Integrand size = 23, antiderivative size = 83 \[ \int \frac {\cos ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {(a-3 b) x}{2 (a-b)^2}+\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 (a-b) d} \]
1/2*(a-3*b)*x/(a-b)^2+1/2*cos(d*x+c)*sin(d*x+c)/(a-b)/d+b^(3/2)*arctan(b^( 1/2)*tan(d*x+c)/a^(1/2))/(a-b)^2/d/a^(1/2)
Time = 0.68 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {4 b^{3/2} \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )+\sqrt {a} (2 (a-3 b) (c+d x)+(a-b) \sin (2 (c+d x)))}{4 \sqrt {a} (a-b)^2 d} \]
(4*b^(3/2)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]] + Sqrt[a]*(2*(a - 3*b)*( c + d*x) + (a - b)*Sin[2*(c + d*x)]))/(4*Sqrt[a]*(a - b)^2*d)
Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.24, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4158, 316, 25, 397, 216, 218}
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.
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (c+d x)^2 \left (a+b \tan (c+d x)^2\right )}dx\) |
| \(\Big \downarrow \) 4158 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(c+d x)+1\right )^2 \left (b \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right )}-\frac {\int -\frac {b \tan ^2(c+d x)+a-2 b}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{2 (a-b)}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {b \tan ^2(c+d x)+a-2 b}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{2 (a-b)}+\frac {\tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right )}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {2 b^2 \int \frac {1}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}+\frac {(a-3 b) \int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)}{a-b}}{2 (a-b)}+\frac {\tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right )}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {2 b^2 \int \frac {1}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}+\frac {(a-3 b) \arctan (\tan (c+d x))}{a-b}}{2 (a-b)}+\frac {\tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right )}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)}+\frac {(a-3 b) \arctan (\tan (c+d x))}{a-b}}{2 (a-b)}+\frac {\tan (c+d x)}{2 (a-b) \left (\tan ^2(c+d x)+1\right )}}{d}\) |
((((a - 3*b)*ArcTan[Tan[c + d*x]])/(a - b) + (2*b^(3/2)*ArcTan[(Sqrt[b]*Ta n[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)))/(2*(a - b)) + Tan[c + d*x]/(2*(a - b)*(1 + Tan[c + d*x]^2)))/d
3.5.60.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[ff/(c^(m - 1)*f) Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)^n)^ p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && I ntegerQ[m/2] && (IntegersQ[n, p] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Time = 1.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\left (\frac {a}{2}-\frac {b}{2}\right ) \tan \left (d x +c \right )}{1+\tan \left (d x +c \right )^{2}}+\frac {\left (a -3 b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{\left (a -b \right )^{2}}+\frac {b^{2} \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{\left (a -b \right )^{2} \sqrt {a b}}}{d}\) | \(85\) |
| default | \(\frac {\frac {\frac {\left (\frac {a}{2}-\frac {b}{2}\right ) \tan \left (d x +c \right )}{1+\tan \left (d x +c \right )^{2}}+\frac {\left (a -3 b \right ) \arctan \left (\tan \left (d x +c \right )\right )}{2}}{\left (a -b \right )^{2}}+\frac {b^{2} \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{\left (a -b \right )^{2} \sqrt {a b}}}{d}\) | \(85\) |
| risch | \(\frac {x a}{2 \left (a -b \right )^{2}}-\frac {3 x b}{2 \left (a -b \right )^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 \left (a -b \right ) d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 \left (a -b \right ) d}+\frac {\sqrt {-a b}\, b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{2 a \left (a -b \right )^{2} d}-\frac {\sqrt {-a b}\, b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{2 a \left (a -b \right )^{2} d}\) | \(179\) |
1/d*(1/(a-b)^2*((1/2*a-1/2*b)*tan(d*x+c)/(1+tan(d*x+c)^2)+1/2*(a-3*b)*arct an(tan(d*x+c)))+b^2/(a-b)^2/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2)))
Time = 0.30 (sec) , antiderivative size = 290, normalized size of antiderivative = 3.49 \[ \int \frac {\cos ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\left [\frac {2 \, {\left (a - 3 \, b\right )} d x + 2 \, {\left (a - b\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \sqrt {-\frac {b}{a}} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (a^{2} + a b\right )} \cos \left (d x + c\right )^{3} - a b \cos \left (d x + c\right )\right )} \sqrt {-\frac {b}{a}} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} d}, \frac {{\left (a - 3 \, b\right )} d x + {\left (a - b\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - b \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {\frac {b}{a}}}{2 \, b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} d}\right ] \]
[1/4*(2*(a - 3*b)*d*x + 2*(a - b)*cos(d*x + c)*sin(d*x + c) + b*sqrt(-b/a) *log(((a^2 + 6*a*b + b^2)*cos(d*x + c)^4 - 2*(3*a*b + b^2)*cos(d*x + c)^2 - 4*((a^2 + a*b)*cos(d*x + c)^3 - a*b*cos(d*x + c))*sqrt(-b/a)*sin(d*x + c ) + b^2)/((a^2 - 2*a*b + b^2)*cos(d*x + c)^4 + 2*(a*b - b^2)*cos(d*x + c)^ 2 + b^2)))/((a^2 - 2*a*b + b^2)*d), 1/2*((a - 3*b)*d*x + (a - b)*cos(d*x + c)*sin(d*x + c) - b*sqrt(b/a)*arctan(1/2*((a + b)*cos(d*x + c)^2 - b)*sqr t(b/a)/(b*cos(d*x + c)*sin(d*x + c))))/((a^2 - 2*a*b + b^2)*d)]
Timed out. \[ \int \frac {\cos ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\text {Timed out} \]
Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.14 \[ \int \frac {\cos ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\frac {2 \, b^{2} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a b}} + \frac {{\left (d x + c\right )} {\left (a - 3 \, b\right )}}{a^{2} - 2 \, a b + b^{2}} + \frac {\tan \left (d x + c\right )}{{\left (a - b\right )} \tan \left (d x + c\right )^{2} + a - b}}{2 \, d} \]
1/2*(2*b^2*arctan(b*tan(d*x + c)/sqrt(a*b))/((a^2 - 2*a*b + b^2)*sqrt(a*b) ) + (d*x + c)*(a - 3*b)/(a^2 - 2*a*b + b^2) + tan(d*x + c)/((a - b)*tan(d* x + c)^2 + a - b))/d
Time = 0.50 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.33 \[ \int \frac {\cos ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )} b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {a b}} + \frac {{\left (d x + c\right )} {\left (a - 3 \, b\right )}}{a^{2} - 2 \, a b + b^{2}} + \frac {\tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )} {\left (a - b\right )}}}{2 \, d} \]
1/2*(2*(pi*floor((d*x + c)/pi + 1/2)*sgn(b) + arctan(b*tan(d*x + c)/sqrt(a *b)))*b^2/((a^2 - 2*a*b + b^2)*sqrt(a*b)) + (d*x + c)*(a - 3*b)/(a^2 - 2*a *b + b^2) + tan(d*x + c)/((tan(d*x + c)^2 + 1)*(a - b)))/d
Time = 13.92 (sec) , antiderivative size = 254, normalized size of antiderivative = 3.06 \[ \int \frac {\cos ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=-\frac {6\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (c+d\,x\right )}{\cos \left (c+d\,x\right )}\right )-a^2\,\sin \left (2\,c+2\,d\,x\right )-2\,a^2\,\mathrm {atan}\left (\frac {\sin \left (c+d\,x\right )}{\cos \left (c+d\,x\right )}\right )+a\,b\,\sin \left (2\,c+2\,d\,x\right )+\mathrm {atan}\left (\frac {a^2\,b^3\,\sin \left (c+d\,x\right )\,\sqrt {-a\,b^3}\,9{}\mathrm {i}-a^3\,b^2\,\sin \left (c+d\,x\right )\,\sqrt {-a\,b^3}\,6{}\mathrm {i}-a\,b^4\,\sin \left (c+d\,x\right )\,\sqrt {-a\,b^3}\,4{}\mathrm {i}+a^4\,b\,\sin \left (c+d\,x\right )\,\sqrt {-a\,b^3}\,1{}\mathrm {i}}{-\cos \left (c+d\,x\right )\,a^5\,b^2+6\,\cos \left (c+d\,x\right )\,a^4\,b^3-9\,\cos \left (c+d\,x\right )\,a^3\,b^4+4\,\cos \left (c+d\,x\right )\,a^2\,b^5}\right )\,\sqrt {-a\,b^3}\,4{}\mathrm {i}}{4\,d\,a^3-8\,d\,a^2\,b+4\,d\,a\,b^2} \]
-(atan((a^2*b^3*sin(c + d*x)*(-a*b^3)^(1/2)*9i - a^3*b^2*sin(c + d*x)*(-a* b^3)^(1/2)*6i - a*b^4*sin(c + d*x)*(-a*b^3)^(1/2)*4i + a^4*b*sin(c + d*x)* (-a*b^3)^(1/2)*1i)/(4*a^2*b^5*cos(c + d*x) - 9*a^3*b^4*cos(c + d*x) + 6*a^ 4*b^3*cos(c + d*x) - a^5*b^2*cos(c + d*x)))*(-a*b^3)^(1/2)*4i - 2*a^2*atan (sin(c + d*x)/cos(c + d*x)) - a^2*sin(2*c + 2*d*x) + 6*a*b*atan(sin(c + d* x)/cos(c + d*x)) + a*b*sin(2*c + 2*d*x))/(4*a^3*d + 4*a*b^2*d - 8*a^2*b*d)