Integrand size = 37, antiderivative size = 50 \[ \int \frac {\sqrt {d+e x^2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} x \sqrt {d+e x^2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}}\right )}{\sqrt {b}} \] Output:
arctanh(b^(1/2)*x*(e*x^2+d)^(1/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2))/b^(1/ 2)
Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt {d+e x^2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b} \sqrt {\left (a+b x^2\right ) \left (d+e x^2\right )}} \] Input:
Integrate[Sqrt[d + e*x^2]/Sqrt[a*d + (b*d + a*e)*x^2 + b*e*x^4],x]
Output:
(Sqrt[a + b*x^2]*Sqrt[d + e*x^2]*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(Sq rt[b]*Sqrt[(a + b*x^2)*(d + e*x^2)])
Time = 0.32 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.44, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {1395, 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.
\(\displaystyle \int \frac {\sqrt {d+e x^2}}{\sqrt {x^2 (a e+b d)+a d+b e x^4}} \, dx\) |
\(\Big \downarrow \) 1395 |
\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \int \frac {1}{\sqrt {b x^2+a}}dx}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b} \sqrt {x^2 (a e+b d)+a d+b e x^4}}\) |
Input:
Int[Sqrt[d + e*x^2]/Sqrt[a*d + (b*d + a*e)*x^2 + b*e*x^4],x]
Output:
(Sqrt[a + b*x^2]*Sqrt[d + e*x^2]*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(Sq rt[b]*Sqrt[a*d + (b*d + a*e)*x^2 + b*e*x^4])
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[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p]) Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && E qQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && !IntegerQ[p] && !(EqQ[q, 1] && EqQ[n, 2])
Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.12
method | result | size |
default | \(\frac {\sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}\, \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {e \,x^{2}+d}\, \sqrt {b \,x^{2}+a}\, \sqrt {b}}\) | \(56\) |
Input:
int((e*x^2+d)^(1/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x,method=_RETURNVERB OSE)
Output:
1/(e*x^2+d)^(1/2)*((e*x^2+d)*(b*x^2+a))^(1/2)/(b*x^2+a)^(1/2)*ln(b^(1/2)*x +(b*x^2+a)^(1/2))/b^(1/2)
Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.56 \[ \int \frac {\sqrt {d+e x^2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\left [\frac {\log \left (\frac {2 \, b e x^{4} + {\left (2 \, b d + a e\right )} x^{2} + 2 \, \sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} \sqrt {e x^{2} + d} \sqrt {b} x + a d}{e x^{2} + d}\right )}{2 \, \sqrt {b}}, -\frac {\sqrt {-b} \arctan \left (\frac {\sqrt {e x^{2} + d} \sqrt {-b} x}{\sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d}}\right )}{b}\right ] \] Input:
integrate((e*x^2+d)^(1/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x, algorithm=" fricas")
Output:
[1/2*log((2*b*e*x^4 + (2*b*d + a*e)*x^2 + 2*sqrt(b*e*x^4 + (b*d + a*e)*x^2 + a*d)*sqrt(e*x^2 + d)*sqrt(b)*x + a*d)/(e*x^2 + d))/sqrt(b), -sqrt(-b)*a rctan(sqrt(e*x^2 + d)*sqrt(-b)*x/sqrt(b*e*x^4 + (b*d + a*e)*x^2 + a*d))/b]
\[ \int \frac {\sqrt {d+e x^2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\int \frac {\sqrt {d + e x^{2}}}{\sqrt {\left (a + b x^{2}\right ) \left (d + e x^{2}\right )}}\, dx \] Input:
integrate((e*x**2+d)**(1/2)/(a*d+(a*e+b*d)*x**2+b*e*x**4)**(1/2),x)
Output:
Integral(sqrt(d + e*x**2)/sqrt((a + b*x**2)*(d + e*x**2)), x)
\[ \int \frac {\sqrt {d+e x^2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\int { \frac {\sqrt {e x^{2} + d}}{\sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d}} \,d x } \] Input:
integrate((e*x^2+d)^(1/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x, algorithm=" maxima")
Output:
integrate(sqrt(e*x^2 + d)/sqrt(b*e*x^4 + (b*d + a*e)*x^2 + a*d), x)
\[ \int \frac {\sqrt {d+e x^2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\int { \frac {\sqrt {e x^{2} + d}}{\sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d}} \,d x } \] Input:
integrate((e*x^2+d)^(1/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x, algorithm=" giac")
Output:
integrate(sqrt(e*x^2 + d)/sqrt(b*e*x^4 + (b*d + a*e)*x^2 + a*d), x)
Timed out. \[ \int \frac {\sqrt {d+e x^2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\int \frac {\sqrt {e\,x^2+d}}{\sqrt {b\,e\,x^4+\left (a\,e+b\,d\right )\,x^2+a\,d}} \,d x \] Input:
int((d + e*x^2)^(1/2)/(a*d + x^2*(a*e + b*d) + b*e*x^4)^(1/2),x)
Output:
int((d + e*x^2)^(1/2)/(a*d + x^2*(a*e + b*d) + b*e*x^4)^(1/2), x)
Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.50 \[ \int \frac {\sqrt {d+e x^2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\frac {\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right )}{b} \] Input:
int((e*x^2+d)^(1/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x)
Output:
(sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a)))/b