Integrand size = 23, antiderivative size = 189 \[ \int \frac {\tan ^6(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}-\frac {2 a \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac {2 a^2 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}+\frac {6 a^3 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}}+\frac {2 a^4 \tan ^9(c+d x)}{9 d (a+a \sec (c+d x))^{9/2}} \]
-2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d/a^(1/2)+2*tan(d*x+c )/d/(a+a*sec(d*x+c))^(1/2)-2/3*a*tan(d*x+c)^3/d/(a+a*sec(d*x+c))^(3/2)+2/5 *a^2*tan(d*x+c)^5/d/(a+a*sec(d*x+c))^(5/2)+6/7*a^3*tan(d*x+c)^7/d/(a+a*sec (d*x+c))^(7/2)+2/9*a^4*tan(d*x+c)^9/d/(a+a*sec(d*x+c))^(9/2)
Time = 14.08 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.88 \[ \int \frac {\tan ^6(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {-5040 \arctan \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {\sec (c+d x)}{(1+\sec (c+d x))^2}} \sqrt {1+\sec (c+d x)}+(901+164 \cos (c+d x)+1004 \cos (2 (c+d x))+68 \cos (3 (c+d x))+383 \cos (4 (c+d x))) \sec ^4(c+d x) \tan (c+d x)}{1260 d \sqrt {a (1+\sec (c+d x))}} \]
(-5040*ArcTan[Tan[(c + d*x)/2]/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]]*Cos[ (c + d*x)/2]^2*Sqrt[Sec[c + d*x]]*Sqrt[Sec[c + d*x]/(1 + Sec[c + d*x])^2]* Sqrt[1 + Sec[c + d*x]] + (901 + 164*Cos[c + d*x] + 1004*Cos[2*(c + d*x)] + 68*Cos[3*(c + d*x)] + 383*Cos[4*(c + d*x)])*Sec[c + d*x]^4*Tan[c + d*x])/ (1260*d*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.30 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4375, 364, 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.
\(\displaystyle \int \frac {\tan ^6(c+d x)}{\sqrt {a \sec (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cot \left (c+d x+\frac {\pi }{2}\right )^6}{\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
\(\Big \downarrow \) 4375 |
\(\displaystyle -\frac {2 a^3 \int \frac {\tan ^6(c+d x) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )^2}{(\sec (c+d x) a+a)^3 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
\(\Big \downarrow \) 364 |
\(\displaystyle -\frac {2 a^3 \int \left (\frac {a \tan ^8(c+d x)}{(\sec (c+d x) a+a)^4}+\frac {3 \tan ^6(c+d x)}{(\sec (c+d x) a+a)^3}+\frac {\tan ^4(c+d x)}{a (\sec (c+d x) a+a)^2}-\frac {\tan ^2(c+d x)}{a^2 (\sec (c+d x) a+a)}-\frac {1}{a^3 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}+\frac {1}{a^3}\right )d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^3 \left (\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{7/2}}-\frac {\tan (c+d x)}{a^3 \sqrt {a \sec (c+d x)+a}}+\frac {\tan ^3(c+d x)}{3 a^2 (a \sec (c+d x)+a)^{3/2}}-\frac {a \tan ^9(c+d x)}{9 (a \sec (c+d x)+a)^{9/2}}-\frac {3 \tan ^7(c+d x)}{7 (a \sec (c+d x)+a)^{7/2}}-\frac {\tan ^5(c+d x)}{5 a (a \sec (c+d x)+a)^{5/2}}\right )}{d}\) |
(-2*a^3*(ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]/a^(7/2) - Tan[c + d*x]/(a^3*Sqrt[a + a*Sec[c + d*x]]) + Tan[c + d*x]^3/(3*a^2*(a + a*Sec[c + d*x])^(3/2)) - Tan[c + d*x]^5/(5*a*(a + a*Sec[c + d*x])^(5/2)) - (3*Tan[c + d*x]^7)/(7*(a + a*Sec[c + d*x])^(7/2)) - (a*Tan[c + d*x]^9)/(9 *(a + a*Sec[c + d*x])^(9/2))))/d
3.2.77.3.1 Defintions of rubi rules used
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _.), x_Symbol] :> Simp[-2*(a^(m/2 + n + 1/2)/d) Subst[Int[x^m*((2 + a*x^2 )^(m/2 + n - 1/2)/(1 + a*x^2)), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x] ]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && I ntegerQ[n - 1/2]
Time = 3.72 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.16
method | result | size |
default | \(-\frac {2 \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (315 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+315 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-383 \sin \left (d x +c \right )-34 \tan \left (d x +c \right )+132 \sec \left (d x +c \right ) \tan \left (d x +c \right )+5 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}-35 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}\right )}{315 d a \left (\cos \left (d x +c \right )+1\right )}\) | \(220\) |
-2/315/d/a*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*(315*(-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)+315*(-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))-383*sin(d*x+c)- 34*tan(d*x+c)+132*sec(d*x+c)*tan(d*x+c)+5*tan(d*x+c)*sec(d*x+c)^2-35*tan(d *x+c)*sec(d*x+c)^3)
Time = 0.30 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.88 \[ \int \frac {\tan ^6(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [-\frac {315 \, {\left (\cos \left (d x + c\right )^{5} + \cos \left (d x + c\right )^{4}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 2 \, {\left (383 \, \cos \left (d x + c\right )^{4} + 34 \, \cos \left (d x + c\right )^{3} - 132 \, \cos \left (d x + c\right )^{2} - 5 \, \cos \left (d x + c\right ) + 35\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}}, \frac {2 \, {\left (315 \, {\left (\cos \left (d x + c\right )^{5} + \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + {\left (383 \, \cos \left (d x + c\right )^{4} + 34 \, \cos \left (d x + c\right )^{3} - 132 \, \cos \left (d x + c\right )^{2} - 5 \, \cos \left (d x + c\right ) + 35\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{315 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}}\right ] \]
[-1/315*(315*(cos(d*x + c)^5 + cos(d*x + c)^4)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*si n(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) - 2*(383*cos(d*x + c) ^4 + 34*cos(d*x + c)^3 - 132*cos(d*x + c)^2 - 5*cos(d*x + c) + 35)*sqrt((a *cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c)^5 + a*d*c os(d*x + c)^4), 2/315*(315*(cos(d*x + c)^5 + cos(d*x + c)^4)*sqrt(a)*arcta n(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + (383*cos(d*x + c)^4 + 34*cos(d*x + c)^3 - 132*cos(d*x + c)^2 - 5*co s(d*x + c) + 35)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a* d*cos(d*x + c)^5 + a*d*cos(d*x + c)^4)]
\[ \int \frac {\tan ^6(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\tan ^{6}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]
\[ \int \frac {\tan ^6(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{6}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \]
-1/630*(12*(105*sin(8*d*x + 8*c) + 280*sin(6*d*x + 6*c) + 546*sin(4*d*x + 4*c) + 312*sin(2*d*x + 2*c))*cos(9/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 4*(315*cos(8*d*x + 8*c) + 840*cos(6*d*x + 6*c) + 1638*cos(4* d*x + 4*c) + 936*cos(2*d*x + 2*c) + 383)*sin(9/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 315*((cos(2*d*x + 2*c)^4 + sin(2*d*x + 2*c)^4 + 4*cos(2*d*x + 2*c)^3 + 2*(cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin (2*d*x + 2*c)^2 + 6*cos(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*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)*si n(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) - (cos(2*d*x + 2*c)^4 + sin(2*d *x + 2*c)^4 + 4*cos(2*d*x + 2*c)^3 + 2*(cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c)^2 + 6*cos(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*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) + 2*(a*d*cos(2*d* x + 2*c)^4 + a*d*sin(2*d*x + 2*c)^4 + 4*a*d*cos(2*d*x + 2*c)^3 + 6*a*d*cos (2*d*x + 2*c)^2 + 4*a*d*cos(2*d*x + 2*c) + 2*(a*d*cos(2*d*x + 2*c)^2 + 2*a *d*cos(2*d*x + 2*c) + a*d)*sin(2*d*x + 2*c)^2 + a*d)*integrate(-(cos(2*...
Time = 2.84 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.41 \[ \int \frac {\tan ^6(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} {\left (\frac {315 \, \sqrt {2} \sqrt {-a} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right )}{{\left | a \right |}} + \frac {4 \, {\left (315 \, a^{4} - {\left (1470 \, a^{4} - {\left (2772 \, a^{4} + {\left (257 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1314 \, a^{4}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )}}{630 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \]
1/630*sqrt(2)*(315*sqrt(2)*sqrt(-a)*log(abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2* c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs( 2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*sqrt(2)*abs(a) - 6*a))/abs(a) + 4*(315*a^4 - (1470*a^4 - (2772*a^4 + ( 257*a^4*tan(1/2*d*x + 1/2*c)^2 - 1314*a^4)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2 *d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2* d*x + 1/2*c)^2 - a)^4*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(d*sgn(cos(d*x + c)))
Timed out. \[ \int \frac {\tan ^6(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^6}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]