Integrand size = 33, antiderivative size = 332 \[ \int \frac {A+B x}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {b (b B d-3 A b e+2 a B e)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (A b-a B)}{2 (b d-a e)^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e (B d-A e) (a+b x)}{2 (b d-a e)^3 (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e (2 b B d-3 A b e+a B e) (a+b x)}{(b d-a e)^4 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 b e (b B d-2 A b e+a B e) (a+b x) \log (a+b x)}{(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b e (b B d-2 A b e+a B e) (a+b x) \log (d+e x)}{(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]
-b*(-3*A*b*e+2*B*a*e+B*b*d)/(-a*e+b*d)^4/((b*x+a)^2)^(1/2)-1/2*b*(A*b-B*a) /(-a*e+b*d)^3/(b*x+a)/((b*x+a)^2)^(1/2)-1/2*e*(-A*e+B*d)*(b*x+a)/(-a*e+b*d )^3/(e*x+d)^2/((b*x+a)^2)^(1/2)-e*(-3*A*b*e+B*a*e+2*B*b*d)*(b*x+a)/(-a*e+b *d)^4/(e*x+d)/((b*x+a)^2)^(1/2)-3*b*e*(-2*A*b*e+B*a*e+B*b*d)*(b*x+a)*ln(b* x+a)/(-a*e+b*d)^5/((b*x+a)^2)^(1/2)+3*b*e*(-2*A*b*e+B*a*e+B*b*d)*(b*x+a)*l n(e*x+d)/(-a*e+b*d)^5/((b*x+a)^2)^(1/2)
Time = 1.13 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {(a+b x) \left (-b (A b-a B) (b d-a e)^2-2 b (b d-a e) (b B d-3 A b e+2 a B e) (a+b x)+\frac {e (b d-a e)^2 (-B d+A e) (a+b x)^2}{(d+e x)^2}+\frac {2 e (b d-a e) (-2 b B d+3 A b e-a B e) (a+b x)^2}{d+e x}-6 b e (b B d-2 A b e+a B e) (a+b x)^2 \log (a+b x)+6 b e (b B d-2 A b e+a B e) (a+b x)^2 \log (d+e x)\right )}{2 (b d-a e)^5 \left ((a+b x)^2\right )^{3/2}} \]
((a + b*x)*(-(b*(A*b - a*B)*(b*d - a*e)^2) - 2*b*(b*d - a*e)*(b*B*d - 3*A* b*e + 2*a*B*e)*(a + b*x) + (e*(b*d - a*e)^2*(-(B*d) + A*e)*(a + b*x)^2)/(d + e*x)^2 + (2*e*(b*d - a*e)*(-2*b*B*d + 3*A*b*e - a*B*e)*(a + b*x)^2)/(d + e*x) - 6*b*e*(b*B*d - 2*A*b*e + a*B*e)*(a + b*x)^2*Log[a + b*x] + 6*b*e* (b*B*d - 2*A*b*e + a*B*e)*(a + b*x)^2*Log[d + e*x]))/(2*(b*d - a*e)^5*((a + b*x)^2)^(3/2))
Time = 0.51 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.68, 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.
| \(\displaystyle \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (d+e x)^3} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^3 (a+b x) \int \frac {A+B x}{b^3 (a+b x)^3 (d+e x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {A+B x}{(a+b x)^3 (d+e x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {(a+b x) \int \left (\frac {3 e (-b B d+2 A b e-a B e) b^2}{(b d-a e)^5 (a+b x)}+\frac {(b B d-3 A b e+2 a B e) b^2}{(b d-a e)^4 (a+b x)^2}+\frac {(A b-a B) b^2}{(b d-a e)^3 (a+b x)^3}-\frac {3 e^2 (-b B d+2 A b e-a B e) b}{(b d-a e)^5 (d+e x)}-\frac {e^2 (-2 b B d+3 A b e-a B e)}{(b d-a e)^4 (d+e x)^2}-\frac {e^2 (A e-B d)}{(b d-a e)^3 (d+e x)^3}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {b (A b-a B)}{2 (a+b x)^2 (b d-a e)^3}-\frac {b (2 a B e-3 A b e+b B d)}{(a+b x) (b d-a e)^4}-\frac {e (a B e-3 A b e+2 b B d)}{(d+e x) (b d-a e)^4}-\frac {e (B d-A e)}{2 (d+e x)^2 (b d-a e)^3}-\frac {3 b e \log (a+b x) (a B e-2 A b e+b B d)}{(b d-a e)^5}+\frac {3 b e \log (d+e x) (a B e-2 A b e+b B d)}{(b d-a e)^5}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-1/2*(b*(A*b - a*B))/((b*d - a*e)^3*(a + b*x)^2) - (b*(b*B*d - 3*A*b*e + 2*a*B*e))/((b*d - a*e)^4*(a + b*x)) - (e*(B*d - A*e))/(2*(b*d - a*e)^3*(d + e*x)^2) - (e*(2*b*B*d - 3*A*b*e + a*B*e))/((b*d - a*e)^4*(d + e*x)) - (3*b*e*(b*B*d - 2*A*b*e + a*B*e)*Log[a + b*x])/(b*d - a*e)^5 + (3 *b*e*(b*B*d - 2*A*b*e + a*B*e)*Log[d + e*x])/(b*d - a*e)^5))/Sqrt[a^2 + 2* a*b*x + b^2*x^2]
3.18.74.3.1 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]
Leaf count of result is larger than twice the leaf count of optimal. \(650\) vs. \(2(262)=524\).
Time = 0.45 (sec) , antiderivative size = 651, normalized size of antiderivative = 1.96
| method | result | size |
| risch | \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {3 b^{2} e^{2} \left (2 A b e -B a e -B b d \right ) x^{3}}{e^{4} a^{4}-4 b d \,e^{3} a^{3}+6 b^{2} d^{2} e^{2} a^{2}-4 b^{3} d^{3} e a +b^{4} d^{4}}+\frac {9 b e \left (a e +b d \right ) \left (2 A b e -B a e -B b d \right ) x^{2}}{2 \left (e^{4} a^{4}-4 b d \,e^{3} a^{3}+6 b^{2} d^{2} e^{2} a^{2}-4 b^{3} d^{3} e a +b^{4} d^{4}\right )}+\frac {\left (2 A \,a^{2} b \,e^{3}+14 A a \,b^{2} d \,e^{2}+2 A \,b^{3} d^{2} e -B \,e^{3} a^{3}-8 B \,a^{2} b d \,e^{2}-8 B a \,b^{2} d^{2} e -B \,b^{3} d^{3}\right ) x}{e^{4} a^{4}-4 b d \,e^{3} a^{3}+6 b^{2} d^{2} e^{2} a^{2}-4 b^{3} d^{3} e a +b^{4} d^{4}}-\frac {A \,a^{3} e^{3}-7 A \,a^{2} b d \,e^{2}-7 A a \,b^{2} d^{2} e +A \,b^{3} d^{3}+B \,a^{3} d \,e^{2}+10 B \,a^{2} b \,d^{2} e +B a \,b^{2} d^{3}}{2 \left (e^{4} a^{4}-4 b d \,e^{3} a^{3}+6 b^{2} d^{2} e^{2} a^{2}-4 b^{3} d^{3} e a +b^{4} d^{4}\right )}\right )}{\left (b x +a \right )^{3} \left (e x +d \right )^{2}}+\frac {3 \sqrt {\left (b x +a \right )^{2}}\, b e \left (2 A b e -B a e -B b d \right ) \ln \left (-e x -d \right )}{\left (b x +a \right ) \left (e^{5} a^{5}-5 b d \,e^{4} a^{4}+10 b^{2} d^{2} e^{3} a^{3}-10 b^{3} d^{3} e^{2} a^{2}+5 b^{4} d^{4} e a -b^{5} d^{5}\right )}-\frac {3 \sqrt {\left (b x +a \right )^{2}}\, b e \left (2 A b e -B a e -B b d \right ) \ln \left (b x +a \right )}{\left (b x +a \right ) \left (e^{5} a^{5}-5 b d \,e^{4} a^{4}+10 b^{2} d^{2} e^{3} a^{3}-10 b^{3} d^{3} e^{2} a^{2}+5 b^{4} d^{4} e a -b^{5} d^{5}\right )}\) | \(651\) |
| default | \(\text {Expression too large to display}\) | \(1271\) |
((b*x+a)^2)^(1/2)/(b*x+a)^3*(3*b^2*e^2*(2*A*b*e-B*a*e-B*b*d)/(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)*x^3+9/2*b*e*(a*e+b*d)*( 2*A*b*e-B*a*e-B*b*d)/(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)*x^2+(2*A*a^2*b*e^3+14*A*a*b^2*d*e^2+2*A*b^3*d^2*e-B*a^3*e^3-8*B *a^2*b*d*e^2-8*B*a*b^2*d^2*e-B*b^3*d^3)/(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)*x-1/2*(A*a^3*e^3-7*A*a^2*b*d*e^2-7*A*a*b^2*d ^2*e+A*b^3*d^3+B*a^3*d*e^2+10*B*a^2*b*d^2*e+B*a*b^2*d^3)/(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)^2+3*((b*x+a)^2)^(1 /2)/(b*x+a)*b*e*(2*A*b*e-B*a*e-B*b*d)/(a^5*e^5-5*a^4*b*d*e^4+10*a^3*b^2*d^ 2*e^3-10*a^2*b^3*d^3*e^2+5*a*b^4*d^4*e-b^5*d^5)*ln(-e*x-d)-3*((b*x+a)^2)^( 1/2)/(b*x+a)*b*e*(2*A*b*e-B*a*e-B*b*d)/(a^5*e^5-5*a^4*b*d*e^4+10*a^3*b^2*d ^2*e^3-10*a^2*b^3*d^3*e^2+5*a*b^4*d^4*e-b^5*d^5)*ln(b*x+a)
Leaf count of result is larger than twice the leaf count of optimal. 1215 vs. \(2 (262) = 524\).
Time = 0.40 (sec) , antiderivative size = 1215, normalized size of antiderivative = 3.66 \[ \int \frac {A+B x}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\text {Too large to display} \]
1/2*(9*B*a^3*b*d^2*e^2 + A*a^4*e^4 - (B*a*b^3 + A*b^4)*d^4 - (9*B*a^2*b^2 - 8*A*a*b^3)*d^3*e + (B*a^4 - 8*A*a^3*b)*d*e^3 - 6*(B*b^4*d^2*e^2 - 2*A*b^ 4*d*e^3 - (B*a^2*b^2 - 2*A*a*b^3)*e^4)*x^3 - 9*(B*b^4*d^3*e - B*a^2*b^2*d* e^3 + (B*a*b^3 - 2*A*b^4)*d^2*e^2 - (B*a^3*b - 2*A*a^2*b^2)*e^4)*x^2 - 2*( B*b^4*d^4 - 12*A*a*b^3*d^2*e^2 + (7*B*a*b^3 - 2*A*b^4)*d^3*e - (7*B*a^3*b - 12*A*a^2*b^2)*d*e^3 - (B*a^4 - 2*A*a^3*b)*e^4)*x - 6*(B*a^2*b^2*d^3*e + (B*a^3*b - 2*A*a^2*b^2)*d^2*e^2 + (B*b^4*d*e^3 + (B*a*b^3 - 2*A*b^4)*e^4)* x^4 + 2*(B*b^4*d^2*e^2 + 2*(B*a*b^3 - A*b^4)*d*e^3 + (B*a^2*b^2 - 2*A*a*b^ 3)*e^4)*x^3 + (B*b^4*d^3*e + (5*B*a*b^3 - 2*A*b^4)*d^2*e^2 + (5*B*a^2*b^2 - 8*A*a*b^3)*d*e^3 + (B*a^3*b - 2*A*a^2*b^2)*e^4)*x^2 + 2*(B*a*b^3*d^3*e + 2*(B*a^2*b^2 - A*a*b^3)*d^2*e^2 + (B*a^3*b - 2*A*a^2*b^2)*d*e^3)*x)*log(b *x + a) + 6*(B*a^2*b^2*d^3*e + (B*a^3*b - 2*A*a^2*b^2)*d^2*e^2 + (B*b^4*d* e^3 + (B*a*b^3 - 2*A*b^4)*e^4)*x^4 + 2*(B*b^4*d^2*e^2 + 2*(B*a*b^3 - A*b^4 )*d*e^3 + (B*a^2*b^2 - 2*A*a*b^3)*e^4)*x^3 + (B*b^4*d^3*e + (5*B*a*b^3 - 2 *A*b^4)*d^2*e^2 + (5*B*a^2*b^2 - 8*A*a*b^3)*d*e^3 + (B*a^3*b - 2*A*a^2*b^2 )*e^4)*x^2 + 2*(B*a*b^3*d^3*e + 2*(B*a^2*b^2 - A*a*b^3)*d^2*e^2 + (B*a^3*b - 2*A*a^2*b^2)*d*e^3)*x)*log(e*x + d))/(a^2*b^5*d^7 - 5*a^3*b^4*d^6*e + 1 0*a^4*b^3*d^5*e^2 - 10*a^5*b^2*d^4*e^3 + 5*a^6*b*d^3*e^4 - a^7*d^2*e^5 + ( b^7*d^5*e^2 - 5*a*b^6*d^4*e^3 + 10*a^2*b^5*d^3*e^4 - 10*a^3*b^4*d^2*e^5 + 5*a^4*b^3*d*e^6 - a^5*b^2*e^7)*x^4 + 2*(b^7*d^6*e - 4*a*b^6*d^5*e^2 + 5...
\[ \int \frac {A+B x}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {A + B x}{\left (d + e x\right )^{3} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {A+B x}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 634 vs. \(2 (262) = 524\).
Time = 0.30 (sec) , antiderivative size = 634, normalized size of antiderivative = 1.91 \[ \int \frac {A+B x}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {3 \, {\left (B b^{3} d e + B a b^{2} e^{2} - 2 \, A b^{3} e^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{6} d^{5} \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{5} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{4} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{3} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b^{2} d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{5} b e^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {3 \, {\left (B b^{2} d e^{2} + B a b e^{3} - 2 \, A b^{2} e^{3}\right )} \log \left ({\left | e x + d \right |}\right )}{b^{5} d^{5} e \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{4} d^{4} e^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} d^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{2} d^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b d e^{5} \mathrm {sgn}\left (b x + a\right ) - a^{5} e^{6} \mathrm {sgn}\left (b x + a\right )} - \frac {6 \, B b^{3} d e^{2} x^{3} + 6 \, B a b^{2} e^{3} x^{3} - 12 \, A b^{3} e^{3} x^{3} + 9 \, B b^{3} d^{2} e x^{2} + 18 \, B a b^{2} d e^{2} x^{2} - 18 \, A b^{3} d e^{2} x^{2} + 9 \, B a^{2} b e^{3} x^{2} - 18 \, A a b^{2} e^{3} x^{2} + 2 \, B b^{3} d^{3} x + 16 \, B a b^{2} d^{2} e x - 4 \, A b^{3} d^{2} e x + 16 \, B a^{2} b d e^{2} x - 28 \, A a b^{2} d e^{2} x + 2 \, B a^{3} e^{3} x - 4 \, A a^{2} b e^{3} x + B a b^{2} d^{3} + A b^{3} d^{3} + 10 \, B a^{2} b d^{2} e - 7 \, A a b^{2} d^{2} e + B a^{3} d e^{2} - 7 \, A a^{2} b d e^{2} + A a^{3} e^{3}}{2 \, {\left (b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} {\left (b e x^{2} + b d x + a e x + a d\right )}^{2}} \]
-3*(B*b^3*d*e + B*a*b^2*e^2 - 2*A*b^3*e^2)*log(abs(b*x + a))/(b^6*d^5*sgn( b*x + a) - 5*a*b^5*d^4*e*sgn(b*x + a) + 10*a^2*b^4*d^3*e^2*sgn(b*x + a) - 10*a^3*b^3*d^2*e^3*sgn(b*x + a) + 5*a^4*b^2*d*e^4*sgn(b*x + a) - a^5*b*e^5 *sgn(b*x + a)) + 3*(B*b^2*d*e^2 + B*a*b*e^3 - 2*A*b^2*e^3)*log(abs(e*x + d ))/(b^5*d^5*e*sgn(b*x + a) - 5*a*b^4*d^4*e^2*sgn(b*x + a) + 10*a^2*b^3*d^3 *e^3*sgn(b*x + a) - 10*a^3*b^2*d^2*e^4*sgn(b*x + a) + 5*a^4*b*d*e^5*sgn(b* x + a) - a^5*e^6*sgn(b*x + a)) - 1/2*(6*B*b^3*d*e^2*x^3 + 6*B*a*b^2*e^3*x^ 3 - 12*A*b^3*e^3*x^3 + 9*B*b^3*d^2*e*x^2 + 18*B*a*b^2*d*e^2*x^2 - 18*A*b^3 *d*e^2*x^2 + 9*B*a^2*b*e^3*x^2 - 18*A*a*b^2*e^3*x^2 + 2*B*b^3*d^3*x + 16*B *a*b^2*d^2*e*x - 4*A*b^3*d^2*e*x + 16*B*a^2*b*d*e^2*x - 28*A*a*b^2*d*e^2*x + 2*B*a^3*e^3*x - 4*A*a^2*b*e^3*x + B*a*b^2*d^3 + A*b^3*d^3 + 10*B*a^2*b* d^2*e - 7*A*a*b^2*d^2*e + B*a^3*d*e^2 - 7*A*a^2*b*d*e^2 + A*a^3*e^3)/((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))*(b*e*x^2 + b*d* x + a*e*x + a*d)^2)
Timed out. \[ \int \frac {A+B x}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {A+B\,x}{{\left (d+e\,x\right )}^3\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]