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]

-1/7*A*(b^2*x^2+2*a*b*x+a^2)^(7/2)/a^2/x^7+1/6*(A*b-B*a)*(b*x+a)^5*((b*x+a)^2)^(1/2)/a^2/x^6

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Rubi [A]  time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.04, size = 122, normalized size = 1.58 \[ -\frac {\sqrt {(a+b x)^2} \left (a^5 (6 A+7 B x)+7 a^4 b x (5 A+6 B x)+21 a^3 b^2 x^2 (4 A+5 B x)+35 a^2 b^3 x^3 (3 A+4 B x)+35 a b^4 x^4 (2 A+3 B x)+21 b^5 x^5 (A+2 B x)\right )}{42 x^7 (a+b x)} \]

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]

-1/42*(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)))/(x^7*(a + b*x))

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fricas [A]  time = 0.95, size = 119, normalized size = 1.55 \[ -\frac {42 \, B b^{5} x^{6} + 6 \, A a^{5} + 21 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 70 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 105 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 42 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 7 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{42 \, x^{7}} \]

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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giac [B]  time = 0.22, size = 221, normalized size = 2.87 \[ -\frac {{\left (7 \, B a b^{6} - A b^{7}\right )} \mathrm {sgn}\left (b x + a\right )}{42 \, a^{2}} - \frac {42 \, B b^{5} x^{6} \mathrm {sgn}\left (b x + a\right ) + 105 \, B a b^{4} x^{5} \mathrm {sgn}\left (b x + a\right ) + 21 \, A b^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + 140 \, B a^{2} b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 70 \, A a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 105 \, B a^{3} b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 105 \, A a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 42 \, B a^{4} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 84 \, A a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 7 \, B a^{5} x \mathrm {sgn}\left (b x + a\right ) + 35 \, A a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 6 \, A a^{5} \mathrm {sgn}\left (b x + a\right )}{42 \, x^{7}} \]

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

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maple [B]  time = 0.08, size = 140, normalized size = 1.82 \[ -\frac {\left (42 B \,b^{5} x^{6}+21 A \,b^{5} x^{5}+105 B a \,b^{4} x^{5}+70 A a \,b^{4} x^{4}+140 B \,a^{2} b^{3} x^{4}+105 A \,a^{2} b^{3} x^{3}+105 B \,a^{3} b^{2} x^{3}+84 A \,a^{3} b^{2} x^{2}+42 B \,a^{4} b \,x^{2}+35 A \,a^{4} b x +7 B \,a^{5} x +6 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{42 \left (b x +a \right )^{5} x^{7}} \]

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 [B]  time = 0.58, size = 435, normalized size = 5.65 \[ \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{6}}{6 \, a^{6}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{7}}{6 \, a^{7}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{5}}{6 \, a^{5} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{6}}{6 \, a^{6} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b^{4}}{6 \, a^{6} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{5}}{6 \, a^{7} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b^{3}}{6 \, a^{5} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{4}}{6 \, a^{6} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b^{2}}{6 \, a^{4} x^{4}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{3}}{6 \, a^{5} x^{4}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b}{6 \, a^{3} x^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{2}}{6 \, a^{4} x^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B}{6 \, a^{2} x^{6}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b}{6 \, a^{3} x^{6}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A}{7 \, a^{2} x^{7}} \]

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]

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

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mupad [B]  time = 1.33, size = 284, normalized size = 3.69 \[ -\frac {\left (\frac {B\,a^5}{6}+\frac {5\,A\,b\,a^4}{6}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^6\,\left (a+b\,x\right )}-\frac {\left (\frac {A\,b^5}{2}+\frac {5\,B\,a\,b^4}{2}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^2\,\left (a+b\,x\right )}-\frac {A\,a^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )}-\frac {B\,b^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x\,\left (a+b\,x\right )}-\frac {5\,a\,b^3\,\left (A\,b+2\,B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{3\,x^3\,\left (a+b\,x\right )}-\frac {a^3\,b\,\left (2\,A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^5\,\left (a+b\,x\right )}-\frac {5\,a^2\,b^2\,\left (A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,x^4\,\left (a+b\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{8}}\, dx \]

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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