\(\int \cos ^6(c+d x) (a+b \tan (c+d x))^3 \, dx\) [543]

Optimal result

Integrand size = 21, antiderivative size = 147 \[ \int \cos ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {1}{16} a \left (5 a^2+3 b^2\right ) x-\frac {b \left (5 a^2+b^2\right ) \cos ^2(c+d x)}{12 d}+\frac {a \left (15 a^2-b^2\right ) \cos (c+d x) \sin (c+d x)}{48 d}-\frac {\cos ^4(c+d x) (2 b-5 a \tan (c+d x)) (a+b \tan (c+d x))^2}{24 d}+\frac {\cos ^5(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{6 d} \] Output:

1/16*a*(5*a^2+3*b^2)*x-1/12*b*(5*a^2+b^2)*cos(d*x+c)^2/d+1/48*a*(15*a^2-b^ 
2)*cos(d*x+c)*sin(d*x+c)/d-1/24*cos(d*x+c)^4*(2*b-5*a*tan(d*x+c))*(a+b*tan 
(d*x+c))^2/d+1/6*cos(d*x+c)^5*sin(d*x+c)*(a+b*tan(d*x+c))^3/d
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(390\) vs. \(2(147)=294\).

Time = 5.63 (sec) , antiderivative size = 390, normalized size of antiderivative = 2.65 \[ \int \cos ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\frac {8 b \left (2 a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^4}{a^2+b^2}+16 \cos ^6(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^4-\frac {\left (-5 a^3-3 a b^2\right ) \left (4 \cos ^4(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^5-\frac {2 \cos ^2(c+d x) (a+b \tan (c+d x))^5 \left (-5 a^2 b+b^3-3 a \left (a^2-b^2\right ) \tan (c+d x)\right )}{a^2+b^2}+\frac {b \left (\frac {3 \left (a^2+b^2\right )^4 \left (\log \left (\sqrt {-b^2}-b \tan (c+d x)\right )-\log \left (\sqrt {-b^2}+b \tan (c+d x)\right )\right )}{\sqrt {-b^2}}-6 b \left (10 a^6+5 a^4 b^2+4 a^2 b^4+b^6\right ) \tan (c+d x)+12 a^3 b^2 \left (-5 a^2+b^2\right ) \tan ^2(c+d x)+2 b^3 \left (-15 a^4+10 a^2 b^2+b^4\right ) \tan ^3(c+d x)+6 a b^4 \left (-a^2+b^2\right ) \tan ^4(c+d x)\right )}{a^2+b^2}\right )}{\left (a^2+b^2\right )^2}}{96 \left (a^2+b^2\right ) d} \] Input:

Integrate[Cos[c + d*x]^6*(a + b*Tan[c + d*x])^3,x]
 

Output:

((8*b*(2*a^2 + b^2)*(a*Cos[c + d*x] + b*Sin[c + d*x])^4)/(a^2 + b^2) + 16* 
Cos[c + d*x]^6*(b + a*Tan[c + d*x])*(a + b*Tan[c + d*x])^4 - ((-5*a^3 - 3* 
a*b^2)*(4*Cos[c + d*x]^4*(b + a*Tan[c + d*x])*(a + b*Tan[c + d*x])^5 - (2* 
Cos[c + d*x]^2*(a + b*Tan[c + d*x])^5*(-5*a^2*b + b^3 - 3*a*(a^2 - b^2)*Ta 
n[c + d*x]))/(a^2 + b^2) + (b*((3*(a^2 + b^2)^4*(Log[Sqrt[-b^2] - b*Tan[c 
+ d*x]] - Log[Sqrt[-b^2] + b*Tan[c + d*x]]))/Sqrt[-b^2] - 6*b*(10*a^6 + 5* 
a^4*b^2 + 4*a^2*b^4 + b^6)*Tan[c + d*x] + 12*a^3*b^2*(-5*a^2 + b^2)*Tan[c 
+ d*x]^2 + 2*b^3*(-15*a^4 + 10*a^2*b^2 + b^4)*Tan[c + d*x]^3 + 6*a*b^4*(-a 
^2 + b^2)*Tan[c + d*x]^4))/(a^2 + b^2)))/(a^2 + b^2)^2)/(96*(a^2 + b^2)*d)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.46, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3987, 27, 494, 25, 685, 675, 216}

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.

\(\displaystyle \int \cos ^6(c+d x) (a+b \tan (c+d x))^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (c+d x))^3}{\sec (c+d x)^6}dx\)

\(\Big \downarrow \) 3987

\(\displaystyle \frac {\int \frac {b^8 (a+b \tan (c+d x))^3}{\left (\tan ^2(c+d x) b^2+b^2\right )^4}d(b \tan (c+d x))}{b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b^7 \int \frac {(a+b \tan (c+d x))^3}{\left (\tan ^2(c+d x) b^2+b^2\right )^4}d(b \tan (c+d x))}{d}\)

\(\Big \downarrow \) 494

\(\displaystyle \frac {b^7 \left (\frac {\tan (c+d x) (a+b \tan (c+d x))^3}{6 b \left (b^2 \tan ^2(c+d x)+b^2\right )^3}-\frac {\int -\frac {(a+b \tan (c+d x))^2 (5 a+2 b \tan (c+d x))}{\left (\tan ^2(c+d x) b^2+b^2\right )^3}d(b \tan (c+d x))}{6 b^2}\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b^7 \left (\frac {\int \frac {(a+b \tan (c+d x))^2 (5 a+2 b \tan (c+d x))}{\left (\tan ^2(c+d x) b^2+b^2\right )^3}d(b \tan (c+d x))}{6 b^2}+\frac {\tan (c+d x) (a+b \tan (c+d x))^3}{6 b \left (b^2 \tan ^2(c+d x)+b^2\right )^3}\right )}{d}\)

\(\Big \downarrow \) 685

\(\displaystyle \frac {b^7 \left (\frac {\frac {\int \frac {(a+b \tan (c+d x)) \left (15 a^2+5 b \tan (c+d x) a+4 b^2\right )}{\left (\tan ^2(c+d x) b^2+b^2\right )^2}d(b \tan (c+d x))}{4 b^2}-\frac {(a+b \tan (c+d x))^2 \left (2 b^2-5 a b \tan (c+d x)\right )}{4 b^2 \left (b^2 \tan ^2(c+d x)+b^2\right )^2}}{6 b^2}+\frac {\tan (c+d x) (a+b \tan (c+d x))^3}{6 b \left (b^2 \tan ^2(c+d x)+b^2\right )^3}\right )}{d}\)

\(\Big \downarrow \) 675

\(\displaystyle \frac {b^7 \left (\frac {\frac {\frac {3}{2} a \left (\frac {5 a^2}{b^2}+3\right ) \int \frac {1}{\tan ^2(c+d x) b^2+b^2}d(b \tan (c+d x))-\frac {2 \left (5 a^2+b^2\right )}{b^2 \tan ^2(c+d x)+b^2}-\frac {a b \left (1-\frac {15 a^2}{b^2}\right ) \tan (c+d x)}{2 \left (b^2 \tan ^2(c+d x)+b^2\right )}}{4 b^2}-\frac {(a+b \tan (c+d x))^2 \left (2 b^2-5 a b \tan (c+d x)\right )}{4 b^2 \left (b^2 \tan ^2(c+d x)+b^2\right )^2}}{6 b^2}+\frac {\tan (c+d x) (a+b \tan (c+d x))^3}{6 b \left (b^2 \tan ^2(c+d x)+b^2\right )^3}\right )}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {b^7 \left (\frac {\frac {\frac {3 a \left (\frac {5 a^2}{b^2}+3\right ) \arctan (\tan (c+d x))}{2 b}-\frac {2 \left (5 a^2+b^2\right )}{b^2 \tan ^2(c+d x)+b^2}-\frac {a b \left (1-\frac {15 a^2}{b^2}\right ) \tan (c+d x)}{2 \left (b^2 \tan ^2(c+d x)+b^2\right )}}{4 b^2}-\frac {(a+b \tan (c+d x))^2 \left (2 b^2-5 a b \tan (c+d x)\right )}{4 b^2 \left (b^2 \tan ^2(c+d x)+b^2\right )^2}}{6 b^2}+\frac {\tan (c+d x) (a+b \tan (c+d x))^3}{6 b \left (b^2 \tan ^2(c+d x)+b^2\right )^3}\right )}{d}\)

Input:

Int[Cos[c + d*x]^6*(a + b*Tan[c + d*x])^3,x]
 

Output:

(b^7*((Tan[c + d*x]*(a + b*Tan[c + d*x])^3)/(6*b*(b^2 + b^2*Tan[c + d*x]^2 
)^3) + (-1/4*((a + b*Tan[c + d*x])^2*(2*b^2 - 5*a*b*Tan[c + d*x]))/(b^2*(b 
^2 + b^2*Tan[c + d*x]^2)^2) + ((3*a*(3 + (5*a^2)/b^2)*ArcTan[Tan[c + d*x]] 
)/(2*b) - (2*(5*a^2 + b^2))/(b^2 + b^2*Tan[c + d*x]^2) - (a*(1 - (15*a^2)/ 
b^2)*b*Tan[c + d*x])/(2*(b^2 + b^2*Tan[c + d*x]^2)))/(4*b^2))/(6*b^2)))/d
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 494

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-x)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[1/(2*a*(p + 
 1))   Int[(c + d*x)^(n - 1)*(a + b*x^2)^(p + 1)*(c*(2*p + 3) + d*(n + 2*p 
+ 3)*x), x], x] /; FreeQ[{a, b, c, d}, x] && LtQ[p, -1] && GtQ[n, 0] && (Lt 
Q[n, 1] || (ILtQ[n + 2*p + 3, 0] && NeQ[n, 2])) && IntQuadraticQ[a, 0, b, c 
, d, n, p, x]
 

rule 675

Int[((d_) + (e_.)*(x_))*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[a*(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + (- 
Simp[(c*d*f - a*e*g)*x*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Simp[(a* 
e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1))   Int[(a + c*x^2)^(p + 1), x], x]) / 
; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1] &&  !(IntegerQ[p] && NiceSqrtQ 
[(-a)*c])
 

rule 685

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> Simp[(d + e*x)^m*(a + c*x^2)^(p + 1)*((a*g - c*f*x)/(2*a*c 
*(p + 1))), x] - Simp[1/(2*a*c*(p + 1))   Int[(d + e*x)^(m - 1)*(a + c*x^2) 
^(p + 1)*Simp[a*e*g*m - c*d*f*(2*p + 3) - c*e*f*(m + 2*p + 3)*x, x], x], x] 
 /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] 
 || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 3042

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

rule 3987

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[1/(b*f)   Subst[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), 
 x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b^2, 
0] && IntegerQ[m/2]
 

Maple [A] (verified)

Time = 66.30 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {b^{3} \left (-\frac {\cos \left (d x +c \right )^{4} \sin \left (d x +c \right )^{2}}{6}-\frac {\cos \left (d x +c \right )^{4}}{12}\right )+3 a \,b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {a^{2} b \cos \left (d x +c \right )^{6}}{2}+a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\)	\(155\)
default	\(\frac {b^{3} \left (-\frac {\cos \left (d x +c \right )^{4} \sin \left (d x +c \right )^{2}}{6}-\frac {\cos \left (d x +c \right )^{4}}{12}\right )+3 a \,b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{5}}{6}+\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {a^{2} b \cos \left (d x +c \right )^{6}}{2}+a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\)	\(155\)
risch	\(\frac {5 a^{3} x}{16}+\frac {3 a \,b^{2} x}{16}-\frac {b \cos \left (6 d x +6 c \right ) a^{2}}{64 d}+\frac {b^{3} \cos \left (6 d x +6 c \right )}{192 d}+\frac {a^{3} \sin \left (6 d x +6 c \right )}{192 d}-\frac {a \sin \left (6 d x +6 c \right ) b^{2}}{64 d}-\frac {3 b \cos \left (4 d x +4 c \right ) a^{2}}{32 d}+\frac {3 a^{3} \sin \left (4 d x +4 c \right )}{64 d}-\frac {3 a \sin \left (4 d x +4 c \right ) b^{2}}{64 d}-\frac {15 b \cos \left (2 d x +2 c \right ) a^{2}}{64 d}-\frac {3 b^{3} \cos \left (2 d x +2 c \right )}{64 d}+\frac {15 a^{3} \sin \left (2 d x +2 c \right )}{64 d}+\frac {3 a \sin \left (2 d x +2 c \right ) b^{2}}{64 d}\)	\(208\)

Input:

int(cos(d*x+c)^6*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(b^3*(-1/6*cos(d*x+c)^4*sin(d*x+c)^2-1/12*cos(d*x+c)^4)+3*a*b^2*(-1/6* 
sin(d*x+c)*cos(d*x+c)^5+1/24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16 
*d*x+1/16*c)-1/2*a^2*b*cos(d*x+c)^6+a^3*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^ 
3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.87 \[ \int \cos ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {12 \, b^{3} \cos \left (d x + c\right )^{4} + 8 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} d x - {\left (8 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d} \] Input:

integrate(cos(d*x+c)^6*(a+b*tan(d*x+c))^3,x, algorithm="fricas")
 

Output:

-1/48*(12*b^3*cos(d*x + c)^4 + 8*(3*a^2*b - b^3)*cos(d*x + c)^6 - 3*(5*a^3 
 + 3*a*b^2)*d*x - (8*(a^3 - 3*a*b^2)*cos(d*x + c)^5 + 2*(5*a^3 + 3*a*b^2)* 
cos(d*x + c)^3 + 3*(5*a^3 + 3*a*b^2)*cos(d*x + c))*sin(d*x + c))/d
 

Sympy [F]

\[ \int \cos ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \cos ^{6}{\left (c + d x \right )}\, dx \] Input:

integrate(cos(d*x+c)**6*(a+b*tan(d*x+c))**3,x)
 

Output:

Integral((a + b*tan(c + d*x))**3*cos(c + d*x)**6, x)
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00 \[ \int \cos ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {3 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} {\left (d x + c\right )} + \frac {3 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{5} - 12 \, b^{3} \tan \left (d x + c\right )^{2} + 8 \, {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{3} - 24 \, a^{2} b - 4 \, b^{3} + 3 \, {\left (11 \, a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{6} + 3 \, \tan \left (d x + c\right )^{4} + 3 \, \tan \left (d x + c\right )^{2} + 1}}{48 \, d} \] Input:

integrate(cos(d*x+c)^6*(a+b*tan(d*x+c))^3,x, algorithm="maxima")
 

Output:

1/48*(3*(5*a^3 + 3*a*b^2)*(d*x + c) + (3*(5*a^3 + 3*a*b^2)*tan(d*x + c)^5 
- 12*b^3*tan(d*x + c)^2 + 8*(5*a^3 + 3*a*b^2)*tan(d*x + c)^3 - 24*a^2*b - 
4*b^3 + 3*(11*a^3 - 3*a*b^2)*tan(d*x + c))/(tan(d*x + c)^6 + 3*tan(d*x + c 
)^4 + 3*tan(d*x + c)^2 + 1))/d
 

Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.97 \[ \int \cos ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {{\left (5 \, a^{3} + 3 \, a b^{2}\right )} {\left (d x + c\right )}}{16 \, d} + \frac {15 \, a^{3} \tan \left (d x + c\right )^{5} + 9 \, a b^{2} \tan \left (d x + c\right )^{5} + 40 \, a^{3} \tan \left (d x + c\right )^{3} + 24 \, a b^{2} \tan \left (d x + c\right )^{3} - 12 \, b^{3} \tan \left (d x + c\right )^{2} + 33 \, a^{3} \tan \left (d x + c\right ) - 9 \, a b^{2} \tan \left (d x + c\right ) - 24 \, a^{2} b - 4 \, b^{3}}{48 \, {\left (\tan \left (d x + c\right )^{2} + 1\right )}^{3} d} \] Input:

integrate(cos(d*x+c)^6*(a+b*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

1/16*(5*a^3 + 3*a*b^2)*(d*x + c)/d + 1/48*(15*a^3*tan(d*x + c)^5 + 9*a*b^2 
*tan(d*x + c)^5 + 40*a^3*tan(d*x + c)^3 + 24*a*b^2*tan(d*x + c)^3 - 12*b^3 
*tan(d*x + c)^2 + 33*a^3*tan(d*x + c) - 9*a*b^2*tan(d*x + c) - 24*a^2*b - 
4*b^3)/((tan(d*x + c)^2 + 1)^3*d)
 

Mupad [B] (verification not implemented)

Time = 1.33 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.96 \[ \int \cos ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {5\,a^3\,x}{16}-\frac {24\,a^2\,b-{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (40\,a^3+24\,a\,b^2\right )-{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (15\,a^3+9\,a\,b^2\right )+4\,b^3+\mathrm {tan}\left (c+d\,x\right )\,\left (9\,a\,b^2-33\,a^3\right )+12\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d\,\left (48\,{\mathrm {tan}\left (c+d\,x\right )}^6+144\,{\mathrm {tan}\left (c+d\,x\right )}^4+144\,{\mathrm {tan}\left (c+d\,x\right )}^2+48\right )}+\frac {3\,a\,b^2\,x}{16} \] Input:

int(cos(c + d*x)^6*(a + b*tan(c + d*x))^3,x)
 

Output:

(5*a^3*x)/16 - (24*a^2*b - tan(c + d*x)^3*(24*a*b^2 + 40*a^3) - tan(c + d* 
x)^5*(9*a*b^2 + 15*a^3) + 4*b^3 + tan(c + d*x)*(9*a*b^2 - 33*a^3) + 12*b^3 
*tan(c + d*x)^2)/(d*(144*tan(c + d*x)^2 + 144*tan(c + d*x)^4 + 48*tan(c + 
d*x)^6 + 48)) + (3*a*b^2*x)/16
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.37 \[ \int \cos ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} a^{3}-24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} a \,b^{2}-26 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{3}+42 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a \,b^{2}+33 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3}-9 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{2}+24 \sin \left (d x +c \right )^{6} a^{2} b -8 \sin \left (d x +c \right )^{6} b^{3}-72 \sin \left (d x +c \right )^{4} a^{2} b +12 \sin \left (d x +c \right )^{4} b^{3}+72 \sin \left (d x +c \right )^{2} a^{2} b +15 a^{3} d x +9 a \,b^{2} d x}{48 d} \] Input:

int(cos(d*x+c)^6*(a+b*tan(d*x+c))^3,x)
 

Output:

(8*cos(c + d*x)*sin(c + d*x)**5*a**3 - 24*cos(c + d*x)*sin(c + d*x)**5*a*b 
**2 - 26*cos(c + d*x)*sin(c + d*x)**3*a**3 + 42*cos(c + d*x)*sin(c + d*x)* 
*3*a*b**2 + 33*cos(c + d*x)*sin(c + d*x)*a**3 - 9*cos(c + d*x)*sin(c + d*x 
)*a*b**2 + 24*sin(c + d*x)**6*a**2*b - 8*sin(c + d*x)**6*b**3 - 72*sin(c + 
 d*x)**4*a**2*b + 12*sin(c + d*x)**4*b**3 + 72*sin(c + d*x)**2*a**2*b + 15 
*a**3*d*x + 9*a*b**2*d*x)/(48*d)