Integrand size = 21, antiderivative size = 85 \[ \int \frac {\sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d} \] Output:
2*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/a^(1/2)/d-2^(1/2)*arc tanh(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(1/2)/d
Time = 0.07 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.76 \[ \int \frac {\sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {2 \left (\text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\sqrt {2} \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \cos \left (\frac {1}{2} (c+d x)\right )}{d \sqrt {a (1+\cos (c+d x))}} \] Input:
Integrate[Sec[c + d*x]/Sqrt[a + a*Cos[c + d*x]],x]
Output:
(-2*(ArcTanh[Sin[(c + d*x)/2]] - Sqrt[2]*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]] )*Cos[(c + d*x)/2])/(d*Sqrt[a*(1 + Cos[c + d*x])])
Time = 0.43 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3259, 3042, 3128, 219, 3252, 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.
| \(\displaystyle \int \frac {\sec (c+d x)}{\sqrt {a \cos (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
| \(\Big \downarrow \) 3259 |
| \(\displaystyle \frac {\int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx}{a}-\int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}-\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {2 \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle -\frac {2 \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {2 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}\) |
Input:
Int[Sec[c + d*x]/Sqrt[a + a*Cos[c + d*x]],x]
Output:
(2*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(Sqrt[a]*d) - (Sqrt[2]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]] )])/(Sqrt[a]*d)
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_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] - Simp[d/(b*c - a*d) Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*S in[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] & & EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(225\) vs. \(2(70)=140\).
Time = 0.96 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.66
| method | result | size |
| default | \(-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right )-\ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right )-\ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right )\right )}{\sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) | \(226\) |
Input:
int(sec(d*x+c)/(a+a*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
-cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2^(1/2)*ln(4*(a^(1/2)* (a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))-ln(-4/(2*cos(1/2*d*x +1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2* c)^2)^(1/2)*a^(1/2)-2*a))-ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*c os(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a)))/a^ (1/2)/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (70) = 140\).
Time = 0.09 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.93 \[ \int \frac {\sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {2} \sqrt {a} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{2 \, a d} \] Input:
integrate(sec(d*x+c)/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")
Output:
1/2*(sqrt(2)*sqrt(a)*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d* x + c) + 1)) + sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt (a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d *x + c)^3 + cos(d*x + c)^2)))/(a*d)
\[ \int \frac {\sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}\, dx \] Input:
integrate(sec(d*x+c)/(a+a*cos(d*x+c))**(1/2),x)
Output:
Integral(sec(c + d*x)/sqrt(a*(cos(c + d*x) + 1)), x)
\[ \int \frac {\sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )}{\sqrt {a \cos \left (d x + c\right ) + a}} \,d x } \] Input:
integrate(sec(d*x+c)/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate(sec(d*x + c)/sqrt(a*cos(d*x + c) + a), x)
Time = 0.39 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.42 \[ \int \frac {\sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} {\left (\frac {\sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )}{\mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {\log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{2 \, \sqrt {a} d} \] Input:
integrate(sec(d*x+c)/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")
Output:
-1/2*sqrt(2)*(sqrt(2)*log(abs(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))/abs(2*s qrt(2) + 4*sin(1/2*d*x + 1/2*c)))/sgn(cos(1/2*d*x + 1/2*c)) + log(sin(1/2* d*x + 1/2*c) + 1)/sgn(cos(1/2*d*x + 1/2*c)) - log(-sin(1/2*d*x + 1/2*c) + 1)/sgn(cos(1/2*d*x + 1/2*c)))/(sqrt(a)*d)
Timed out. \[ \int \frac {\sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \] Input:
int(1/(cos(c + d*x)*(a + a*cos(c + d*x))^(1/2)),x)
Output:
int(1/(cos(c + d*x)*(a + a*cos(c + d*x))^(1/2)), x)
\[ \int \frac {\sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )+1}d x \right )}{a} \] Input:
int(sec(d*x+c)/(a+a*cos(d*x+c))^(1/2),x)
Output:
(sqrt(a)*int((sqrt(cos(c + d*x) + 1)*sec(c + d*x))/(cos(c + d*x) + 1),x))/ a