Integrand size = 23, antiderivative size = 157 \[ \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{3/2} d}+\frac {2 \tan (c+d x)}{a d \sqrt {a+a \sec (c+d x)}}-\frac {2 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac {2 a \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}+\frac {2 a^2 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}} \]
-2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d+2*tan(d*x+c )/a/d/(a+a*sec(d*x+c))^(1/2)-2/3*tan(d*x+c)^3/d/(a+a*sec(d*x+c))^(3/2)+2/5 *a*tan(d*x+c)^5/d/(a+a*sec(d*x+c))^(5/2)+2/7*a^2*tan(d*x+c)^7/d/(a+a*sec(d *x+c))^(7/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.93 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.58 \[ \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {32 \sqrt {2} \left (\frac {1}{1+\sec (c+d x)}\right )^{11/2} \left (\frac {\cos (c+d x) (11+7 \cos (c+d x)) \csc ^8\left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (105 \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \cos ^3(c+d x)+(76-198 \cos (c+d x)+61 \cos (2 (c+d x))-44 \cos (3 (c+d x))) \sqrt {1-\sec (c+d x)}\right )}{3360 \sqrt {1-\sec (c+d x)}}-\frac {4}{11} \operatorname {Hypergeometric2F1}\left (2,\frac {11}{2},\frac {13}{2},-2 \sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec (c+d x) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^7(c+d x)}{7 d (a (1+\sec (c+d x)))^{3/2} \left (1-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{9/2}} \]
(32*Sqrt[2]*((1 + Sec[c + d*x])^(-1))^(11/2)*((Cos[c + d*x]*(11 + 7*Cos[c + d*x])*Csc[(c + d*x)/2]^8*Sec[(c + d*x)/2]^2*(105*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Cos[c + d*x]^3 + (76 - 198*Cos[c + d*x] + 61*Cos[2*(c + d*x)] - 4 4*Cos[3*(c + d*x)])*Sqrt[1 - Sec[c + d*x]]))/(3360*Sqrt[1 - Sec[c + d*x]]) - (4*Hypergeometric2F1[2, 11/2, 13/2, -2*Sec[c + d*x]*Sin[(c + d*x)/2]^2] *Sec[c + d*x]*Tan[(c + d*x)/2]^2)/11)*Tan[c + d*x]^7)/(7*d*(a*(1 + Sec[c + d*x]))^(3/2)*(1 - Tan[(c + d*x)/2]^2)^(9/2))
Time = 0.29 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4375, 363, 254, 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.
| \(\displaystyle \int \frac {\tan ^6(c+d x)}{(a \sec (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cot \left (c+d x+\frac {\pi }{2}\right )^6}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4375 |
| \(\displaystyle -\frac {2 a^2 \int \frac {\tan ^6(c+d x) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )}{(\sec (c+d x) a+a)^3 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle -\frac {2 a^2 \left (\int \frac {\tan ^6(c+d x)}{(\sec (c+d x) a+a)^3 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )-\frac {\tan ^7(c+d x)}{7 (a \sec (c+d x)+a)^{7/2}}\right )}{d}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle -\frac {2 a^2 \left (\int \left (\frac {\tan ^4(c+d x)}{a (\sec (c+d x) a+a)^2}-\frac {\tan ^2(c+d x)}{a^2 (\sec (c+d x) a+a)}-\frac {1}{a^3 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}+\frac {1}{a^3}\right )d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )-\frac {\tan ^7(c+d x)}{7 (a \sec (c+d x)+a)^{7/2}}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a^2 \left (\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{7/2}}-\frac {\tan (c+d x)}{a^3 \sqrt {a \sec (c+d x)+a}}+\frac {\tan ^3(c+d x)}{3 a^2 (a \sec (c+d x)+a)^{3/2}}-\frac {\tan ^7(c+d x)}{7 (a \sec (c+d x)+a)^{7/2}}-\frac {\tan ^5(c+d x)}{5 a (a \sec (c+d x)+a)^{5/2}}\right )}{d}\) |
(-2*a^2*(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)) - Tan[c + d*x]^7/(7*(a + a*Sec[c + d*x])^(7/2))))/d
3.2.89.3.1 Defintions of rubi rules used
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
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 = 3.74 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.48
| method | result | size |
| default | \(-\frac {\left (105 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {7}{2}}-278 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+1078 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-770 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+210 \csc \left (d x +c \right )-210 \cot \left (d x +c \right )\right ) \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}}{105 d \,a^{2} \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right )^{3} \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+1\right )^{3}}\) | \(233\) |
-1/105/d/a^2*(105*2^(1/2)*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1 )^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(7/2)- 278*(1-cos(d*x+c))^7*csc(d*x+c)^7+1078*(1-cos(d*x+c))^5*csc(d*x+c)^5-770*( 1-cos(d*x+c))^3*csc(d*x+c)^3+210*csc(d*x+c)-210*cot(d*x+c))*(-2*a/((1-cos( d*x+c))^2*csc(d*x+c)^2-1))^(1/2)/(-cot(d*x+c)+csc(d*x+c)-1)^3/(csc(d*x+c)- cot(d*x+c)+1)^3
Time = 0.30 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.18 \[ \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {105 \, {\left (\cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{3}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 2 \, {\left (146 \, \cos \left (d x + c\right )^{3} - 32 \, \cos \left (d x + c\right )^{2} - 24 \, \cos \left (d x + c\right ) + 15\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}}, \frac {2 \, {\left (105 \, {\left (\cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + {\left (146 \, \cos \left (d x + c\right )^{3} - 32 \, \cos \left (d x + c\right )^{2} - 24 \, \cos \left (d x + c\right ) + 15\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{105 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}}\right ] \]
[-1/105*(105*(cos(d*x + c)^4 + cos(d*x + c)^3)*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*(146*cos(d*x + c) ^3 - 32*cos(d*x + c)^2 - 24*cos(d*x + c) + 15)*sqrt((a*cos(d*x + c) + a)/c os(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3), 2/105*(105*(cos(d*x + c)^4 + cos(d*x + c)^3)*sqrt(a)*arctan(sqrt((a*cos(d* x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + (146*cos( d*x + c)^3 - 32*cos(d*x + c)^2 - 24*cos(d*x + c) + 15)*sqrt((a*cos(d*x + c ) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3)]
\[ \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\tan ^{6}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{6}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
1/210*(105*((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*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) - (cos(2*d*x + 2* c)^2 + sin(2*d*x + 2*c)^2 + 2*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^2*d*cos(2*d*x + 2*c)^2 + a^2*d*sin(2* d*x + 2*c)^2 + 2*a^2*d*cos(2*d*x + 2*c) + a^2*d)*integrate(-(cos(2*d*x + 2 *c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*(((cos(14*d*x + 14*c)*cos(2*d*x + 2*c) + 6*cos(12*d*x + 12*c)*cos(2*d*x + 2*c) + 15*cos(1 0*d*x + 10*c)*cos(2*d*x + 2*c) + 20*cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 15 *cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 6*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(14*d*x + 14*c)*sin(2*d*x + 2*c) + 6*sin(12*d*x + 12*c)*sin(2*d*x + 2*c) + 15*sin(10*d*x + 10*c)*sin(2*d*x + 2*c) + 20*sin( 8*d*x + 8*c)*sin(2*d*x + 2*c) + 15*sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 6*s in(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(9/2*arctan2(sin (2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)*sin(14*d*x + 14*c...
Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (137) = 274\).
Time = 3.75 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.89 \[ \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {105 \, \sqrt {-a} {\left (\frac {\log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {\log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} + \frac {2 \, {\left ({\left ({\left (\frac {139 \, \sqrt {2} a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {539 \, \sqrt {2} a^{2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {385 \, \sqrt {2} a^{2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {105 \, \sqrt {2} a^{2}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{105 \, d} \]
1/105*(105*sqrt(-a)*(log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan( 1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2) + 3)))/(a^2*sgn(cos(d*x + c))) - log(abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3)))/(a^2*sgn(cos(d*x + c)))) + 2*(((139*sqrt(2)*a ^2*tan(1/2*d*x + 1/2*c)^2/sgn(cos(d*x + c)) - 539*sqrt(2)*a^2/sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2 + 385*sqrt(2)*a^2/sgn(cos(d*x + c)))*tan(1/2 *d*x + 1/2*c)^2 - 105*sqrt(2)*a^2/sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)/ ((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d
Timed out. \[ \int \frac {\tan ^6(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^6}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]