\(\int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx\) [272]

Optimal result

Integrand size = 28, antiderivative size = 264 \[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {\left (a^2-2 b^2\right ) \cos ^{-1+m}(c+d x) \sin (c+d x)}{d m (2+m)}-\frac {2 a b \cos ^m(c+d x) \sin (c+d x)}{d \left (2+3 m+m^2\right )}-\frac {\cos ^{-1+m}(c+d x) (b+a \cos (c+d x))^2 \sin (c+d x)}{d (2+m)}-\frac {\left (a^2 (1-m)-b^2 (2+m)\right ) \cos ^{-1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+m),\frac {1+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1-m) m (2+m) \sqrt {\sin ^2(c+d x)}}-\frac {2 a b \cos ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d m (1+m) \sqrt {\sin ^2(c+d x)}} \] Output:

(a^2-2*b^2)*cos(d*x+c)^(-1+m)*sin(d*x+c)/d/m/(2+m)-2*a*b*cos(d*x+c)^m*sin( 
d*x+c)/d/(m^2+3*m+2)-cos(d*x+c)^(-1+m)*(b+a*cos(d*x+c))^2*sin(d*x+c)/d/(2+ 
m)-(a^2*(1-m)-b^2*(2+m))*cos(d*x+c)^(-1+m)*hypergeom([1/2, -1/2+1/2*m],[1/ 
2+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(1-m)/m/(2+m)/(sin(d*x+c)^2)^(1/2)-2*a 
*b*cos(d*x+c)^m*hypergeom([1/2, 1/2*m],[1+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/ 
d/m/(1+m)/(sin(d*x+c)^2)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 37.61 (sec) , antiderivative size = 6404, normalized size of antiderivative = 24.26 \[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\text {Result too large to show} \] Input:

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

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.01, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.536, Rules used = {3042, 4897, 3042, 3368, 3042, 3529, 3042, 3512, 3042, 3502, 25, 3042, 3227, 3042, 3122}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4897

\(\displaystyle \int \sin ^2(c+d x) \cos ^{m-2}(c+d x) (a \cos (c+d x)+b)^2dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos \left (c+d x+\frac {\pi }{2}\right )^2 \sin \left (c+d x+\frac {\pi }{2}\right )^{m-2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2dx\)

\(\Big \downarrow \) 3368

\(\displaystyle \int \left (1-\cos ^2(c+d x)\right ) \cos ^{m-2}(c+d x) (a \cos (c+d x)+b)^2dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (1-\sin \left (c+d x+\frac {\pi }{2}\right )^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^{m-2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2dx\)

\(\Big \downarrow \) 3529

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3512

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\frac {\frac {(m+1) \left (a^2-2 b^2\right ) \sin (c+d x) \cos ^{m-1}(c+d x)}{d m}-\frac {\frac {(m+1) \left (a^2 (1-m)-b^2 (m+2)\right ) \sin (c+d x) \cos ^{m-1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m-1}{2},\frac {m+1}{2},\cos ^2(c+d x)\right )}{d (1-m) \sqrt {\sin ^2(c+d x)}}+\frac {2 a b (m+2) \sin (c+d x) \cos ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {m+2}{2},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}}{m}}{m+1}-\frac {2 a b \sin (c+d x) \cos ^m(c+d x)}{d (m+1)}}{m+2}-\frac {\sin (c+d x) \cos ^{m-1}(c+d x) (a \cos (c+d x)+b)^2}{d (m+2)}\)

Input:

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

Output:

-((Cos[c + d*x]^(-1 + m)*(b + a*Cos[c + d*x])^2*Sin[c + d*x])/(d*(2 + m))) 
 + ((-2*a*b*Cos[c + d*x]^m*Sin[c + d*x])/(d*(1 + m)) + (((a^2 - 2*b^2)*(1 
+ m)*Cos[c + d*x]^(-1 + m)*Sin[c + d*x])/(d*m) - (((1 + m)*(a^2*(1 - m) - 
b^2*(2 + m))*Cos[c + d*x]^(-1 + m)*Hypergeometric2F1[1/2, (-1 + m)/2, (1 + 
 m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 - m)*Sqrt[Sin[c + d*x]^2]) + (2 
*a*b*(2 + m)*Cos[c + d*x]^m*Hypergeometric2F1[1/2, m/2, (2 + m)/2, Cos[c + 
 d*x]^2]*Sin[c + d*x])/(d*Sqrt[Sin[c + d*x]^2]))/m)/(1 + m))/(2 + m)
 

Defintions of rubi rules used

rule 25

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

rule 3042

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

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3368

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + 
(b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a 
 + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, m, n 
}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

rule 3512

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si 
n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3))   Int[(a + b*Si 
n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + 
A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 
0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

rule 3529

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] : 
> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 
1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2))   Int[(a + b*Sin[e + f*x 
])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*( 
n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + C* 
(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f 
, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 
0])))
 

rule 4897

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]
 

Maple [F]

\[\int \cos \left (d x +c \right )^{m} \left (a \sin \left (d x +c \right )+b \tan \left (d x +c \right )\right )^{2}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\int { {\left (a \sin \left (d x + c\right ) + b \tan \left (d x + c\right )\right )}^{2} \cos \left (d x + c\right )^{m} \,d x } \] Input:

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

Output:

integral(-(a^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c)*tan(d*x + c) - b^2*tan( 
d*x + c)^2 - a^2)*cos(d*x + c)^m, x)
 

Sympy [F]

\[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{2} \cos ^{m}{\left (c + d x \right )}\, dx \] Input:

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

Output:

Integral((a*sin(c + d*x) + b*tan(c + d*x))**2*cos(c + d*x)**m, x)
 

Maxima [F]

\[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\int { {\left (a \sin \left (d x + c\right ) + b \tan \left (d x + c\right )\right )}^{2} \cos \left (d x + c\right )^{m} \,d x } \] Input:

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

Output:

integrate((a*sin(d*x + c) + b*tan(d*x + c))^2*cos(d*x + c)^m, x)
 

Giac [F]

\[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\int { {\left (a \sin \left (d x + c\right ) + b \tan \left (d x + c\right )\right )}^{2} \cos \left (d x + c\right )^{m} \,d x } \] Input:

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

Output:

integrate((a*sin(d*x + c) + b*tan(d*x + c))^2*cos(d*x + c)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\int {\cos \left (c+d\,x\right )}^m\,{\left (a\,\sin \left (c+d\,x\right )+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^2 \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\left (\int \cos \left (d x +c \right )^{m} \sin \left (d x +c \right )^{2}d x \right ) a^{2}+\left (\int \cos \left (d x +c \right )^{m} \tan \left (d x +c \right )^{2}d x \right ) b^{2}+2 \left (\int \cos \left (d x +c \right )^{m} \sin \left (d x +c \right ) \tan \left (d x +c \right )d x \right ) a b \] Input:

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

Output:

int(cos(c + d*x)**m*sin(c + d*x)**2,x)*a**2 + int(cos(c + d*x)**m*tan(c + 
d*x)**2,x)*b**2 + 2*int(cos(c + d*x)**m*sin(c + d*x)*tan(c + d*x),x)*a*b