Integrand size = 24, antiderivative size = 304 \[ \int \sqrt {a+b x} (A+B x) (d+e x)^{5/2} \, dx=-\frac {(b d-a e)^3 (3 b B d-10 A b e+7 a B e) \sqrt {a+b x} \sqrt {d+e x}}{128 b^4 e^2}-\frac {(b d-a e)^2 (3 b B d-10 A b e+7 a B e) (a+b x)^{3/2} \sqrt {d+e x}}{64 b^4 e}-\frac {(b d-a e) (3 b B d-10 A b e+7 a B e) (a+b x)^{3/2} (d+e x)^{3/2}}{48 b^3 e}-\frac {(3 b B d-10 A b e+7 a B e) (a+b x)^{3/2} (d+e x)^{5/2}}{40 b^2 e}+\frac {B (a+b x)^{3/2} (d+e x)^{7/2}}{5 b e}+\frac {(b d-a e)^4 (3 b B d-10 A b e+7 a B e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{128 b^{9/2} e^{5/2}} \]
-1/48*(-a*e+b*d)*(-10*A*b*e+7*B*a*e+3*B*b*d)*(b*x+a)^(3/2)*(e*x+d)^(3/2)/b ^3/e-1/40*(-10*A*b*e+7*B*a*e+3*B*b*d)*(b*x+a)^(3/2)*(e*x+d)^(5/2)/b^2/e+1/ 5*B*(b*x+a)^(3/2)*(e*x+d)^(7/2)/b/e+1/128*(-a*e+b*d)^4*(-10*A*b*e+7*B*a*e+ 3*B*b*d)*arctanh(e^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(e*x+d)^(1/2))/b^(9/2)/e^(5 /2)-1/64*(-a*e+b*d)^2*(-10*A*b*e+7*B*a*e+3*B*b*d)*(b*x+a)^(3/2)*(e*x+d)^(1 /2)/b^4/e-1/128*(-a*e+b*d)^3*(-10*A*b*e+7*B*a*e+3*B*b*d)*(b*x+a)^(1/2)*(e* x+d)^(1/2)/b^4/e^2
Time = 0.90 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.06 \[ \int \sqrt {a+b x} (A+B x) (d+e x)^{5/2} \, dx=\frac {\sqrt {a+b x} \sqrt {d+e x} \left (-105 a^4 B e^4+10 a^3 b e^3 (34 B d+15 A e+7 B e x)-2 a^2 b^2 e^2 \left (25 A e (11 d+2 e x)+B \left (173 d^2+111 d e x+28 e^2 x^2\right )\right )+2 a b^3 e \left (5 A e \left (73 d^2+36 d e x+8 e^2 x^2\right )+B \left (30 d^3+109 d^2 e x+88 d e^2 x^2+24 e^3 x^3\right )\right )+b^4 \left (10 A e \left (15 d^3+118 d^2 e x+136 d e^2 x^2+48 e^3 x^3\right )+B \left (-45 d^4+30 d^3 e x+744 d^2 e^2 x^2+1008 d e^3 x^3+384 e^4 x^4\right )\right )\right )}{1920 b^4 e^2}+\frac {(b d-a e)^4 (3 b B d-10 A b e+7 a B e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{128 b^{9/2} e^{5/2}} \]
(Sqrt[a + b*x]*Sqrt[d + e*x]*(-105*a^4*B*e^4 + 10*a^3*b*e^3*(34*B*d + 15*A *e + 7*B*e*x) - 2*a^2*b^2*e^2*(25*A*e*(11*d + 2*e*x) + B*(173*d^2 + 111*d* e*x + 28*e^2*x^2)) + 2*a*b^3*e*(5*A*e*(73*d^2 + 36*d*e*x + 8*e^2*x^2) + B* (30*d^3 + 109*d^2*e*x + 88*d*e^2*x^2 + 24*e^3*x^3)) + b^4*(10*A*e*(15*d^3 + 118*d^2*e*x + 136*d*e^2*x^2 + 48*e^3*x^3) + B*(-45*d^4 + 30*d^3*e*x + 74 4*d^2*e^2*x^2 + 1008*d*e^3*x^3 + 384*e^4*x^4))))/(1920*b^4*e^2) + ((b*d - a*e)^4*(3*b*B*d - 10*A*b*e + 7*a*B*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqr t[b]*Sqrt[d + e*x])])/(128*b^(9/2)*e^(5/2))
Time = 0.30 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.83, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {90, 60, 60, 60, 60, 66, 221}
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.
| \(\displaystyle \int \sqrt {a+b x} (A+B x) (d+e x)^{5/2} \, dx\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {B (a+b x)^{3/2} (d+e x)^{7/2}}{5 b e}-\frac {(7 a B e-10 A b e+3 b B d) \int \sqrt {a+b x} (d+e x)^{5/2}dx}{10 b e}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {B (a+b x)^{3/2} (d+e x)^{7/2}}{5 b e}-\frac {(7 a B e-10 A b e+3 b B d) \left (\frac {5 (b d-a e) \int \sqrt {a+b x} (d+e x)^{3/2}dx}{8 b}+\frac {(a+b x)^{3/2} (d+e x)^{5/2}}{4 b}\right )}{10 b e}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {B (a+b x)^{3/2} (d+e x)^{7/2}}{5 b e}-\frac {(7 a B e-10 A b e+3 b B d) \left (\frac {5 (b d-a e) \left (\frac {(b d-a e) \int \sqrt {a+b x} \sqrt {d+e x}dx}{2 b}+\frac {(a+b x)^{3/2} (d+e x)^{3/2}}{3 b}\right )}{8 b}+\frac {(a+b x)^{3/2} (d+e x)^{5/2}}{4 b}\right )}{10 b e}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {B (a+b x)^{3/2} (d+e x)^{7/2}}{5 b e}-\frac {(7 a B e-10 A b e+3 b B d) \left (\frac {5 (b d-a e) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {\sqrt {a+b x}}{\sqrt {d+e x}}dx}{4 b}+\frac {(a+b x)^{3/2} \sqrt {d+e x}}{2 b}\right )}{2 b}+\frac {(a+b x)^{3/2} (d+e x)^{3/2}}{3 b}\right )}{8 b}+\frac {(a+b x)^{3/2} (d+e x)^{5/2}}{4 b}\right )}{10 b e}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {B (a+b x)^{3/2} (d+e x)^{7/2}}{5 b e}-\frac {(7 a B e-10 A b e+3 b B d) \left (\frac {5 (b d-a e) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {\sqrt {a+b x} \sqrt {d+e x}}{e}-\frac {(b d-a e) \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}}dx}{2 e}\right )}{4 b}+\frac {(a+b x)^{3/2} \sqrt {d+e x}}{2 b}\right )}{2 b}+\frac {(a+b x)^{3/2} (d+e x)^{3/2}}{3 b}\right )}{8 b}+\frac {(a+b x)^{3/2} (d+e x)^{5/2}}{4 b}\right )}{10 b e}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {B (a+b x)^{3/2} (d+e x)^{7/2}}{5 b e}-\frac {(7 a B e-10 A b e+3 b B d) \left (\frac {5 (b d-a e) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {\sqrt {a+b x} \sqrt {d+e x}}{e}-\frac {(b d-a e) \int \frac {1}{b-\frac {e (a+b x)}{d+e x}}d\frac {\sqrt {a+b x}}{\sqrt {d+e x}}}{e}\right )}{4 b}+\frac {(a+b x)^{3/2} \sqrt {d+e x}}{2 b}\right )}{2 b}+\frac {(a+b x)^{3/2} (d+e x)^{3/2}}{3 b}\right )}{8 b}+\frac {(a+b x)^{3/2} (d+e x)^{5/2}}{4 b}\right )}{10 b e}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {B (a+b x)^{3/2} (d+e x)^{7/2}}{5 b e}-\frac {(7 a B e-10 A b e+3 b B d) \left (\frac {5 (b d-a e) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {\sqrt {a+b x} \sqrt {d+e x}}{e}-\frac {(b d-a e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{\sqrt {b} e^{3/2}}\right )}{4 b}+\frac {(a+b x)^{3/2} \sqrt {d+e x}}{2 b}\right )}{2 b}+\frac {(a+b x)^{3/2} (d+e x)^{3/2}}{3 b}\right )}{8 b}+\frac {(a+b x)^{3/2} (d+e x)^{5/2}}{4 b}\right )}{10 b e}\) |
(B*(a + b*x)^(3/2)*(d + e*x)^(7/2))/(5*b*e) - ((3*b*B*d - 10*A*b*e + 7*a*B *e)*(((a + b*x)^(3/2)*(d + e*x)^(5/2))/(4*b) + (5*(b*d - a*e)*(((a + b*x)^ (3/2)*(d + e*x)^(3/2))/(3*b) + ((b*d - a*e)*(((a + b*x)^(3/2)*Sqrt[d + e*x ])/(2*b) + ((b*d - a*e)*((Sqrt[a + b*x]*Sqrt[d + e*x])/e - ((b*d - a*e)*Ar cTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(Sqrt[b]*e^(3/2))) )/(4*b)))/(2*b)))/(8*b)))/(10*b*e)
3.22.100.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Leaf count of result is larger than twice the leaf count of optimal. \(1371\) vs. \(2(260)=520\).
Time = 1.08 (sec) , antiderivative size = 1372, normalized size of antiderivative = 4.51
-1/3840*(e*x+d)^(1/2)*(b*x+a)^(1/2)*(-352*B*a*b^3*d*e^3*x^2*((b*x+a)*(e*x+ d))^(1/2)*(b*e)^(1/2)-720*A*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*a*b^3*d*e^ 3*x+444*B*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*a^2*b^2*d*e^3*x-436*B*((b*x+ a)*(e*x+d))^(1/2)*(b*e)^(1/2)*a*b^3*d^2*e^2*x+900*A*ln(1/2*(2*b*e*x+2*((b* x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*b^3*d^2*e^3+375* B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/ 2))*a^4*b*d*e^4-450*B*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2 )+a*e+b*d)/(b*e)^(1/2))*a^3*b^2*d^2*e^3+150*B*ln(1/2*(2*b*e*x+2*((b*x+a)*( e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*b^3*d^3*e^2-600*A*ln(1 /2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^ 3*b^2*d*e^4-300*A*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*b^4*d^3*e+75*B*ln(1/ 2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b ^4*d^4*e-600*A*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b *d)/(b*e)^(1/2))*a*b^4*d^3*e^2-300*A*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*a ^3*b*e^4-960*A*b^4*e^4*x^3*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+150*A*ln(1/ 2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^4 *b*e^5+150*A*ln(1/2*(2*b*e*x+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d )/(b*e)^(1/2))*b^5*d^4*e+210*B*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*a^4*e^4 +90*B*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)*b^4*d^4-45*B*ln(1/2*(2*b*e*x+2*( (b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^5*d^5-96*B*a...
Time = 0.28 (sec) , antiderivative size = 1044, normalized size of antiderivative = 3.43 \[ \int \sqrt {a+b x} (A+B x) (d+e x)^{5/2} \, dx=\text {Too large to display} \]
[-1/7680*(15*(3*B*b^5*d^5 - 5*(B*a*b^4 + 2*A*b^5)*d^4*e - 10*(B*a^2*b^3 - 4*A*a*b^4)*d^3*e^2 + 30*(B*a^3*b^2 - 2*A*a^2*b^3)*d^2*e^3 - 5*(5*B*a^4*b - 8*A*a^3*b^2)*d*e^4 + (7*B*a^5 - 10*A*a^4*b)*e^5)*sqrt(b*e)*log(8*b^2*e^2* x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 - 4*(2*b*e*x + b*d + a*e)*sqrt(b*e)*sq rt(b*x + a)*sqrt(e*x + d) + 8*(b^2*d*e + a*b*e^2)*x) - 4*(384*B*b^5*e^5*x^ 4 - 45*B*b^5*d^4*e + 30*(2*B*a*b^4 + 5*A*b^5)*d^3*e^2 - 2*(173*B*a^2*b^3 - 365*A*a*b^4)*d^2*e^3 + 10*(34*B*a^3*b^2 - 55*A*a^2*b^3)*d*e^4 - 15*(7*B*a ^4*b - 10*A*a^3*b^2)*e^5 + 48*(21*B*b^5*d*e^4 + (B*a*b^4 + 10*A*b^5)*e^5)* x^3 + 8*(93*B*b^5*d^2*e^3 + 2*(11*B*a*b^4 + 85*A*b^5)*d*e^4 - (7*B*a^2*b^3 - 10*A*a*b^4)*e^5)*x^2 + 2*(15*B*b^5*d^3*e^2 + (109*B*a*b^4 + 590*A*b^5)* d^2*e^3 - 3*(37*B*a^2*b^3 - 60*A*a*b^4)*d*e^4 + 5*(7*B*a^3*b^2 - 10*A*a^2* b^3)*e^5)*x)*sqrt(b*x + a)*sqrt(e*x + d))/(b^5*e^3), -1/3840*(15*(3*B*b^5* d^5 - 5*(B*a*b^4 + 2*A*b^5)*d^4*e - 10*(B*a^2*b^3 - 4*A*a*b^4)*d^3*e^2 + 3 0*(B*a^3*b^2 - 2*A*a^2*b^3)*d^2*e^3 - 5*(5*B*a^4*b - 8*A*a^3*b^2)*d*e^4 + (7*B*a^5 - 10*A*a^4*b)*e^5)*sqrt(-b*e)*arctan(1/2*(2*b*e*x + b*d + a*e)*sq rt(-b*e)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*e^2*x^2 + a*b*d*e + (b^2*d*e + a *b*e^2)*x)) - 2*(384*B*b^5*e^5*x^4 - 45*B*b^5*d^4*e + 30*(2*B*a*b^4 + 5*A* b^5)*d^3*e^2 - 2*(173*B*a^2*b^3 - 365*A*a*b^4)*d^2*e^3 + 10*(34*B*a^3*b^2 - 55*A*a^2*b^3)*d*e^4 - 15*(7*B*a^4*b - 10*A*a^3*b^2)*e^5 + 48*(21*B*b^5*d *e^4 + (B*a*b^4 + 10*A*b^5)*e^5)*x^3 + 8*(93*B*b^5*d^2*e^3 + 2*(11*B*a*...
\[ \int \sqrt {a+b x} (A+B x) (d+e x)^{5/2} \, dx=\int \left (A + B x\right ) \sqrt {a + b x} \left (d + e x\right )^{\frac {5}{2}}\, dx \]
Exception generated. \[ \int \sqrt {a+b x} (A+B x) (d+e x)^{5/2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Leaf count of result is larger than twice the leaf count of optimal. 2612 vs. \(2 (260) = 520\).
Time = 0.63 (sec) , antiderivative size = 2612, normalized size of antiderivative = 8.59 \[ \int \sqrt {a+b x} (A+B x) (d+e x)^{5/2} \, dx=\text {Too large to display} \]
-1/1920*(1920*((b^2*d - a*b*e)*log(abs(-sqrt(b*e)*sqrt(b*x + a) + sqrt(b^2 *d + (b*x + a)*b*e - a*b*e)))/sqrt(b*e) - sqrt(b^2*d + (b*x + a)*b*e - a*b *e)*sqrt(b*x + a))*A*a*d^2*abs(b)/b^2 - 80*(sqrt(b^2*d + (b*x + a)*b*e - a *b*e)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/b^2 + (b^6*d*e^3 - 13*a*b^5* e^4)/(b^7*e^4)) - 3*(b^7*d^2*e^2 + 2*a*b^6*d*e^3 - 11*a^2*b^5*e^4)/(b^7*e^ 4)) - 3*(b^3*d^3 + a*b^2*d^2*e + 3*a^2*b*d*e^2 - 5*a^3*e^3)*log(abs(-sqrt( b*e)*sqrt(b*x + a) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/(sqrt(b*e)*b*e^ 2))*B*d^2*abs(b)/b - 160*(sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a )*(2*(b*x + a)*(4*(b*x + a)/b^2 + (b^6*d*e^3 - 13*a*b^5*e^4)/(b^7*e^4)) - 3*(b^7*d^2*e^2 + 2*a*b^6*d*e^3 - 11*a^2*b^5*e^4)/(b^7*e^4)) - 3*(b^3*d^3 + a*b^2*d^2*e + 3*a^2*b*d*e^2 - 5*a^3*e^3)*log(abs(-sqrt(b*e)*sqrt(b*x + a) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/(sqrt(b*e)*b*e^2))*B*a*d*e*abs(b) /b^2 - 160*(sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a) *(4*(b*x + a)/b^2 + (b^6*d*e^3 - 13*a*b^5*e^4)/(b^7*e^4)) - 3*(b^7*d^2*e^2 + 2*a*b^6*d*e^3 - 11*a^2*b^5*e^4)/(b^7*e^4)) - 3*(b^3*d^3 + a*b^2*d^2*e + 3*a^2*b*d*e^2 - 5*a^3*e^3)*log(abs(-sqrt(b*e)*sqrt(b*x + a) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/(sqrt(b*e)*b*e^2))*A*d*e*abs(b)/b - 20*(sqrt(b^ 2*d + (b*x + a)*b*e - a*b*e)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/b^3 + (b^12*d*e^5 - 25*a*b^11*e^6)/(b^14*e^6)) - (5*b^13*d^2*e^4 + 14*a*b^12*d*e ^5 - 163*a^2*b^11*e^6)/(b^14*e^6)) + 3*(5*b^14*d^3*e^3 + 9*a*b^13*d^2*e...
Timed out. \[ \int \sqrt {a+b x} (A+B x) (d+e x)^{5/2} \, dx=\int \left (A+B\,x\right )\,\sqrt {a+b\,x}\,{\left (d+e\,x\right )}^{5/2} \,d x \]