Integrand size = 29, antiderivative size = 212 \[ \int \frac {x^3 (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {a^2 (A b-a B) x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a (A b-a B) x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^3 (a+b x)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x^4 (a+b x)}{4 b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^3 (A b-a B) (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]
a^2*(A*b-B*a)*x*(b*x+a)/b^4/((b*x+a)^2)^(1/2)-1/2*a*(A*b-B*a)*x^2*(b*x+a)/ b^3/((b*x+a)^2)^(1/2)+1/3*(A*b-B*a)*x^3*(b*x+a)/b^2/((b*x+a)^2)^(1/2)+1/4* B*x^4*(b*x+a)/b/((b*x+a)^2)^(1/2)-a^3*(A*b-B*a)*(b*x+a)*ln(b*x+a)/b^5/((b* x+a)^2)^(1/2)
Time = 1.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.45 \[ \int \frac {x^3 (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {(a+b x) \left (b x \left (-12 a^3 B+6 a^2 b (2 A+B x)-2 a b^2 x (3 A+2 B x)+b^3 x^2 (4 A+3 B x)\right )+12 a^3 (-A b+a B) \log (a+b x)\right )}{12 b^5 \sqrt {(a+b x)^2}} \]
((a + b*x)*(b*x*(-12*a^3*B + 6*a^2*b*(2*A + B*x) - 2*a*b^2*x*(3*A + 2*B*x) + b^3*x^2*(4*A + 3*B*x)) + 12*a^3*(-(A*b) + a*B)*Log[a + b*x]))/(12*b^5*S qrt[(a + b*x)^2])
Time = 0.28 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.53, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1187, 27, 86, 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.
\(\displaystyle \int \frac {x^3 (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b (a+b x) \int \frac {x^3 (A+B x)}{b (a+b x)}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {x^3 (A+B x)}{a+b x}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {(a+b x) \int \left (\frac {(a B-A b) a^3}{b^4 (a+b x)}-\frac {(a B-A b) a^2}{b^4}+\frac {(a B-A b) x a}{b^3}+\frac {B x^3}{b}+\frac {(A b-a B) x^2}{b^2}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x) \left (-\frac {a^3 (A b-a B) \log (a+b x)}{b^5}+\frac {a^2 x (A b-a B)}{b^4}-\frac {a x^2 (A b-a B)}{2 b^3}+\frac {x^3 (A b-a B)}{3 b^2}+\frac {B x^4}{4 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*((a^2*(A*b - a*B)*x)/b^4 - (a*(A*b - a*B)*x^2)/(2*b^3) + ((A*b - a*B)*x^3)/(3*b^2) + (B*x^4)/(4*b) - (a^3*(A*b - a*B)*Log[a + b*x])/b^5)) /Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.8.5.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 0.32 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.54
method | result | size |
default | \(-\frac {\left (b x +a \right ) \left (-3 b^{4} B \,x^{4}-4 A \,b^{4} x^{3}+4 B a \,b^{3} x^{3}+6 A a \,b^{3} x^{2}-6 B \,a^{2} b^{2} x^{2}+12 A \ln \left (b x +a \right ) a^{3} b -12 A \,a^{2} b^{2} x -12 B \ln \left (b x +a \right ) a^{4}+12 B \,a^{3} b x \right )}{12 \sqrt {\left (b x +a \right )^{2}}\, b^{5}}\) | \(114\) |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {1}{4} x^{4} B \,b^{3}+\frac {1}{3} A \,b^{3} x^{3}-\frac {1}{3} B a \,b^{2} x^{3}-\frac {1}{2} A a \,b^{2} x^{2}+\frac {1}{2} B \,a^{2} b \,x^{2}+A \,a^{2} b x -a^{3} B x \right )}{\left (b x +a \right ) b^{4}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, a^{3} \left (A b -B a \right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{5}}\) | \(123\) |
-1/12*(b*x+a)*(-3*b^4*B*x^4-4*A*b^4*x^3+4*B*a*b^3*x^3+6*A*a*b^3*x^2-6*B*a^ 2*b^2*x^2+12*A*ln(b*x+a)*a^3*b-12*A*a^2*b^2*x-12*B*ln(b*x+a)*a^4+12*B*a^3* b*x)/((b*x+a)^2)^(1/2)/b^5
Time = 0.47 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.44 \[ \int \frac {x^3 (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {3 \, B b^{4} x^{4} - 4 \, {\left (B a b^{3} - A b^{4}\right )} x^{3} + 6 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 12 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x + 12 \, {\left (B a^{4} - A a^{3} b\right )} \log \left (b x + a\right )}{12 \, b^{5}} \]
1/12*(3*B*b^4*x^4 - 4*(B*a*b^3 - A*b^4)*x^3 + 6*(B*a^2*b^2 - A*a*b^3)*x^2 - 12*(B*a^3*b - A*a^2*b^2)*x + 12*(B*a^4 - A*a^3*b)*log(b*x + a))/b^5
Leaf count of result is larger than twice the leaf count of optimal. 488 vs. \(2 (153) = 306\).
Time = 1.47 (sec) , antiderivative size = 488, normalized size of antiderivative = 2.30 \[ \int \frac {x^3 (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (\frac {B x^{3}}{4 b^{2}} + \frac {x^{2} \left (A - \frac {7 B a}{4 b}\right )}{3 b^{2}} + \frac {x \left (- \frac {3 B a^{2}}{4 b^{2}} - \frac {5 a \left (A - \frac {7 B a}{4 b}\right )}{3 b}\right )}{2 b^{2}} + \frac {- \frac {2 a^{2} \left (A - \frac {7 B a}{4 b}\right )}{3 b^{2}} - \frac {3 a \left (- \frac {3 B a^{2}}{4 b^{2}} - \frac {5 a \left (A - \frac {7 B a}{4 b}\right )}{3 b}\right )}{2 b}}{b^{2}}\right ) + \frac {\left (\frac {a}{b} + x\right ) \left (- \frac {a^{2} \left (- \frac {3 B a^{2}}{4 b^{2}} - \frac {5 a \left (A - \frac {7 B a}{4 b}\right )}{3 b}\right )}{2 b^{2}} - \frac {a \left (- \frac {2 a^{2} \left (A - \frac {7 B a}{4 b}\right )}{3 b^{2}} - \frac {3 a \left (- \frac {3 B a^{2}}{4 b^{2}} - \frac {5 a \left (A - \frac {7 B a}{4 b}\right )}{3 b}\right )}{2 b}\right )}{b}\right ) \log {\left (\frac {a}{b} + x \right )}}{\sqrt {b^{2} \left (\frac {a}{b} + x\right )^{2}}} & \text {for}\: b^{2} \neq 0 \\\frac {\frac {A \left (- a^{6} \sqrt {a^{2} + 2 a b x} + a^{4} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}} - \frac {3 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7}\right )}{4 a^{3} b^{3}} + \frac {B \left (a^{8} \sqrt {a^{2} + 2 a b x} - \frac {4 a^{6} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} + \frac {6 a^{4} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} - \frac {4 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {9}{2}}}{9}\right )}{8 a^{4} b^{4}}}{2 a b} & \text {for}\: a b \neq 0 \\\frac {\frac {A x^{4}}{4} + \frac {B x^{5}}{5}}{\sqrt {a^{2}}} & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(B*x**3/(4*b**2) + x**2*(A - 7 *B*a/(4*b))/(3*b**2) + x*(-3*B*a**2/(4*b**2) - 5*a*(A - 7*B*a/(4*b))/(3*b) )/(2*b**2) + (-2*a**2*(A - 7*B*a/(4*b))/(3*b**2) - 3*a*(-3*B*a**2/(4*b**2) - 5*a*(A - 7*B*a/(4*b))/(3*b))/(2*b))/b**2) + (a/b + x)*(-a**2*(-3*B*a**2 /(4*b**2) - 5*a*(A - 7*B*a/(4*b))/(3*b))/(2*b**2) - a*(-2*a**2*(A - 7*B*a/ (4*b))/(3*b**2) - 3*a*(-3*B*a**2/(4*b**2) - 5*a*(A - 7*B*a/(4*b))/(3*b))/( 2*b))/b)*log(a/b + x)/sqrt(b**2*(a/b + x)**2), Ne(b**2, 0)), ((A*(-a**6*sq rt(a**2 + 2*a*b*x) + a**4*(a**2 + 2*a*b*x)**(3/2) - 3*a**2*(a**2 + 2*a*b*x )**(5/2)/5 + (a**2 + 2*a*b*x)**(7/2)/7)/(4*a**3*b**3) + B*(a**8*sqrt(a**2 + 2*a*b*x) - 4*a**6*(a**2 + 2*a*b*x)**(3/2)/3 + 6*a**4*(a**2 + 2*a*b*x)**( 5/2)/5 - 4*a**2*(a**2 + 2*a*b*x)**(7/2)/7 + (a**2 + 2*a*b*x)**(9/2)/9)/(8* a**4*b**4))/(2*a*b), Ne(a*b, 0)), ((A*x**4/4 + B*x**5/5)/sqrt(a**2), True) )
Time = 0.21 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00 \[ \int \frac {x^3 (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B x^{3}}{4 \, b^{2}} + \frac {13 \, B a^{2} x^{2}}{12 \, b^{3}} - \frac {5 \, A a x^{2}}{6 \, b^{2}} - \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a x^{2}}{12 \, b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A x^{2}}{3 \, b^{2}} - \frac {13 \, B a^{3} x}{6 \, b^{4}} + \frac {5 \, A a^{2} x}{3 \, b^{3}} + \frac {B a^{4} \log \left (x + \frac {a}{b}\right )}{b^{5}} - \frac {A a^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{3}}{6 \, b^{5}} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a^{2}}{3 \, b^{4}} \]
1/4*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*x^3/b^2 + 13/12*B*a^2*x^2/b^3 - 5/6*A* a*x^2/b^2 - 7/12*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a*x^2/b^3 + 1/3*sqrt(b^2* x^2 + 2*a*b*x + a^2)*A*x^2/b^2 - 13/6*B*a^3*x/b^4 + 5/3*A*a^2*x/b^3 + B*a^ 4*log(x + a/b)/b^5 - A*a^3*log(x + a/b)/b^4 + 7/6*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^3/b^5 - 2/3*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a^2/b^4
Time = 0.26 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.70 \[ \int \frac {x^3 (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {3 \, B b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 4 \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) - 6 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) - 12 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 12 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right )}{12 \, b^{4}} + \frac {{\left (B a^{4} \mathrm {sgn}\left (b x + a\right ) - A a^{3} b \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} \]
1/12*(3*B*b^3*x^4*sgn(b*x + a) - 4*B*a*b^2*x^3*sgn(b*x + a) + 4*A*b^3*x^3* sgn(b*x + a) + 6*B*a^2*b*x^2*sgn(b*x + a) - 6*A*a*b^2*x^2*sgn(b*x + a) - 1 2*B*a^3*x*sgn(b*x + a) + 12*A*a^2*b*x*sgn(b*x + a))/b^4 + (B*a^4*sgn(b*x + a) - A*a^3*b*sgn(b*x + a))*log(abs(b*x + a))/b^5
Timed out. \[ \int \frac {x^3 (A+B x)}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {x^3\,\left (A+B\,x\right )}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]