\(\int (a+b \cos (e+f x))^{3/2} (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx\) [641]

Optimal result

Integrand size = 35, antiderivative size = 35 \[ \int (a+b \cos (e+f x))^{3/2} (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=\frac {2 b B \cos (e+f x) \sqrt {a+b \cos (e+f x)} (c \sec (e+f x))^m \sin (e+f x)}{f (5-2 m)}+\frac {2 (c \cos (e+f x))^m (c \sec (e+f x))^m \text {Int}\left (\frac {(c \cos (e+f x))^{-m} \left (\frac {1}{2} a c \left (2 b B (1-m)+2 a A \left (\frac {5}{2}-m\right )\right )+\frac {1}{2} c \left (b^2 B (3-2 m)+a (2 A b+a B) (5-2 m)\right ) \cos (e+f x)+\frac {1}{2} b c (A b (5-2 m)+2 a B (3-m)) \cos ^2(e+f x)\right )}{\sqrt {a+b \cos (e+f x)}},x\right )}{c (5-2 m)} \] Output:

2*b*B*cos(f*x+e)*(a+b*cos(f*x+e))^(1/2)*(c*sec(f*x+e))^m*sin(f*x+e)/f/(5-2 
*m)+2*(c*cos(f*x+e))^m*(c*sec(f*x+e))^m*Defer(Int)((1/2*a*c*(2*b*B*(1-m)+2 
*a*A*(5/2-m))+1/2*c*(b^2*B*(3-2*m)+a*(2*A*b+B*a)*(5-2*m))*cos(f*x+e)+1/2*b 
*c*(A*b*(5-2*m)+2*a*B*(3-m))*cos(f*x+e)^2)/((c*cos(f*x+e))^m)/(a+b*cos(f*x 
+e))^(1/2),x)/c/(5-2*m)
 

Mathematica [F(-1)]

Timed out. \[ \int (a+b \cos (e+f x))^{3/2} (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=\text {\$Aborted} \] Input:

Integrate[(a + b*Cos[e + f*x])^(3/2)*(A + B*Cos[e + f*x])*(c*Sec[e + f*x]) 
^m,x]
 

Output:

$Aborted
 

Rubi [N/A]

Not integrable

Time = 1.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {3042, 3440, 3042, 3469, 27, 3042, 3544}

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.

\(\displaystyle \int (a+b \cos (e+f x))^{3/2} (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a+b \sin \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2} \left (A+B \sin \left (e+f x+\frac {\pi }{2}\right )\right ) \left (c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^mdx\)

\(\Big \downarrow \) 3440

\(\displaystyle (c \cos (e+f x))^m (c \sec (e+f x))^m \int (c \cos (e+f x))^{-m} (a+b \cos (e+f x))^{3/2} (A+B \cos (e+f x))dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (c \cos (e+f x))^m (c \sec (e+f x))^m \int \left (c \sin \left (e+f x+\frac {\pi }{2}\right )\right )^{-m} \left (a+b \sin \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2} \left (A+B \sin \left (e+f x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 3469

\(\displaystyle (c \cos (e+f x))^m (c \sec (e+f x))^m \left (\frac {2 \int \frac {(c \cos (e+f x))^{-m} \left (b c (A b (5-2 m)+2 a B (3-m)) \cos ^2(e+f x)+c \left (B (3-2 m) b^2+a (2 A b+a B) (5-2 m)\right ) \cos (e+f x)+a c (a A (5-2 m)+2 b B (1-m))\right )}{2 \sqrt {a+b \cos (e+f x)}}dx}{c (5-2 m)}+\frac {2 b B \sin (e+f x) \sqrt {a+b \cos (e+f x)} (c \cos (e+f x))^{1-m}}{c f (5-2 m)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle (c \cos (e+f x))^m (c \sec (e+f x))^m \left (\frac {\int \frac {(c \cos (e+f x))^{-m} \left (b c (A b (5-2 m)+2 a B (3-m)) \cos ^2(e+f x)+c \left (B (3-2 m) b^2+a (2 A b+a B) (5-2 m)\right ) \cos (e+f x)+a c (a A (5-2 m)+2 b B (1-m))\right )}{\sqrt {a+b \cos (e+f x)}}dx}{c (5-2 m)}+\frac {2 b B \sin (e+f x) \sqrt {a+b \cos (e+f x)} (c \cos (e+f x))^{1-m}}{c f (5-2 m)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle (c \cos (e+f x))^m (c \sec (e+f x))^m \left (\frac {\int \frac {\left (c \sin \left (e+f x+\frac {\pi }{2}\right )\right )^{-m} \left (b c (A b (5-2 m)+2 a B (3-m)) \sin \left (e+f x+\frac {\pi }{2}\right )^2+c \left (B (3-2 m) b^2+a (2 A b+a B) (5-2 m)\right ) \sin \left (e+f x+\frac {\pi }{2}\right )+a c (a A (5-2 m)+2 b B (1-m))\right )}{\sqrt {a+b \sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{c (5-2 m)}+\frac {2 b B \sin (e+f x) \sqrt {a+b \cos (e+f x)} (c \cos (e+f x))^{1-m}}{c f (5-2 m)}\right )\)

\(\Big \downarrow \) 3544

\(\displaystyle (c \cos (e+f x))^m (c \sec (e+f x))^m \left (\frac {\int \frac {(c \cos (e+f x))^{-m} \left (b c (A b (5-2 m)+2 a B (3-m)) \cos ^2(e+f x)+c \left (B (3-2 m) b^2+a (2 A b+a B) (5-2 m)\right ) \cos (e+f x)+a c (a A (5-2 m)+2 b B (1-m))\right )}{\sqrt {a+b \cos (e+f x)}}dx}{c (5-2 m)}+\frac {2 b B \sin (e+f x) \sqrt {a+b \cos (e+f x)} (c \cos (e+f x))^{1-m}}{c f (5-2 m)}\right )\)

Input:

Int[(a + b*Cos[e + f*x])^(3/2)*(A + B*Cos[e + f*x])*(c*Sec[e + f*x])^m,x]
 

Output:

$Aborted
 

Defintions of rubi rules used

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 3042

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

rule 3440

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p   Int[(a + b*Sin[e + f*x])^m*((c + 
d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g 
, m, n, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && I 
ntegerQ[n])
 

rule 3469

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

rule 3544

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

Maple [N/A]

Not integrable

Time = 0.23 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94

\[\int \left (a +b \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (A +B \cos \left (f x +e \right )\right ) \left (c \sec \left (f x +e \right )\right )^{m}d x\]

Input:

int((a+b*cos(f*x+e))^(3/2)*(A+B*cos(f*x+e))*(c*sec(f*x+e))^m,x)
 

Output:

int((a+b*cos(f*x+e))^(3/2)*(A+B*cos(f*x+e))*(c*sec(f*x+e))^m,x)
 

Fricas [N/A]

Not integrable

Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.54 \[ \int (a+b \cos (e+f x))^{3/2} (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=\int { {\left (B \cos \left (f x + e\right ) + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (c \sec \left (f x + e\right )\right )^{m} \,d x } \] Input:

integrate((a+b*cos(f*x+e))^(3/2)*(A+B*cos(f*x+e))*(c*sec(f*x+e))^m,x, algo 
rithm="fricas")
 

Output:

integral((B*b*cos(f*x + e)^2 + A*a + (B*a + A*b)*cos(f*x + e))*sqrt(b*cos( 
f*x + e) + a)*(c*sec(f*x + e))^m, x)
 

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (e+f x))^{3/2} (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(f*x+e))**(3/2)*(A+B*cos(f*x+e))*(c*sec(f*x+e))**m,x)
 

Output:

Timed out
 

Maxima [N/A]

Not integrable

Time = 2.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int (a+b \cos (e+f x))^{3/2} (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=\int { {\left (B \cos \left (f x + e\right ) + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (c \sec \left (f x + e\right )\right )^{m} \,d x } \] Input:

integrate((a+b*cos(f*x+e))^(3/2)*(A+B*cos(f*x+e))*(c*sec(f*x+e))^m,x, algo 
rithm="maxima")
 

Output:

integrate((B*cos(f*x + e) + A)*(b*cos(f*x + e) + a)^(3/2)*(c*sec(f*x + e)) 
^m, x)
 

Giac [N/A]

Not integrable

Time = 1.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int (a+b \cos (e+f x))^{3/2} (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=\int { {\left (B \cos \left (f x + e\right ) + A\right )} {\left (b \cos \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (c \sec \left (f x + e\right )\right )^{m} \,d x } \] Input:

integrate((a+b*cos(f*x+e))^(3/2)*(A+B*cos(f*x+e))*(c*sec(f*x+e))^m,x, algo 
rithm="giac")
 

Output:

integrate((B*cos(f*x + e) + A)*(b*cos(f*x + e) + a)^(3/2)*(c*sec(f*x + e)) 
^m, x)
 

Mupad [N/A]

Not integrable

Time = 45.44 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int (a+b \cos (e+f x))^{3/2} (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=\int {\left (\frac {c}{\cos \left (e+f\,x\right )}\right )}^m\,\left (A+B\,\cos \left (e+f\,x\right )\right )\,{\left (a+b\,\cos \left (e+f\,x\right )\right )}^{3/2} \,d x \] Input:

int((c/cos(e + f*x))^m*(A + B*cos(e + f*x))*(a + b*cos(e + f*x))^(3/2),x)
 

Output:

int((c/cos(e + f*x))^m*(A + B*cos(e + f*x))*(a + b*cos(e + f*x))^(3/2), x)
 

Reduce [N/A]

Not integrable

Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.77 \[ \int (a+b \cos (e+f x))^{3/2} (A+B \cos (e+f x)) (c \sec (e+f x))^m \, dx=c^{m} \left (2 \left (\int \sec \left (f x +e \right )^{m} \sqrt {\cos \left (f x +e \right ) b +a}\, \cos \left (f x +e \right )d x \right ) a b +\left (\int \sec \left (f x +e \right )^{m} \sqrt {\cos \left (f x +e \right ) b +a}\, \cos \left (f x +e \right )^{2}d x \right ) b^{2}+\left (\int \sec \left (f x +e \right )^{m} \sqrt {\cos \left (f x +e \right ) b +a}d x \right ) a^{2}\right ) \] Input:

int((a+b*cos(f*x+e))^(3/2)*(A+B*cos(f*x+e))*(c*sec(f*x+e))^m,x)
 

Output:

c**m*(2*int(sec(e + f*x)**m*sqrt(cos(e + f*x)*b + a)*cos(e + f*x),x)*a*b + 
 int(sec(e + f*x)**m*sqrt(cos(e + f*x)*b + a)*cos(e + f*x)**2,x)*b**2 + in 
t(sec(e + f*x)**m*sqrt(cos(e + f*x)*b + a),x)*a**2)