Integrand size = 35, antiderivative size = 164 \[ \int (A+B x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 (b d-a e) (B d-A e) (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^3 (a+b x)}-\frac {2 (2 b B d-A b e-a B e) (d+e x)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 e^3 (a+b x)}+\frac {2 b B (d+e x)^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 e^3 (a+b x)} \]
2/7*(-a*e+b*d)*(-A*e+B*d)*(e*x+d)^(7/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)-2/9* (-A*b*e-B*a*e+2*B*b*d)*(e*x+d)^(9/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)+2/11*b* B*(e*x+d)^(11/2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.54 \[ \int (A+B x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \sqrt {(a+b x)^2} (d+e x)^{7/2} \left (11 A b e (-2 d+7 e x)+11 a e (-2 B d+9 A e+7 B e x)+b B \left (8 d^2-28 d e x+63 e^2 x^2\right )\right )}{693 e^3 (a+b x)} \]
(2*Sqrt[(a + b*x)^2]*(d + e*x)^(7/2)*(11*A*b*e*(-2*d + 7*e*x) + 11*a*e*(-2 *B*d + 9*A*e + 7*B*e*x) + b*B*(8*d^2 - 28*d*e*x + 63*e^2*x^2)))/(693*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.
\(\displaystyle \int \sqrt {a^2+2 a b x+b^2 x^2} (A+B x) (d+e x)^{5/2} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b (a+b x) (A+B x) (d+e x)^{5/2}dx}{b (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a+b x) (A+B x) (d+e x)^{5/2}dx}{a+b x}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b B (d+e x)^{9/2}}{e^2}+\frac {(-2 b B d+A b e+a B e) (d+e x)^{7/2}}{e^2}+\frac {(a e-b d) (A e-B d) (d+e x)^{5/2}}{e^2}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {2 (d+e x)^{9/2} (-a B e-A b e+2 b B d)}{9 e^3}+\frac {2 (d+e x)^{7/2} (b d-a e) (B d-A e)}{7 e^3}+\frac {2 b B (d+e x)^{11/2}}{11 e^3}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((2*(b*d - a*e)*(B*d - A*e)*(d + e*x)^(7/2) )/(7*e^3) - (2*(2*b*B*d - A*b*e - a*B*e)*(d + e*x)^(9/2))/(9*e^3) + (2*b*B *(d + e*x)^(11/2))/(11*e^3)))/(a + b*x)
3.19.36.3.1 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 = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.48
method | result | size |
default | \(\frac {2 \,\operatorname {csgn}\left (b x +a \right ) \left (e x +d \right )^{\frac {7}{2}} \left (63 B b \,e^{2} x^{2}+77 A b \,e^{2} x +77 B a \,e^{2} x -28 B b d e x +99 A a \,e^{2}-22 A b d e -22 B a d e +8 B b \,d^{2}\right )}{693 e^{3}}\) | \(79\) |
gosper | \(\frac {2 \left (e x +d \right )^{\frac {7}{2}} \left (63 B b \,e^{2} x^{2}+77 A b \,e^{2} x +77 B a \,e^{2} x -28 B b d e x +99 A a \,e^{2}-22 A b d e -22 B a d e +8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{693 e^{3} \left (b x +a \right )}\) | \(89\) |
risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (63 B b \,e^{5} x^{5}+77 A b \,e^{5} x^{4}+77 B a \,e^{5} x^{4}+161 B b d \,e^{4} x^{4}+99 A a \,e^{5} x^{3}+209 A b d \,e^{4} x^{3}+209 B a d \,e^{4} x^{3}+113 B b \,d^{2} e^{3} x^{3}+297 A a d \,e^{4} x^{2}+165 A b \,d^{2} e^{3} x^{2}+165 B a \,d^{2} e^{3} x^{2}+3 B b \,d^{3} e^{2} x^{2}+297 A a \,d^{2} e^{3} x +11 A b \,d^{3} e^{2} x +11 B a \,d^{3} e^{2} x -4 B b \,d^{4} e x +99 A a \,d^{3} e^{2}-22 A b \,d^{4} e -22 B a \,d^{4} e +8 B b \,d^{5}\right ) \sqrt {e x +d}}{693 \left (b x +a \right ) e^{3}}\) | \(241\) |
2/693*csgn(b*x+a)*(e*x+d)^(7/2)*(63*B*b*e^2*x^2+77*A*b*e^2*x+77*B*a*e^2*x- 28*B*b*d*e*x+99*A*a*e^2-22*A*b*d*e-22*B*a*d*e+8*B*b*d^2)/e^3
Time = 0.45 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.15 \[ \int (A+B x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (63 \, B b e^{5} x^{5} + 8 \, B b d^{5} + 99 \, A a d^{3} e^{2} - 22 \, {\left (B a + A b\right )} d^{4} e + 7 \, {\left (23 \, B b d e^{4} + 11 \, {\left (B a + A b\right )} e^{5}\right )} x^{4} + {\left (113 \, B b d^{2} e^{3} + 99 \, A a e^{5} + 209 \, {\left (B a + A b\right )} d e^{4}\right )} x^{3} + 3 \, {\left (B b d^{3} e^{2} + 99 \, A a d e^{4} + 55 \, {\left (B a + A b\right )} d^{2} e^{3}\right )} x^{2} - {\left (4 \, B b d^{4} e - 297 \, A a d^{2} e^{3} - 11 \, {\left (B a + A b\right )} d^{3} e^{2}\right )} x\right )} \sqrt {e x + d}}{693 \, e^{3}} \]
2/693*(63*B*b*e^5*x^5 + 8*B*b*d^5 + 99*A*a*d^3*e^2 - 22*(B*a + A*b)*d^4*e + 7*(23*B*b*d*e^4 + 11*(B*a + A*b)*e^5)*x^4 + (113*B*b*d^2*e^3 + 99*A*a*e^ 5 + 209*(B*a + A*b)*d*e^4)*x^3 + 3*(B*b*d^3*e^2 + 99*A*a*d*e^4 + 55*(B*a + A*b)*d^2*e^3)*x^2 - (4*B*b*d^4*e - 297*A*a*d^2*e^3 - 11*(B*a + A*b)*d^3*e ^2)*x)*sqrt(e*x + d)/e^3
Timed out. \[ \int (A+B x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.30 \[ \int (A+B x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (7 \, b e^{4} x^{4} - 2 \, b d^{4} + 9 \, a d^{3} e + {\left (19 \, b d e^{3} + 9 \, a e^{4}\right )} x^{3} + 3 \, {\left (5 \, b d^{2} e^{2} + 9 \, a d e^{3}\right )} x^{2} + {\left (b d^{3} e + 27 \, a d^{2} e^{2}\right )} x\right )} \sqrt {e x + d} A}{63 \, e^{2}} + \frac {2 \, {\left (63 \, b e^{5} x^{5} + 8 \, b d^{5} - 22 \, a d^{4} e + 7 \, {\left (23 \, b d e^{4} + 11 \, a e^{5}\right )} x^{4} + {\left (113 \, b d^{2} e^{3} + 209 \, a d e^{4}\right )} x^{3} + 3 \, {\left (b d^{3} e^{2} + 55 \, a d^{2} e^{3}\right )} x^{2} - {\left (4 \, b d^{4} e - 11 \, a d^{3} e^{2}\right )} x\right )} \sqrt {e x + d} B}{693 \, e^{3}} \]
2/63*(7*b*e^4*x^4 - 2*b*d^4 + 9*a*d^3*e + (19*b*d*e^3 + 9*a*e^4)*x^3 + 3*( 5*b*d^2*e^2 + 9*a*d*e^3)*x^2 + (b*d^3*e + 27*a*d^2*e^2)*x)*sqrt(e*x + d)*A /e^2 + 2/693*(63*b*e^5*x^5 + 8*b*d^5 - 22*a*d^4*e + 7*(23*b*d*e^4 + 11*a*e ^5)*x^4 + (113*b*d^2*e^3 + 209*a*d*e^4)*x^3 + 3*(b*d^3*e^2 + 55*a*d^2*e^3) *x^2 - (4*b*d^4*e - 11*a*d^3*e^2)*x)*sqrt(e*x + d)*B/e^3
Leaf count of result is larger than twice the leaf count of optimal. 831 vs. \(2 (119) = 238\).
Time = 0.29 (sec) , antiderivative size = 831, normalized size of antiderivative = 5.07 \[ \int (A+B x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\text {Too large to display} \]
2/3465*(3465*sqrt(e*x + d)*A*a*d^3*sgn(b*x + a) + 3465*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*A*a*d^2*sgn(b*x + a) + 1155*((e*x + d)^(3/2) - 3*sqrt(e *x + d)*d)*B*a*d^3*sgn(b*x + a)/e + 1155*((e*x + d)^(3/2) - 3*sqrt(e*x + d )*d)*A*b*d^3*sgn(b*x + a)/e + 693*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)* d + 15*sqrt(e*x + d)*d^2)*A*a*d*sgn(b*x + a) + 231*(3*(e*x + d)^(5/2) - 10 *(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*b*d^3*sgn(b*x + a)/e^2 + 693* (3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*B*a*d^2* sgn(b*x + a)/e + 693*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e *x + d)*d^2)*A*b*d^2*sgn(b*x + a)/e + 99*(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*a*sgn(b*x + a) + 297*(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^2*sgn(b*x + a)/e^2 + 297*(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*d*sgn(b*x + a)/e + 297*(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*d*sgn(b*x + a)/e + 33*(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*b*d*sgn(b*x + a)/e^2 + 11*(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*a*sgn(b*x + a)/e + 11*(35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*...
Timed out. \[ \int (A+B x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\int \sqrt {{\left (a+b\,x\right )}^2}\,\left (A+B\,x\right )\,{\left (d+e\,x\right )}^{5/2} \,d x \]