Integrand size = 33, antiderivative size = 239 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx=\frac {6 b^2 (b d-a e)^2 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)}-\frac {(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) (d+e x)}-\frac {2 b^3 (b d-a e) (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}+\frac {b^4 (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x)}-\frac {4 b (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^5 (a+b x)} \]
6*b^2*(-a*e+b*d)^2*x*((b*x+a)^2)^(1/2)/e^4/(b*x+a)-(-a*e+b*d)^4*((b*x+a)^2 )^(1/2)/e^5/(b*x+a)/(e*x+d)-2*b^3*(-a*e+b*d)*(e*x+d)^2*((b*x+a)^2)^(1/2)/e ^5/(b*x+a)+1/3*b^4*(e*x+d)^3*((b*x+a)^2)^(1/2)/e^5/(b*x+a)-4*b*(-a*e+b*d)^ 3*ln(e*x+d)*((b*x+a)^2)^(1/2)/e^5/(b*x+a)
Time = 1.09 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx=\frac {\sqrt {(a+b x)^2} \left (12 a^3 b d e^3-3 a^4 e^4+18 a^2 b^2 e^2 \left (-d^2+d e x+e^2 x^2\right )+6 a b^3 e \left (2 d^3-4 d^2 e x-3 d e^2 x^2+e^3 x^3\right )+b^4 \left (-3 d^4+9 d^3 e x+6 d^2 e^2 x^2-2 d e^3 x^3+e^4 x^4\right )-12 b (b d-a e)^3 (d+e x) \log (d+e x)\right )}{3 e^5 (a+b x) (d+e x)} \]
(Sqrt[(a + b*x)^2]*(12*a^3*b*d*e^3 - 3*a^4*e^4 + 18*a^2*b^2*e^2*(-d^2 + d* e*x + e^2*x^2) + 6*a*b^3*e*(2*d^3 - 4*d^2*e*x - 3*d*e^2*x^2 + e^3*x^3) + b ^4*(-3*d^4 + 9*d^3*e*x + 6*d^2*e^2*x^2 - 2*d*e^3*x^3 + e^4*x^4) - 12*b*(b* d - a*e)^3*(d + e*x)*Log[d + e*x]))/(3*e^5*(a + b*x)*(d + e*x))
Time = 0.33 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.55, 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.
\(\displaystyle \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^4}{(d+e x)^2}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^4}{(d+e x)^2}dx}{a+b x}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {(d+e x)^2 b^4}{e^4}-\frac {4 (b d-a e) (d+e x) b^3}{e^4}+\frac {6 (b d-a e)^2 b^2}{e^4}-\frac {4 (b d-a e)^3 b}{e^4 (d+e x)}+\frac {(a e-b d)^4}{e^4 (d+e x)^2}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {2 b^3 (d+e x)^2 (b d-a e)}{e^5}+\frac {6 b^2 x (b d-a e)^2}{e^4}-\frac {(b d-a e)^4}{e^5 (d+e x)}-\frac {4 b (b d-a e)^3 \log (d+e x)}{e^5}+\frac {b^4 (d+e x)^3}{3 e^5}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((6*b^2*(b*d - a*e)^2*x)/e^4 - (b*d - a*e)^ 4/(e^5*(d + e*x)) - (2*b^3*(b*d - a*e)*(d + e*x)^2)/e^5 + (b^4*(d + e*x)^3 )/(3*e^5) - (4*b*(b*d - a*e)^3*Log[d + e*x])/e^5))/(a + b*x)
3.20.77.3.1 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 = 0.35 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.93
method | result | size |
risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{2} \left (\frac {1}{3} b^{2} e^{2} x^{3}+2 a b \,e^{2} x^{2}-b^{2} d e \,x^{2}+6 e^{2} a^{2} x -8 a b d e x +3 b^{2} d^{2} x \right )}{\left (b x +a \right ) e^{4}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (e^{4} a^{4}-4 b d \,e^{3} a^{3}+6 b^{2} d^{2} e^{2} a^{2}-4 b^{3} d^{3} e a +b^{4} d^{4}\right )}{\left (b x +a \right ) e^{5} \left (e x +d \right )}+\frac {4 \sqrt {\left (b x +a \right )^{2}}\, b \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \ln \left (e x +d \right )}{\left (b x +a \right ) e^{5}}\) | \(223\) |
default | \(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (e^{4} x^{4} b^{4}+6 x^{3} a \,b^{3} e^{4}-2 x^{3} b^{4} d \,e^{3}+12 \ln \left (e x +d \right ) a^{3} b \,e^{4} x -36 \ln \left (e x +d \right ) a^{2} b^{2} d \,e^{3} x +36 \ln \left (e x +d \right ) a \,b^{3} d^{2} e^{2} x -12 \ln \left (e x +d \right ) b^{4} d^{3} e x +18 x^{2} a^{2} b^{2} e^{4}-18 x^{2} a \,b^{3} d \,e^{3}+6 x^{2} b^{4} d^{2} e^{2}+12 \ln \left (e x +d \right ) a^{3} b d \,e^{3}-36 \ln \left (e x +d \right ) a^{2} b^{2} d^{2} e^{2}+36 \ln \left (e x +d \right ) a \,b^{3} d^{3} e -12 \ln \left (e x +d \right ) b^{4} d^{4}+18 x \,a^{2} b^{2} d \,e^{3}-24 x a \,b^{3} d^{2} e^{2}+9 x \,b^{4} d^{3} e -3 e^{4} a^{4}+12 b d \,e^{3} a^{3}-18 b^{2} d^{2} e^{2} a^{2}+12 b^{3} d^{3} e a -3 b^{4} d^{4}\right )}{3 \left (b x +a \right )^{3} e^{5} \left (e x +d \right )}\) | \(327\) |
((b*x+a)^2)^(1/2)/(b*x+a)*b^2/e^4*(1/3*b^2*e^2*x^3+2*a*b*e^2*x^2-b^2*d*e*x ^2+6*e^2*a^2*x-8*a*b*d*e*x+3*b^2*d^2*x)-((b*x+a)^2)^(1/2)/(b*x+a)/e^5*(a^4 *e^4-4*a^3*b*d*e^3+6*a^2*b^2*d^2*e^2-4*a*b^3*d^3*e+b^4*d^4)/(e*x+d)+4*((b* x+a)^2)^(1/2)/(b*x+a)*b/e^5*(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)* ln(e*x+d)
Time = 0.29 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx=\frac {b^{4} e^{4} x^{4} - 3 \, b^{4} d^{4} + 12 \, a b^{3} d^{3} e - 18 \, a^{2} b^{2} d^{2} e^{2} + 12 \, a^{3} b d e^{3} - 3 \, a^{4} e^{4} - 2 \, {\left (b^{4} d e^{3} - 3 \, a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (b^{4} d^{2} e^{2} - 3 \, a b^{3} d e^{3} + 3 \, a^{2} b^{2} e^{4}\right )} x^{2} + 3 \, {\left (3 \, b^{4} d^{3} e - 8 \, a b^{3} d^{2} e^{2} + 6 \, a^{2} b^{2} d e^{3}\right )} x - 12 \, {\left (b^{4} d^{4} - 3 \, a b^{3} d^{3} e + 3 \, a^{2} b^{2} d^{2} e^{2} - a^{3} b d e^{3} + {\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x\right )} \log \left (e x + d\right )}{3 \, {\left (e^{6} x + d e^{5}\right )}} \]
1/3*(b^4*e^4*x^4 - 3*b^4*d^4 + 12*a*b^3*d^3*e - 18*a^2*b^2*d^2*e^2 + 12*a^ 3*b*d*e^3 - 3*a^4*e^4 - 2*(b^4*d*e^3 - 3*a*b^3*e^4)*x^3 + 6*(b^4*d^2*e^2 - 3*a*b^3*d*e^3 + 3*a^2*b^2*e^4)*x^2 + 3*(3*b^4*d^3*e - 8*a*b^3*d^2*e^2 + 6 *a^2*b^2*d*e^3)*x - 12*(b^4*d^4 - 3*a*b^3*d^3*e + 3*a^2*b^2*d^2*e^2 - a^3* b*d*e^3 + (b^4*d^3*e - 3*a*b^3*d^2*e^2 + 3*a^2*b^2*d*e^3 - a^3*b*e^4)*x)*l og(e*x + d))/(e^6*x + d*e^5)
\[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx=\int \frac {\left (a + b x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for m ore detail
Time = 0.27 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.16 \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx=-\frac {4 \, {\left (b^{4} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{3} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} b e^{3} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | e x + d \right |}\right )}{e^{5}} + \frac {b^{4} e^{4} x^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, b^{4} d e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, a b^{3} e^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) + 9 \, b^{4} d^{2} e^{2} x \mathrm {sgn}\left (b x + a\right ) - 24 \, a b^{3} d e^{3} x \mathrm {sgn}\left (b x + a\right ) + 18 \, a^{2} b^{2} e^{4} x \mathrm {sgn}\left (b x + a\right )}{3 \, e^{6}} - \frac {b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )}{{\left (e x + d\right )} e^{5}} \]
-4*(b^4*d^3*sgn(b*x + a) - 3*a*b^3*d^2*e*sgn(b*x + a) + 3*a^2*b^2*d*e^2*sg n(b*x + a) - a^3*b*e^3*sgn(b*x + a))*log(abs(e*x + d))/e^5 + 1/3*(b^4*e^4* x^3*sgn(b*x + a) - 3*b^4*d*e^3*x^2*sgn(b*x + a) + 6*a*b^3*e^4*x^2*sgn(b*x + a) + 9*b^4*d^2*e^2*x*sgn(b*x + a) - 24*a*b^3*d*e^3*x*sgn(b*x + a) + 18*a ^2*b^2*e^4*x*sgn(b*x + a))/e^6 - (b^4*d^4*sgn(b*x + a) - 4*a*b^3*d^3*e*sgn (b*x + a) + 6*a^2*b^2*d^2*e^2*sgn(b*x + a) - 4*a^3*b*d*e^3*sgn(b*x + a) + a^4*e^4*sgn(b*x + a))/((e*x + d)*e^5)
Timed out. \[ \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx=\int \frac {\left (a+b\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^2} \,d x \]