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

[Out]

-1/8*A*(b*x+a)^5*((b*x+a)^2)^(1/2)/a/x^8+1/28*(A*b-4*B*a)*(b*x+a)^5*((b*x+a)^2)^(1/2)/a^2/x^7-1/168*b*(A*b-4*B
*a)*(b*x+a)^5*((b*x+a)^2)^(1/2)/a^3/x^6

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Rubi [A]  time = 0.07, antiderivative size = 130, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.04, size = 125, normalized size = 0.96 \[ -\frac {\sqrt {(a+b x)^2} \left (3 a^5 (7 A+8 B x)+20 a^4 b x (6 A+7 B x)+56 a^3 b^2 x^2 (5 A+6 B x)+84 a^2 b^3 x^3 (4 A+5 B x)+70 a b^4 x^4 (3 A+4 B x)+28 b^5 x^5 (2 A+3 B x)\right )}{168 x^8 (a+b x)} \]

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]

-1/168*(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)))/(x^8*(a + b*x))

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fricas [A]  time = 0.72, size = 119, normalized size = 0.92 \[ -\frac {84 \, B b^{5} x^{6} + 21 \, A a^{5} + 56 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 210 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 336 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} + 140 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} + 24 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{168 \, x^{8}} \]

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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giac [B]  time = 0.17, size = 221, normalized size = 1.70 \[ \frac {{\left (4 \, B a b^{7} - A b^{8}\right )} \mathrm {sgn}\left (b x + a\right )}{168 \, a^{3}} - \frac {84 \, B b^{5} x^{6} \mathrm {sgn}\left (b x + a\right ) + 280 \, B a b^{4} x^{5} \mathrm {sgn}\left (b x + a\right ) + 56 \, A b^{5} x^{5} \mathrm {sgn}\left (b x + a\right ) + 420 \, B a^{2} b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 210 \, A a b^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 336 \, B a^{3} b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 336 \, A a^{2} b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 140 \, B a^{4} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 280 \, A a^{3} b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 24 \, B a^{5} x \mathrm {sgn}\left (b x + a\right ) + 120 \, A a^{4} b x \mathrm {sgn}\left (b x + a\right ) + 21 \, A a^{5} \mathrm {sgn}\left (b x + a\right )}{168 \, x^{8}} \]

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

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maple [A]  time = 0.05, size = 140, normalized size = 1.08 \[ -\frac {\left (84 B \,b^{5} x^{6}+56 A \,b^{5} x^{5}+280 B a \,b^{4} x^{5}+210 A a \,b^{4} x^{4}+420 B \,a^{2} b^{3} x^{4}+336 A \,a^{2} b^{3} x^{3}+336 B \,a^{3} b^{2} x^{3}+280 A \,a^{3} b^{2} x^{2}+140 B \,a^{4} b \,x^{2}+120 A \,a^{4} b x +24 B \,a^{5} x +21 A \,a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{168 \left (b x +a \right )^{5} x^{8}} \]

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 [B]  time = 0.63, size = 495, normalized size = 3.81 \[ -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{7}}{6 \, a^{7}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{8}}{6 \, a^{8}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{6}}{6 \, a^{6} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{7}}{6 \, a^{7} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b^{5}}{6 \, a^{7} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{6}}{6 \, a^{8} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b^{4}}{6 \, a^{6} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{5}}{6 \, a^{7} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b^{3}}{6 \, a^{5} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{4}}{6 \, a^{6} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b^{2}}{6 \, a^{4} x^{5}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{3}}{6 \, a^{5} x^{5}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b}{6 \, a^{3} x^{6}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{2}}{6 \, a^{4} x^{6}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B}{7 \, a^{2} x^{7}} + \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b}{56 \, a^{3} x^{7}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A}{8 \, a^{2} x^{8}} \]

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]

-1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^7/a^7 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^8/a^8 - 1/6*(b^2*x^2
+ 2*a*b*x + a^2)^(5/2)*B*b^6/(a^6*x) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^7/(a^7*x) + 1/6*(b^2*x^2 + 2*a*
b*x + a^2)^(7/2)*B*b^5/(a^7*x^2) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^6/(a^8*x^2) - 1/6*(b^2*x^2 + 2*a*b*
x + a^2)^(7/2)*B*b^4/(a^6*x^3) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^5/(a^7*x^3) + 1/6*(b^2*x^2 + 2*a*b*x
+ a^2)^(7/2)*B*b^3/(a^5*x^4) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^4/(a^6*x^4) - 1/6*(b^2*x^2 + 2*a*b*x +
a^2)^(7/2)*B*b^2/(a^4*x^5) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^3/(a^5*x^5) + 1/6*(b^2*x^2 + 2*a*b*x + a^
2)^(7/2)*B*b/(a^3*x^6) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^2/(a^4*x^6) - 1/7*(b^2*x^2 + 2*a*b*x + a^2)^(
7/2)*B/(a^2*x^7) + 9/56*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b/(a^3*x^7) - 1/8*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A/
(a^2*x^8)

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mupad [B]  time = 1.31, size = 284, normalized size = 2.18 \[ -\frac {\left (\frac {B\,a^5}{7}+\frac {5\,A\,b\,a^4}{7}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^7\,\left (a+b\,x\right )}-\frac {\left (\frac {A\,b^5}{3}+\frac {5\,B\,a\,b^4}{3}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^3\,\left (a+b\,x\right )}-\frac {A\,a^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{8\,x^8\,\left (a+b\,x\right )}-\frac {B\,b^5\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,x^2\,\left (a+b\,x\right )}-\frac {5\,a\,b^3\,\left (A\,b+2\,B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,x^4\,\left (a+b\,x\right )}-\frac {5\,a^3\,b\,\left (2\,A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{6\,x^6\,\left (a+b\,x\right )}-\frac {2\,a^2\,b^2\,\left (A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^5\,\left (a+b\,x\right )} \]

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^9,x)

[Out]

- (((B*a^5)/7 + (5*A*a^4*b)/7)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^7*(a + b*x)) - (((A*b^5)/3 + (5*B*a*b^4)/3)
*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^3*(a + b*x)) - (A*a^5*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(8*x^8*(a + b*x))
- (B*b^5*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(2*x^2*(a + b*x)) - (5*a*b^3*(A*b + 2*B*a)*(a^2 + b^2*x^2 + 2*a*b*x)
^(1/2))/(4*x^4*(a + b*x)) - (5*a^3*b*(2*A*b + B*a)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(6*x^6*(a + b*x)) - (2*a^2
*b^2*(A*b + B*a)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^5*(a + b*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{x^{9}}\, dx \]

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**9,x)

[Out]

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

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