Integrand size = 13, antiderivative size = 79 \[ \int \frac {\sec ^3(x)}{a+b \cot (x)} \, dx=\frac {\text {arctanh}(\sin (x))}{2 a}+\frac {b^2 \text {arctanh}(\sin (x))}{a^3}+\frac {b \sqrt {a^2+b^2} \text {arctanh}\left (\frac {a \cos (x)-b \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^3}-\frac {b \sec (x)}{a^2}+\frac {\sec (x) \tan (x)}{2 a} \] Output:
1/2*arctanh(sin(x))/a+b^2*arctanh(sin(x))/a^3+b*(a^2+b^2)^(1/2)*arctanh((a *cos(x)-b*sin(x))/(a^2+b^2)^(1/2))/a^3-b*sec(x)/a^2+1/2*sec(x)*tan(x)/a
Leaf count is larger than twice the leaf count of optimal. \(192\) vs. \(2(79)=158\).
Time = 0.48 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.43 \[ \int \frac {\sec ^3(x)}{a+b \cot (x)} \, dx=-\frac {8 b \sqrt {a^2+b^2} \text {arctanh}\left (\frac {-a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )+\sec ^2(x) \left (4 a b \cos (x)+a^2 \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+2 b^2 \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+\left (a^2+2 b^2\right ) \cos (2 x) \left (\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )\right )-a^2 \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )-2 b^2 \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )-2 a^2 \sin (x)\right )}{4 a^3} \] Input:
Integrate[Sec[x]^3/(a + b*Cot[x]),x]
Output:
-1/4*(8*b*Sqrt[a^2 + b^2]*ArcTanh[(-a + b*Tan[x/2])/Sqrt[a^2 + b^2]] + Sec [x]^2*(4*a*b*Cos[x] + a^2*Log[Cos[x/2] - Sin[x/2]] + 2*b^2*Log[Cos[x/2] - Sin[x/2]] + (a^2 + 2*b^2)*Cos[2*x]*(Log[Cos[x/2] - Sin[x/2]] - Log[Cos[x/2 ] + Sin[x/2]]) - a^2*Log[Cos[x/2] + Sin[x/2]] - 2*b^2*Log[Cos[x/2] + Sin[x /2]] - 2*a^2*Sin[x]))/a^3
Time = 0.45 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3042, 4001, 25, 25, 3042, 3589, 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.
| \(\displaystyle \int \frac {\sec ^3(x)}{a+b \cot (x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (x+\frac {\pi }{2}\right )^3 \left (a-b \tan \left (x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4001 |
| \(\displaystyle \int -\frac {\tan (x) \sec ^2(x)}{-a \sin (x)-b \cos (x)}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -\frac {\sec ^2(x) \tan (x)}{b \cos (x)+a \sin (x)}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\tan (x) \sec ^2(x)}{a \sin (x)+b \cos (x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (x)}{\cos (x)^3 (a \sin (x)+b \cos (x))}dx\) |
| \(\Big \downarrow \) 3589 |
| \(\displaystyle \int \left (\frac {\sec ^3(x)}{a}-\frac {b \sec ^2(x)}{a (a \sin (x)+b \cos (x))}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^2 \text {arctanh}(\sin (x))}{a^3}-\frac {b \sec (x)}{a^2}+\frac {b \sqrt {a^2+b^2} \text {arctanh}\left (\frac {a \cos (x)-b \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^3}+\frac {\text {arctanh}(\sin (x))}{2 a}+\frac {\tan (x) \sec (x)}{2 a}\) |
Input:
Int[Sec[x]^3/(a + b*Cot[x]),x]
Output:
ArcTanh[Sin[x]]/(2*a) + (b^2*ArcTanh[Sin[x]])/a^3 + (b*Sqrt[a^2 + b^2]*Arc Tanh[(a*Cos[x] - b*Sin[x])/Sqrt[a^2 + b^2]])/a^3 - (b*Sec[x])/a^2 + (Sec[x ]*Tan[x])/(2*a)
Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_. ) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[Ex pandTrig[cos[c + d*x]^m*(sin[c + d*x]^n/(a*cos[c + d*x] + b*sin[c + d*x])), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegersQ[m, n]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n _.), x_Symbol] :> Int[Sin[e + f*x]^m*((a*Cos[e + f*x] + b*Sin[e + f*x])^n/C os[e + f*x]^n), x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && ILtQ [n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))
Leaf count of result is larger than twice the leaf count of optimal. \(149\) vs. \(2(71)=142\).
Time = 1.49 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.90
| method | result | size |
| default | \(\frac {2 b \sqrt {a^{2}+b^{2}}\, \operatorname {arctanh}\left (\frac {-2 b \tan \left (\frac {x}{2}\right )+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{3}}+\frac {1}{2 a \left (\tan \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {-a -2 b}{2 a^{2} \left (\tan \left (\frac {x}{2}\right )-1\right )}+\frac {\left (-a^{2}-2 b^{2}\right ) \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{2 a^{3}}-\frac {1}{2 a \left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {-a +2 b}{2 a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )}+\frac {\left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{2 a^{3}}\) | \(150\) |
| risch | \(-\frac {{\mathrm e}^{i x} \left (i a \,{\mathrm e}^{2 i x}+2 b \,{\mathrm e}^{2 i x}-i a +2 b \right )}{\left ({\mathrm e}^{2 i x}+1\right )^{2} a^{2}}-\frac {\ln \left ({\mathrm e}^{i x}-i\right )}{2 a}-\frac {\ln \left ({\mathrm e}^{i x}-i\right ) b^{2}}{a^{3}}+\frac {\ln \left ({\mathrm e}^{i x}+i\right )}{2 a}+\frac {\ln \left ({\mathrm e}^{i x}+i\right ) b^{2}}{a^{3}}+\frac {\sqrt {a^{2}+b^{2}}\, b \ln \left ({\mathrm e}^{i x}-\frac {i b -a}{\sqrt {a^{2}+b^{2}}}\right )}{a^{3}}-\frac {\sqrt {a^{2}+b^{2}}\, b \ln \left ({\mathrm e}^{i x}+\frac {i b -a}{\sqrt {a^{2}+b^{2}}}\right )}{a^{3}}\) | \(193\) |
Input:
int(sec(x)^3/(a+b*cot(x)),x,method=_RETURNVERBOSE)
Output:
2*b*(a^2+b^2)^(1/2)/a^3*arctanh(1/2*(-2*b*tan(1/2*x)+2*a)/(a^2+b^2)^(1/2)) +1/2/a/(tan(1/2*x)-1)^2-1/2*(-a-2*b)/a^2/(tan(1/2*x)-1)+1/2/a^3*(-a^2-2*b^ 2)*ln(tan(1/2*x)-1)-1/2/a/(tan(1/2*x)+1)^2-1/2*(-a+2*b)/a^2/(tan(1/2*x)+1) +1/2*(a^2+2*b^2)/a^3*ln(tan(1/2*x)+1)
Leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (71) = 142\).
Time = 0.14 (sec) , antiderivative size = 166, normalized size of antiderivative = 2.10 \[ \int \frac {\sec ^3(x)}{a+b \cot (x)} \, dx=\frac {2 \, \sqrt {a^{2} + b^{2}} b \cos \left (x\right )^{2} \log \left (\frac {2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} - 2 \, b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cos \left (x\right ) - b \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2}}\right ) + {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) - {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) - 4 \, a b \cos \left (x\right ) + 2 \, a^{2} \sin \left (x\right )}{4 \, a^{3} \cos \left (x\right )^{2}} \] Input:
integrate(sec(x)^3/(a+b*cot(x)),x, algorithm="fricas")
Output:
1/4*(2*sqrt(a^2 + b^2)*b*cos(x)^2*log((2*a*b*cos(x)*sin(x) - (a^2 - b^2)*c os(x)^2 - a^2 - 2*b^2 - 2*sqrt(a^2 + b^2)*(a*cos(x) - b*sin(x)))/(2*a*b*co s(x)*sin(x) - (a^2 - b^2)*cos(x)^2 + a^2)) + (a^2 + 2*b^2)*cos(x)^2*log(si n(x) + 1) - (a^2 + 2*b^2)*cos(x)^2*log(-sin(x) + 1) - 4*a*b*cos(x) + 2*a^2 *sin(x))/(a^3*cos(x)^2)
\[ \int \frac {\sec ^3(x)}{a+b \cot (x)} \, dx=\int \frac {\sec ^{3}{\left (x \right )}}{a + b \cot {\left (x \right )}}\, dx \] Input:
integrate(sec(x)**3/(a+b*cot(x)),x)
Output:
Integral(sec(x)**3/(a + b*cot(x)), x)
Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (71) = 142\).
Time = 0.11 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.57 \[ \int \frac {\sec ^3(x)}{a+b \cot (x)} \, dx=-\frac {2 \, b - \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac {2 \, b \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac {a \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}}}{a^{2} - \frac {2 \, a^{2} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}}} + \frac {{\left (a^{2} + 2 \, b^{2}\right )} \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}{2 \, a^{3}} - \frac {{\left (a^{2} + 2 \, b^{2}\right )} \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right )}{2 \, a^{3}} + \frac {{\left (a^{2} b + b^{3}\right )} \log \left (\frac {a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {a^{2} + b^{2}}}{a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} \] Input:
integrate(sec(x)^3/(a+b*cot(x)),x, algorithm="maxima")
Output:
-(2*b - a*sin(x)/(cos(x) + 1) - 2*b*sin(x)^2/(cos(x) + 1)^2 - a*sin(x)^3/( cos(x) + 1)^3)/(a^2 - 2*a^2*sin(x)^2/(cos(x) + 1)^2 + a^2*sin(x)^4/(cos(x) + 1)^4) + 1/2*(a^2 + 2*b^2)*log(sin(x)/(cos(x) + 1) + 1)/a^3 - 1/2*(a^2 + 2*b^2)*log(sin(x)/(cos(x) + 1) - 1)/a^3 + (a^2*b + b^3)*log((a - b*sin(x) /(cos(x) + 1) + sqrt(a^2 + b^2))/(a - b*sin(x)/(cos(x) + 1) - sqrt(a^2 + b ^2)))/(sqrt(a^2 + b^2)*a^3)
Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (71) = 142\).
Time = 0.15 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.00 \[ \int \frac {\sec ^3(x)}{a+b \cot (x)} \, dx=\frac {{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right )}{2 \, a^{3}} - \frac {{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right )}{2 \, a^{3}} + \frac {{\left (a^{2} b + b^{3}\right )} \log \left (\frac {{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{3}} + \frac {a \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, x\right )^{2} + a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} - 1\right )}^{2} a^{2}} \] Input:
integrate(sec(x)^3/(a+b*cot(x)),x, algorithm="giac")
Output:
1/2*(a^2 + 2*b^2)*log(abs(tan(1/2*x) + 1))/a^3 - 1/2*(a^2 + 2*b^2)*log(abs (tan(1/2*x) - 1))/a^3 + (a^2*b + b^3)*log(abs(2*b*tan(1/2*x) - 2*a - 2*sqr t(a^2 + b^2))/abs(2*b*tan(1/2*x) - 2*a + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b ^2)*a^3) + (a*tan(1/2*x)^3 + 2*b*tan(1/2*x)^2 + a*tan(1/2*x) - 2*b)/((tan( 1/2*x)^2 - 1)^2*a^2)
Time = 9.27 (sec) , antiderivative size = 346, normalized size of antiderivative = 4.38 \[ \int \frac {\sec ^3(x)}{a+b \cot (x)} \, dx=\frac {\frac {\mathrm {tan}\left (\frac {x}{2}\right )}{a}-\frac {2\,b}{a^2}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{a}+\frac {2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}+\frac {\mathrm {atanh}\left (\frac {24\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a^2\,b+24\,b^3+\frac {16\,b^5}{a^2}}+\frac {16\,b^5\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a^4\,b+24\,a^2\,b^3+16\,b^5}+\frac {8\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a\,b+\frac {24\,b^3}{a}+\frac {16\,b^5}{a^3}}\right )\,\left (a^2+2\,b^2\right )}{a^3}-\frac {2\,b\,\mathrm {atanh}\left (\frac {16\,b^3\,\sqrt {a^2+b^2}}{32\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )+16\,a\,b^3+\frac {16\,b^5}{a}+32\,a^2\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}+\frac {32\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2+b^2}}{16\,b^3+\frac {16\,b^5}{a^2}+\frac {32\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a}+32\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}+\frac {16\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2+b^2}}{32\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^3\,b^2+16\,a^2\,b^3+32\,\mathrm {tan}\left (\frac {x}{2}\right )\,a\,b^4+16\,b^5}\right )\,\sqrt {a^2+b^2}}{a^3} \] Input:
int(1/(cos(x)^3*(a + b*cot(x))),x)
Output:
(tan(x/2)/a - (2*b)/a^2 + tan(x/2)^3/a + (2*b*tan(x/2)^2)/a^2)/(tan(x/2)^4 - 2*tan(x/2)^2 + 1) + (atanh((24*b^3*tan(x/2))/(8*a^2*b + 24*b^3 + (16*b^ 5)/a^2) + (16*b^5*tan(x/2))/(8*a^4*b + 16*b^5 + 24*a^2*b^3) + (8*a*b*tan(x /2))/(8*a*b + (24*b^3)/a + (16*b^5)/a^3))*(a^2 + 2*b^2))/a^3 - (2*b*atanh( (16*b^3*(a^2 + b^2)^(1/2))/(32*b^4*tan(x/2) + 16*a*b^3 + (16*b^5)/a + 32*a ^2*b^2*tan(x/2)) + (32*b^2*tan(x/2)*(a^2 + b^2)^(1/2))/(16*b^3 + (16*b^5)/ a^2 + (32*b^4*tan(x/2))/a + 32*a*b^2*tan(x/2)) + (16*b^4*tan(x/2)*(a^2 + b ^2)^(1/2))/(16*b^5 + 16*a^2*b^3 + 32*a^3*b^2*tan(x/2) + 32*a*b^4*tan(x/2)) )*(a^2 + b^2)^(1/2))/a^3
Time = 0.17 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.70 \[ \int \frac {\sec ^3(x)}{a+b \cot (x)} \, dx=\frac {4 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) b i -a i}{\sqrt {a^{2}+b^{2}}}\right ) \sin \left (x \right )^{2} b i -4 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) b i -a i}{\sqrt {a^{2}+b^{2}}}\right ) b i +2 \cos \left (x \right ) a b -\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )^{2} a^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )^{2} b^{2}+\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) a^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) b^{2}+\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )^{2} a^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )^{2} b^{2}-\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) a^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) b^{2}-\sin \left (x \right ) a^{2}}{2 a^{3} \left (\sin \left (x \right )^{2}-1\right )} \] Input:
int(sec(x)^3/(a+b*cot(x)),x)
Output:
(4*sqrt(a**2 + b**2)*atan((tan(x/2)*b*i - a*i)/sqrt(a**2 + b**2))*sin(x)** 2*b*i - 4*sqrt(a**2 + b**2)*atan((tan(x/2)*b*i - a*i)/sqrt(a**2 + b**2))*b *i + 2*cos(x)*a*b - log(tan(x/2) - 1)*sin(x)**2*a**2 - 2*log(tan(x/2) - 1) *sin(x)**2*b**2 + log(tan(x/2) - 1)*a**2 + 2*log(tan(x/2) - 1)*b**2 + log( tan(x/2) + 1)*sin(x)**2*a**2 + 2*log(tan(x/2) + 1)*sin(x)**2*b**2 - log(ta n(x/2) + 1)*a**2 - 2*log(tan(x/2) + 1)*b**2 - sin(x)*a**2)/(2*a**3*(sin(x) **2 - 1))