\(\int (a+b \cos (c+d x)) \sec ^m(c+d x) \, dx\) [777]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 19, antiderivative size = 143 \[ \int (a+b \cos (c+d x)) \sec ^m(c+d x) \, dx=-\frac {b \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-m}{2},\frac {4-m}{2},\cos ^2(c+d x)\right ) \sec ^{-2+m}(c+d x) \sin (c+d x)}{d (2-m) \sqrt {\sin ^2(c+d x)}}-\frac {a \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d (1-m) \sqrt {\sin ^2(c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.75 \[ \int (a+b \cos (c+d x)) \sec ^m(c+d x) \, dx=\frac {\csc (c+d x) \left (b m \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+m),\frac {1+m}{2},\sec ^2(c+d x)\right )+a (-1+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(c+d x)\right )\right ) \sec ^{-1+m}(c+d x) \sqrt {-\tan ^2(c+d x)}}{d (-1+m) m} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 3717, 3042, 4274, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3717

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

Input:

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

Output:

-((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])) - (a*Hyperge 
ometric2F1[1/2, (1 - m)/2, (3 - m)/2, Cos[c + d*x]^2]*Sec[c + d*x]^(-1 + m 
)*Sin[c + d*x])/(d*(1 - 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 4274
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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