\(\int \cos ^9(c+d x) (a+b \tan (c+d x))^2 \, dx\) [535]

Optimal result

Integrand size = 21, antiderivative size = 133 \[ \int \cos ^9(c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {2 a b \cos ^9(c+d x)}{9 d}+\frac {a^2 \sin (c+d x)}{d}-\frac {\left (4 a^2-b^2\right ) \sin ^3(c+d x)}{3 d}+\frac {3 \left (2 a^2-b^2\right ) \sin ^5(c+d x)}{5 d}-\frac {\left (4 a^2-3 b^2\right ) \sin ^7(c+d x)}{7 d}+\frac {\left (a^2-b^2\right ) \sin ^9(c+d x)}{9 d} \] Output:

-2/9*a*b*cos(d*x+c)^9/d+a^2*sin(d*x+c)/d-1/3*(4*a^2-b^2)*sin(d*x+c)^3/d+3/ 
5*(2*a^2-b^2)*sin(d*x+c)^5/d-1/7*(4*a^2-3*b^2)*sin(d*x+c)^7/d+1/9*(a^2-b^2 
)*sin(d*x+c)^9/d
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.85 \[ \int \cos ^9(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {-70 a b \cos ^9(c+d x)+315 a^2 \sin (c+d x)-105 \left (4 a^2-b^2\right ) \sin ^3(c+d x)+189 \left (2 a^2-b^2\right ) \sin ^5(c+d x)-45 \left (4 a^2-3 b^2\right ) \sin ^7(c+d x)+35 \left (a^2-b^2\right ) \sin ^9(c+d x)}{315 d} \] Input:

Integrate[Cos[c + d*x]^9*(a + b*Tan[c + d*x])^2,x]
 

Output:

(-70*a*b*Cos[c + d*x]^9 + 315*a^2*Sin[c + d*x] - 105*(4*a^2 - b^2)*Sin[c + 
 d*x]^3 + 189*(2*a^2 - b^2)*Sin[c + d*x]^5 - 45*(4*a^2 - 3*b^2)*Sin[c + d* 
x]^7 + 35*(a^2 - b^2)*Sin[c + d*x]^9)/(315*d)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3991, 27, 3042, 3045, 15, 4159, 290, 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.

\(\displaystyle \int \cos ^9(c+d x) (a+b \tan (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (c+d x))^2}{\sec (c+d x)^9}dx\)

\(\Big \downarrow \) 3991

\(\displaystyle \int \cos ^9(c+d x) \left (a^2+b^2 \tan ^2(c+d x)\right )dx+\int 2 a b \cos ^8(c+d x) \sin (c+d x)dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \cos ^9(c+d x) \left (a^2+b^2 \tan ^2(c+d x)\right )dx+2 a b \int \cos ^8(c+d x) \sin (c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {a^2+b^2 \tan (c+d x)^2}{\sec (c+d x)^9}dx+2 a b \int \cos (c+d x)^8 \sin (c+d x)dx\)

\(\Big \downarrow \) 3045

\(\displaystyle \int \frac {a^2+b^2 \tan (c+d x)^2}{\sec (c+d x)^9}dx-\frac {2 a b \int \cos ^8(c+d x)d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 15

\(\displaystyle \int \frac {a^2+b^2 \tan (c+d x)^2}{\sec (c+d x)^9}dx-\frac {2 a b \cos ^9(c+d x)}{9 d}\)

\(\Big \downarrow \) 4159

\(\displaystyle \frac {\int \left (1-\sin ^2(c+d x)\right )^3 \left (a^2-\left (a^2-b^2\right ) \sin ^2(c+d x)\right )d\sin (c+d x)}{d}-\frac {2 a b \cos ^9(c+d x)}{9 d}\)

\(\Big \downarrow \) 290

\(\displaystyle \frac {\int \left ((a-b) (a+b) \sin ^8(c+d x)-\left (4 a^2-3 b^2\right ) \sin ^6(c+d x)+3 \left (2 a^2-b^2\right ) \sin ^4(c+d x)-\left (4 a^2-b^2\right ) \sin ^2(c+d x)+a^2\right )d\sin (c+d x)}{d}-\frac {2 a b \cos ^9(c+d x)}{9 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{9} \left (a^2-b^2\right ) \sin ^9(c+d x)-\frac {1}{7} \left (4 a^2-3 b^2\right ) \sin ^7(c+d x)+\frac {3}{5} \left (2 a^2-b^2\right ) \sin ^5(c+d x)-\frac {1}{3} \left (4 a^2-b^2\right ) \sin ^3(c+d x)+a^2 \sin (c+d x)}{d}-\frac {2 a b \cos ^9(c+d x)}{9 d}\)

Input:

Int[Cos[c + d*x]^9*(a + b*Tan[c + d*x])^2,x]
 

Output:

(-2*a*b*Cos[c + d*x]^9)/(9*d) + (a^2*Sin[c + d*x] - ((4*a^2 - b^2)*Sin[c + 
 d*x]^3)/3 + (3*(2*a^2 - b^2)*Sin[c + d*x]^5)/5 - ((4*a^2 - 3*b^2)*Sin[c + 
 d*x]^7)/7 + ((a^2 - b^2)*Sin[c + d*x]^9)/9)/d
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

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 290

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I 
nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d 
}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
 

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 3045

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ 
Symbol] :> Simp[-(a*f)^(-1)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], 
x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && 
 !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
 

rule 3991

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

rule 4159

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_ 
))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f 
  Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2 
*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f} 
, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
 

Maple [A] (verified)

Time = 186.21 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {\frac {a^{2} \left (\frac {128}{35}+\cos \left (d x +c \right )^{8}+\frac {8 \cos \left (d x +c \right )^{6}}{7}+\frac {48 \cos \left (d x +c \right )^{4}}{35}+\frac {64 \cos \left (d x +c \right )^{2}}{35}\right ) \sin \left (d x +c \right )}{9}-\frac {2 a b \cos \left (d x +c \right )^{9}}{9}+b^{2} \left (-\frac {\cos \left (d x +c \right )^{8} \sin \left (d x +c \right )}{9}+\frac {\left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{63}\right )}{d}\)	\(128\)
default	\(\frac {\frac {a^{2} \left (\frac {128}{35}+\cos \left (d x +c \right )^{8}+\frac {8 \cos \left (d x +c \right )^{6}}{7}+\frac {48 \cos \left (d x +c \right )^{4}}{35}+\frac {64 \cos \left (d x +c \right )^{2}}{35}\right ) \sin \left (d x +c \right )}{9}-\frac {2 a b \cos \left (d x +c \right )^{9}}{9}+b^{2} \left (-\frac {\cos \left (d x +c \right )^{8} \sin \left (d x +c \right )}{9}+\frac {\left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{63}\right )}{d}\)	\(128\)
risch	\(-\frac {7 a b \cos \left (d x +c \right )}{64 d}+\frac {63 a^{2} \sin \left (d x +c \right )}{128 d}+\frac {7 \sin \left (d x +c \right ) b^{2}}{128 d}-\frac {a b \cos \left (9 d x +9 c \right )}{1152 d}+\frac {\sin \left (9 d x +9 c \right ) a^{2}}{2304 d}-\frac {\sin \left (9 d x +9 c \right ) b^{2}}{2304 d}-\frac {a b \cos \left (7 d x +7 c \right )}{128 d}+\frac {9 \sin \left (7 d x +7 c \right ) a^{2}}{1792 d}-\frac {5 \sin \left (7 d x +7 c \right ) b^{2}}{1792 d}-\frac {a b \cos \left (5 d x +5 c \right )}{32 d}+\frac {9 \sin \left (5 d x +5 c \right ) a^{2}}{320 d}-\frac {\sin \left (5 d x +5 c \right ) b^{2}}{160 d}-\frac {7 a b \cos \left (3 d x +3 c \right )}{96 d}+\frac {7 \sin \left (3 d x +3 c \right ) a^{2}}{64 d}\)	\(226\)

Input:

int(cos(d*x+c)^9*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/9*a^2*(128/35+cos(d*x+c)^8+8/7*cos(d*x+c)^6+48/35*cos(d*x+c)^4+64/3 
5*cos(d*x+c)^2)*sin(d*x+c)-2/9*a*b*cos(d*x+c)^9+b^2*(-1/9*cos(d*x+c)^8*sin 
(d*x+c)+1/63*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x 
+c)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.85 \[ \int \cos ^9(c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {70 \, a b \cos \left (d x + c\right )^{9} - {\left (35 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{8} + 5 \, {\left (8 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{6} + 6 \, {\left (8 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} + 8 \, {\left (8 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} + 128 \, a^{2} + 16 \, b^{2}\right )} \sin \left (d x + c\right )}{315 \, d} \] Input:

integrate(cos(d*x+c)^9*(a+b*tan(d*x+c))^2,x, algorithm="fricas")
 

Output:

-1/315*(70*a*b*cos(d*x + c)^9 - (35*(a^2 - b^2)*cos(d*x + c)^8 + 5*(8*a^2 
+ b^2)*cos(d*x + c)^6 + 6*(8*a^2 + b^2)*cos(d*x + c)^4 + 8*(8*a^2 + b^2)*c 
os(d*x + c)^2 + 128*a^2 + 16*b^2)*sin(d*x + c))/d
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^9(c+d x) (a+b \tan (c+d x))^2 \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**9*(a+b*tan(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.88 \[ \int \cos ^9(c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {70 \, a b \cos \left (d x + c\right )^{9} - {\left (35 \, \sin \left (d x + c\right )^{9} - 180 \, \sin \left (d x + c\right )^{7} + 378 \, \sin \left (d x + c\right )^{5} - 420 \, \sin \left (d x + c\right )^{3} + 315 \, \sin \left (d x + c\right )\right )} a^{2} + {\left (35 \, \sin \left (d x + c\right )^{9} - 135 \, \sin \left (d x + c\right )^{7} + 189 \, \sin \left (d x + c\right )^{5} - 105 \, \sin \left (d x + c\right )^{3}\right )} b^{2}}{315 \, d} \] Input:

integrate(cos(d*x+c)^9*(a+b*tan(d*x+c))^2,x, algorithm="maxima")
 

Output:

-1/315*(70*a*b*cos(d*x + c)^9 - (35*sin(d*x + c)^9 - 180*sin(d*x + c)^7 + 
378*sin(d*x + c)^5 - 420*sin(d*x + c)^3 + 315*sin(d*x + c))*a^2 + (35*sin( 
d*x + c)^9 - 135*sin(d*x + c)^7 + 189*sin(d*x + c)^5 - 105*sin(d*x + c)^3) 
*b^2)/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80368 vs. \(2 (123) = 246\).

Time = 79.01 (sec) , antiderivative size = 80368, normalized size of antiderivative = 604.27 \[ \int \cos ^9(c+d x) (a+b \tan (c+d x))^2 \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^9*(a+b*tan(d*x+c))^2,x, algorithm="giac")
 

Output:

-1/80640*(2205*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*t 
an(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 + 2*tan(1/2*c) - 1)*sgn(tan(1/2* 
d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1 
/2*c)^2 + 2*tan(1/2*d*x) - 1)*tan(1/2*d*x)^18*tan(1/2*c)^18 + 2205*pi*a*b* 
sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d* 
x)^2 - tan(1/2*c)^2 - 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 
2*tan(1/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d* 
x) - 1)*tan(1/2*d*x)^18*tan(1/2*c)^18 + 2205*pi*a*b*sgn(tan(1/2*d*x)^2*tan 
(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 + 
2*tan(1/2*c) - 1)*tan(1/2*d*x)^18*tan(1/2*c)^18 + 2205*pi*a*b*sgn(tan(1/2* 
d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1 
/2*c)^2 - 2*tan(1/2*c) - 1)*tan(1/2*d*x)^18*tan(1/2*c)^18 - 19530*pi*a*b*s 
gn(tan(1/2*d*x)^2*tan(1/2*c)^2 - tan(1/2*d*x)^2 - 4*tan(1/2*d*x)*tan(1/2*c 
) - tan(1/2*c)^2 + 1)*tan(1/2*d*x)^18*tan(1/2*c)^18 + 19845*pi*a*b*sgn(tan 
(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - 
tan(1/2*c)^2 + 2*tan(1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1 
/2*d*x)*tan(1/2*c)^2 - tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) - 1) 
*tan(1/2*d*x)^18*tan(1/2*c)^16 + 19845*pi*a*b*sgn(tan(1/2*d*x)^2*tan(1/2*c 
)^2 - 2*tan(1/2*d*x)^2*tan(1/2*c) + tan(1/2*d*x)^2 - tan(1/2*c)^2 - 2*tan( 
1/2*c) - 1)*sgn(tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)*tan(1/2*c)...
 

Mupad [B] (verification not implemented)

Time = 1.20 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.44 \[ \int \cos ^9(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {2\,\left (\frac {35\,\sin \left (c+d\,x\right )\,a^2\,{\cos \left (c+d\,x\right )}^8}{2}+20\,\sin \left (c+d\,x\right )\,a^2\,{\cos \left (c+d\,x\right )}^6+24\,\sin \left (c+d\,x\right )\,a^2\,{\cos \left (c+d\,x\right )}^4+32\,\sin \left (c+d\,x\right )\,a^2\,{\cos \left (c+d\,x\right )}^2+64\,\sin \left (c+d\,x\right )\,a^2-35\,a\,b\,{\cos \left (c+d\,x\right )}^9-\frac {35\,\sin \left (c+d\,x\right )\,b^2\,{\cos \left (c+d\,x\right )}^8}{2}+\frac {5\,\sin \left (c+d\,x\right )\,b^2\,{\cos \left (c+d\,x\right )}^6}{2}+3\,\sin \left (c+d\,x\right )\,b^2\,{\cos \left (c+d\,x\right )}^4+4\,\sin \left (c+d\,x\right )\,b^2\,{\cos \left (c+d\,x\right )}^2+8\,\sin \left (c+d\,x\right )\,b^2\right )}{315\,d} \] Input:

int(cos(c + d*x)^9*(a + b*tan(c + d*x))^2,x)
 

Output:

(2*(64*a^2*sin(c + d*x) + 8*b^2*sin(c + d*x) + 32*a^2*cos(c + d*x)^2*sin(c 
 + d*x) + 24*a^2*cos(c + d*x)^4*sin(c + d*x) + 20*a^2*cos(c + d*x)^6*sin(c 
 + d*x) + (35*a^2*cos(c + d*x)^8*sin(c + d*x))/2 + 4*b^2*cos(c + d*x)^2*si 
n(c + d*x) + 3*b^2*cos(c + d*x)^4*sin(c + d*x) + (5*b^2*cos(c + d*x)^6*sin 
(c + d*x))/2 - (35*b^2*cos(c + d*x)^8*sin(c + d*x))/2 - 35*a*b*cos(c + d*x 
)^9))/(315*d)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.56 \[ \int \cos ^9(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {-70 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8} a b +280 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} a b -420 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a b +280 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a b -70 \cos \left (d x +c \right ) a b +35 \sin \left (d x +c \right )^{9} a^{2}-35 \sin \left (d x +c \right )^{9} b^{2}-180 \sin \left (d x +c \right )^{7} a^{2}+135 \sin \left (d x +c \right )^{7} b^{2}+378 \sin \left (d x +c \right )^{5} a^{2}-189 \sin \left (d x +c \right )^{5} b^{2}-420 \sin \left (d x +c \right )^{3} a^{2}+105 \sin \left (d x +c \right )^{3} b^{2}+315 \sin \left (d x +c \right ) a^{2}+70 a b}{315 d} \] Input:

int(cos(d*x+c)^9*(a+b*tan(d*x+c))^2,x)
 

Output:

( - 70*cos(c + d*x)*sin(c + d*x)**8*a*b + 280*cos(c + d*x)*sin(c + d*x)**6 
*a*b - 420*cos(c + d*x)*sin(c + d*x)**4*a*b + 280*cos(c + d*x)*sin(c + d*x 
)**2*a*b - 70*cos(c + d*x)*a*b + 35*sin(c + d*x)**9*a**2 - 35*sin(c + d*x) 
**9*b**2 - 180*sin(c + d*x)**7*a**2 + 135*sin(c + d*x)**7*b**2 + 378*sin(c 
 + d*x)**5*a**2 - 189*sin(c + d*x)**5*b**2 - 420*sin(c + d*x)**3*a**2 + 10 
5*sin(c + d*x)**3*b**2 + 315*sin(c + d*x)*a**2 + 70*a*b)/(315*d)