Integrand size = 33, antiderivative size = 280 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx=-\frac {b^2 (3 b B d-A b e-3 a B e) x \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)}+\frac {b^3 B x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^3 (a+b x)}-\frac {(b d-a e)^3 (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^5 (a+b x) (d+e x)^2}+\frac {(b d-a e)^2 (4 b B d-3 A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) (d+e x)}+\frac {3 b (b d-a e) (2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^5 (a+b x)} \]
-b^2*(-A*b*e-3*B*a*e+3*B*b*d)*x*((b*x+a)^2)^(1/2)/e^4/(b*x+a)+1/2*b^3*B*x^ 2*((b*x+a)^2)^(1/2)/e^3/(b*x+a)-1/2*(-a*e+b*d)^3*(-A*e+B*d)*((b*x+a)^2)^(1 /2)/e^5/(b*x+a)/(e*x+d)^2+(-a*e+b*d)^2*(-3*A*b*e-B*a*e+4*B*b*d)*((b*x+a)^2 )^(1/2)/e^5/(b*x+a)/(e*x+d)+3*b*(-a*e+b*d)*(-A*b*e-B*a*e+2*B*b*d)*ln(e*x+d )*((b*x+a)^2)^(1/2)/e^5/(b*x+a)
Time = 1.10 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx=\frac {\sqrt {(a+b x)^2} \left (-a^3 e^3 (A e+B (d+2 e x))-3 a^2 b e^2 (A e (d+2 e x)-B d (3 d+4 e x))+3 a b^2 e \left (A d e (3 d+4 e x)+B \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )\right )+b^3 \left (A e \left (-5 d^3-4 d^2 e x+4 d e^2 x^2+2 e^3 x^3\right )+B \left (7 d^4+2 d^3 e x-11 d^2 e^2 x^2-4 d e^3 x^3+e^4 x^4\right )\right )+6 b (b d-a e) (2 b B d-A b e-a B e) (d+e x)^2 \log (d+e x)\right )}{2 e^5 (a+b x) (d+e x)^2} \]
(Sqrt[(a + b*x)^2]*(-(a^3*e^3*(A*e + B*(d + 2*e*x))) - 3*a^2*b*e^2*(A*e*(d + 2*e*x) - B*d*(3*d + 4*e*x)) + 3*a*b^2*e*(A*d*e*(3*d + 4*e*x) + B*(-5*d^ 3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*e^3*x^3)) + b^3*(A*e*(-5*d^3 - 4*d^2*e*x + 4*d*e^2*x^2 + 2*e^3*x^3) + B*(7*d^4 + 2*d^3*e*x - 11*d^2*e^2*x^2 - 4*d*e^ 3*x^3 + e^4*x^4)) + 6*b*(b*d - a*e)*(2*b*B*d - A*b*e - a*B*e)*(d + e*x)^2* Log[d + e*x]))/(2*e^5*(a + b*x)*(d + e*x)^2)
Time = 0.41 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.62, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, 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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (A+B x)}{(d+e x)^3} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^3 (a+b x)^3 (A+B x)}{(d+e x)^3}dx}{b^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^3}dx}{a+b x}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {B x b^3}{e^3}+\frac {(-3 b B d+A b e+3 a B e) b^2}{e^4}-\frac {3 (b d-a e) (-2 b B d+A b e+a B e) b}{e^4 (d+e x)}+\frac {(a e-b d)^2 (-4 b B d+3 A b e+a B e)}{e^4 (d+e x)^2}+\frac {(a e-b d)^3 (A e-B d)}{e^4 (d+e x)^3}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {b^2 x (-3 a B e-A b e+3 b B d)}{e^4}+\frac {(b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 (d+e x)}-\frac {(b d-a e)^3 (B d-A e)}{2 e^5 (d+e x)^2}+\frac {3 b (b d-a e) \log (d+e x) (-a B e-A b e+2 b B d)}{e^5}+\frac {b^3 B x^2}{2 e^3}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-((b^2*(3*b*B*d - A*b*e - 3*a*B*e)*x)/e^4) + (b^3*B*x^2)/(2*e^3) - ((b*d - a*e)^3*(B*d - A*e))/(2*e^5*(d + e*x)^2) + ((b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a*B*e))/(e^5*(d + e*x)) + (3*b*(b*d - a*e)*(2*b*B*d - A*b*e - a*B*e)*Log[d + e*x])/e^5))/(a + b*x)
3.18.28.3.1 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 = 0.40 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.11
| method | result | size |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{2} \left (\frac {1}{2} B b e \,x^{2}+A b e x +3 B a e x -3 B b d x \right )}{\left (b x +a \right ) e^{4}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-3 A \,a^{2} b \,e^{3}+6 A a \,b^{2} d \,e^{2}-3 A \,b^{3} d^{2} e -B \,e^{3} a^{3}+6 B \,a^{2} b d \,e^{2}-9 B a \,b^{2} d^{2} e +4 B \,b^{3} d^{3}\right ) x -\frac {A \,a^{3} e^{4}+3 A \,a^{2} b d \,e^{3}-9 A a \,b^{2} d^{2} e^{2}+5 A \,b^{3} d^{3} e +B \,a^{3} d \,e^{3}-9 B \,a^{2} b \,d^{2} e^{2}+15 B a \,b^{2} d^{3} e -7 B \,b^{3} d^{4}}{2 e}\right )}{\left (b x +a \right ) e^{4} \left (e x +d \right )^{2}}+\frac {3 \sqrt {\left (b x +a \right )^{2}}\, b \left (A a b \,e^{2}-A \,b^{2} d e +a^{2} B \,e^{2}-3 B a b d e +2 B \,b^{2} d^{2}\right ) \ln \left (e x +d \right )}{\left (b x +a \right ) e^{5}}\) | \(311\) |
| default | \(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (7 B \,b^{3} d^{4}-A \,a^{3} e^{4}+2 A \,b^{3} e^{4} x^{3}+6 A \ln \left (e x +d \right ) a \,b^{2} e^{4} x^{2}-6 A \ln \left (e x +d \right ) b^{3} d \,e^{3} x^{2}+6 B \ln \left (e x +d \right ) a^{2} b \,e^{4} x^{2}+12 B \ln \left (e x +d \right ) b^{3} d^{2} e^{2} x^{2}-12 A \ln \left (e x +d \right ) b^{3} d^{2} e^{2} x +24 B \ln \left (e x +d \right ) b^{3} d^{3} e x +6 A \ln \left (e x +d \right ) a \,b^{2} d^{2} e^{2}+6 B \ln \left (e x +d \right ) a^{2} b \,d^{2} e^{2}-18 B \ln \left (e x +d \right ) a \,b^{2} d^{3} e +12 B \ln \left (e x +d \right ) b^{3} d^{4}+B \,b^{3} e^{4} x^{4}-2 B \,a^{3} e^{4} x +12 B \,a^{2} b d \,e^{3} x -12 B a \,b^{2} d^{2} e^{2} x +12 B a \,b^{2} d \,e^{3} x^{2}+12 A a \,b^{2} d \,e^{3} x -B \,a^{3} d \,e^{3}-5 A \,b^{3} d^{3} e -4 B \,b^{3} d \,e^{3} x^{3}+4 A \,b^{3} d \,e^{3} x^{2}-11 B \,b^{3} d^{2} e^{2} x^{2}-4 A \,b^{3} d^{2} e^{2} x +2 B \,b^{3} d^{3} e x -6 A \ln \left (e x +d \right ) b^{3} d^{3} e +12 A \ln \left (e x +d \right ) a \,b^{2} d \,e^{3} x +12 B \ln \left (e x +d \right ) a^{2} b d \,e^{3} x -36 B \ln \left (e x +d \right ) a \,b^{2} d^{2} e^{2} x -18 B \ln \left (e x +d \right ) a \,b^{2} d \,e^{3} x^{2}+6 B a \,b^{2} e^{4} x^{3}-6 A \,a^{2} b \,e^{4} x +9 B \,a^{2} b \,d^{2} e^{2}-15 B a \,b^{2} d^{3} e -3 A \,a^{2} b d \,e^{3}+9 A a \,b^{2} d^{2} e^{2}\right )}{2 \left (b x +a \right )^{3} e^{5} \left (e x +d \right )^{2}}\) | \(566\) |
((b*x+a)^2)^(1/2)/(b*x+a)*b^2/e^4*(1/2*B*b*e*x^2+A*b*e*x+3*B*a*e*x-3*B*b*d *x)+((b*x+a)^2)^(1/2)/(b*x+a)*((-3*A*a^2*b*e^3+6*A*a*b^2*d*e^2-3*A*b^3*d^2 *e-B*a^3*e^3+6*B*a^2*b*d*e^2-9*B*a*b^2*d^2*e+4*B*b^3*d^3)*x-1/2*(A*a^3*e^4 +3*A*a^2*b*d*e^3-9*A*a*b^2*d^2*e^2+5*A*b^3*d^3*e+B*a^3*d*e^3-9*B*a^2*b*d^2 *e^2+15*B*a*b^2*d^3*e-7*B*b^3*d^4)/e)/e^4/(e*x+d)^2+3*((b*x+a)^2)^(1/2)/(b *x+a)*b/e^5*(A*a*b*e^2-A*b^2*d*e+B*a^2*e^2-3*B*a*b*d*e+2*B*b^2*d^2)*ln(e*x +d)
Time = 0.66 (sec) , antiderivative size = 420, normalized size of antiderivative = 1.50 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx=\frac {B b^{3} e^{4} x^{4} + 7 \, B b^{3} d^{4} - A a^{3} e^{4} - 5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 9 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - 2 \, {\left (2 \, B b^{3} d e^{3} - {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} - {\left (11 \, B b^{3} d^{2} e^{2} - 4 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3}\right )} x^{2} + 2 \, {\left (B b^{3} d^{3} e - 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 6 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x + 6 \, {\left (2 \, B b^{3} d^{4} - {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + {\left (2 \, B b^{3} d^{2} e^{2} - {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 2 \, {\left (2 \, B b^{3} d^{3} e - {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + {\left (B a^{2} b + A a b^{2}\right )} d e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \]
1/2*(B*b^3*e^4*x^4 + 7*B*b^3*d^4 - A*a^3*e^4 - 5*(3*B*a*b^2 + A*b^3)*d^3*e + 9*(B*a^2*b + A*a*b^2)*d^2*e^2 - (B*a^3 + 3*A*a^2*b)*d*e^3 - 2*(2*B*b^3* d*e^3 - (3*B*a*b^2 + A*b^3)*e^4)*x^3 - (11*B*b^3*d^2*e^2 - 4*(3*B*a*b^2 + A*b^3)*d*e^3)*x^2 + 2*(B*b^3*d^3*e - 2*(3*B*a*b^2 + A*b^3)*d^2*e^2 + 6*(B* a^2*b + A*a*b^2)*d*e^3 - (B*a^3 + 3*A*a^2*b)*e^4)*x + 6*(2*B*b^3*d^4 - (3* B*a*b^2 + A*b^3)*d^3*e + (B*a^2*b + A*a*b^2)*d^2*e^2 + (2*B*b^3*d^2*e^2 - (3*B*a*b^2 + A*b^3)*d*e^3 + (B*a^2*b + A*a*b^2)*e^4)*x^2 + 2*(2*B*b^3*d^3* e - (3*B*a*b^2 + A*b^3)*d^2*e^2 + (B*a^2*b + A*a*b^2)*d*e^3)*x)*log(e*x + d))/(e^7*x^2 + 2*d*e^6*x + d^2*e^5)
\[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx=\int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{3}}\, dx \]
Exception generated. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx=\text {Exception raised: ValueError} \]
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 = 0.29 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.54 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx=\frac {3 \, {\left (2 \, B b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, B a b^{2} d e \mathrm {sgn}\left (b x + a\right ) - A b^{3} d e \mathrm {sgn}\left (b x + a\right ) + B a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right ) + A a b^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | e x + d \right |}\right )}{e^{5}} + \frac {B b^{3} e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) - 6 \, B b^{3} d e^{2} x \mathrm {sgn}\left (b x + a\right ) + 6 \, B a b^{2} e^{3} x \mathrm {sgn}\left (b x + a\right ) + 2 \, A b^{3} e^{3} x \mathrm {sgn}\left (b x + a\right )}{2 \, e^{6}} + \frac {7 \, B b^{3} d^{4} \mathrm {sgn}\left (b x + a\right ) - 15 \, B a b^{2} d^{3} e \mathrm {sgn}\left (b x + a\right ) - 5 \, A b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 9 \, B a^{2} b d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 9 \, A a b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - B a^{3} d e^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, A a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) - A a^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) + 2 \, {\left (4 \, B b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) - 9 \, B a b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, A b^{3} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, B a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, A a b^{2} d e^{3} \mathrm {sgn}\left (b x + a\right ) - B a^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) - 3 \, A a^{2} b e^{4} \mathrm {sgn}\left (b x + a\right )\right )} x}{2 \, {\left (e x + d\right )}^{2} e^{5}} \]
3*(2*B*b^3*d^2*sgn(b*x + a) - 3*B*a*b^2*d*e*sgn(b*x + a) - A*b^3*d*e*sgn(b *x + a) + B*a^2*b*e^2*sgn(b*x + a) + A*a*b^2*e^2*sgn(b*x + a))*log(abs(e*x + d))/e^5 + 1/2*(B*b^3*e^3*x^2*sgn(b*x + a) - 6*B*b^3*d*e^2*x*sgn(b*x + a ) + 6*B*a*b^2*e^3*x*sgn(b*x + a) + 2*A*b^3*e^3*x*sgn(b*x + a))/e^6 + 1/2*( 7*B*b^3*d^4*sgn(b*x + a) - 15*B*a*b^2*d^3*e*sgn(b*x + a) - 5*A*b^3*d^3*e*s gn(b*x + a) + 9*B*a^2*b*d^2*e^2*sgn(b*x + a) + 9*A*a*b^2*d^2*e^2*sgn(b*x + a) - B*a^3*d*e^3*sgn(b*x + a) - 3*A*a^2*b*d*e^3*sgn(b*x + a) - A*a^3*e^4* sgn(b*x + a) + 2*(4*B*b^3*d^3*e*sgn(b*x + a) - 9*B*a*b^2*d^2*e^2*sgn(b*x + a) - 3*A*b^3*d^2*e^2*sgn(b*x + a) + 6*B*a^2*b*d*e^3*sgn(b*x + a) + 6*A*a* b^2*d*e^3*sgn(b*x + a) - B*a^3*e^4*sgn(b*x + a) - 3*A*a^2*b*e^4*sgn(b*x + a))*x)/((e*x + d)^2*e^5)
Timed out. \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^3} \,d x \]