\(\int (a+a \sec (c+d x))^{5/2} \tan ^6(c+d x) \, dx\) [165]

Optimal result

Integrand size = 23, antiderivative size = 290 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^6(c+d x) \, dx=-\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a^3 \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}-\frac {2 a^4 \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac {2 a^5 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}+\frac {62 a^6 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}}+\frac {98 a^7 \tan ^9(c+d x)}{9 d (a+a \sec (c+d x))^{9/2}}+\frac {62 a^8 \tan ^{11}(c+d x)}{11 d (a+a \sec (c+d x))^{11/2}}+\frac {18 a^9 \tan ^{13}(c+d x)}{13 d (a+a \sec (c+d x))^{13/2}}+\frac {2 a^{10} \tan ^{15}(c+d x)}{15 d (a+a \sec (c+d x))^{15/2}} \] Output:

-2*a^(5/2)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2*a^3*tan(d 
*x+c)/d/(a+a*sec(d*x+c))^(1/2)-2/3*a^4*tan(d*x+c)^3/d/(a+a*sec(d*x+c))^(3/ 
2)+2/5*a^5*tan(d*x+c)^5/d/(a+a*sec(d*x+c))^(5/2)+62/7*a^6*tan(d*x+c)^7/d/( 
a+a*sec(d*x+c))^(7/2)+98/9*a^7*tan(d*x+c)^9/d/(a+a*sec(d*x+c))^(9/2)+62/11 
*a^8*tan(d*x+c)^11/d/(a+a*sec(d*x+c))^(11/2)+18/13*a^9*tan(d*x+c)^13/d/(a+ 
a*sec(d*x+c))^(13/2)+2/15*a^10*tan(d*x+c)^15/d/(a+a*sec(d*x+c))^(15/2)
 

Mathematica [A] (verified)

Time = 8.49 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.60 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^6(c+d x) \, dx=\frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^7(c+d x) \sqrt {a (1+\sec (c+d x))} \left (-2882880 \sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {15}{2}}(c+d x)+604890 \sin \left (\frac {1}{2} (c+d x)\right )-87230 \sin \left (\frac {3}{2} (c+d x)\right )+450450 \sin \left (\frac {5}{2} (c+d x)\right )-137670 \sin \left (\frac {7}{2} (c+d x)\right )+210210 \sin \left (\frac {9}{2} (c+d x)\right )+75450 \sin \left (\frac {11}{2} (c+d x)\right )+90090 \sin \left (\frac {13}{2} (c+d x)\right )+16066 \sin \left (\frac {15}{2} (c+d x)\right )\right )}{2882880 d} \] Input:

Integrate[(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x]^6,x]
 

Output:

(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]^7*Sqrt[a*(1 + Sec[c + d*x])]*(-2882880* 
Sqrt[2]*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[c + d*x]^(15/2) + 604890*Sin[ 
(c + d*x)/2] - 87230*Sin[(3*(c + d*x))/2] + 450450*Sin[(5*(c + d*x))/2] - 
137670*Sin[(7*(c + d*x))/2] + 210210*Sin[(9*(c + d*x))/2] + 75450*Sin[(11* 
(c + d*x))/2] + 90090*Sin[(13*(c + d*x))/2] + 16066*Sin[(15*(c + d*x))/2]) 
)/(2882880*d)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.91, 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.

\(\displaystyle \int \tan ^6(c+d x) (a \sec (c+d x)+a)^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cot \left (c+d x+\frac {\pi }{2}\right )^6 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}dx\)

\(\Big \downarrow \) 4375

\(\displaystyle -\frac {2 a^6 \int \frac {\tan ^6(c+d x) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )^5}{(\sec (c+d x) a+a)^3 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\)

\(\Big \downarrow \) 364

\(\displaystyle -\frac {2 a^6 \int \left (\frac {a^4 \tan ^{14}(c+d x)}{(\sec (c+d x) a+a)^7}+\frac {9 a^3 \tan ^{12}(c+d x)}{(\sec (c+d x) a+a)^6}+\frac {31 a^2 \tan ^{10}(c+d x)}{(\sec (c+d x) a+a)^5}+\frac {49 a \tan ^8(c+d x)}{(\sec (c+d x) a+a)^4}+\frac {31 \tan ^6(c+d x)}{(\sec (c+d x) a+a)^3}+\frac {\tan ^4(c+d x)}{a (\sec (c+d x) a+a)^2}-\frac {\tan ^2(c+d x)}{a^2 (\sec (c+d x) a+a)}-\frac {1}{a^3 \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right )}+\frac {1}{a^3}\right )d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 a^6 \left (\frac {\arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{7/2}}-\frac {a^4 \tan ^{15}(c+d x)}{15 (a \sec (c+d x)+a)^{15/2}}-\frac {9 a^3 \tan ^{13}(c+d x)}{13 (a \sec (c+d x)+a)^{13/2}}-\frac {\tan (c+d x)}{a^3 \sqrt {a \sec (c+d x)+a}}-\frac {31 a^2 \tan ^{11}(c+d x)}{11 (a \sec (c+d x)+a)^{11/2}}+\frac {\tan ^3(c+d x)}{3 a^2 (a \sec (c+d x)+a)^{3/2}}-\frac {49 a \tan ^9(c+d x)}{9 (a \sec (c+d x)+a)^{9/2}}-\frac {31 \tan ^7(c+d x)}{7 (a \sec (c+d x)+a)^{7/2}}-\frac {\tan ^5(c+d x)}{5 a (a \sec (c+d x)+a)^{5/2}}\right )}{d}\)

Input:

Int[(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x]^6,x]
 

Output:

(-2*a^6*(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)) - 
 (31*Tan[c + d*x]^7)/(7*(a + a*Sec[c + d*x])^(7/2)) - (49*a*Tan[c + d*x]^9 
)/(9*(a + a*Sec[c + d*x])^(9/2)) - (31*a^2*Tan[c + d*x]^11)/(11*(a + a*Sec 
[c + d*x])^(11/2)) - (9*a^3*Tan[c + d*x]^13)/(13*(a + a*Sec[c + d*x])^(13/ 
2)) - (a^4*Tan[c + d*x]^15)/(15*(a + a*Sec[c + d*x])^(15/2))))/d
 

Defintions of rubi rules used

rule 364

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])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4375

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]
 

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.69

\[\frac {2 a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (16066 \cos \left (d x +c \right )^{7}+53078 \cos \left (d x +c \right )^{6}+17286 \cos \left (d x +c \right )^{5}-30640 \cos \left (d x +c \right )^{4}-26810 \cos \left (d x +c \right )^{3}+2898 \cos \left (d x +c \right )^{2}+10164 \cos \left (d x +c \right )+3003\right ) \tan \left (d x +c \right ) \sec \left (d x +c \right )^{6}-45045 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )\right )}{45045 d \left (1+\cos \left (d x +c \right )\right )}\]

Input:

int((a+a*sec(d*x+c))^(5/2)*tan(d*x+c)^6,x)
 

Output:

2/45045/d*a^2*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))*((16066*cos(d*x+c)^7 
+53078*cos(d*x+c)^6+17286*cos(d*x+c)^5-30640*cos(d*x+c)^4-26810*cos(d*x+c) 
^3+2898*cos(d*x+c)^2+10164*cos(d*x+c)+3003)*tan(d*x+c)*sec(d*x+c)^6-45045* 
(1+cos(d*x+c))*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(2^(1/2)/(cot(d*x 
+c)^2-2*csc(d*x+c)*cot(d*x+c)+csc(d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c) 
)))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 477, normalized size of antiderivative = 1.64 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^6(c+d x) \, dx=\left [\frac {45045 \, {\left (a^{2} \cos \left (d x + c\right )^{8} + a^{2} \cos \left (d x + c\right )^{7}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (16066 \, a^{2} \cos \left (d x + c\right )^{7} + 53078 \, a^{2} \cos \left (d x + c\right )^{6} + 17286 \, a^{2} \cos \left (d x + c\right )^{5} - 30640 \, a^{2} \cos \left (d x + c\right )^{4} - 26810 \, a^{2} \cos \left (d x + c\right )^{3} + 2898 \, a^{2} \cos \left (d x + c\right )^{2} + 10164 \, a^{2} \cos \left (d x + c\right ) + 3003 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{45045 \, {\left (d \cos \left (d x + c\right )^{8} + d \cos \left (d x + c\right )^{7}\right )}}, \frac {2 \, {\left (45045 \, {\left (a^{2} \cos \left (d x + c\right )^{8} + a^{2} \cos \left (d x + c\right )^{7}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + {\left (16066 \, a^{2} \cos \left (d x + c\right )^{7} + 53078 \, a^{2} \cos \left (d x + c\right )^{6} + 17286 \, a^{2} \cos \left (d x + c\right )^{5} - 30640 \, a^{2} \cos \left (d x + c\right )^{4} - 26810 \, a^{2} \cos \left (d x + c\right )^{3} + 2898 \, a^{2} \cos \left (d x + c\right )^{2} + 10164 \, a^{2} \cos \left (d x + c\right ) + 3003 \, a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{45045 \, {\left (d \cos \left (d x + c\right )^{8} + d \cos \left (d x + c\right )^{7}\right )}}\right ] \] Input:

integrate((a+a*sec(d*x+c))^(5/2)*tan(d*x+c)^6,x, algorithm="fricas")
 

Output:

[1/45045*(45045*(a^2*cos(d*x + c)^8 + a^2*cos(d*x + c)^7)*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)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(16066 
*a^2*cos(d*x + c)^7 + 53078*a^2*cos(d*x + c)^6 + 17286*a^2*cos(d*x + c)^5 
- 30640*a^2*cos(d*x + c)^4 - 26810*a^2*cos(d*x + c)^3 + 2898*a^2*cos(d*x + 
 c)^2 + 10164*a^2*cos(d*x + c) + 3003*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d 
*x + c))*sin(d*x + c))/(d*cos(d*x + c)^8 + d*cos(d*x + c)^7), 2/45045*(450 
45*(a^2*cos(d*x + c)^8 + a^2*cos(d*x + c)^7)*sqrt(a)*arctan(sqrt((a*cos(d* 
x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + (16066*a^ 
2*cos(d*x + c)^7 + 53078*a^2*cos(d*x + c)^6 + 17286*a^2*cos(d*x + c)^5 - 3 
0640*a^2*cos(d*x + c)^4 - 26810*a^2*cos(d*x + c)^3 + 2898*a^2*cos(d*x + c) 
^2 + 10164*a^2*cos(d*x + c) + 3003*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x 
+ c))*sin(d*x + c))/(d*cos(d*x + c)^8 + d*cos(d*x + c)^7)]
 

Sympy [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^{5/2} \tan ^6(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*sec(d*x+c))**(5/2)*tan(d*x+c)**6,x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^{5/2} \tan ^6(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+a*sec(d*x+c))^(5/2)*tan(d*x+c)^6,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 397, normalized size of antiderivative = 1.37 \[ \int (a+a \sec (c+d x))^{5/2} \tan ^6(c+d x) \, dx=\frac {\frac {45045 \, \sqrt {-a} a^{3} \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 ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{{\left | a \right |}} - \frac {2 \, {\left (45045 \, \sqrt {2} a^{10} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (345345 \, \sqrt {2} a^{10} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (1162161 \, \sqrt {2} a^{10} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (611325 \, \sqrt {2} a^{10} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (77935 \, \sqrt {2} a^{10} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (109005 \, \sqrt {2} a^{10} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (11633 \, \sqrt {2} a^{10} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 64725 \, \sqrt {2} a^{10} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\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 )^{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 )}^{7} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{45045 \, d} \] Input:

integrate((a+a*sec(d*x+c))^(5/2)*tan(d*x+c)^6,x, algorithm="giac")
 

Output:

1/45045*(45045*sqrt(-a)*a^3*log(abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqr 
t(-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))*sgn(cos(d*x + c))/abs(a) - 2*(45045*sqrt(2)*a^10*sgn(co 
s(d*x + c)) - (345345*sqrt(2)*a^10*sgn(cos(d*x + c)) - (1162161*sqrt(2)*a^ 
10*sgn(cos(d*x + c)) - (611325*sqrt(2)*a^10*sgn(cos(d*x + c)) - (77935*sqr 
t(2)*a^10*sgn(cos(d*x + c)) + (109005*sqrt(2)*a^10*sgn(cos(d*x + c)) + (11 
633*sqrt(2)*a^10*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)^2 - 64725*sqrt(2)* 
a^10*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*ta 
n(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)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^7 
*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^{5/2} \tan ^6(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^6\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \] Input:

int(tan(c + d*x)^6*(a + a/cos(c + d*x))^(5/2),x)
 

Output:

int(tan(c + d*x)^6*(a + a/cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int (a+a \sec (c+d x))^{5/2} \tan ^6(c+d x) \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2} \tan \left (d x +c \right )^{6}d x +\int \sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{6}d x +2 \left (\int \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )^{6}d x \right )\right ) \] Input:

int((a+a*sec(d*x+c))^(5/2)*tan(d*x+c)^6,x)
 

Output:

sqrt(a)*a**2*(int(sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2*tan(c + d*x)**6,x 
) + int(sqrt(sec(c + d*x) + 1)*tan(c + d*x)**6,x) + 2*int(sqrt(sec(c + d*x 
) + 1)*sec(c + d*x)*tan(c + d*x)**6,x))