(b*x + a) + 6*A*a^5*sgn(b*x + a))/x^7