3.7.6 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{x^9} \, dx\)

Optimal. Leaf size=210 \[ -\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{7 x^7 (a+b x)}-\frac {a b \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{2 x^6 (a+b x)}-\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{5 x^5 (a+b x)}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {770, 76} \begin {gather*} -\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{7 x^7 (a+b x)}-\frac {a b \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{2 x^6 (a+b x)}-\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{5 x^5 (a+b x)}-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^9,x]

[Out]

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

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 87, normalized size = 0.41 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (5 a^3 (7 A+8 B x)+20 a^2 b x (6 A+7 B x)+28 a b^2 x^2 (5 A+6 B x)+14 b^3 x^3 (4 A+5 B x)\right )}{280 x^8 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^9,x]

[Out]

-1/280*(Sqrt[(a + b*x)^2]*(14*b^3*x^3*(4*A + 5*B*x) + 28*a*b^2*x^2*(5*A + 6*B*x) + 20*a^2*b*x*(6*A + 7*B*x) +
5*a^3*(7*A + 8*B*x)))/(x^8*(a + b*x))

________________________________________________________________________________________

IntegrateAlgebraic [B]  time = 2.85, size = 754, normalized size = 3.59 \begin {gather*} \frac {16 b^7 \sqrt {a^2+2 a b x+b^2 x^2} \left (-35 a^{10} A b-40 a^{10} b B x-365 a^9 A b^2 x-420 a^9 b^2 B x^2-1715 a^8 A b^3 x^2-1988 a^8 b^3 B x^3-4781 a^7 A b^4 x^3-5586 a^7 b^4 B x^4-8757 a^6 A b^5 x^4-10318 a^6 b^5 B x^5-11011 a^5 A b^6 x^5-13090 a^5 b^6 B x^6-9625 a^4 A b^7 x^6-11550 a^4 b^7 B x^7-5775 a^3 A b^8 x^7-6998 a^3 b^8 B x^8-2276 a^2 A b^9 x^8-2786 a^2 b^9 B x^9-532 a A b^{10} x^9-658 a b^{10} B x^{10}-56 A b^{11} x^{10}-70 b^{11} B x^{11}\right )+16 \sqrt {b^2} b^7 \left (35 a^{11} A+40 a^{11} B x+400 a^{10} A b x+460 a^{10} b B x^2+2080 a^9 A b^2 x^2+2408 a^9 b^2 B x^3+6496 a^8 A b^3 x^3+7574 a^8 b^3 B x^4+13538 a^7 A b^4 x^4+15904 a^7 b^4 B x^5+19768 a^6 A b^5 x^5+23408 a^6 b^5 B x^6+20636 a^5 A b^6 x^6+24640 a^5 b^6 B x^7+15400 a^4 A b^7 x^7+18548 a^4 b^7 B x^8+8051 a^3 A b^8 x^8+9784 a^3 b^8 B x^9+2808 a^2 A b^9 x^9+3444 a^2 b^9 B x^{10}+588 a A b^{10} x^{10}+728 a b^{10} B x^{11}+56 A b^{11} x^{11}+70 b^{11} B x^{12}\right )}{35 \sqrt {b^2} x^8 \sqrt {a^2+2 a b x+b^2 x^2} \left (-128 a^7 b^7-896 a^6 b^8 x-2688 a^5 b^9 x^2-4480 a^4 b^{10} x^3-4480 a^3 b^{11} x^4-2688 a^2 b^{12} x^5-896 a b^{13} x^6-128 b^{14} x^7\right )+35 x^8 \left (128 a^8 b^8+1024 a^7 b^9 x+3584 a^6 b^{10} x^2+7168 a^5 b^{11} x^3+8960 a^4 b^{12} x^4+7168 a^3 b^{13} x^5+3584 a^2 b^{14} x^6+1024 a b^{15} x^7+128 b^{16} x^8\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^9,x]

[Out]

(16*b^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-35*a^10*A*b - 365*a^9*A*b^2*x - 40*a^10*b*B*x - 1715*a^8*A*b^3*x^2 - 4
20*a^9*b^2*B*x^2 - 4781*a^7*A*b^4*x^3 - 1988*a^8*b^3*B*x^3 - 8757*a^6*A*b^5*x^4 - 5586*a^7*b^4*B*x^4 - 11011*a
^5*A*b^6*x^5 - 10318*a^6*b^5*B*x^5 - 9625*a^4*A*b^7*x^6 - 13090*a^5*b^6*B*x^6 - 5775*a^3*A*b^8*x^7 - 11550*a^4
*b^7*B*x^7 - 2276*a^2*A*b^9*x^8 - 6998*a^3*b^8*B*x^8 - 532*a*A*b^10*x^9 - 2786*a^2*b^9*B*x^9 - 56*A*b^11*x^10
- 658*a*b^10*B*x^10 - 70*b^11*B*x^11) + 16*b^7*Sqrt[b^2]*(35*a^11*A + 400*a^10*A*b*x + 40*a^11*B*x + 2080*a^9*
A*b^2*x^2 + 460*a^10*b*B*x^2 + 6496*a^8*A*b^3*x^3 + 2408*a^9*b^2*B*x^3 + 13538*a^7*A*b^4*x^4 + 7574*a^8*b^3*B*
x^4 + 19768*a^6*A*b^5*x^5 + 15904*a^7*b^4*B*x^5 + 20636*a^5*A*b^6*x^6 + 23408*a^6*b^5*B*x^6 + 15400*a^4*A*b^7*
x^7 + 24640*a^5*b^6*B*x^7 + 8051*a^3*A*b^8*x^8 + 18548*a^4*b^7*B*x^8 + 2808*a^2*A*b^9*x^9 + 9784*a^3*b^8*B*x^9
+ 588*a*A*b^10*x^10 + 3444*a^2*b^9*B*x^10 + 56*A*b^11*x^11 + 728*a*b^10*B*x^11 + 70*b^11*B*x^12))/(35*Sqrt[b^
2]*x^8*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-128*a^7*b^7 - 896*a^6*b^8*x - 2688*a^5*b^9*x^2 - 4480*a^4*b^10*x^3 - 44
80*a^3*b^11*x^4 - 2688*a^2*b^12*x^5 - 896*a*b^13*x^6 - 128*b^14*x^7) + 35*x^8*(128*a^8*b^8 + 1024*a^7*b^9*x +
3584*a^6*b^10*x^2 + 7168*a^5*b^11*x^3 + 8960*a^4*b^12*x^4 + 7168*a^3*b^13*x^5 + 3584*a^2*b^14*x^6 + 1024*a*b^1
5*x^7 + 128*b^16*x^8))

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 73, normalized size = 0.35 \begin {gather*} -\frac {70 \, B b^{3} x^{4} + 35 \, A a^{3} + 56 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 140 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 40 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{280 \, x^{8}} \end {gather*}

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)^(3/2)/x^9,x, algorithm="fricas")

[Out]

-1/280*(70*B*b^3*x^4 + 35*A*a^3 + 56*(3*B*a*b^2 + A*b^3)*x^3 + 140*(B*a^2*b + A*a*b^2)*x^2 + 40*(B*a^3 + 3*A*a
^2*b)*x)/x^8

________________________________________________________________________________________

giac [A]  time = 0.17, size = 149, normalized size = 0.71 \begin {gather*} \frac {{\left (2 \, B a b^{7} - A b^{8}\right )} \mathrm {sgn}\left (b x + a\right )}{280 \, a^{5}} - \frac {70 \, B b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 168 \, B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 56 \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 140 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 140 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 40 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 120 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 35 \, A a^{3} \mathrm {sgn}\left (b x + a\right )}{280 \, x^{8}} \end {gather*}

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)^(3/2)/x^9,x, algorithm="giac")

[Out]

1/280*(2*B*a*b^7 - A*b^8)*sgn(b*x + a)/a^5 - 1/280*(70*B*b^3*x^4*sgn(b*x + a) + 168*B*a*b^2*x^3*sgn(b*x + a) +
56*A*b^3*x^3*sgn(b*x + a) + 140*B*a^2*b*x^2*sgn(b*x + a) + 140*A*a*b^2*x^2*sgn(b*x + a) + 40*B*a^3*x*sgn(b*x
+ a) + 120*A*a^2*b*x*sgn(b*x + a) + 35*A*a^3*sgn(b*x + a))/x^8

________________________________________________________________________________________

maple [A]  time = 0.05, size = 92, normalized size = 0.44 \begin {gather*} -\frac {\left (70 B \,b^{3} x^{4}+56 A \,b^{3} x^{3}+168 B a \,b^{2} x^{3}+140 A a \,b^{2} x^{2}+140 B \,a^{2} b \,x^{2}+120 A \,a^{2} b x +40 B \,a^{3} x +35 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{280 \left (b x +a \right )^{3} x^{8}} \end {gather*}

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)^(3/2)/x^9,x)

[Out]

-1/280*(70*B*b^3*x^4+56*A*b^3*x^3+168*B*a*b^2*x^3+140*A*a*b^2*x^2+140*B*a^2*b*x^2+120*A*a^2*b*x+40*B*a^3*x+35*
A*a^3)*((b*x+a)^2)^(3/2)/x^8/(b*x+a)^3

________________________________________________________________________________________

maxima [B]  time = 0.52, size = 495, normalized size = 2.36 \begin {gather*} -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{7}}{4 \, a^{7}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{8}}{4 \, a^{8}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{6}}{4 \, a^{6} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{7}}{4 \, a^{7} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{5}}{4 \, a^{7} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{6}}{4 \, a^{8} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{4}}{4 \, a^{6} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{5}}{4 \, a^{7} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{3}}{4 \, a^{5} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{4}}{4 \, a^{6} x^{4}} - \frac {17 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{2}}{70 \, a^{4} x^{5}} + \frac {69 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{3}}{280 \, a^{5} x^{5}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b}{14 \, a^{3} x^{6}} - \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{2}}{56 \, a^{4} x^{6}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{7 \, a^{2} x^{7}} + \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b}{56 \, a^{3} x^{7}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{8 \, a^{2} x^{8}} \end {gather*}

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)^(3/2)/x^9,x, algorithm="maxima")

[Out]

-1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^7/a^7 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^8/a^8 - 1/4*(b^2*x^2
+ 2*a*b*x + a^2)^(3/2)*B*b^6/(a^6*x) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^7/(a^7*x) + 1/4*(b^2*x^2 + 2*a*
b*x + a^2)^(5/2)*B*b^5/(a^7*x^2) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^6/(a^8*x^2) - 1/4*(b^2*x^2 + 2*a*b*
x + a^2)^(5/2)*B*b^4/(a^6*x^3) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^5/(a^7*x^3) + 1/4*(b^2*x^2 + 2*a*b*x
+ a^2)^(5/2)*B*b^3/(a^5*x^4) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^4/(a^6*x^4) - 17/70*(b^2*x^2 + 2*a*b*x
+ a^2)^(5/2)*B*b^2/(a^4*x^5) + 69/280*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^3/(a^5*x^5) + 3/14*(b^2*x^2 + 2*a*b*
x + a^2)^(5/2)*B*b/(a^3*x^6) - 13/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^2/(a^4*x^6) - 1/7*(b^2*x^2 + 2*a*b*x
+ a^2)^(5/2)*B/(a^2*x^7) + 11/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b/(a^3*x^7) - 1/8*(b^2*x^2 + 2*a*b*x + a^2)
^(5/2)*A/(a^2*x^8)

________________________________________________________________________________________

mupad [B]  time = 1.17, size = 196, normalized size = 0.93 \begin {gather*} -\frac {\left (\frac {B\,a^3}{7}+\frac {3\,A\,b\,a^2}{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^3}{5}+\frac {3\,B\,a\,b^2}{5}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^5\,\left (a+b\,x\right )}-\frac {A\,a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{8\,x^8\,\left (a+b\,x\right )}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,x^4\,\left (a+b\,x\right )}-\frac {a\,b\,\left (A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,x^6\,\left (a+b\,x\right )} \end {gather*}

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)^(3/2))/x^9,x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{9}}\, dx \end {gather*}

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)**(3/2)/x**9,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**9, x)

________________________________________________________________________________________