Integrand size = 33, antiderivative size = 154 \[ \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {(d+e x)^3}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {2 e^2 (b d-a e)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e (b d-a e)^2}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^3 (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]
-1/3*(e*x+d)^3/b/(b^2*x^2+2*a*b*x+a^2)^(3/2)-2*e^2*(-a*e+b*d)/b^4/((b*x+a) ^2)^(1/2)-1/2*e*(-a*e+b*d)^2/b^4/(b*x+a)/((b*x+a)^2)^(1/2)+e^3*(b*x+a)*ln( b*x+a)/b^4/((b*x+a)^2)^(1/2)
Time = 1.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.59 \[ \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-\left ((b d-a e) \left (11 a^2 e^2+a b e (5 d+27 e x)+b^2 \left (2 d^2+9 d e x+18 e^2 x^2\right )\right )\right )+6 e^3 (a+b x)^3 \log (a+b x)}{6 b^4 \left ((a+b x)^2\right )^{3/2}} \]
(-((b*d - a*e)*(11*a^2*e^2 + a*b*e*(5*d + 27*e*x) + b^2*(2*d^2 + 9*d*e*x + 18*e^2*x^2))) + 6*e^3*(a + b*x)^3*Log[a + b*x])/(6*b^4*((a + b*x)^2)^(3/2 ))
Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.73, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1187, 27, 49, 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.
\(\displaystyle \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b^5 (a+b x) \int \frac {(d+e x)^3}{b^5 (a+b x)^4}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {(d+e x)^3}{(a+b x)^4}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {(a+b x) \int \left (\frac {e^3}{b^3 (a+b x)}+\frac {3 (b d-a e) e^2}{b^3 (a+b x)^2}+\frac {3 (b d-a e)^2 e}{b^3 (a+b x)^3}+\frac {(b d-a e)^3}{b^3 (a+b x)^4}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x) \left (-\frac {3 e^2 (b d-a e)}{b^4 (a+b x)}-\frac {3 e (b d-a e)^2}{2 b^4 (a+b x)^2}-\frac {(b d-a e)^3}{3 b^4 (a+b x)^3}+\frac {e^3 \log (a+b x)}{b^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-1/3*(b*d - a*e)^3/(b^4*(a + b*x)^3) - (3*e*(b*d - a*e)^2)/(2* b^4*(a + b*x)^2) - (3*e^2*(b*d - a*e))/(b^4*(a + b*x)) + (e^3*Log[a + b*x] )/b^4))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.21.35.3.1 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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
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 = 0.49 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {3 e^{2} \left (a e -b d \right ) x^{2}}{b^{2}}+\frac {3 e \left (3 e^{2} a^{2}-2 a b d e -b^{2} d^{2}\right ) x}{2 b^{3}}+\frac {11 a^{3} e^{3}-6 a^{2} b d \,e^{2}-3 a \,b^{2} d^{2} e -2 b^{3} d^{3}}{6 b^{4}}\right )}{\left (b x +a \right )^{4}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{3} \ln \left (b x +a \right )}{\left (b x +a \right ) b^{4}}\) | \(140\) |
default | \(\frac {\left (6 \ln \left (b x +a \right ) b^{3} e^{3} x^{3}+18 \ln \left (b x +a \right ) x^{2} a \,b^{2} e^{3}+18 \ln \left (b x +a \right ) x \,a^{2} b \,e^{3}+18 x^{2} a \,b^{2} e^{3}-18 x^{2} b^{3} d \,e^{2}+6 \ln \left (b x +a \right ) a^{3} e^{3}+27 x \,a^{2} b \,e^{3}-18 x a \,b^{2} d \,e^{2}-9 x \,b^{3} d^{2} e +11 a^{3} e^{3}-6 a^{2} b d \,e^{2}-3 a \,b^{2} d^{2} e -2 b^{3} d^{3}\right ) \left (b x +a \right )^{2}}{6 b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(179\) |
((b*x+a)^2)^(1/2)/(b*x+a)^4*(3*e^2*(a*e-b*d)/b^2*x^2+3/2*e*(3*a^2*e^2-2*a* b*d*e-b^2*d^2)/b^3*x+1/6*(11*a^3*e^3-6*a^2*b*d*e^2-3*a*b^2*d^2*e-2*b^3*d^3 )/b^4)+((b*x+a)^2)^(1/2)/(b*x+a)*e^3/b^4*ln(b*x+a)
Time = 0.29 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.14 \[ \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, b^{3} d^{3} + 3 \, a b^{2} d^{2} e + 6 \, a^{2} b d e^{2} - 11 \, a^{3} e^{3} + 18 \, {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 9 \, {\left (b^{3} d^{2} e + 2 \, a b^{2} d e^{2} - 3 \, a^{2} b e^{3}\right )} x - 6 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \log \left (b x + a\right )}{6 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}} \]
-1/6*(2*b^3*d^3 + 3*a*b^2*d^2*e + 6*a^2*b*d*e^2 - 11*a^3*e^3 + 18*(b^3*d*e ^2 - a*b^2*e^3)*x^2 + 9*(b^3*d^2*e + 2*a*b^2*d*e^2 - 3*a^2*b*e^3)*x - 6*(b ^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*log(b*x + a))/(b^7 *x^3 + 3*a*b^6*x^2 + 3*a^2*b^5*x + a^3*b^4)
\[ \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (a + b x\right ) \left (d + e x\right )^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 533 vs. \(2 (115) = 230\).
Time = 0.21 (sec) , antiderivative size = 533, normalized size of antiderivative = 3.46 \[ \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {1}{12} \, b e^{3} {\left (\frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4}}{b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}} + \frac {12 \, \log \left (b x + a\right )}{b^{5}}\right )} - \frac {1}{4} \, b d e^{2} {\left (\frac {12 \, x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} + \frac {6 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{3}}{b^{8} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{12} \, a e^{3} {\left (\frac {12 \, x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} + \frac {6 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{3}}{b^{8} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{12} \, b d^{3} {\left (\frac {4}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {3 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{4} \, a d^{2} e {\left (\frac {4}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {3 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{4} \, b d^{2} e {\left (\frac {6}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {3 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{4} \, a d e^{2} {\left (\frac {6}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {3 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {a d^{3}}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} \]
1/12*b*e^3*((48*a*b^3*x^3 + 108*a^2*b^2*x^2 + 88*a^3*b*x + 25*a^4)/(b^9*x^ 4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*b^5) + 12*log(b*x + a) /b^5) - 1/4*b*d*e^2*(12*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) + 8*a^2/ ((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) + 6*a/(b^6*(x + a/b)^2) - 8*a^2/(b^7 *(x + a/b)^3) - 3*a^3/(b^8*(x + a/b)^4)) - 1/12*a*e^3*(12*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) + 8*a^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) + 6*a/(b^6*(x + a/b)^2) - 8*a^2/(b^7*(x + a/b)^3) - 3*a^3/(b^8*(x + a/b)^4)) - 1/12*b*d^3*(4/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 3*a/(b^6*(x + a/b )^4)) - 1/4*a*d^2*e*(4/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 3*a/(b^6*(x + a/b)^4)) - 1/4*b*d^2*e*(6/(b^5*(x + a/b)^2) - 8*a/(b^6*(x + a/b)^3) + 3 *a^2/(b^7*(x + a/b)^4)) - 1/4*a*d*e^2*(6/(b^5*(x + a/b)^2) - 8*a/(b^6*(x + a/b)^3) + 3*a^2/(b^7*(x + a/b)^4)) - 1/4*a*d^3/(b^5*(x + a/b)^4)
Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^{3} \log \left ({\left | b x + a \right |}\right )}{b^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {18 \, {\left (b^{2} d e^{2} - a b e^{3}\right )} x^{2} + 9 \, {\left (b^{2} d^{2} e + 2 \, a b d e^{2} - 3 \, a^{2} e^{3}\right )} x + \frac {2 \, b^{3} d^{3} + 3 \, a b^{2} d^{2} e + 6 \, a^{2} b d e^{2} - 11 \, a^{3} e^{3}}{b}}{6 \, {\left (b x + a\right )}^{3} b^{3} \mathrm {sgn}\left (b x + a\right )} \]
e^3*log(abs(b*x + a))/(b^4*sgn(b*x + a)) - 1/6*(18*(b^2*d*e^2 - a*b*e^3)*x ^2 + 9*(b^2*d^2*e + 2*a*b*d*e^2 - 3*a^2*e^3)*x + (2*b^3*d^3 + 3*a*b^2*d^2* e + 6*a^2*b*d*e^2 - 11*a^3*e^3)/b)/((b*x + a)^3*b^3*sgn(b*x + a))
Timed out. \[ \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]