Integrand size = 29, antiderivative size = 210 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{11}} \, dx=-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{10 x^{10} (a+b x)}-\frac {a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{9 x^9 (a+b x)}-\frac {3 a b (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac {b^2 (A b+3 a B) \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)} \] Output:
-1/10*a^3*A*((b*x+a)^2)^(1/2)/x^10/(b*x+a)-1/9*a^2*(3*A*b+B*a)*((b*x+a)^2) ^(1/2)/x^9/(b*x+a)-3/8*a*b*(A*b+B*a)*((b*x+a)^2)^(1/2)/x^8/(b*x+a)-1/7*b^2 *(A*b+3*B*a)*((b*x+a)^2)^(1/2)/x^7/(b*x+a)-1/6*b^3*B*((b*x+a)^2)^(1/2)/x^6 /(b*x+a)
Time = 1.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.41 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{11}} \, dx=-\frac {\sqrt {(a+b x)^2} \left (60 b^3 x^3 (6 A+7 B x)+135 a b^2 x^2 (7 A+8 B x)+105 a^2 b x (8 A+9 B x)+28 a^3 (9 A+10 B x)\right )}{2520 x^{10} (a+b x)} \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^11,x]
Output:
-1/2520*(Sqrt[(a + b*x)^2]*(60*b^3*x^3*(6*A + 7*B*x) + 135*a*b^2*x^2*(7*A + 8*B*x) + 105*a^2*b*x*(8*A + 9*B*x) + 28*a^3*(9*A + 10*B*x)))/(x^10*(a + b*x))
Time = 0.42 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.49, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1187, 27, 85, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{x^{11}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{x^{11}}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{x^{11}}dx}{a+b x}\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {A a^3}{x^{11}}+\frac {(3 A b+a B) a^2}{x^{10}}+\frac {3 b (A b+a B) a}{x^9}+\frac {b^3 B}{x^7}+\frac {b^2 (A b+3 a B)}{x^8}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {a^3 A}{10 x^{10}}-\frac {a^2 (a B+3 A b)}{9 x^9}-\frac {b^2 (3 a B+A b)}{7 x^7}-\frac {3 a b (a B+A b)}{8 x^8}-\frac {b^3 B}{6 x^6}\right )}{a+b x}\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^11,x]
Output:
((-1/10*(a^3*A)/x^10 - (a^2*(3*A*b + a*B))/(9*x^9) - (3*a*b*(A*b + a*B))/( 8*x^8) - (b^2*(A*b + 3*a*B))/(7*x^7) - (b^3*B)/(6*x^6))*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > 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]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(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)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Time = 1.61 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.43
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,b^{3} x^{4}}{6}+\left (-\frac {1}{7} A \,b^{3}-\frac {3}{7} B a \,b^{2}\right ) x^{3}+\left (-\frac {3}{8} A a \,b^{2}-\frac {3}{8} B \,a^{2} b \right ) x^{2}+\left (-\frac {1}{3} A \,a^{2} b -\frac {1}{9} B \,a^{3}\right ) x -\frac {a^{3} A}{10}\right )}{\left (b x +a \right ) x^{10}}\) | \(90\) |
gosper | \(-\frac {\left (420 B \,b^{3} x^{4}+360 A \,b^{3} x^{3}+1080 B a \,b^{2} x^{3}+945 A a \,b^{2} x^{2}+945 B \,a^{2} b \,x^{2}+840 A \,a^{2} b x +280 B \,a^{3} x +252 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{2520 x^{10} \left (b x +a \right )^{3}}\) | \(92\) |
default | \(-\frac {\left (420 B \,b^{3} x^{4}+360 A \,b^{3} x^{3}+1080 B a \,b^{2} x^{3}+945 A a \,b^{2} x^{2}+945 B \,a^{2} b \,x^{2}+840 A \,a^{2} b x +280 B \,a^{3} x +252 a^{3} A \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{2520 x^{10} \left (b x +a \right )^{3}}\) | \(92\) |
orering | \(-\frac {\left (420 B \,b^{3} x^{4}+360 A \,b^{3} x^{3}+1080 B a \,b^{2} x^{3}+945 A a \,b^{2} x^{2}+945 B \,a^{2} b \,x^{2}+840 A \,a^{2} b x +280 B \,a^{3} x +252 a^{3} A \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}{2520 x^{10} \left (b x +a \right )^{3}}\) | \(101\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^11,x,method=_RETURNVERBOSE)
Output:
((b*x+a)^2)^(1/2)/(b*x+a)*(-1/6*B*b^3*x^4+(-1/7*A*b^3-3/7*B*a*b^2)*x^3+(-3 /8*A*a*b^2-3/8*B*a^2*b)*x^2+(-1/3*A*a^2*b-1/9*B*a^3)*x-1/10*a^3*A)/x^10
Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.35 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{11}} \, dx=-\frac {420 \, B b^{3} x^{4} + 252 \, A a^{3} + 360 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 945 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 280 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{2520 \, x^{10}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^11,x, algorithm="fricas")
Output:
-1/2520*(420*B*b^3*x^4 + 252*A*a^3 + 360*(3*B*a*b^2 + A*b^3)*x^3 + 945*(B* a^2*b + A*a*b^2)*x^2 + 280*(B*a^3 + 3*A*a^2*b)*x)/x^10
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{11}} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{11}}\, dx \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/x**11,x)
Output:
Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**11, x)
Leaf count of result is larger than twice the leaf count of optimal. 615 vs. \(2 (145) = 290\).
Time = 0.04 (sec) , antiderivative size = 615, normalized size of antiderivative = 2.93 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{11}} \, dx =\text {Too large to display} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^11,x, algorithm="maxima")
Output:
-1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^9/a^9 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^10/a^10 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^8/(a^8*x) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^9/(a^9*x) + 1/4*(b^2*x^2 + 2*a* b*x + a^2)^(5/2)*B*b^7/(a^9*x^2) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b ^8/(a^10*x^2) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^6/(a^8*x^3) + 1/4* (b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^7/(a^9*x^3) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^5/(a^7*x^4) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^6/(a ^8*x^4) - 125/504*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^4/(a^6*x^5) + 209/84 0*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^5/(a^7*x^5) + 121/504*(b^2*x^2 + 2*a *b*x + a^2)^(5/2)*B*b^3/(a^5*x^6) - 41/168*(b^2*x^2 + 2*a*b*x + a^2)^(5/2) *A*b^4/(a^6*x^6) - 37/168*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^2/(a^4*x^7) + 13/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^3/(a^5*x^7) + 13/72*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b/(a^3*x^8) - 5/24*(b^2*x^2 + 2*a*b*x + a^2)^(5/2) *A*b^2/(a^4*x^8) - 1/9*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B/(a^2*x^9) + 1/6*( b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b/(a^3*x^9) - 1/10*(b^2*x^2 + 2*a*b*x + a ^2)^(5/2)*A/(a^2*x^10)
Time = 0.27 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.71 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{11}} \, dx=\frac {{\left (5 \, B a b^{9} - 3 \, A b^{10}\right )} \mathrm {sgn}\left (b x + a\right )}{2520 \, a^{7}} - \frac {420 \, B b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 1080 \, B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 360 \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 945 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 945 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 280 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 840 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 252 \, A a^{3} \mathrm {sgn}\left (b x + a\right )}{2520 \, x^{10}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^11,x, algorithm="giac")
Output:
1/2520*(5*B*a*b^9 - 3*A*b^10)*sgn(b*x + a)/a^7 - 1/2520*(420*B*b^3*x^4*sgn (b*x + a) + 1080*B*a*b^2*x^3*sgn(b*x + a) + 360*A*b^3*x^3*sgn(b*x + a) + 9 45*B*a^2*b*x^2*sgn(b*x + a) + 945*A*a*b^2*x^2*sgn(b*x + a) + 280*B*a^3*x*s gn(b*x + a) + 840*A*a^2*b*x*sgn(b*x + a) + 252*A*a^3*sgn(b*x + a))/x^10
Time = 10.80 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{11}} \, dx=-\frac {\left (\frac {B\,a^3}{9}+\frac {A\,b\,a^2}{3}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^9\,\left (a+b\,x\right )}-\frac {\left (\frac {A\,b^3}{7}+\frac {3\,B\,a\,b^2}{7}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^7\,\left (a+b\,x\right )}-\frac {A\,a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{10\,x^{10}\,\left (a+b\,x\right )}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{6\,x^6\,\left (a+b\,x\right )}-\frac {3\,a\,b\,\left (A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{8\,x^8\,\left (a+b\,x\right )} \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/x^11,x)
Output:
- (((B*a^3)/9 + (A*a^2*b)/3)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^9*(a + b* x)) - (((A*b^3)/7 + (3*B*a*b^2)/7)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^7*( a + b*x)) - (A*a^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(10*x^10*(a + b*x)) - (B*b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(6*x^6*(a + b*x)) - (3*a*b*(A*b + B*a)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(8*x^8*(a + b*x))
Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.22 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^{11}} \, dx=\frac {-210 b^{4} x^{4}-720 a \,b^{3} x^{3}-945 a^{2} b^{2} x^{2}-560 a^{3} b x -126 a^{4}}{1260 x^{10}} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^11,x)
Output:
( - 126*a**4 - 560*a**3*b*x - 945*a**2*b**2*x**2 - 720*a*b**3*x**3 - 210*b **4*x**4)/(1260*x**10)