Integrand size = 37, antiderivative size = 177 \[ \int \frac {\sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx=\frac {(b c+2 a d) x \sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 c \left (a+b x^2\right )}-\frac {a \left (c+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{c x \left (a+b x^2\right )}+\frac {(b c+2 a d) \sqrt {a^2+2 a b x^2+b^2 x^4} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} \left (a+b x^2\right )} \]
-a*(d*x^2+c)^(3/2)*((b*x^2+a)^2)^(1/2)/c/x/(b*x^2+a)+1/2*(2*a*d+b*c)*arcta nh(x*d^(1/2)/(d*x^2+c)^(1/2))*((b*x^2+a)^2)^(1/2)/(b*x^2+a)/d^(1/2)+1/2*(2 *a*d+b*c)*x*(d*x^2+c)^(1/2)*((b*x^2+a)^2)^(1/2)/c/(b*x^2+a)
Time = 0.17 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx=\frac {\sqrt {\left (a+b x^2\right )^2} \left (\sqrt {d} \left (-2 a+b x^2\right ) \sqrt {c+d x^2}+2 (b c+2 a d) x \text {arctanh}\left (\frac {\sqrt {d} x}{-\sqrt {c}+\sqrt {c+d x^2}}\right )\right )}{2 \sqrt {d} x \left (a+b x^2\right )} \]
(Sqrt[(a + b*x^2)^2]*(Sqrt[d]*(-2*a + b*x^2)*Sqrt[c + d*x^2] + 2*(b*c + 2* a*d)*x*ArcTanh[(Sqrt[d]*x)/(-Sqrt[c] + Sqrt[c + d*x^2])]))/(2*Sqrt[d]*x*(a + b*x^2))
Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.63, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {1384, 27, 359, 211, 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.
\(\displaystyle \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \sqrt {c+d x^2}}{x^2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b \left (b x^2+a\right ) \sqrt {d x^2+c}}{x^2}dx}{b \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right ) \sqrt {d x^2+c}}{x^2}dx}{a+b x^2}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {(2 a d+b c) \int \sqrt {d x^2+c}dx}{c}-\frac {a \left (c+d x^2\right )^{3/2}}{c x}\right )}{a+b x^2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {(2 a d+b c) \left (\frac {1}{2} c \int \frac {1}{\sqrt {d x^2+c}}dx+\frac {1}{2} x \sqrt {c+d x^2}\right )}{c}-\frac {a \left (c+d x^2\right )^{3/2}}{c x}\right )}{a+b x^2}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {(2 a d+b c) \left (\frac {1}{2} c \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}+\frac {1}{2} x \sqrt {c+d x^2}\right )}{c}-\frac {a \left (c+d x^2\right )^{3/2}}{c x}\right )}{a+b x^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {(2 a d+b c) \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}}+\frac {1}{2} x \sqrt {c+d x^2}\right )}{c}-\frac {a \left (c+d x^2\right )^{3/2}}{c x}\right )}{a+b x^2}\) |
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-((a*(c + d*x^2)^(3/2))/(c*x)) + ((b*c + 2*a*d)*((x*Sqrt[c + d*x^2])/2 + (c*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/ (2*Sqrt[d])))/c))/(a + b*x^2)
3.3.71.3.1 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)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac Part[p])) Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)] && !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
Time = 0.33 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.53
method | result | size |
risch | \(-\frac {\sqrt {d \,x^{2}+c}\, \left (-b \,x^{2}+2 a \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{2 x \left (b \,x^{2}+a \right )}+\frac {\left (d a +\frac {b c}{2}\right ) \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{\sqrt {d}\, \left (b \,x^{2}+a \right )}\) | \(94\) |
default | \(-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-2 \sqrt {d \,x^{2}+c}\, d^{\frac {3}{2}} a \,x^{2}-\sqrt {d \,x^{2}+c}\, \sqrt {d}\, b c \,x^{2}+2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {d}\, a -2 \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) a c d x -\ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) b \,c^{2} x \right )}{2 \left (b \,x^{2}+a \right ) c x \sqrt {d}}\) | \(130\) |
-1/2*(d*x^2+c)^(1/2)*(-b*x^2+2*a)/x*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+(d*a+1/2 *b*c)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))/d^(1/2)*((b*x^2+a)^2)^(1/2)/(b*x^2+a)
Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx=\left [\frac {{\left (b c + 2 \, a d\right )} \sqrt {d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (b d x^{2} - 2 \, a d\right )} \sqrt {d x^{2} + c}}{4 \, d x}, -\frac {{\left (b c + 2 \, a d\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (b d x^{2} - 2 \, a d\right )} \sqrt {d x^{2} + c}}{2 \, d x}\right ] \]
[1/4*((b*c + 2*a*d)*sqrt(d)*x*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(b*d*x^2 - 2*a*d)*sqrt(d*x^2 + c))/(d*x), -1/2*((b*c + 2*a*d)*sqrt (-d)*x*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (b*d*x^2 - 2*a*d)*sqrt(d*x^2 + c))/(d*x)]
\[ \int \frac {\sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx=\int \frac {\sqrt {c + d x^{2}} \sqrt {\left (a + b x^{2}\right )^{2}}}{x^{2}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.33 \[ \int \frac {\sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx=\frac {1}{2} \, \sqrt {d x^{2} + c} b x + \frac {b c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, \sqrt {d}} + a \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {\sqrt {d x^{2} + c} a}{x} \]
1/2*sqrt(d*x^2 + c)*b*x + 1/2*b*c*arcsinh(d*x/sqrt(c*d))/sqrt(d) + a*sqrt( d)*arcsinh(d*x/sqrt(c*d)) - sqrt(d*x^2 + c)*a/x
Time = 0.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.63 \[ \int \frac {\sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx=\frac {1}{2} \, \sqrt {d x^{2} + c} b x \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {2 \, a c \sqrt {d} \mathrm {sgn}\left (b x^{2} + a\right )}{{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c} - \frac {{\left (b c \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, a d \mathrm {sgn}\left (b x^{2} + a\right )\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{4 \, \sqrt {d}} \]
1/2*sqrt(d*x^2 + c)*b*x*sgn(b*x^2 + a) + 2*a*c*sqrt(d)*sgn(b*x^2 + a)/((sq rt(d)*x - sqrt(d*x^2 + c))^2 - c) - 1/4*(b*c*sgn(b*x^2 + a) + 2*a*d*sgn(b* x^2 + a))*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/sqrt(d)
Timed out. \[ \int \frac {\sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx=\int \frac {\sqrt {d\,x^2+c}\,\sqrt {{\left (b\,x^2+a\right )}^2}}{x^2} \,d x \]