Integrand size = 25, antiderivative size = 39 \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{\sqrt {b} f} \] Output:
arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/b^(1/2)/f
Leaf count is larger than twice the leaf count of optimal. \(87\) vs. \(2(39)=78\).
Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.23 \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right ) \sqrt {a+2 b+a \cos (2 e+2 f x)} \sec (e+f x)}{\sqrt {2} \sqrt {b} f \sqrt {a+b \sec ^2(e+f x)}} \] Input:
Integrate[Sec[e + f*x]^2/Sqrt[a + b*Sec[e + f*x]^2],x]
Output:
(ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]]*Sqrt[a + 2 *b + a*Cos[2*e + 2*f*x]]*Sec[e + f*x])/(Sqrt[2]*Sqrt[b]*f*Sqrt[a + b*Sec[e + f*x]^2])
Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 4634, 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.
\(\displaystyle \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (e+f x)^2}{\sqrt {a+b \sec (e+f x)^2}}dx\) |
\(\Big \downarrow \) 4634 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {b \tan ^2(e+f x)+a+b}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\int \frac {1}{1-\frac {b \tan ^2(e+f x)}{b \tan ^2(e+f x)+a+b}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a+b}}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{\sqrt {b} f}\) |
Input:
Int[Sec[e + f*x]^2/Sqrt[a + b*Sec[e + f*x]^2],x]
Output:
ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(Sqrt[b]*f)
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[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) )^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ [m/2] && IntegerQ[n/2]
Leaf count of result is larger than twice the leaf count of optimal. \(240\) vs. \(2(33)=66\).
Time = 9.14 (sec) , antiderivative size = 241, normalized size of antiderivative = 6.18
method | result | size |
default | \(\frac {\left (\ln \left (\frac {-4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sin \left (f x +e \right ) a -4 \sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 a -4 b}{\sin \left (f x +e \right )-1}\right )+\ln \left (-\frac {4 \left (-\sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+\sin \left (f x +e \right ) a -\sqrt {b}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+a +b \right )}{\sin \left (f x +e \right )+1}\right )\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \left (\sec \left (f x +e \right )+1\right )}{2 f \sqrt {b}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}\) | \(241\) |
Input:
int(sec(f*x+e)^2/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/2/f/b^(1/2)*(ln(4*(-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)* cos(f*x+e)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2 )-a-b)/(sin(f*x+e)-1))+ln(-4*(-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^ 2)^(1/2)*cos(f*x+e)+sin(f*x+e)*a-b^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e) )^2)^(1/2)+a+b)/(sin(f*x+e)+1)))*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/ 2)/(a+b*sec(f*x+e)^2)^(1/2)*(sec(f*x+e)+1)
Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (33) = 66\).
Time = 0.13 (sec) , antiderivative size = 215, normalized size of antiderivative = 5.51 \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [\frac {\log \left (\frac {{\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right )}{4 \, \sqrt {b} f}, \frac {\sqrt {-b} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left (a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right )}{2 \, b f}\right ] \] Input:
integrate(sec(f*x+e)^2/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")
Output:
[1/4*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^ 2 + 4*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4)/(sqrt(b) *f), 1/2*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))* sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e)))/(b*f)]
\[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\sec ^{2}{\left (e + f x \right )}}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \] Input:
integrate(sec(f*x+e)**2/(a+b*sec(f*x+e)**2)**(1/2),x)
Output:
Integral(sec(e + f*x)**2/sqrt(a + b*sec(e + f*x)**2), x)
Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.59 \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\frac {\operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b} f} \] Input:
integrate(sec(f*x+e)^2/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")
Output:
arcsinh(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt(b)*f)
\[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\sec \left (f x + e\right )^{2}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \] Input:
integrate(sec(f*x+e)^2/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")
Output:
integrate(sec(f*x + e)^2/sqrt(b*sec(f*x + e)^2 + a), x)
Timed out. \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {1}{{\cos \left (e+f\,x\right )}^2\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \] Input:
int(1/(cos(e + f*x)^2*(a + b/cos(e + f*x)^2)^(1/2)),x)
Output:
int(1/(cos(e + f*x)^2*(a + b/cos(e + f*x)^2)^(1/2)), x)
\[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right )^{2} b +a}\, \sec \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{2} b +a}d x \] Input:
int(sec(f*x+e)^2/(a+b*sec(f*x+e)^2)^(1/2),x)
Output:
int((sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**2)/(sec(e + f*x)**2*b + a), x)