Integrand size = 35, antiderivative size = 164 \[ \int (A+B x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 (b d-a e) (B d-A e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)}-\frac {2 (2 b B d-A b e-a B e) (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^3 (a+b x)}+\frac {2 b B (d+e x)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 e^3 (a+b x)} \]
2/5*(-a*e+b*d)*(-A*e+B*d)*(e*x+d)^(5/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)-2/7* (-A*b*e-B*a*e+2*B*b*d)*(e*x+d)^(7/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)+2/9*b*B *(e*x+d)^(9/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.54 \[ \int (A+B x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \sqrt {(a+b x)^2} (d+e x)^{5/2} \left (9 A b e (-2 d+5 e x)+9 a e (-2 B d+7 A e+5 B e x)+b B \left (8 d^2-20 d e x+35 e^2 x^2\right )\right )}{315 e^3 (a+b x)} \]
(2*Sqrt[(a + b*x)^2]*(d + e*x)^(5/2)*(9*A*b*e*(-2*d + 5*e*x) + 9*a*e*(-2*B *d + 7*A*e + 5*B*e*x) + b*B*(8*d^2 - 20*d*e*x + 35*e^2*x^2)))/(315*e^3*(a + b*x))
Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.68, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, 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.
\(\displaystyle \int \sqrt {a^2+2 a b x+b^2 x^2} (A+B x) (d+e x)^{3/2} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b (a+b x) (A+B x) (d+e x)^{3/2}dx}{b (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a+b x) (A+B x) (d+e x)^{3/2}dx}{a+b x}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b B (d+e x)^{7/2}}{e^2}+\frac {(-2 b B d+A b e+a B e) (d+e x)^{5/2}}{e^2}+\frac {(a e-b d) (A e-B d) (d+e x)^{3/2}}{e^2}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {2 (d+e x)^{7/2} (-a B e-A b e+2 b B d)}{7 e^3}+\frac {2 (d+e x)^{5/2} (b d-a e) (B d-A e)}{5 e^3}+\frac {2 b B (d+e x)^{9/2}}{9 e^3}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((2*(b*d - a*e)*(B*d - A*e)*(d + e*x)^(5/2) )/(5*e^3) - (2*(2*b*B*d - A*b*e - a*B*e)*(d + e*x)^(7/2))/(7*e^3) + (2*b*B *(d + e*x)^(9/2))/(9*e^3)))/(a + b*x)
3.19.37.3.1 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]
Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.22 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.48
method | result | size |
default | \(\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (e x +d \right )^{\frac {5}{2}} \left (35 B b \,e^{2} x^{2}+45 A b \,e^{2} x +45 B a \,e^{2} x -20 B b d e x +63 A a \,e^{2}-18 A b d e -18 B a d e +8 B b \,d^{2}\right )}{315 e^{3}}\) | \(79\) |
gosper | \(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (35 B b \,e^{2} x^{2}+45 A b \,e^{2} x +45 B a \,e^{2} x -20 B b d e x +63 A a \,e^{2}-18 A b d e -18 B a d e +8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{315 e^{3} \left (b x +a \right )}\) | \(89\) |
risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (35 B b \,x^{4} e^{4}+45 A b \,e^{4} x^{3}+45 B a \,e^{4} x^{3}+50 B b d \,e^{3} x^{3}+63 A a \,e^{4} x^{2}+72 A b d \,e^{3} x^{2}+72 B a d \,e^{3} x^{2}+3 B b \,d^{2} e^{2} x^{2}+126 A a d \,e^{3} x +9 A b \,d^{2} e^{2} x +9 B a \,d^{2} e^{2} x -4 B b \,d^{3} e x +63 A a \,d^{2} e^{2}-18 A b \,d^{3} e -18 B a \,d^{3} e +8 B b \,d^{4}\right ) \sqrt {e x +d}}{315 \left (b x +a \right ) e^{3}}\) | \(189\) |
2/315*csgn(b*x+a)*(e*x+d)^(5/2)*(35*B*b*e^2*x^2+45*A*b*e^2*x+45*B*a*e^2*x- 20*B*b*d*e*x+63*A*a*e^2-18*A*b*d*e-18*B*a*d*e+8*B*b*d^2)/e^3
Time = 0.33 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.91 \[ \int (A+B x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (35 \, B b e^{4} x^{4} + 8 \, B b d^{4} + 63 \, A a d^{2} e^{2} - 18 \, {\left (B a + A b\right )} d^{3} e + 5 \, {\left (10 \, B b d e^{3} + 9 \, {\left (B a + A b\right )} e^{4}\right )} x^{3} + 3 \, {\left (B b d^{2} e^{2} + 21 \, A a e^{4} + 24 \, {\left (B a + A b\right )} d e^{3}\right )} x^{2} - {\left (4 \, B b d^{3} e - 126 \, A a d e^{3} - 9 \, {\left (B a + A b\right )} d^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{3}} \]
2/315*(35*B*b*e^4*x^4 + 8*B*b*d^4 + 63*A*a*d^2*e^2 - 18*(B*a + A*b)*d^3*e + 5*(10*B*b*d*e^3 + 9*(B*a + A*b)*e^4)*x^3 + 3*(B*b*d^2*e^2 + 21*A*a*e^4 + 24*(B*a + A*b)*d*e^3)*x^2 - (4*B*b*d^3*e - 126*A*a*d*e^3 - 9*(B*a + A*b)* d^2*e^2)*x)*sqrt(e*x + d)/e^3
\[ \int (A+B x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\int \left (A + B x\right ) \left (d + e x\right )^{\frac {3}{2}} \sqrt {\left (a + b x\right )^{2}}\, dx \]
Time = 0.23 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.02 \[ \int (A+B x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (5 \, b e^{3} x^{3} - 2 \, b d^{3} + 7 \, a d^{2} e + {\left (8 \, b d e^{2} + 7 \, a e^{3}\right )} x^{2} + {\left (b d^{2} e + 14 \, a d e^{2}\right )} x\right )} \sqrt {e x + d} A}{35 \, e^{2}} + \frac {2 \, {\left (35 \, b e^{4} x^{4} + 8 \, b d^{4} - 18 \, a d^{3} e + 5 \, {\left (10 \, b d e^{3} + 9 \, a e^{4}\right )} x^{3} + 3 \, {\left (b d^{2} e^{2} + 24 \, a d e^{3}\right )} x^{2} - {\left (4 \, b d^{3} e - 9 \, a d^{2} e^{2}\right )} x\right )} \sqrt {e x + d} B}{315 \, e^{3}} \]
2/35*(5*b*e^3*x^3 - 2*b*d^3 + 7*a*d^2*e + (8*b*d*e^2 + 7*a*e^3)*x^2 + (b*d ^2*e + 14*a*d*e^2)*x)*sqrt(e*x + d)*A/e^2 + 2/315*(35*b*e^4*x^4 + 8*b*d^4 - 18*a*d^3*e + 5*(10*b*d*e^3 + 9*a*e^4)*x^3 + 3*(b*d^2*e^2 + 24*a*d*e^3)*x ^2 - (4*b*d^3*e - 9*a*d^2*e^2)*x)*sqrt(e*x + d)*B/e^3
Leaf count of result is larger than twice the leaf count of optimal. 545 vs. \(2 (119) = 238\).
Time = 0.28 (sec) , antiderivative size = 545, normalized size of antiderivative = 3.32 \[ \int (A+B x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} A a d^{2} \mathrm {sgn}\left (b x + a\right ) + 210 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A a d \mathrm {sgn}\left (b x + a\right ) + \frac {105 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} B a d^{2} \mathrm {sgn}\left (b x + a\right )}{e} + \frac {105 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} A b d^{2} \mathrm {sgn}\left (b x + a\right )}{e} + 21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} A a \mathrm {sgn}\left (b x + a\right ) + \frac {21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} B b d^{2} \mathrm {sgn}\left (b x + a\right )}{e^{2}} + \frac {42 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} B a d \mathrm {sgn}\left (b x + a\right )}{e} + \frac {42 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} A b d \mathrm {sgn}\left (b x + a\right )}{e} + \frac {18 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} B b d \mathrm {sgn}\left (b x + a\right )}{e^{2}} + \frac {9 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} B a \mathrm {sgn}\left (b x + a\right )}{e} + \frac {9 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} A b \mathrm {sgn}\left (b x + a\right )}{e} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} B b \mathrm {sgn}\left (b x + a\right )}{e^{2}}\right )}}{315 \, e} \]
2/315*(315*sqrt(e*x + d)*A*a*d^2*sgn(b*x + a) + 210*((e*x + d)^(3/2) - 3*s qrt(e*x + d)*d)*A*a*d*sgn(b*x + a) + 105*((e*x + d)^(3/2) - 3*sqrt(e*x + d )*d)*B*a*d^2*sgn(b*x + a)/e + 105*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A* b*d^2*sgn(b*x + a)/e + 21*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*s qrt(e*x + d)*d^2)*A*a*sgn(b*x + a) + 21*(3*(e*x + d)^(5/2) - 10*(e*x + d)^ (3/2)*d + 15*sqrt(e*x + d)*d^2)*B*b*d^2*sgn(b*x + a)/e^2 + 42*(3*(e*x + d) ^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*a*d*sgn(b*x + a)/e + 42*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*A* b*d*sgn(b*x + a)/e + 18*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e* x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*B*b*d*sgn(b*x + a)/e^2 + 9*(5*(e* x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*B*a*sgn(b*x + a)/e + 9*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)* d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*A*b*sgn(b*x + a)/e + (3 5*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420* (e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*B*b*sgn(b*x + a)/e^2)/e
Timed out. \[ \int (A+B x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\int \sqrt {{\left (a+b\,x\right )}^2}\,\left (A+B\,x\right )\,{\left (d+e\,x\right )}^{3/2} \,d x \]