3.682 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{x^{12}} \, dx\)

Optimal. Leaf size=210 \[ -\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{10 x^{10} (a+b x)}-\frac {a b \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{3 x^9 (a+b x)}-\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{8 x^8 (a+b x)}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{11 x^{11} (a+b x)} \]

[Out]

-1/11*a^3*A*((b*x+a)^2)^(1/2)/x^11/(b*x+a)-1/10*a^2*(3*A*b+B*a)*((b*x+a)^2)^(1/2)/x^10/(b*x+a)-1/3*a*b*(A*b+B*
a)*((b*x+a)^2)^(1/2)/x^9/(b*x+a)-1/8*b^2*(A*b+3*B*a)*((b*x+a)^2)^(1/2)/x^8/(b*x+a)-1/7*b^3*B*((b*x+a)^2)^(1/2)
/x^7/(b*x+a)

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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} \[ -\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{10 x^{10} (a+b x)}-\frac {a b \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{3 x^9 (a+b x)}-\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{8 x^8 (a+b x)}-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{11 x^{11} (a+b x)}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^12,x]

[Out]

-(a^3*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(11*x^11*(a + b*x)) - (a^2*(3*A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
/(10*x^10*(a + b*x)) - (a*b*(A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*x^9*(a + b*x)) - (b^2*(A*b + 3*a*B)*
Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*x^8*(a + b*x)) - (b^3*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*x^7*(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^{12}} \, 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^{12}} \, 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^{12}}+\frac {a^2 b^3 (3 A b+a B)}{x^{11}}+\frac {3 a b^4 (A b+a B)}{x^{10}}+\frac {b^5 (A b+3 a B)}{x^9}+\frac {b^6 B}{x^8}\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}}{11 x^{11} (a+b x)}-\frac {a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{10 x^{10} (a+b x)}-\frac {a b (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^9 (a+b x)}-\frac {b^2 (A b+3 a B) \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}}{7 x^7 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 87, normalized size = 0.41 \[ -\frac {\sqrt {(a+b x)^2} \left (84 a^3 (10 A+11 B x)+308 a^2 b x (9 A+10 B x)+385 a b^2 x^2 (8 A+9 B x)+165 b^3 x^3 (7 A+8 B x)\right )}{9240 x^{11} (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^12,x]

[Out]

-1/9240*(Sqrt[(a + b*x)^2]*(165*b^3*x^3*(7*A + 8*B*x) + 385*a*b^2*x^2*(8*A + 9*B*x) + 308*a^2*b*x*(9*A + 10*B*
x) + 84*a^3*(10*A + 11*B*x)))/(x^11*(a + b*x))

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fricas [A]  time = 0.82, size = 73, normalized size = 0.35 \[ -\frac {1320 \, B b^{3} x^{4} + 840 \, A a^{3} + 1155 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 3080 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 924 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{9240 \, x^{11}} \]

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^12,x, algorithm="fricas")

[Out]

-1/9240*(1320*B*b^3*x^4 + 840*A*a^3 + 1155*(3*B*a*b^2 + A*b^3)*x^3 + 3080*(B*a^2*b + A*a*b^2)*x^2 + 924*(B*a^3
+ 3*A*a^2*b)*x)/x^11

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giac [A]  time = 0.20, size = 149, normalized size = 0.71 \[ -\frac {{\left (11 \, B a b^{10} - 7 \, A b^{11}\right )} \mathrm {sgn}\left (b x + a\right )}{9240 \, a^{8}} - \frac {1320 \, B b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 3465 \, B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 1155 \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 3080 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 3080 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 924 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 2772 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 840 \, A a^{3} \mathrm {sgn}\left (b x + a\right )}{9240 \, x^{11}} \]

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^12,x, algorithm="giac")

[Out]

-1/9240*(11*B*a*b^10 - 7*A*b^11)*sgn(b*x + a)/a^8 - 1/9240*(1320*B*b^3*x^4*sgn(b*x + a) + 3465*B*a*b^2*x^3*sgn
(b*x + a) + 1155*A*b^3*x^3*sgn(b*x + a) + 3080*B*a^2*b*x^2*sgn(b*x + a) + 3080*A*a*b^2*x^2*sgn(b*x + a) + 924*
B*a^3*x*sgn(b*x + a) + 2772*A*a^2*b*x*sgn(b*x + a) + 840*A*a^3*sgn(b*x + a))/x^11

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maple [A]  time = 0.06, size = 92, normalized size = 0.44 \[ -\frac {\left (1320 B \,b^{3} x^{4}+1155 A \,b^{3} x^{3}+3465 B a \,b^{2} x^{3}+3080 A a \,b^{2} x^{2}+3080 B \,a^{2} b \,x^{2}+2772 A \,a^{2} b x +924 B \,a^{3} x +840 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{9240 \left (b x +a \right )^{3} x^{11}} \]

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^12,x)

[Out]

-1/9240*(1320*B*b^3*x^4+1155*A*b^3*x^3+3465*B*a*b^2*x^3+3080*A*a*b^2*x^2+3080*B*a^2*b*x^2+2772*A*a^2*b*x+924*B
*a^3*x+840*A*a^3)*((b*x+a)^2)^(3/2)/x^11/(b*x+a)^3

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maxima [B]  time = 0.65, size = 675, normalized size = 3.21 \[ \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{10}}{4 \, a^{10}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{11}}{4 \, a^{11}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{9}}{4 \, a^{9} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{10}}{4 \, a^{10} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{8}}{4 \, a^{10} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{9}}{4 \, a^{11} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{7}}{4 \, a^{9} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{8}}{4 \, a^{10} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{6}}{4 \, a^{8} x^{4}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{7}}{4 \, a^{9} x^{4}} + \frac {209 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{5}}{840 \, a^{7} x^{5}} - \frac {329 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{6}}{1320 \, a^{8} x^{5}} - \frac {41 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{4}}{168 \, a^{6} x^{6}} + \frac {65 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{5}}{264 \, a^{7} x^{6}} + \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{3}}{56 \, a^{5} x^{7}} - \frac {21 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{4}}{88 \, a^{6} x^{7}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{2}}{24 \, a^{4} x^{8}} + \frac {59 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{3}}{264 \, a^{5} x^{8}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b}{6 \, a^{3} x^{9}} - \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{2}}{66 \, a^{4} x^{9}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{10 \, a^{2} x^{10}} + \frac {17 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b}{110 \, a^{3} x^{10}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{11 \, a^{2} x^{11}} \]

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^12,x, algorithm="maxima")

[Out]

1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^10/a^10 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^11/a^11 + 1/4*(b^2*x
^2 + 2*a*b*x + a^2)^(3/2)*B*b^9/(a^9*x) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^10/(a^10*x) - 1/4*(b^2*x^2 +
2*a*b*x + a^2)^(5/2)*B*b^8/(a^10*x^2) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^9/(a^11*x^2) + 1/4*(b^2*x^2 +
2*a*b*x + a^2)^(5/2)*B*b^7/(a^9*x^3) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^8/(a^10*x^3) - 1/4*(b^2*x^2 +
2*a*b*x + a^2)^(5/2)*B*b^6/(a^8*x^4) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^7/(a^9*x^4) + 209/840*(b^2*x^2
+ 2*a*b*x + a^2)^(5/2)*B*b^5/(a^7*x^5) - 329/1320*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^6/(a^8*x^5) - 41/168*(b^
2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^4/(a^6*x^6) + 65/264*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^5/(a^7*x^6) + 13/56*
(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^3/(a^5*x^7) - 21/88*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^4/(a^6*x^7) - 5/24
*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^2/(a^4*x^8) + 59/264*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^3/(a^5*x^8) + 1/
6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b/(a^3*x^9) - 13/66*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^2/(a^4*x^9) - 1/10
*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B/(a^2*x^10) + 17/110*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b/(a^3*x^10) - 1/11*(
b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A/(a^2*x^11)

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mupad [B]  time = 1.16, size = 196, normalized size = 0.93 \[ -\frac {\left (\frac {B\,a^3}{10}+\frac {3\,A\,b\,a^2}{10}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^{10}\,\left (a+b\,x\right )}-\frac {\left (\frac {A\,b^3}{8}+\frac {3\,B\,a\,b^2}{8}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^8\,\left (a+b\,x\right )}-\frac {A\,a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{11\,x^{11}\,\left (a+b\,x\right )}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{7\,x^7\,\left (a+b\,x\right )}-\frac {a\,b\,\left (A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{3\,x^9\,\left (a+b\,x\right )} \]

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)^(3/2))/x^12,x)

[Out]

- (((B*a^3)/10 + (3*A*a^2*b)/10)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^10*(a + b*x)) - (((A*b^3)/8 + (3*B*a*b^2)
/8)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^8*(a + b*x)) - (A*a^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(11*x^11*(a + b
*x)) - (B*b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(7*x^7*(a + b*x)) - (a*b*(A*b + B*a)*(a^2 + b^2*x^2 + 2*a*b*x)^
(1/2))/(3*x^9*(a + b*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{12}}\, dx \]

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)**(3/2)/x**12,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**12, x)

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