Integrand size = 26, antiderivative size = 196 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {i a^{7/2} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{8 \sqrt {2} d}-\frac {i a^3 \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {i a^2 \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{12 d}-\frac {i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{10 d}-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2}}{7 d} \]
1/16*I*a^(7/2)*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^( 1/2))/d*2^(1/2)-1/8*I*a^3*cos(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-1/12*I*a^2 *cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(3/2)/d-1/10*I*a*cos(d*x+c)^5*(a+I*a*tan( d*x+c))^(5/2)/d-1/7*I*cos(d*x+c)^7*(a+I*a*tan(d*x+c))^(7/2)/d
Time = 2.41 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.67 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {i a^3 e^{-i (c+d x)} \left (176+298 e^{2 i (c+d x)}+188 e^{4 i (c+d x)}+81 e^{6 i (c+d x)}+15 e^{8 i (c+d x)}-105 \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {a+i a \tan (c+d x)}}{1680 d} \]
((-1/1680*I)*a^3*(176 + 298*E^((2*I)*(c + d*x)) + 188*E^((4*I)*(c + d*x)) + 81*E^((6*I)*(c + d*x)) + 15*E^((8*I)*(c + d*x)) - 105*Sqrt[1 + E^((2*I)* (c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])*Sqrt[a + I*a*Tan[c + d *x]])/(d*E^(I*(c + d*x)))
Time = 0.85 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {3042, 3971, 3042, 3971, 3042, 3971, 3042, 3971, 3042, 3970, 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.
\(\displaystyle \int \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{7/2}}{\sec (c+d x)^7}dx\) |
\(\Big \downarrow \) 3971 |
\(\displaystyle \frac {1}{2} a \int \cos ^5(c+d x) (i \tan (c+d x) a+a)^{5/2}dx-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} a \int \frac {(i \tan (c+d x) a+a)^{5/2}}{\sec (c+d x)^5}dx-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2}}{7 d}\) |
\(\Big \downarrow \) 3971 |
\(\displaystyle \frac {1}{2} a \left (\frac {1}{2} a \int \cos ^3(c+d x) (i \tan (c+d x) a+a)^{3/2}dx-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} a \left (\frac {1}{2} a \int \frac {(i \tan (c+d x) a+a)^{3/2}}{\sec (c+d x)^3}dx-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2}}{7 d}\) |
\(\Big \downarrow \) 3971 |
\(\displaystyle \frac {1}{2} a \left (\frac {1}{2} a \left (\frac {1}{2} a \int \cos (c+d x) \sqrt {i \tan (c+d x) a+a}dx-\frac {i \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} a \left (\frac {1}{2} a \left (\frac {1}{2} a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sec (c+d x)}dx-\frac {i \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2}}{7 d}\) |
\(\Big \downarrow \) 3971 |
\(\displaystyle \frac {1}{2} a \left (\frac {1}{2} a \left (\frac {1}{2} a \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {i \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} a \left (\frac {1}{2} a \left (\frac {1}{2} a \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {i \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2}}{7 d}\) |
\(\Big \downarrow \) 3970 |
\(\displaystyle \frac {1}{2} a \left (\frac {1}{2} a \left (\frac {1}{2} a \left (\frac {i a \int \frac {1}{2-\frac {a \sec ^2(c+d x)}{i \tan (c+d x) a+a}}d\frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {i \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2}}{7 d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} a \left (\frac {1}{2} a \left (\frac {1}{2} a \left (\frac {i \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {2} d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {i \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}\right )-\frac {i \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2}}{7 d}\) |
((-1/7*I)*Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^(7/2))/d + (a*(((-1/5*I)*C os[c + d*x]^5*(a + I*a*Tan[c + d*x])^(5/2))/d + (a*(((-1/3*I)*Cos[c + d*x] ^3*(a + I*a*Tan[c + d*x])^(3/2))/d + (a*((I*Sqrt[a]*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) - (I*Cos[c + d *x]*Sqrt[a + I*a*Tan[c + d*x]])/d))/2))/2))/2
3.4.30.3.1 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[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_S ymbol] :> Simp[-2*(a/(b*f)) Subst[Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/ Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^2, 0 ]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/ (a*f*m)), x] + Simp[a/(2*d^2) Int[(d*Sec[e + f*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && EqQ[m/2 + n, 0] && GtQ[n, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1184 vs. \(2 (159 ) = 318\).
Time = 3.88 (sec) , antiderivative size = 1185, normalized size of antiderivative = 6.05
\[\text {Expression too large to display}\]
1/1680*I/d*(tan(d*x+c)-I)^3*(a*(1+I*tan(d*x+c)))^(1/2)*a^3*cos(d*x+c)^3*(- 315*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1 ))^(1/2))*cos(d*x+c)-420*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin( d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)*sin(d *x+c)-770*I*cos(d*x+c)^2+840*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh( sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^3 *sin(d*x+c)-840*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos (d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^4+840*(-cos(d*x+ c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d* x+c)^3*sin(d*x+c)+105*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d* x+c)/(cos(d*x+c)+1))^(1/2))+1528*I*cos(d*x+c)^4-105*I*(-cos(d*x+c)/(cos(d* x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+ 1))^(1/2))*sin(d*x+c)-420*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d *x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^3+420* (-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/ 2))*cos(d*x+c)^2*sin(d*x+c)+420*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arcta nh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c )^2*sin(d*x+c)-840*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos(d*x+c )/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2+840*arctanh(sin(d*x+c)/(cos(d*x+c)+1 )/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2...
Time = 0.26 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.32 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {105 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d \log \left (\frac {{\left (i \, a^{4} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, d}\right ) - 105 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d \log \left (\frac {{\left (i \, a^{4} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, d}\right ) + \sqrt {2} {\left (-15 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 81 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 188 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 298 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 176 i \, a^{3}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{1680 \, d} \]
1/1680*(105*sqrt(1/2)*sqrt(-a^7/d^2)*d*log(1/4*(I*a^4 + sqrt(2)*sqrt(1/2)* sqrt(-a^7/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1 )))*e^(-I*d*x - I*c)/d) - 105*sqrt(1/2)*sqrt(-a^7/d^2)*d*log(1/4*(I*a^4 - sqrt(2)*sqrt(1/2)*sqrt(-a^7/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2* I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/d) + sqrt(2)*(-15*I*a^3*e^(8*I*d*x + 8*I*c) - 81*I*a^3*e^(6*I*d*x + 6*I*c) - 188*I*a^3*e^(4*I*d*x + 4*I*c) - 298*I*a^3*e^(2*I*d*x + 2*I*c) - 176*I*a^3)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1 )))/d
Timed out. \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\text {Timed out} \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1253 vs. \(2 (149) = 298\).
Time = 0.70 (sec) , antiderivative size = 1253, normalized size of antiderivative = 6.39 \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\text {Too large to display} \]
-1/6720*(20*(7*I*sqrt(2)*a^3*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 7*sqrt(2)*a^3*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 3*(I*sqrt(2)*a^3*cos(2*d*x + 2*c)^2 + I*sqrt(2)*a^3*sin(2*d*x + 2*c)^2 + 2*I*sqrt(2)*a^3*cos(2*d*x + 2*c) + I*sqrt(2)*a^3)*cos(7/2*arct an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 3*(sqrt(2)*a^3*cos(2*d*x + 2*c)^2 + sqrt(2)*a^3*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*a^3*cos(2*d*x + 2*c) + sqrt(2)*a^3)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*(c os(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*sqr t(a) + 84*(5*I*sqrt(2)*a^3*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2 *c) + 1)) - 5*sqrt(2)*a^3*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2* c) + 1)) + (I*sqrt(2)*a^3*cos(2*d*x + 2*c)^2 + I*sqrt(2)*a^3*sin(2*d*x + 2 *c)^2 + 2*I*sqrt(2)*a^3*cos(2*d*x + 2*c) + I*sqrt(2)*a^3)*cos(5/2*arctan2( sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (sqrt(2)*a^3*cos(2*d*x + 2*c)^2 + sqrt(2)*a^3*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*a^3*cos(2*d*x + 2*c) + sqrt( 2)*a^3)*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*(cos(2*d *x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + 105*(2*sqrt(2)*a^3*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*c os(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2 *c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1)...
\[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \cos \left (d x + c\right )^{7} \,d x } \]
Timed out. \[ \int \cos ^7(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\int {\cos \left (c+d\,x\right )}^7\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2} \,d x \]