Integrand size = 24, antiderivative size = 82 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx=-\frac {a^2 \sqrt {c+d x^2}}{c x}+\frac {b^2 x \sqrt {c+d x^2}}{2 d}-\frac {b (b c-4 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{3/2}} \]
-1/2*b*(-4*a*d+b*c)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/d^(3/2)-a^2*(d*x^2+ c)^(1/2)/c/x+1/2*b^2*x*(d*x^2+c)^(1/2)/d
Time = 0.11 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx=\frac {\left (-2 a^2 d+b^2 c x^2\right ) \sqrt {c+d x^2}}{2 c d x}+\frac {b (b c-4 a d) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{2 d^{3/2}} \]
((-2*a^2*d + b^2*c*x^2)*Sqrt[c + d*x^2])/(2*c*d*x) + (b*(b*c - 4*a*d)*Log[ -(Sqrt[d]*x) + Sqrt[c + d*x^2]])/(2*d^(3/2))
Time = 0.20 (sec) , antiderivative size = 82, 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 = {365, 27, 299, 224, 219}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx\) |
\(\Big \downarrow \) 365 |
\(\displaystyle \frac {\int \frac {b c \left (b x^2+2 a\right )}{\sqrt {d x^2+c}}dx}{c}-\frac {a^2 \sqrt {c+d x^2}}{c x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle b \int \frac {b x^2+2 a}{\sqrt {d x^2+c}}dx-\frac {a^2 \sqrt {c+d x^2}}{c x}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle b \left (\frac {b x \sqrt {c+d x^2}}{2 d}-\frac {(b c-4 a d) \int \frac {1}{\sqrt {d x^2+c}}dx}{2 d}\right )-\frac {a^2 \sqrt {c+d x^2}}{c x}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle b \left (\frac {b x \sqrt {c+d x^2}}{2 d}-\frac {(b c-4 a d) \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}}{2 d}\right )-\frac {a^2 \sqrt {c+d x^2}}{c x}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle b \left (\frac {b x \sqrt {c+d x^2}}{2 d}-\frac {(b c-4 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{3/2}}\right )-\frac {a^2 \sqrt {c+d x^2}}{c x}\) |
-((a^2*Sqrt[c + d*x^2])/(c*x)) + b*((b*x*Sqrt[c + d*x^2])/(2*d) - ((b*c - 4*a*d)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(2*d^(3/2)))
3.7.42.3.1 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_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x _Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
Time = 2.97 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84
method | result | size |
risch | \(-\frac {\sqrt {d \,x^{2}+c}\, \left (-b^{2} c \,x^{2}+2 a^{2} d \right )}{2 d c x}+\frac {\left (4 a d -b c \right ) b \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}\) | \(69\) |
default | \(b^{2} \left (\frac {x \sqrt {d \,x^{2}+c}}{2 d}-\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}\right )-\frac {a^{2} \sqrt {d \,x^{2}+c}}{c x}+\frac {2 a b \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{\sqrt {d}}\) | \(87\) |
pseudoelliptic | \(\frac {b^{2} c \,x^{2} \sqrt {d \,x^{2}+c}\, \sqrt {d}+4 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right ) a b c d x -\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right ) b^{2} c^{2} x -2 \sqrt {d \,x^{2}+c}\, a^{2} d^{\frac {3}{2}}}{2 d^{\frac {3}{2}} x c}\) | \(100\) |
-1/2*(d*x^2+c)^(1/2)*(-b^2*c*x^2+2*a^2*d)/d/c/x+1/2*(4*a*d-b*c)*b/d^(3/2)* ln(x*d^(1/2)+(d*x^2+c)^(1/2))
Time = 0.26 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.01 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx=\left [-\frac {{\left (b^{2} c^{2} - 4 \, a b c d\right )} \sqrt {d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left (b^{2} c d x^{2} - 2 \, a^{2} d^{2}\right )} \sqrt {d x^{2} + c}}{4 \, c d^{2} x}, \frac {{\left (b^{2} c^{2} - 4 \, a b c d\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (b^{2} c d x^{2} - 2 \, a^{2} d^{2}\right )} \sqrt {d x^{2} + c}}{2 \, c d^{2} x}\right ] \]
[-1/4*((b^2*c^2 - 4*a*b*c*d)*sqrt(d)*x*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sq rt(d)*x - c) - 2*(b^2*c*d*x^2 - 2*a^2*d^2)*sqrt(d*x^2 + c))/(c*d^2*x), 1/2 *((b^2*c^2 - 4*a*b*c*d)*sqrt(-d)*x*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (b ^2*c*d*x^2 - 2*a^2*d^2)*sqrt(d*x^2 + c))/(c*d^2*x)]
Time = 0.83 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.84 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx=- \frac {a^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{c} + 2 a b \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \wedge d \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {for}\: d \neq 0 \\\frac {x}{\sqrt {c}} & \text {otherwise} \end {cases}\right ) + b^{2} \left (\begin {cases} - \frac {c \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{2 d} + \frac {x \sqrt {c + d x^{2}}}{2 d} & \text {for}\: d \neq 0 \\\frac {x^{3}}{3 \sqrt {c}} & \text {otherwise} \end {cases}\right ) \]
-a**2*sqrt(d)*sqrt(c/(d*x**2) + 1)/c + 2*a*b*Piecewise((log(2*sqrt(d)*sqrt (c + d*x**2) + 2*d*x)/sqrt(d), Ne(c, 0) & Ne(d, 0)), (x*log(x)/sqrt(d*x**2 ), Ne(d, 0)), (x/sqrt(c), True)) + b**2*Piecewise((-c*Piecewise((log(2*sqr t(d)*sqrt(c + d*x**2) + 2*d*x)/sqrt(d), Ne(c, 0)), (x*log(x)/sqrt(d*x**2), True))/(2*d) + x*sqrt(c + d*x**2)/(2*d), Ne(d, 0)), (x**3/(3*sqrt(c)), Tr ue))
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx=\frac {\sqrt {d x^{2} + c} b^{2} x}{2 \, d} - \frac {b^{2} c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, d^{\frac {3}{2}}} + \frac {2 \, a b \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{\sqrt {d}} - \frac {\sqrt {d x^{2} + c} a^{2}}{c x} \]
1/2*sqrt(d*x^2 + c)*b^2*x/d - 1/2*b^2*c*arcsinh(d*x/sqrt(c*d))/d^(3/2) + 2 *a*b*arcsinh(d*x/sqrt(c*d))/sqrt(d) - sqrt(d*x^2 + c)*a^2/(c*x)
Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx=\frac {\sqrt {d x^{2} + c} b^{2} x}{2 \, d} + \frac {2 \, a^{2} \sqrt {d}}{{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c} + \frac {{\left (b^{2} c - 4 \, a b d\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{4 \, d^{\frac {3}{2}}} \]
1/2*sqrt(d*x^2 + c)*b^2*x/d + 2*a^2*sqrt(d)/((sqrt(d)*x - sqrt(d*x^2 + c)) ^2 - c) + 1/4*(b^2*c - 4*a*b*d)*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/d^(3/ 2)
Time = 6.21 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.52 \[ \int \frac {\left (a+b x^2\right )^2}{x^2 \sqrt {c+d x^2}} \, dx=\left \{\begin {array}{cl} \frac {-a^2+2\,a\,b\,x^2+\frac {b^2\,x^4}{3}}{\sqrt {c}\,x} & \text {\ if\ \ }d=0\\ \frac {2\,a\,b\,\ln \left (\sqrt {d}\,x+\sqrt {d\,x^2+c}\right )}{\sqrt {d}}+\frac {b^2\,x\,\sqrt {d\,x^2+c}}{2\,d}-\frac {a^2\,\sqrt {d\,x^2+c}}{c\,x}-\frac {b^2\,c\,\ln \left (2\,\sqrt {d}\,x+2\,\sqrt {d\,x^2+c}\right )}{2\,d^{3/2}} & \text {\ if\ \ }d\neq 0 \end {array}\right . \]