Integrand size = 26, antiderivative size = 160 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {8 i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{a^{7/2} d}-\frac {2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}}-\frac {4 i \sec ^3(c+d x)}{3 a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac {8 i \sec (c+d x)}{a^3 d \sqrt {a+i a \tan (c+d x)}} \]
8*I*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*2^(1/ 2)/a^(7/2)/d-8*I*sec(d*x+c)/a^3/d/(a+I*a*tan(d*x+c))^(1/2)-2/5*I*sec(d*x+c )^5/a/d/(a+I*a*tan(d*x+c))^(5/2)-4/3*I*sec(d*x+c)^3/a^2/d/(a+I*a*tan(d*x+c ))^(3/2)
Time = 2.05 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.81 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {128 e^{7 i (c+d x)} \left (-23-35 e^{2 i (c+d x)}-15 e^{4 i (c+d x)}+15 \left (1+e^{2 i (c+d x)}\right )^{5/2} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{15 a^3 d \left (1+e^{2 i (c+d x)}\right )^6 (-i+\tan (c+d x))^3 \sqrt {a+i a \tan (c+d x)}} \]
(-128*E^((7*I)*(c + d*x))*(-23 - 35*E^((2*I)*(c + d*x)) - 15*E^((4*I)*(c + d*x)) + 15*(1 + E^((2*I)*(c + d*x)))^(5/2)*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))/(15*a^3*d*(1 + E^((2*I)*(c + d*x)))^6*(-I + Tan[c + d*x])^3*Sqr t[a + I*a*Tan[c + d*x]])
Time = 0.73 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 3972, 3042, 3972, 3042, 3972, 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.
\(\displaystyle \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^7}{(a+i a \tan (c+d x))^{7/2}}dx\) |
\(\Big \downarrow \) 3972 |
\(\displaystyle \frac {2 \int \frac {\sec ^5(c+d x)}{(i \tan (c+d x) a+a)^{5/2}}dx}{a}-\frac {2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \int \frac {\sec (c+d x)^5}{(i \tan (c+d x) a+a)^{5/2}}dx}{a}-\frac {2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3972 |
\(\displaystyle \frac {2 \left (\frac {2 \int \frac {\sec ^3(c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx}{a}-\frac {2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^{3/2}}\right )}{a}-\frac {2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \left (\frac {2 \int \frac {\sec (c+d x)^3}{(i \tan (c+d x) a+a)^{3/2}}dx}{a}-\frac {2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^{3/2}}\right )}{a}-\frac {2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3972 |
\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {2 \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{a}-\frac {2 i \sec (c+d x)}{a d \sqrt {a+i a \tan (c+d x)}}\right )}{a}-\frac {2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^{3/2}}\right )}{a}-\frac {2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {2 \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx}{a}-\frac {2 i \sec (c+d x)}{a d \sqrt {a+i a \tan (c+d x)}}\right )}{a}-\frac {2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^{3/2}}\right )}{a}-\frac {2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3970 |
\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {4 i \int \frac {1}{2-\frac {a \sec ^2(c+d x)}{i \tan (c+d x) a+a}}d\frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}}{a d}-\frac {2 i \sec (c+d x)}{a d \sqrt {a+i a \tan (c+d x)}}\right )}{a}-\frac {2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^{3/2}}\right )}{a}-\frac {2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 \left (\frac {2 \left (\frac {2 i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac {2 i \sec (c+d x)}{a d \sqrt {a+i a \tan (c+d x)}}\right )}{a}-\frac {2 i \sec ^3(c+d x)}{3 a d (a+i a \tan (c+d x))^{3/2}}\right )}{a}-\frac {2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}}\) |
(((-2*I)/5)*Sec[c + d*x]^5)/(a*d*(a + I*a*Tan[c + d*x])^(5/2)) + (2*((((-2 *I)/3)*Sec[c + d*x]^3)/(a*d*(a + I*a*Tan[c + d*x])^(3/2)) + (2*(((2*I)*Sqr t[2]*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])]) /(a^(3/2)*d) - ((2*I)*Sec[c + d*x])/(a*d*Sqrt[a + I*a*Tan[c + d*x]])))/a)) /a
3.4.88.3.1 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[2*d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m - 2))), x] + Simp[2*(d^2/a) 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] && LtQ[n, -1]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (133 ) = 266\).
Time = 10.79 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.96
method | result | size |
default | \(-\frac {2 \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )+i\right )^{7} \left (-60 \sqrt {2}\, \arctan \left (\frac {\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-1\right ) \sqrt {2}}{2 \sqrt {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}}\right ) {\left (\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1\right )}^{\frac {5}{2}}+73 i \left (\csc ^{5}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{5}-190 i \left (\csc ^{3}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{3}-105 \left (\csc ^{4}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{4}+105 i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+190 \left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-73\right )}{15 d {\left (-\frac {a \left (2 i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1\right )}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}\right )}^{\frac {7}{2}} {\left (\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1\right )}^{6}}\) | \(313\) |
-2/15/d*(-csc(d*x+c)+cot(d*x+c)+I)^7*(-60*2^(1/2)*arctan(1/2*(I*(csc(d*x+c )-cot(d*x+c))-1)*2^(1/2)/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)^(1/2))*(csc(d*x +c)^2*(1-cos(d*x+c))^2-1)^(5/2)+73*I*csc(d*x+c)^5*(1-cos(d*x+c))^5-190*I*c sc(d*x+c)^3*(1-cos(d*x+c))^3-105*csc(d*x+c)^4*(1-cos(d*x+c))^4+105*I*(csc( d*x+c)-cot(d*x+c))+190*csc(d*x+c)^2*(1-cos(d*x+c))^2-73)/(-a*(2*I*(csc(d*x +c)-cot(d*x+c))-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)/(csc(d*x+c)^2*(1-cos(d*x+ c))^2-1))^(7/2)/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)^6
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (125) = 250\).
Time = 0.26 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.04 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {4 \, {\left (15 \, \sqrt {2} {\left (i \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{4} d\right )} \sqrt {\frac {1}{a^{7} d^{2}}} \log \left (-\frac {32 \, {\left ({\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3} d}\right ) + 15 \, \sqrt {2} {\left (-i \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{4} d\right )} \sqrt {\frac {1}{a^{7} d^{2}}} \log \left (-\frac {32 \, {\left ({\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3} d}\right ) + 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (15 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 35 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 23 i\right )}\right )}}{15 \, {\left (a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )}} \]
-4/15*(15*sqrt(2)*(I*a^4*d*e^(4*I*d*x + 4*I*c) + 2*I*a^4*d*e^(2*I*d*x + 2* I*c) + I*a^4*d)*sqrt(1/(a^7*d^2))*log(-32*((I*a^3*d*e^(2*I*d*x + 2*I*c) + I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) - I)*e^(-I*d* x - I*c)/(a^3*d)) + 15*sqrt(2)*(-I*a^4*d*e^(4*I*d*x + 4*I*c) - 2*I*a^4*d*e ^(2*I*d*x + 2*I*c) - I*a^4*d)*sqrt(1/(a^7*d^2))*log(-32*((-I*a^3*d*e^(2*I* d*x + 2*I*c) - I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2) ) - I)*e^(-I*d*x - I*c)/(a^3*d)) + 2*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(15*I*e^(4*I*d*x + 4*I*c) + 35*I*e^(2*I*d*x + 2*I*c) + 23*I))/(a^4*d* e^(4*I*d*x + 4*I*c) + 2*a^4*d*e^(2*I*d*x + 2*I*c) + a^4*d)
\[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int \frac {\sec ^{7}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {7}{2}}}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1164 vs. \(2 (125) = 250\).
Time = 0.72 (sec) , antiderivative size = 1164, normalized size of antiderivative = 7.28 \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Too large to display} \]
-2/15*(15*(2*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2* sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d* x + 2*c)^2 + 2*cos(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*co s(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) - 2*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*arctan2((cos(2*d*x + 2*c)^2 + sin (2*d*x + 2*c)^2 + 2*cos(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) + (-I*sqrt(2)*cos(2*d*x + 2*c)^2 - I*sqrt(2)*sin(2*d*x + 2*c)^2 - 2*I*sqrt(2)*cos(2*d*x + 2*c) - I*sqrt(2))*log(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(1/2*arctan2(sin( 2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d *x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), co s(2*d*x + 2*c) + 1))^2 + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*co s(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) + (I*sqrt(2)*cos(2*d*x + 2*c)^2 + I*sqrt(2)*sin(2*d*x + 2*c) ^2 + 2*I*sqrt(2)*cos(2*d*x + 2*c) + I*sqrt(2))*log(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(1/2*arctan2(sin(2*d...
\[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{7}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^7\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}} \,d x \]