Integrand size = 23, antiderivative size = 88 \[ \int x^2 \sqrt {a-b x} \sqrt {a+b x} \, dx=-\frac {a^2 x \sqrt {a-b x} \sqrt {a+b x}}{8 b^2}+\frac {1}{4} x^3 \sqrt {a-b x} \sqrt {a+b x}+\frac {a^4 \arctan \left (\frac {\sqrt {a+b x}}{\sqrt {a-b x}}\right )}{4 b^3} \] Output:
-1/8*a^2*x*(-b*x+a)^(1/2)*(b*x+a)^(1/2)/b^2+1/4*x^3*(-b*x+a)^(1/2)*(b*x+a) ^(1/2)+1/4*a^4*arctan((b*x+a)^(1/2)/(-b*x+a)^(1/2))/b^3
Time = 0.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.97 \[ \int x^2 \sqrt {a-b x} \sqrt {a+b x} \, dx=\frac {\frac {b x \sqrt {a+b x} \left (-a^3+a^2 b x+2 a b^2 x^2-2 b^3 x^3\right )}{\sqrt {a-b x}}+2 a^4 \arctan \left (\frac {\sqrt {a+b x}}{\sqrt {a-b x}}\right )}{8 b^3} \] Input:
Integrate[x^2*Sqrt[a - b*x]*Sqrt[a + b*x],x]
Output:
((b*x*Sqrt[a + b*x]*(-a^3 + a^2*b*x + 2*a*b^2*x^2 - 2*b^3*x^3))/Sqrt[a - b *x] + 2*a^4*ArcTan[Sqrt[a + b*x]/Sqrt[a - b*x]])/(8*b^3)
Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {101, 25, 27, 40, 45, 218}
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.
\(\displaystyle \int x^2 \sqrt {a-b x} \sqrt {a+b x} \, dx\) |
\(\Big \downarrow \) 101 |
\(\displaystyle -\frac {\int -a^2 \sqrt {a-b x} \sqrt {a+b x}dx}{4 b^2}-\frac {x (a-b x)^{3/2} (a+b x)^{3/2}}{4 b^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int a^2 \sqrt {a-b x} \sqrt {a+b x}dx}{4 b^2}-\frac {x (a-b x)^{3/2} (a+b x)^{3/2}}{4 b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 \int \sqrt {a-b x} \sqrt {a+b x}dx}{4 b^2}-\frac {x (a-b x)^{3/2} (a+b x)^{3/2}}{4 b^2}\) |
\(\Big \downarrow \) 40 |
\(\displaystyle \frac {a^2 \left (\frac {1}{2} a^2 \int \frac {1}{\sqrt {a-b x} \sqrt {a+b x}}dx+\frac {1}{2} x \sqrt {a-b x} \sqrt {a+b x}\right )}{4 b^2}-\frac {x (a-b x)^{3/2} (a+b x)^{3/2}}{4 b^2}\) |
\(\Big \downarrow \) 45 |
\(\displaystyle \frac {a^2 \left (a^2 \int \frac {1}{-\frac {(a-b x) b}{a+b x}-b}d\frac {\sqrt {a-b x}}{\sqrt {a+b x}}+\frac {1}{2} x \sqrt {a-b x} \sqrt {a+b x}\right )}{4 b^2}-\frac {x (a-b x)^{3/2} (a+b x)^{3/2}}{4 b^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {a^2 \left (\frac {1}{2} x \sqrt {a-b x} \sqrt {a+b x}-\frac {a^2 \arctan \left (\frac {\sqrt {a-b x}}{\sqrt {a+b x}}\right )}{b}\right )}{4 b^2}-\frac {x (a-b x)^{3/2} (a+b x)^{3/2}}{4 b^2}\) |
Input:
Int[x^2*Sqrt[a - b*x]*Sqrt[a + b*x],x]
Output:
-1/4*(x*(a - b*x)^(3/2)*(a + b*x)^(3/2))/b^2 + (a^2*((x*Sqrt[a - b*x]*Sqrt [a + b*x])/2 - (a^2*ArcTan[Sqrt[a - b*x]/Sqrt[a + b*x]])/b))/(4*b^2)
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_))^(m_), x_Symbol] :> Simp[x* (a + b*x)^m*((c + d*x)^m/(2*m + 1)), x] + Simp[2*a*c*(m/(2*m + 1)) Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ b*c + a*d, 0] && IGtQ[m + 1/2, 0]
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] && EqQ[b*c + a*d, 0] && !GtQ[c, 0]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Simp[1/(d*f*(n + p + 3)) Int[(c + d*x)^n*(e + f*x)^p*Simp [a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f *(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Time = 0.23 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.12
method | result | size |
risch | \(-\frac {x \left (-2 b^{2} x^{2}+a^{2}\right ) \sqrt {-b x +a}\, \sqrt {b x +a}}{8 b^{2}}+\frac {a^{4} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right ) \sqrt {\left (-b x +a \right ) \left (b x +a \right )}}{8 b^{2} \sqrt {b^{2}}\, \sqrt {-b x +a}\, \sqrt {b x +a}}\) | \(99\) |
default | \(-\frac {\sqrt {-b x +a}\, \sqrt {b x +a}\, \left (-2 \,\operatorname {csgn}\left (b \right ) b^{3} x^{3} \sqrt {-b^{2} x^{2}+a^{2}}+\sqrt {-b^{2} x^{2}+a^{2}}\, \operatorname {csgn}\left (b \right ) b \,a^{2} x -\arctan \left (\frac {\operatorname {csgn}\left (b \right ) b x}{\sqrt {-b^{2} x^{2}+a^{2}}}\right ) a^{4}\right ) \operatorname {csgn}\left (b \right )}{8 \sqrt {-b^{2} x^{2}+a^{2}}\, b^{3}}\) | \(109\) |
Input:
int(x^2*(-b*x+a)^(1/2)*(b*x+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/8*x*(-2*b^2*x^2+a^2)/b^2*(-b*x+a)^(1/2)*(b*x+a)^(1/2)+1/8*a^4/b^2/(b^2) ^(1/2)*arctan((b^2)^(1/2)*x/(-b^2*x^2+a^2)^(1/2))*((-b*x+a)*(b*x+a))^(1/2) /(-b*x+a)^(1/2)/(b*x+a)^(1/2)
Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.82 \[ \int x^2 \sqrt {a-b x} \sqrt {a+b x} \, dx=-\frac {2 \, a^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b x + a} - a}{b x}\right ) - {\left (2 \, b^{3} x^{3} - a^{2} b x\right )} \sqrt {b x + a} \sqrt {-b x + a}}{8 \, b^{3}} \] Input:
integrate(x^2*(-b*x+a)^(1/2)*(b*x+a)^(1/2),x, algorithm="fricas")
Output:
-1/8*(2*a^4*arctan((sqrt(b*x + a)*sqrt(-b*x + a) - a)/(b*x)) - (2*b^3*x^3 - a^2*b*x)*sqrt(b*x + a)*sqrt(-b*x + a))/b^3
\[ \int x^2 \sqrt {a-b x} \sqrt {a+b x} \, dx=\int x^{2} \sqrt {a - b x} \sqrt {a + b x}\, dx \] Input:
integrate(x**2*(-b*x+a)**(1/2)*(b*x+a)**(1/2),x)
Output:
Integral(x**2*sqrt(a - b*x)*sqrt(a + b*x), x)
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.67 \[ \int x^2 \sqrt {a-b x} \sqrt {a+b x} \, dx=\frac {a^{4} \arcsin \left (\frac {b x}{a}\right )}{8 \, b^{3}} + \frac {\sqrt {-b^{2} x^{2} + a^{2}} a^{2} x}{8 \, b^{2}} - \frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}} x}{4 \, b^{2}} \] Input:
integrate(x^2*(-b*x+a)^(1/2)*(b*x+a)^(1/2),x, algorithm="maxima")
Output:
1/8*a^4*arcsin(b*x/a)/b^3 + 1/8*sqrt(-b^2*x^2 + a^2)*a^2*x/b^2 - 1/4*(-b^2 *x^2 + a^2)^(3/2)*x/b^2
Time = 0.19 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.57 \[ \int x^2 \sqrt {a-b x} \sqrt {a+b x} \, dx=-\frac {18 \, a^{4} \arcsin \left (\frac {\sqrt {2} \sqrt {b x + a}}{2 \, \sqrt {a}}\right ) + {\left (39 \, a^{3} - {\left (2 \, {\left (3 \, b x - 10 \, a\right )} {\left (b x + a\right )} + 43 \, a^{2}\right )} {\left (b x + a\right )}\right )} \sqrt {b x + a} \sqrt {-b x + a} - 4 \, {\left (6 \, a^{3} \arcsin \left (\frac {\sqrt {2} \sqrt {b x + a}}{2 \, \sqrt {a}}\right ) + {\left ({\left (2 \, b x - 5 \, a\right )} {\left (b x + a\right )} + 9 \, a^{2}\right )} \sqrt {b x + a} \sqrt {-b x + a}\right )} a}{24 \, b^{3}} \] Input:
integrate(x^2*(-b*x+a)^(1/2)*(b*x+a)^(1/2),x, algorithm="giac")
Output:
-1/24*(18*a^4*arcsin(1/2*sqrt(2)*sqrt(b*x + a)/sqrt(a)) + (39*a^3 - (2*(3* b*x - 10*a)*(b*x + a) + 43*a^2)*(b*x + a))*sqrt(b*x + a)*sqrt(-b*x + a) - 4*(6*a^3*arcsin(1/2*sqrt(2)*sqrt(b*x + a)/sqrt(a)) + ((2*b*x - 5*a)*(b*x + a) + 9*a^2)*sqrt(b*x + a)*sqrt(-b*x + a))*a)/b^3
Time = 4.92 (sec) , antiderivative size = 369, normalized size of antiderivative = 4.19 \[ \int x^2 \sqrt {a-b x} \sqrt {a+b x} \, dx=\frac {a^4\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {a-b\,x}-\sqrt {a}}\right )}{2\,b^3}-\frac {\frac {a^4\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{2\,\left (\sqrt {a-b\,x}-\sqrt {a}\right )}-\frac {35\,a^4\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}{2\,{\left (\sqrt {a-b\,x}-\sqrt {a}\right )}^3}+\frac {273\,a^4\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5}{2\,{\left (\sqrt {a-b\,x}-\sqrt {a}\right )}^5}-\frac {715\,a^4\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7}{2\,{\left (\sqrt {a-b\,x}-\sqrt {a}\right )}^7}+\frac {715\,a^4\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^9}{2\,{\left (\sqrt {a-b\,x}-\sqrt {a}\right )}^9}-\frac {273\,a^4\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{11}}{2\,{\left (\sqrt {a-b\,x}-\sqrt {a}\right )}^{11}}+\frac {35\,a^4\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{13}}{2\,{\left (\sqrt {a-b\,x}-\sqrt {a}\right )}^{13}}-\frac {a^4\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{15}}{2\,{\left (\sqrt {a-b\,x}-\sqrt {a}\right )}^{15}}}{b^3\,{\left (\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{{\left (\sqrt {a-b\,x}-\sqrt {a}\right )}^2}+1\right )}^8} \] Input:
int(x^2*(a + b*x)^(1/2)*(a - b*x)^(1/2),x)
Output:
(a^4*atan(((a + b*x)^(1/2) - a^(1/2))/((a - b*x)^(1/2) - a^(1/2))))/(2*b^3 ) - ((a^4*((a + b*x)^(1/2) - a^(1/2)))/(2*((a - b*x)^(1/2) - a^(1/2))) - ( 35*a^4*((a + b*x)^(1/2) - a^(1/2))^3)/(2*((a - b*x)^(1/2) - a^(1/2))^3) + (273*a^4*((a + b*x)^(1/2) - a^(1/2))^5)/(2*((a - b*x)^(1/2) - a^(1/2))^5) - (715*a^4*((a + b*x)^(1/2) - a^(1/2))^7)/(2*((a - b*x)^(1/2) - a^(1/2))^7 ) + (715*a^4*((a + b*x)^(1/2) - a^(1/2))^9)/(2*((a - b*x)^(1/2) - a^(1/2)) ^9) - (273*a^4*((a + b*x)^(1/2) - a^(1/2))^11)/(2*((a - b*x)^(1/2) - a^(1/ 2))^11) + (35*a^4*((a + b*x)^(1/2) - a^(1/2))^13)/(2*((a - b*x)^(1/2) - a^ (1/2))^13) - (a^4*((a + b*x)^(1/2) - a^(1/2))^15)/(2*((a - b*x)^(1/2) - a^ (1/2))^15))/(b^3*(((a + b*x)^(1/2) - a^(1/2))^2/((a - b*x)^(1/2) - a^(1/2) )^2 + 1)^8)
Time = 0.15 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78 \[ \int x^2 \sqrt {a-b x} \sqrt {a+b x} \, dx=\frac {-2 \mathit {asin} \left (\frac {\sqrt {-b x +a}}{\sqrt {a}\, \sqrt {2}}\right ) a^{4}-\sqrt {b x +a}\, \sqrt {-b x +a}\, a^{2} b x +2 \sqrt {b x +a}\, \sqrt {-b x +a}\, b^{3} x^{3}}{8 b^{3}} \] Input:
int(x^2*(-b*x+a)^(1/2)*(b*x+a)^(1/2),x)
Output:
( - 2*asin(sqrt(a - b*x)/(sqrt(a)*sqrt(2)))*a**4 - sqrt(a + b*x)*sqrt(a - b*x)*a**2*b*x + 2*sqrt(a + b*x)*sqrt(a - b*x)*b**3*x**3)/(8*b**3)