\(\int (a+b \cos (c+d x))^2 \sec ^m(c+d x) \, dx\) [776]

Optimal result

Integrand size = 21, antiderivative size = 200 \[ \int (a+b \cos (c+d x))^2 \sec ^m(c+d x) \, dx=-\frac {a^2 \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (1-m)}-\frac {\left (b^2 (1-m)+a^2 (2-m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-m}{2},\frac {5-m}{2},\cos ^2(c+d x)\right ) \sec ^{-3+m}(c+d x) \sin (c+d x)}{d (1-m) (3-m) \sqrt {\sin ^2(c+d x)}}-\frac {2 a b \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-m}{2},\frac {4-m}{2},\cos ^2(c+d x)\right ) \sec ^{-2+m}(c+d x) \sin (c+d x)}{d (2-m) \sqrt {\sin ^2(c+d x)}} \] Output:

-a^2*sec(d*x+c)^(-1+m)*sin(d*x+c)/d/(1-m)-(b^2*(1-m)+a^2*(2-m))*hypergeom( 
[1/2, 3/2-1/2*m],[5/2-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-3+m)*sin(d*x+c)/d/ 
(1-m)/(3-m)/(sin(d*x+c)^2)^(1/2)-2*a*b*hypergeom([1/2, 1-1/2*m],[2-1/2*m], 
cos(d*x+c)^2)*sec(d*x+c)^(-2+m)*sin(d*x+c)/d/(2-m)/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.80 \[ \int (a+b \cos (c+d x))^2 \sec ^m(c+d x) \, dx=\frac {\csc (c+d x) \sec ^{-3+m}(c+d x) \left (b^2 (-1+m) m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2+m),\frac {m}{2},\sec ^2(c+d x)\right )+a (-2+m) \left (2 b m \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+m),\frac {1+m}{2},\sec ^2(c+d x)\right )+a (-1+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(c+d x)\right )\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (-2+m) (-1+m) m} \] Input:

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

Output:

(Csc[c + d*x]*Sec[c + d*x]^(-3 + m)*(b^2*(-1 + m)*m*Hypergeometric2F1[1/2, 
 (-2 + m)/2, m/2, Sec[c + d*x]^2] + a*(-2 + m)*(2*b*m*Cos[c + d*x]*Hyperge 
ometric2F1[1/2, (-1 + m)/2, (1 + m)/2, Sec[c + d*x]^2] + a*(-1 + m)*Hyperg 
eometric2F1[1/2, m/2, (2 + m)/2, Sec[c + d*x]^2])*Sec[c + d*x]^2)*Sqrt[-Ta 
n[c + d*x]^2])/(d*(-2 + m)*(-1 + m)*m)
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.97, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 3717, 3042, 4275, 3042, 4259, 3042, 3122, 4534, 3042, 4259, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3717

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4275

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-2} \left (b^2+a^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx-\frac {2 a b \sin (c+d x) \sec ^{m-2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-m}{2},\frac {4-m}{2},\cos ^2(c+d x)\right )}{d (2-m) \sqrt {\sin ^2(c+d x)}}\)

\(\Big \downarrow \) 4534

\(\displaystyle \left (\frac {a^2 (2-m)}{1-m}+b^2\right ) \int \sec ^{m-2}(c+d x)dx-\frac {a^2 \sin (c+d x) \sec ^{m-1}(c+d x)}{d (1-m)}-\frac {2 a b \sin (c+d x) \sec ^{m-2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-m}{2},\frac {4-m}{2},\cos ^2(c+d x)\right )}{d (2-m) \sqrt {\sin ^2(c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \left (\frac {a^2 (2-m)}{1-m}+b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-2}dx-\frac {a^2 \sin (c+d x) \sec ^{m-1}(c+d x)}{d (1-m)}-\frac {2 a b \sin (c+d x) \sec ^{m-2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-m}{2},\frac {4-m}{2},\cos ^2(c+d x)\right )}{d (2-m) \sqrt {\sin ^2(c+d x)}}\)

\(\Big \downarrow \) 4259

\(\displaystyle \left (\frac {a^2 (2-m)}{1-m}+b^2\right ) \cos ^m(c+d x) \sec ^m(c+d x) \int \cos ^{2-m}(c+d x)dx-\frac {a^2 \sin (c+d x) \sec ^{m-1}(c+d x)}{d (1-m)}-\frac {2 a b \sin (c+d x) \sec ^{m-2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-m}{2},\frac {4-m}{2},\cos ^2(c+d x)\right )}{d (2-m) \sqrt {\sin ^2(c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \left (\frac {a^2 (2-m)}{1-m}+b^2\right ) \cos ^m(c+d x) \sec ^m(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{2-m}dx-\frac {a^2 \sin (c+d x) \sec ^{m-1}(c+d x)}{d (1-m)}-\frac {2 a b \sin (c+d x) \sec ^{m-2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-m}{2},\frac {4-m}{2},\cos ^2(c+d x)\right )}{d (2-m) \sqrt {\sin ^2(c+d x)}}\)

\(\Big \downarrow \) 3122

\(\displaystyle -\frac {\left (\frac {a^2 (2-m)}{1-m}+b^2\right ) \sin (c+d x) \sec ^{m-3}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-m}{2},\frac {5-m}{2},\cos ^2(c+d x)\right )}{d (3-m) \sqrt {\sin ^2(c+d x)}}-\frac {a^2 \sin (c+d x) \sec ^{m-1}(c+d x)}{d (1-m)}-\frac {2 a b \sin (c+d x) \sec ^{m-2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-m}{2},\frac {4-m}{2},\cos ^2(c+d x)\right )}{d (2-m) \sqrt {\sin ^2(c+d x)}}\)

Input:

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

Output:

-((a^2*Sec[c + d*x]^(-1 + m)*Sin[c + d*x])/(d*(1 - m))) - ((b^2 + (a^2*(2 
- m))/(1 - m))*Hypergeometric2F1[1/2, (3 - m)/2, (5 - m)/2, Cos[c + d*x]^2 
]*Sec[c + d*x]^(-3 + m)*Sin[c + d*x])/(d*(3 - m)*Sqrt[Sin[c + d*x]^2]) - ( 
2*a*b*Hypergeometric2F1[1/2, (2 - m)/2, (4 - m)/2, Cos[c + d*x]^2]*Sec[c + 
 d*x]^(-2 + m)*Sin[c + d*x])/(d*(2 - m)*Sqrt[Sin[c + d*x]^2])
 

Defintions of rubi rules used

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 3717

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p)   Int[(d*Csc[e + f*x])^(m - n*p 
)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && 
  !IntegerQ[m] && IntegersQ[n, p]
 

rule 4259

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^(n - 1)*((Sin[c + d*x]/b)^(n - 1)   Int[1/(Sin[c + d*x]/b)^n, x]), x] /; 
FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]
 

rule 4275

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^2, x_Symbol] :> Simp[2*a*(b/d)   Int[(d*Csc[e + f*x])^(n + 1), x], x] 
 + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b, d, 
 e, f, n}, x]
 

rule 4534

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) 
)), x] + Simp[(C*m + A*(m + 1))/(m + 1)   Int[(b*Csc[e + f*x])^m, x], x] /; 
 FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*sec(d*x + c)^m, x 
)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^2 \sec ^m(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2 \,d x \] Input:

int((1/cos(c + d*x))^m*(a + b*cos(c + d*x))^2,x)
 

Output:

int((1/cos(c + d*x))^m*(a + b*cos(c + d*x))^2, x)
 

Reduce [F]

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

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

Output:

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