Integrand size = 35, antiderivative size = 152 \[ \int (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)}-\frac {4 b (b d-a e) (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^3 (a+b x)}+\frac {2 b^2 (d+e x)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 e^3 (a+b x)} \]
2/5*(-a*e+b*d)^2*(e*x+d)^(5/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)-4/7*b*(-a*e+b *d)*(e*x+d)^(7/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)+2/9*b^2*(e*x+d)^(9/2)*((b* x+a)^2)^(1/2)/e^3/(b*x+a)
Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.52 \[ \int (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \sqrt {(a+b x)^2} (d+e x)^{5/2} \left (63 a^2 e^2+18 a b e (-2 d+5 e x)+b^2 \left (8 d^2-20 d e x+35 e^2 x^2\right )\right )}{315 e^3 (a+b x)} \]
(2*Sqrt[(a + b*x)^2]*(d + e*x)^(5/2)*(63*a^2*e^2 + 18*a*b*e*(-2*d + 5*e*x) + b^2*(8*d^2 - 20*d*e*x + 35*e^2*x^2)))/(315*e^3*(a + b*x))
Time = 0.25 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.65, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {1187, 27, 53, 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.
\(\displaystyle \int (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b (a+b x)^2 (d+e x)^{3/2}dx}{b (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a+b x)^2 (d+e x)^{3/2}dx}{a+b x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b^2 (d+e x)^{7/2}}{e^2}-\frac {2 b (b d-a e) (d+e x)^{5/2}}{e^2}+\frac {(a e-b d)^2 (d+e x)^{3/2}}{e^2}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {4 b (d+e x)^{7/2} (b d-a e)}{7 e^3}+\frac {2 (d+e x)^{5/2} (b d-a e)^2}{5 e^3}+\frac {2 b^2 (d+e x)^{9/2}}{9 e^3}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((2*(b*d - a*e)^2*(d + e*x)^(5/2))/(5*e^3) - (4*b*(b*d - a*e)*(d + e*x)^(7/2))/(7*e^3) + (2*b^2*(d + e*x)^(9/2))/(9*e ^3)))/(a + b*x)
3.21.94.3.1 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, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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]
Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.45
method | result | size |
default | \(\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (e x +d \right )^{\frac {5}{2}} \left (35 b^{2} e^{2} x^{2}+90 a b \,e^{2} x -20 b^{2} d e x +63 e^{2} a^{2}-36 a b d e +8 b^{2} d^{2}\right )}{315 e^{3}}\) | \(69\) |
gosper | \(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (35 b^{2} e^{2} x^{2}+90 a b \,e^{2} x -20 b^{2} d e x +63 e^{2} a^{2}-36 a b d e +8 b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{315 e^{3} \left (b x +a \right )}\) | \(79\) |
risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (35 b^{2} x^{4} e^{4}+90 a b \,e^{4} x^{3}+50 b^{2} d \,e^{3} x^{3}+63 a^{2} e^{4} x^{2}+144 a b d \,e^{3} x^{2}+3 b^{2} d^{2} e^{2} x^{2}+126 a^{2} d \,e^{3} x +18 a b \,d^{2} e^{2} x -4 b^{2} d^{3} e x +63 a^{2} d^{2} e^{2}-36 a b \,d^{3} e +8 b^{2} d^{4}\right ) \sqrt {e x +d}}{315 \left (b x +a \right ) e^{3}}\) | \(157\) |
2/315*csgn(b*x+a)*(e*x+d)^(5/2)*(35*b^2*e^2*x^2+90*a*b*e^2*x-20*b^2*d*e*x+ 63*a^2*e^2-36*a*b*d*e+8*b^2*d^2)/e^3
Time = 0.30 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.90 \[ \int (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (35 \, b^{2} e^{4} x^{4} + 8 \, b^{2} d^{4} - 36 \, a b d^{3} e + 63 \, a^{2} d^{2} e^{2} + 10 \, {\left (5 \, b^{2} d e^{3} + 9 \, a b e^{4}\right )} x^{3} + 3 \, {\left (b^{2} d^{2} e^{2} + 48 \, a b d e^{3} + 21 \, a^{2} e^{4}\right )} x^{2} - 2 \, {\left (2 \, b^{2} d^{3} e - 9 \, a b d^{2} e^{2} - 63 \, a^{2} d e^{3}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{3}} \]
2/315*(35*b^2*e^4*x^4 + 8*b^2*d^4 - 36*a*b*d^3*e + 63*a^2*d^2*e^2 + 10*(5* b^2*d*e^3 + 9*a*b*e^4)*x^3 + 3*(b^2*d^2*e^2 + 48*a*b*d*e^3 + 21*a^2*e^4)*x ^2 - 2*(2*b^2*d^3*e - 9*a*b*d^2*e^2 - 63*a^2*d*e^3)*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 = 0.21 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.10 \[ \int (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (5 \, b e^{3} x^{3} - 2 \, b d^{3} + 7 \, a d^{2} e + {\left (8 \, b d e^{2} + 7 \, a e^{3}\right )} x^{2} + {\left (b d^{2} e + 14 \, a d e^{2}\right )} x\right )} \sqrt {e x + d} a}{35 \, e^{2}} + \frac {2 \, {\left (35 \, b e^{4} x^{4} + 8 \, b d^{4} - 18 \, a d^{3} e + 5 \, {\left (10 \, b d e^{3} + 9 \, a e^{4}\right )} x^{3} + 3 \, {\left (b d^{2} e^{2} + 24 \, a d e^{3}\right )} x^{2} - {\left (4 \, b d^{3} e - 9 \, a d^{2} e^{2}\right )} x\right )} \sqrt {e x + d} b}{315 \, e^{3}} \]
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. 414 vs. \(2 (107) = 214\).
Time = 0.28 (sec) , antiderivative size = 414, normalized size of antiderivative = 2.72 \[ \int (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} a^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) + 210 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{2} d \mathrm {sgn}\left (b x + a\right ) + \frac {210 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a b d^{2} \mathrm {sgn}\left (b x + a\right )}{e} + 21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} b^{2} d^{2} \mathrm {sgn}\left (b x + a\right )}{e^{2}} + \frac {84 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a b d \mathrm {sgn}\left (b x + a\right )}{e} + \frac {18 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} b^{2} d \mathrm {sgn}\left (b x + a\right )}{e^{2}} + \frac {18 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} a b \mathrm {sgn}\left (b x + a\right )}{e} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} b^{2} \mathrm {sgn}\left (b x + a\right )}{e^{2}}\right )}}{315 \, e} \]
2/315*(315*sqrt(e*x + d)*a^2*d^2*sgn(b*x + a) + 210*((e*x + d)^(3/2) - 3*s qrt(e*x + d)*d)*a^2*d*sgn(b*x + a) + 210*((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*sqrt(e*x + d)*d^2)*a^2*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^2*d^2*sgn(b*x + a)/e^2 + 84*(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^2*d*sgn(b*x + a)/e^2 + 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) *a*b*sgn(b*x + a)/e + (35*(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^2*s gn(b*x + a)/e^2)/e
Timed out. \[ \int (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\int \sqrt {{\left (a+b\,x\right )}^2}\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^{3/2} \,d x \]