Integrand size = 30, antiderivative size = 208 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=-\frac {2 (b d-a e)^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^4 (a+b x)}+\frac {6 b (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^4 (a+b x)}-\frac {6 b^2 (b d-a e) (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^4 (a+b x)}+\frac {2 b^3 (d+e x)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 e^4 (a+b x)} \]
-2/3*(-a*e+b*d)^3*(e*x+d)^(3/2)*((b*x+a)^2)^(1/2)/e^4/(b*x+a)+6/5*b*(-a*e+ b*d)^2*(e*x+d)^(5/2)*((b*x+a)^2)^(1/2)/e^4/(b*x+a)-6/7*b^2*(-a*e+b*d)*(e*x +d)^(7/2)*((b*x+a)^2)^(1/2)/e^4/(b*x+a)+2/9*b^3*(e*x+d)^(9/2)*((b*x+a)^2)^ (1/2)/e^4/(b*x+a)
Time = 0.07 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.58 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 \sqrt {(a+b x)^2} (d+e x)^{3/2} \left (105 a^3 e^3+63 a^2 b e^2 (-2 d+3 e x)+9 a b^2 e \left (8 d^2-12 d e x+15 e^2 x^2\right )+b^3 \left (-16 d^3+24 d^2 e x-30 d e^2 x^2+35 e^3 x^3\right )\right )}{315 e^4 (a+b x)} \]
(2*Sqrt[(a + b*x)^2]*(d + e*x)^(3/2)*(105*a^3*e^3 + 63*a^2*b*e^2*(-2*d + 3 *e*x) + 9*a*b^2*e*(8*d^2 - 12*d*e*x + 15*e^2*x^2) + b^3*(-16*d^3 + 24*d^2* e*x - 30*d*e^2*x^2 + 35*e^3*x^3)))/(315*e^4*(a + b*x))
Time = 0.26 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.62, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1102, 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.
\(\displaystyle \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \sqrt {d+e x} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^3 (a+b x)^3 \sqrt {d+e x}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a+b x)^3 \sqrt {d+e x}dx}{a+b x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b^3 (d+e x)^{7/2}}{e^3}-\frac {3 b^2 (b d-a e) (d+e x)^{5/2}}{e^3}+\frac {3 b (b d-a e)^2 (d+e x)^{3/2}}{e^3}+\frac {(a e-b d)^3 \sqrt {d+e x}}{e^3}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {6 b^2 (d+e x)^{7/2} (b d-a e)}{7 e^4}+\frac {6 b (d+e x)^{5/2} (b d-a e)^2}{5 e^4}-\frac {2 (d+e x)^{3/2} (b d-a e)^3}{3 e^4}+\frac {2 b^3 (d+e x)^{9/2}}{9 e^4}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-2*(b*d - a*e)^3*(d + e*x)^(3/2))/(3*e^4) + (6*b*(b*d - a*e)^2*(d + e*x)^(5/2))/(5*e^4) - (6*b^2*(b*d - a*e)*(d + e *x)^(7/2))/(7*e^4) + (2*b^3*(d + e*x)^(9/2))/(9*e^4)))/(a + b*x)
3.17.85.3.1 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_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 2.25 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.63
method | result | size |
gosper | \(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (35 e^{3} x^{3} b^{3}+135 x^{2} a \,b^{2} e^{3}-30 x^{2} b^{3} d \,e^{2}+189 a^{2} b \,e^{3} x -108 x a \,b^{2} d \,e^{2}+24 b^{3} d^{2} e x +105 a^{3} e^{3}-126 a^{2} b d \,e^{2}+72 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{315 e^{4} \left (b x +a \right )^{3}}\) | \(132\) |
default | \(\frac {2 \left (e x +d \right )^{\frac {3}{2}} \left (35 e^{3} x^{3} b^{3}+135 x^{2} a \,b^{2} e^{3}-30 x^{2} b^{3} d \,e^{2}+189 a^{2} b \,e^{3} x -108 x a \,b^{2} d \,e^{2}+24 b^{3} d^{2} e x +105 a^{3} e^{3}-126 a^{2} b d \,e^{2}+72 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{315 e^{4} \left (b x +a \right )^{3}}\) | \(132\) |
risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (35 b^{3} x^{4} e^{4}+135 a \,b^{2} e^{4} x^{3}+5 b^{3} d \,e^{3} x^{3}+189 a^{2} b \,e^{4} x^{2}+27 a \,b^{2} d \,e^{3} x^{2}-6 b^{3} d^{2} e^{2} x^{2}+105 e^{4} a^{3} x +63 a^{2} b d \,e^{3} x -36 a \,b^{2} d^{2} e^{2} x +8 b^{3} d^{3} e x +105 a^{3} d \,e^{3}-126 a^{2} b \,d^{2} e^{2}+72 a \,b^{2} d^{3} e -16 b^{3} d^{4}\right ) \sqrt {e x +d}}{315 \left (b x +a \right ) e^{4}}\) | \(186\) |
2/315*(e*x+d)^(3/2)*(35*b^3*e^3*x^3+135*a*b^2*e^3*x^2-30*b^3*d*e^2*x^2+189 *a^2*b*e^3*x-108*a*b^2*d*e^2*x+24*b^3*d^2*e*x+105*a^3*e^3-126*a^2*b*d*e^2+ 72*a*b^2*d^2*e-16*b^3*d^3)*((b*x+a)^2)^(3/2)/e^4/(b*x+a)^3
Time = 0.27 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.79 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 \, {\left (35 \, b^{3} e^{4} x^{4} - 16 \, b^{3} d^{4} + 72 \, a b^{2} d^{3} e - 126 \, a^{2} b d^{2} e^{2} + 105 \, a^{3} d e^{3} + 5 \, {\left (b^{3} d e^{3} + 27 \, a b^{2} e^{4}\right )} x^{3} - 3 \, {\left (2 \, b^{3} d^{2} e^{2} - 9 \, a b^{2} d e^{3} - 63 \, a^{2} b e^{4}\right )} x^{2} + {\left (8 \, b^{3} d^{3} e - 36 \, a b^{2} d^{2} e^{2} + 63 \, a^{2} b d e^{3} + 105 \, a^{3} e^{4}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{4}} \]
2/315*(35*b^3*e^4*x^4 - 16*b^3*d^4 + 72*a*b^2*d^3*e - 126*a^2*b*d^2*e^2 + 105*a^3*d*e^3 + 5*(b^3*d*e^3 + 27*a*b^2*e^4)*x^3 - 3*(2*b^3*d^2*e^2 - 9*a* b^2*d*e^3 - 63*a^2*b*e^4)*x^2 + (8*b^3*d^3*e - 36*a*b^2*d^2*e^2 + 63*a^2*b *d*e^3 + 105*a^3*e^4)*x)*sqrt(e*x + d)/e^4
\[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int \sqrt {d + e x} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.79 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 \, {\left (35 \, b^{3} e^{4} x^{4} - 16 \, b^{3} d^{4} + 72 \, a b^{2} d^{3} e - 126 \, a^{2} b d^{2} e^{2} + 105 \, a^{3} d e^{3} + 5 \, {\left (b^{3} d e^{3} + 27 \, a b^{2} e^{4}\right )} x^{3} - 3 \, {\left (2 \, b^{3} d^{2} e^{2} - 9 \, a b^{2} d e^{3} - 63 \, a^{2} b e^{4}\right )} x^{2} + {\left (8 \, b^{3} d^{3} e - 36 \, a b^{2} d^{2} e^{2} + 63 \, a^{2} b d e^{3} + 105 \, a^{3} e^{4}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{4}} \]
2/315*(35*b^3*e^4*x^4 - 16*b^3*d^4 + 72*a*b^2*d^3*e - 126*a^2*b*d^2*e^2 + 105*a^3*d*e^3 + 5*(b^3*d*e^3 + 27*a*b^2*e^4)*x^3 - 3*(2*b^3*d^2*e^2 - 9*a* b^2*d*e^3 - 63*a^2*b*e^4)*x^2 + (8*b^3*d^3*e - 36*a*b^2*d^2*e^2 + 63*a^2*b *d*e^3 + 105*a^3*e^4)*x)*sqrt(e*x + d)/e^4
Leaf count of result is larger than twice the leaf count of optimal. 370 vs. \(2 (148) = 296\).
Time = 0.46 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.78 \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} a^{3} d \mathrm {sgn}\left (b x + a\right ) + 105 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {315 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{2} b d \mathrm {sgn}\left (b x + a\right )}{e} + \frac {63 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a b^{2} d \mathrm {sgn}\left (b x + a\right )}{e^{2}} + \frac {63 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a^{2} b \mathrm {sgn}\left (b x + a\right )}{e} + \frac {9 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} b^{3} d \mathrm {sgn}\left (b x + a\right )}{e^{3}} + \frac {27 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} a b^{2} \mathrm {sgn}\left (b x + a\right )}{e^{2}} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} b^{3} \mathrm {sgn}\left (b x + a\right )}{e^{3}}\right )}}{315 \, e} \]
2/315*(315*sqrt(e*x + d)*a^3*d*sgn(b*x + a) + 105*((e*x + d)^(3/2) - 3*sqr t(e*x + d)*d)*a^3*sgn(b*x + a) + 315*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d) *a^2*b*d*sgn(b*x + a)/e + 63*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 1 5*sqrt(e*x + d)*d^2)*a*b^2*d*sgn(b*x + a)/e^2 + 63*(3*(e*x + d)^(5/2) - 10 *(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a^2*b*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)*b^3*d*sgn(b*x + a)/e^3 + 27*(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^2*sgn(b*x + a) /e^2 + (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^3*sgn(b*x + a)/e^3 )/e
Timed out. \[ \int \sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int \sqrt {d+e\,x}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \]