Integrand size = 23, antiderivative size = 108 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {\tan (c+d x)}{d \sqrt {a+a \cos (c+d x)}} \]
-arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d/a^(1/2)+arctanh(1/2* sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)+tan(d *x+c)/d/(a+a*cos(d*x+c))^(1/2)
Result contains complex when optimal does not.
Time = 0.85 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.78 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {2} \log \left (i-\sqrt {2} e^{\frac {1}{2} i (c+d x)}-i e^{i (c+d x)}\right )-\sqrt {2} \log \left (i+\sqrt {2} e^{\frac {1}{2} i (c+d x)}-i e^{i (c+d x)}\right )-4 \log \left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )+4 \log \left (\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )\right )+4 \sec (c+d x) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d \sqrt {a (1+\cos (c+d x))}} \]
(Cos[(c + d*x)/2]*(Sqrt[2]*Log[I - Sqrt[2]*E^((I/2)*(c + d*x)) - I*E^(I*(c + d*x))] - Sqrt[2]*Log[I + Sqrt[2]*E^((I/2)*(c + d*x)) - I*E^(I*(c + d*x) )] - 4*Log[Cos[(c + d*x)/4] - Sin[(c + d*x)/4]] + 4*Log[Cos[(c + d*x)/4] + Sin[(c + d*x)/4]] + 4*Sec[c + d*x]*Sin[(c + d*x)/2]))/(2*d*Sqrt[a*(1 + Co s[c + d*x])])
Time = 0.63 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3258, 3042, 3464, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x)}{\sqrt {a \cos (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
| \(\Big \downarrow \) 3258 |
| \(\displaystyle \frac {\tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {(a-a \cos (c+d x)) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\) |
| \(\Big \downarrow \) 3464 |
| \(\displaystyle \frac {\tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-2 a \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{2 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-2 a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {4 a \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}+\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {-\frac {2 a \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {2 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\tan (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {2 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{2 a}\) |
-1/2*((2*Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]) /d - (2*Sqrt[2]*Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a *Cos[c + d*x]])])/d)/a + Tan[c + d*x]/(d*Sqrt[a + a*Cos[c + d*x]])
3.2.28.3.1 Defintions of rubi rules used
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[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. ) + (f_.)*(x_)]], x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((c + d*Sin[e + f*x]) ^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] - Simp[1/(2* b*(n + 1)*(c^2 - d^2)) Int[(c + d*Sin[e + f*x])^(n + 1)*(Simp[a*d - 2*b*c *(n + 1) + b*d*(2*n + 3)*Sin[e + f*x], x]/Sqrt[a + b*Sin[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] && LtQ[n, -1] && IntegerQ[2*n]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A *b - a*B)/(b*c - a*d) Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Simp[(B*c - A*d)/(b*c - a*d) Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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. \(465\) vs. \(2(91)=182\).
Time = 1.75 (sec) , antiderivative size = 466, normalized size of antiderivative = 4.31
| method | result | size |
| default | \(\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (2 a \left (-2 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right )+\ln \left (\frac {4 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right )+\ln \left (-\frac {4 \left (\sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a -\ln \left (\frac {4 \sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a -\ln \left (-\frac {4 \left (\sqrt {2}\, a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a +2 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{a^{\frac {3}{2}} \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\right ) \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(466\) |
cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*a*(-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))*(2^(1/2)*a*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))*(2^(1 /2)*a*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)))*sin(1/2*d*x+1/2*c)^2+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))*a-ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/ 2)*a*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 ))*a-ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*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))*a+2*2^(1/2)*(a*sin(1/2* d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(3/2)/(2*cos(1/2*d*x+1/2*c)-2^(1/2))/(2*cos (1/2*d*x+1/2*c)+2^(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. 236 vs. \(2 (91) = 182\).
Time = 0.27 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.19 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {2 \, \sqrt {2} {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}} + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{4 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}} \]
1/4*((cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*c os(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)) + 2*sqrt(2)*(a*cos(d*x + c)^2 + a*cos(d*x + c))*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*co s(d*x + c) + 1))/sqrt(a) + 4*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a*d*c os(d*x + c)^2 + a*d*cos(d*x + c))
\[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 18435 vs. \(2 (91) = 182\).
Time = 0.55 (sec) , antiderivative size = 18435, normalized size of antiderivative = 170.69 \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Too large to display} \]
1/4*((2*sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*si n(1/2*d*x + 1/2*c) + 1) - 2*sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d *x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*s in(1/2*d*x + 1/2*c) + 2) + log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2 )*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + log(2*cos(1 /2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2 *c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(d*x + c)^4 + (2*sqrt(2)*log (cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 2*sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*s in(1/2*d*x + 1/2*c) + 1) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2 )*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos(1 /2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2 *c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin (1/2*d*x + 1/2*c) + 2))*sin(d*x + c)^4 + 4*sqrt(2)*cos(1/2*d*x + 1/2*c)...
Exception generated. \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^2\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \]