Integrand size = 24, antiderivative size = 83 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx=\frac {2 \sqrt {a^2-b^2 x^2}}{b (a+b x)}-\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{3 b (a+b x)^3}+\frac {\arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{b} \]
-2/3*(-b^2*x^2+a^2)^(3/2)/b/(b*x+a)^3+arctan(b*x/(-b^2*x^2+a^2)^(1/2))/b+2 *(-b^2*x^2+a^2)^(1/2)/b/(b*x+a)
Time = 0.45 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx=\frac {4 (a+2 b x) \sqrt {a^2-b^2 x^2}}{3 b (a+b x)^2}-\frac {2 \arctan \left (\frac {b x}{\sqrt {a^2}-\sqrt {a^2-b^2 x^2}}\right )}{b} \]
(4*(a + 2*b*x)*Sqrt[a^2 - b^2*x^2])/(3*b*(a + b*x)^2) - (2*ArcTan[(b*x)/(S qrt[a^2] - Sqrt[a^2 - b^2*x^2])])/b
Time = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {465, 463, 25, 224, 216}
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.
\(\displaystyle \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx\) |
\(\Big \downarrow \) 465 |
\(\displaystyle -\int \frac {\sqrt {a^2-b^2 x^2}}{(a+b x)^2}dx-\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{3 b (a+b x)^3}\) |
\(\Big \downarrow \) 463 |
\(\displaystyle -\int -\frac {1}{\sqrt {a^2-b^2 x^2}}dx-\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{3 b (a+b x)^3}+\frac {2 \sqrt {a^2-b^2 x^2}}{b (a+b x)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {1}{\sqrt {a^2-b^2 x^2}}dx-\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{3 b (a+b x)^3}+\frac {2 \sqrt {a^2-b^2 x^2}}{b (a+b x)}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \int \frac {1}{\frac {b^2 x^2}{a^2-b^2 x^2}+1}d\frac {x}{\sqrt {a^2-b^2 x^2}}-\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{3 b (a+b x)^3}+\frac {2 \sqrt {a^2-b^2 x^2}}{b (a+b x)}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\arctan \left (\frac {b x}{\sqrt {a^2-b^2 x^2}}\right )}{b}-\frac {2 \left (a^2-b^2 x^2\right )^{3/2}}{3 b (a+b x)^3}+\frac {2 \sqrt {a^2-b^2 x^2}}{b (a+b x)}\) |
(2*Sqrt[a^2 - b^2*x^2])/(b*(a + b*x)) - (2*(a^2 - b^2*x^2)^(3/2))/(3*b*(a + b*x)^3) + ArcTan[(b*x)/Sqrt[a^2 - b^2*x^2]]/b
3.8.94.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-(-c)^(-n - 2))*d^(2*n + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)*b^(n + 2)*(c + d*x ))), x] - Simp[d^(2*n + 2)/b^(n + 1) Int[(1/Sqrt[a + b*x^2])*ExpandToSum[ (2^(-n - 1)*(-c)^(-n - 1) - (-c + d*x)^(-n - 1))/(c + d*x), x], x], x] /; F reeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + p + 1))), x] - Simp[b*(p/(d^2*(n + p + 1))) Int[(c + d*x)^(n + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LtQ[n, -2] || EqQ[n + 2*p + 1, 0]) && NeQ[n + p + 1, 0] && IntegerQ[2*p]
Leaf count of result is larger than twice the leaf count of optimal. \(290\) vs. \(2(75)=150\).
Time = 2.34 (sec) , antiderivative size = 291, normalized size of antiderivative = 3.51
method | result | size |
default | \(\frac {-\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {5}{2}}}{3 a b \left (x +\frac {a}{b}\right )^{4}}-\frac {b \left (-\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {5}{2}}}{a b \left (x +\frac {a}{b}\right )^{3}}-\frac {2 b \left (\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {5}{2}}}{a b \left (x +\frac {a}{b}\right )^{2}}+\frac {3 b \left (\frac {\left (-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )\right )^{\frac {3}{2}}}{3}+a b \left (-\frac {\left (-2 b^{2} \left (x +\frac {a}{b}\right )+2 a b \right ) \sqrt {-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )}}{4 b^{2}}+\frac {a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-b^{2} \left (x +\frac {a}{b}\right )^{2}+2 a b \left (x +\frac {a}{b}\right )}}\right )}{2 \sqrt {b^{2}}}\right )\right )}{a}\right )}{a}\right )}{3 a}}{b^{4}}\) | \(291\) |
1/b^4*(-1/3/a/b/(x+a/b)^4*(-b^2*(x+a/b)^2+2*a*b*(x+a/b))^(5/2)-1/3*b/a*(-1 /a/b/(x+a/b)^3*(-b^2*(x+a/b)^2+2*a*b*(x+a/b))^(5/2)-2*b/a*(1/a/b/(x+a/b)^2 *(-b^2*(x+a/b)^2+2*a*b*(x+a/b))^(5/2)+3*b/a*(1/3*(-b^2*(x+a/b)^2+2*a*b*(x+ a/b))^(3/2)+a*b*(-1/4*(-2*b^2*(x+a/b)+2*a*b)/b^2*(-b^2*(x+a/b)^2+2*a*b*(x+ a/b))^(1/2)+1/2*a^2/(b^2)^(1/2)*arctan((b^2)^(1/2)*x/(-b^2*(x+a/b)^2+2*a*b *(x+a/b))^(1/2)))))))
Time = 0.30 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx=\frac {2 \, {\left (2 \, b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} - 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \arctan \left (-\frac {a - \sqrt {-b^{2} x^{2} + a^{2}}}{b x}\right ) + 2 \, \sqrt {-b^{2} x^{2} + a^{2}} {\left (2 \, b x + a\right )}\right )}}{3 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} \]
2/3*(2*b^2*x^2 + 4*a*b*x + 2*a^2 - 3*(b^2*x^2 + 2*a*b*x + a^2)*arctan(-(a - sqrt(-b^2*x^2 + a^2))/(b*x)) + 2*sqrt(-b^2*x^2 + a^2)*(2*b*x + a))/(b^3* x^2 + 2*a*b^2*x + a^2*b)
\[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx=\int \frac {\left (- \left (- a + b x\right ) \left (a + b x\right )\right )^{\frac {3}{2}}}{\left (a + b x\right )^{4}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.53 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx=-\frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{\frac {3}{2}}}{3 \, {\left (b^{4} x^{3} + 3 \, a b^{3} x^{2} + 3 \, a^{2} b^{2} x + a^{3} b\right )}} - \frac {2 \, \sqrt {-b^{2} x^{2} + a^{2}} a}{3 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} + \frac {\arcsin \left (\frac {b x}{a}\right )}{b} + \frac {7 \, \sqrt {-b^{2} x^{2} + a^{2}}}{3 \, {\left (b^{2} x + a b\right )}} \]
-1/3*(-b^2*x^2 + a^2)^(3/2)/(b^4*x^3 + 3*a*b^3*x^2 + 3*a^2*b^2*x + a^3*b) - 2/3*sqrt(-b^2*x^2 + a^2)*a/(b^3*x^2 + 2*a*b^2*x + a^2*b) + arcsin(b*x/a) /b + 7/3*sqrt(-b^2*x^2 + a^2)/(b^2*x + a*b)
Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx=\frac {\arcsin \left (\frac {b x}{a}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (b\right )}{{\left | b \right |}} - \frac {8 \, {\left (\frac {3 \, {\left (a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}\right )}}{b^{2} x} + 1\right )}}{3 \, {\left (\frac {a b + \sqrt {-b^{2} x^{2} + a^{2}} {\left | b \right |}}{b^{2} x} + 1\right )}^{3} {\left | b \right |}} \]
arcsin(b*x/a)*sgn(a)*sgn(b)/abs(b) - 8/3*(3*(a*b + sqrt(-b^2*x^2 + a^2)*ab s(b))/(b^2*x) + 1)/(((a*b + sqrt(-b^2*x^2 + a^2)*abs(b))/(b^2*x) + 1)^3*ab s(b))
Timed out. \[ \int \frac {\left (a^2-b^2 x^2\right )^{3/2}}{(a+b x)^4} \, dx=\int \frac {{\left (a^2-b^2\,x^2\right )}^{3/2}}{{\left (a+b\,x\right )}^4} \,d x \]