Integrand size = 21, antiderivative size = 106 \[ \int \frac {\cot (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b} d} \]
2*arctanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))/d/a^(1/2)-arctanh((a+b*sec(d*x+c ))^(1/2)/(a-b)^(1/2))/d/(a-b)^(1/2)-arctanh((a+b*sec(d*x+c))^(1/2)/(a+b)^( 1/2))/d/(a+b)^(1/2)
Time = 0.29 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.94 \[ \int \frac {\cot (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{\sqrt {a-b}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{\sqrt {a+b}}}{d} \]
-(((-2*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/Sqrt[a] + ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]]/Sqrt[a - b] + ArcTanh[Sqrt[a + b*Sec[c + d *x]]/Sqrt[a + b]]/Sqrt[a + b])/d)
Time = 0.36 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 25, 4373, 561, 1484, 2009}
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.
| \(\displaystyle \int \frac {\cot (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {1}{\cot \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {1}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right ) \sqrt {a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )}}dx\) |
| \(\Big \downarrow \) 4373 |
| \(\displaystyle -\frac {b^2 \int \frac {\cos (c+d x)}{b \sqrt {a+b \sec (c+d x)} \left (b^2-b^2 \sec ^2(c+d x)\right )}d(b \sec (c+d x))}{d}\) |
| \(\Big \downarrow \) 561 |
| \(\displaystyle -\frac {2 b^2 \int \frac {1}{\left (a-b^2 \sec ^2(c+d x)\right ) \left (b^4 \sec ^4(c+d x)-2 a b^2 \sec ^2(c+d x)+a^2-b^2\right )}d\sqrt {a+b \sec (c+d x)}}{d}\) |
| \(\Big \downarrow \) 1484 |
| \(\displaystyle -\frac {2 b^2 \int \left (\frac {1}{2 b^2 \left (-b^2 \sec ^2(c+d x)+a+b\right )}-\frac {1}{2 b^2 \left (b^2 \sec ^2(c+d x)-a+b\right )}-\frac {1}{b^2 \left (a-b^2 \sec ^2(c+d x)\right )}\right )d\sqrt {a+b \sec (c+d x)}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 b^2 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} b^2}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a-b}}\right )}{2 b^2 \sqrt {a-b}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )}{2 b^2 \sqrt {a+b}}\right )}{d}\) |
(-2*b^2*(-(ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]]/(Sqrt[a]*b^2)) + ArcT anh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]]/(2*Sqrt[a - b]*b^2) + ArcTanh[Sq rt[a + b*Sec[c + d*x]]/Sqrt[a + b]]/(2*b^2*Sqrt[a + b])))/d
3.4.29.3.1 Defintions of rubi rules used
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{k = Denominator[n]}, Simp[k/d Subst[Int[x^(k*(n + 1) - 1)*(-c /d + x^k/d)^m*Simp[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^k/d^2) + b*(x^(2*k)/d^2), x]^p, x], x, (c + d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, m, p}, x] && Frac tionQ[n] && IntegerQ[p] && IntegerQ[m]
Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symb ol] :> Int[ExpandIntegrand[(d + e*x^2)^q/(a + b*x^2 + c*x^4), x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[q]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1)) Subst[Int[(b^2 - x^ 2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(657\) vs. \(2(88)=176\).
Time = 2.29 (sec) , antiderivative size = 658, normalized size of antiderivative = 6.21
| method | result | size |
| default | \(-\frac {\sqrt {a +b \sec \left (d x +c \right )}\, \left (-2 a^{\frac {3}{2}} \left (a -b \right )^{\frac {3}{2}} \ln \left (4 \cos \left (d x +c \right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \sqrt {a}+4 a \cos \left (d x +c \right )+4 \sqrt {a}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}+2 b \right )+\sqrt {a +b}\, \left (a -b \right )^{\frac {3}{2}} \ln \left (-\frac {2 \left (2 \cos \left (d x +c \right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \sqrt {a +b}+2 a \cos \left (d x +c \right )+\cos \left (d x +c \right ) b +2 \sqrt {a +b}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}+b \right )}{\cos \left (d x +c \right )-1}\right ) a -2 \sqrt {a}\, \left (a -b \right )^{\frac {3}{2}} \ln \left (4 \cos \left (d x +c \right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \sqrt {a}+4 a \cos \left (d x +c \right )+4 \sqrt {a}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}+2 b \right ) b -a^{3} \ln \left (\frac {2 \sqrt {a -b}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \cos \left (d x +c \right )+2 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \sqrt {a -b}-2 a \cos \left (d x +c \right )+\cos \left (d x +c \right ) b -b}{\sqrt {a -b}\, \left (\cos \left (d x +c \right )+1\right )}\right )+\ln \left (\frac {2 \sqrt {a -b}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \cos \left (d x +c \right )+2 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \sqrt {a -b}-2 a \cos \left (d x +c \right )+\cos \left (d x +c \right ) b -b}{\sqrt {a -b}\, \left (\cos \left (d x +c \right )+1\right )}\right ) a \,b^{2}\right ) \cos \left (d x +c \right )}{2 d \left (a -b \right )^{\frac {3}{2}} \left (a +b \right ) a \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}}\) | \(658\) |
-1/2/d/(a-b)^(3/2)/(a+b)/a*(a+b*sec(d*x+c))^(1/2)*(-2*a^(3/2)*(a-b)^(3/2)* ln(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/ 2)+4*a*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2) ^(1/2)+2*b)+(a+b)^(1/2)*(a-b)^(3/2)*ln(-2*(2*cos(d*x+c)*((b+a*cos(d*x+c))* cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a+b)^(1/2)+2*a*cos(d*x+c)+cos(d*x+c)*b +2*(a+b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+b)/(co s(d*x+c)-1))*a-2*a^(1/2)*(a-b)^(3/2)*ln(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos (d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d*x+c)+4*a^(1/2)*((b+a*cos (d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)*b-a^3*ln(1/(a-b)^(1/2)*(2 *(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+ c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a* cos(d*x+c)+cos(d*x+c)*b-b)/(cos(d*x+c)+1))+ln(1/(a-b)^(1/2)*(2*(a-b)^(1/2) *((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)+2*((b+a*c os(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+c os(d*x+c)*b-b)/(cos(d*x+c)+1))*a*b^2)*cos(d*x+c)/(cos(d*x+c)+1)/((b+a*cos( d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (88) = 176\).
Time = 1.42 (sec) , antiderivative size = 2420, normalized size of antiderivative = 22.83 \[ \int \frac {\cot (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\text {Too large to display} \]
[1/4*(2*(a^2 - b^2)*sqrt(a)*log(-8*a^2*cos(d*x + c)^2 - 8*a*b*cos(d*x + c) - b^2 - 4*(2*a*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))) + (a^2 + a*b)*sqrt(a - b)*log(-((8*a^2 - 8*a*b + b ^2)*cos(d*x + c)^2 + b^2 - 4*((2*a - b)*cos(d*x + c)^2 + b*cos(d*x + c))*s qrt(a - b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)) + 2*(4*a*b - 3*b^2)*cos (d*x + c))/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + (a^2 - a*b)*sqrt(a + b )*log(-((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^2 + b^2 - 4*((2*a + b)*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a + b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c )) + 2*(4*a*b + 3*b^2)*cos(d*x + c))/(cos(d*x + c)^2 - 2*cos(d*x + c) + 1) ))/((a^3 - a*b^2)*d), -1/4*(4*(a^2 - b^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt( (a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b)) - (a^2 + a*b)*sqrt(a - b)*log(-((8*a^2 - 8*a*b + b^2)*cos(d*x + c)^2 + b^2 - 4*((2*a - b)*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a - b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)) + 2*(4*a*b - 3*b^2)*cos(d*x + c))/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - (a^2 - a*b)*sqrt(a + b)*log(-((8*a^2 + 8*a*b + b ^2)*cos(d*x + c)^2 + b^2 - 4*((2*a + b)*cos(d*x + c)^2 + b*cos(d*x + c))*s qrt(a + b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)) + 2*(4*a*b + 3*b^2)*cos (d*x + c))/(cos(d*x + c)^2 - 2*cos(d*x + c) + 1)))/((a^3 - a*b^2)*d), -1/4 *(2*(a^2 + a*b)*sqrt(-a + b)*arctan(-2*sqrt(-a + b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/((2*a - b)*cos(d*x + c) + b)) - 2*(a^2 -...
\[ \int \frac {\cot (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {\cot {\left (c + d x \right )}}{\sqrt {a + b \sec {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\cot (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {\cot \left (d x + c\right )}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
Exception generated. \[ \int \frac {\cot (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {\cot (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {\mathrm {cot}\left (c+d\,x\right )}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]