Integrand size = 26, antiderivative size = 146 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {5 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} \sqrt {a} d}+\frac {5 i a}{12 d (a+i a \tan (c+d x))^{3/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac {5 i}{8 d \sqrt {a+i a \tan (c+d x)}} \]
-5/16*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2)/a^ (1/2)+5/8*I/d/(a+I*a*tan(d*x+c))^(1/2)+5/12*I*a/d/(a+I*a*tan(d*x+c))^(3/2) -1/2*I*a^2/d/(a-I*a*tan(d*x+c))/(a+I*a*tan(d*x+c))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.14 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.35 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i a \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2,-\frac {1}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{6 d (a+i a \tan (c+d x))^{3/2}} \]
((I/6)*a*Hypergeometric2F1[-3/2, 2, -1/2, (1 + I*Tan[c + d*x])/2])/(d*(a + I*a*Tan[c + d*x])^(3/2))
Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3042, 3968, 52, 61, 61, 73, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (c+d x)^2 \sqrt {a+i a \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 3968 |
| \(\displaystyle -\frac {i a^3 \int \frac {1}{(a-i a \tan (c+d x))^2 (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {i a^3 \left (\frac {5 \int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}\right )}{d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {i a^3 \left (\frac {5 \left (\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{3/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}\right )}{d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {i a^3 \left (\frac {5 \left (\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}\right )}{d}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {i a^3 \left (\frac {5 \left (\frac {\frac {\int \frac {1}{a^2 \tan ^2(c+d x)+2 a}d\sqrt {i \tan (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}\right )}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {i a^3 \left (\frac {5 \left (\frac {\frac {i \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}\right )}{d}\) |
((-I)*a^3*(1/(2*a*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(3/2)) + ( 5*(-1/3*1/(a*(a + I*a*Tan[c + d*x])^(3/2)) + ((I*ArcTan[(Sqrt[a]*Tan[c + d *x])/Sqrt[2]])/(Sqrt[2]*a^(3/2)) - 1/(a*Sqrt[a + I*a*Tan[c + d*x]]))/(2*a) ))/(4*a)))/d
3.4.37.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 405 vs. \(2 (114 ) = 228\).
Time = 10.67 (sec) , antiderivative size = 406, normalized size of antiderivative = 2.78
| method | result | size |
| default | \(-\frac {4 i \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+4 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-20 \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-15 i \cos \left (d x +c \right ) \arctan \left (\frac {\cos \left (d x +c \right )+1+i \sin \left (d x +c \right )}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-30 i \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-20 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )-15 i \arctan \left (\frac {\cos \left (d x +c \right )+1+i \sin \left (d x +c \right )}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-30 i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+15 \arctan \left (\frac {\cos \left (d x +c \right )+1+i \sin \left (d x +c \right )}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sin \left (d x +c \right )}{48 d \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\) | \(406\) |
-1/48/d*(4*I*cos(d*x+c)^3*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+4*I*cos(d*x+c )^2*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-20*sin(d*x+c)*cos(d*x+c)^2*(-cos(d* x+c)/(cos(d*x+c)+1))^(1/2)-15*I*cos(d*x+c)*arctan(1/2*(cos(d*x+c)+1+I*sin( d*x+c))/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-30*I*cos(d*x+c) *(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-20*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)* cos(d*x+c)*sin(d*x+c)-15*I*arctan(1/2*(cos(d*x+c)+1+I*sin(d*x+c))/(cos(d*x +c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-30*I*(-cos(d*x+c)/(cos(d*x+c)+1 ))^(1/2)+15*arctan(1/2*(cos(d*x+c)+1+I*sin(d*x+c))/(cos(d*x+c)+1)/(-cos(d* x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c))/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2) /(cos(d*x+c)+1)/(a*(1+I*tan(d*x+c)))^(1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 271 vs. \(2 (105) = 210\).
Time = 0.25 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.86 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\left (-15 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 15 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-3 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 11 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{48 \, a d} \]
1/48*(-15*I*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(3*I*d*x + 3*I*c)*log(4*(sqrt( 2)*sqrt(1/2)*(a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 15*I*sqrt(1/ 2)*a*d*sqrt(1/(a*d^2))*e^(3*I*d*x + 3*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(a*d* e^(2*I*d*x + 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2 )) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2 *I*c) + 1))*(-3*I*e^(6*I*d*x + 6*I*c) + 11*I*e^(4*I*d*x + 4*I*c) + 16*I*e^ (2*I*d*x + 2*I*c) + 2*I))*e^(-3*I*d*x - 3*I*c)/(a*d)
\[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
Time = 0.32 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i \, {\left (15 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a - 20 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - 8 \, a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a}\right )}}{96 \, a d} \]
1/96*I*(15*sqrt(2)*sqrt(a)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) + 4*(15*(I*a*tan(d*x + c) + a)^2*a - 20*(I*a*tan(d*x + c) + a)*a^2 - 8*a^3)/((I*a*tan(d*x + c) + a)^(5/2) - 2*(I*a*tan(d*x + c) + a)^(3/2)*a))/(a*d)
\[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]