\(\int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx\) [426]

Optimal result

Integrand size = 21, antiderivative size = 110 \[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\frac {\left (3 a^2+4 b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {2 a b \tan (c+d x)}{d}+\frac {\left (3 a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {2 a b \tan ^3(c+d x)}{3 d} \] Output:

1/8*(3*a^2+4*b^2)*arctanh(sin(d*x+c))/d+2*a*b*tan(d*x+c)/d+1/8*(3*a^2+4*b^ 
2)*sec(d*x+c)*tan(d*x+c)/d+1/4*a^2*sec(d*x+c)^3*tan(d*x+c)/d+2/3*a*b*tan(d 
*x+c)^3/d
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.75 \[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\frac {3 \left (3 a^2+4 b^2\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (3 \left (3 a^2+4 b^2\right ) \sec (c+d x)+6 a^2 \sec ^3(c+d x)+16 a b \left (3+\tan ^2(c+d x)\right )\right )}{24 d} \] Input:

Integrate[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^5,x]
 

Output:

(3*(3*a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*(3*a^2 + 4*b^2) 
*Sec[c + d*x] + 6*a^2*Sec[c + d*x]^3 + 16*a*b*(3 + Tan[c + d*x]^2)))/(24*d 
)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.93, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3268, 3042, 3491, 3042, 4254, 2009, 4255, 3042, 4257}

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.

\(\displaystyle \int \sec ^5(c+d x) (a+b \cos (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\)

\(\Big \downarrow \) 3268

\(\displaystyle \int \left (a^2+b^2 \cos ^2(c+d x)\right ) \sec ^5(c+d x)dx+2 a b \int \sec ^4(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {a^2+b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx+2 a b \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\)

\(\Big \downarrow \) 3491

\(\displaystyle \frac {1}{4} \left (3 a^2+4 b^2\right ) \int \sec ^3(c+d x)dx+2 a b \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (3 a^2+4 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+2 a b \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {1}{4} \left (3 a^2+4 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {2 a b \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} \left (3 a^2+4 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {2 a b \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\)

\(\Big \downarrow \) 4255

\(\displaystyle \frac {1}{4} \left (3 a^2+4 b^2\right ) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {2 a b \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (3 a^2+4 b^2\right ) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {2 a b \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{4} \left (3 a^2+4 b^2\right ) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {2 a b \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\)

Input:

Int[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^5,x]
 

Output:

(a^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((3*a^2 + 4*b^2)*(ArcTanh[Sin[c 
+ d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4 - (2*a*b*(-Tan[c + d 
*x] - Tan[c + d*x]^3/3))/d
 

Defintions of rubi rules used

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 3268

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)])^2, x_Symbol] :> Simp[2*c*(d/b)   Int[(b*Sin[e + f*x])^(m + 1), x], x] 
+ Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c, d, 
e, f, m}, x]
 

rule 3491

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

rule 4254

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 5.72 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.01

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-2 a b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(111\)
default	\(\frac {a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-2 a b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(111\)
parts	\(\frac {a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {2 a b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(116\)
parallelrisch	\(\frac {-36 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+\frac {4 b^{2}}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+36 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2}+\frac {4 b^{2}}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (18 a^{2}+24 b^{2}\right ) \sin \left (3 d x +3 c \right )+128 a b \sin \left (2 d x +2 c \right )+32 a b \sin \left (4 d x +4 c \right )+\left (66 a^{2}+24 b^{2}\right ) \sin \left (d x +c \right )}{24 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(186\)
risch	\(-\frac {i \left (9 \,{\mathrm e}^{7 i \left (d x +c \right )} a^{2}+12 b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+33 \,{\mathrm e}^{5 i \left (d x +c \right )} a^{2}+12 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-96 \,{\mathrm e}^{4 i \left (d x +c \right )} a b -33 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-12 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-128 \,{\mathrm e}^{2 i \left (d x +c \right )} a b -9 a^{2} {\mathrm e}^{i \left (d x +c \right )}-12 b^{2} {\mathrm e}^{i \left (d x +c \right )}-32 a b \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{2 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{2 d}\)	\(248\)
norman	\(\frac {\frac {\left (5 a^{2}-16 a b +4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {\left (5 a^{2}+16 a b +4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (21 a^{2}-16 a b -12 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 d}+\frac {\left (21 a^{2}+16 a b -12 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {\left (39 a^{2}-16 a b +12 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}+\frac {\left (39 a^{2}+16 a b +12 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {\left (3 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (3 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(274\)

Input:

int((a+cos(d*x+c)*b)^2*sec(d*x+c)^5,x,method=_RETURNVERBOSE)
 

Output:

1/d*(a^2*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c) 
+tan(d*x+c)))-2*a*b*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+b^2*(1/2*sec(d*x+c) 
*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c))))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.21 \[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\frac {3 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (32 \, a b \cos \left (d x + c\right )^{3} + 16 \, a b \cos \left (d x + c\right ) + 3 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \, a^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:

integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^5,x, algorithm="fricas")
 

Output:

1/48*(3*(3*a^2 + 4*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(3*a^2 + 
4*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(32*a*b*cos(d*x + c)^3 + 
16*a*b*cos(d*x + c) + 3*(3*a^2 + 4*b^2)*cos(d*x + c)^2 + 6*a^2)*sin(d*x + 
c))/(d*cos(d*x + c)^4)
 

Sympy [F]

\[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right )^{2} \sec ^{5}{\left (c + d x \right )}\, dx \] Input:

integrate((a+b*cos(d*x+c))**2*sec(d*x+c)**5,x)
 

Output:

Integral((a + b*cos(c + d*x))**2*sec(c + d*x)**5, x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.31 \[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b - 3 \, a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \] Input:

integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^5,x, algorithm="maxima")
 

Output:

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*a*b - 3*a^2*(2*(3*sin(d*x + c)^ 
3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d* 
x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 12*b^2*(2*sin(d*x + c)/(sin(d*x + 
 c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (102) = 204\).

Time = 0.54 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.35 \[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\frac {3 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 80 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \] Input:

integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^5,x, algorithm="giac")
 

Output:

1/24*(3*(3*a^2 + 4*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*a^2 + 4* 
b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(15*a^2*tan(1/2*d*x + 1/2*c)^7 
 - 48*a*b*tan(1/2*d*x + 1/2*c)^7 + 12*b^2*tan(1/2*d*x + 1/2*c)^7 + 9*a^2*t 
an(1/2*d*x + 1/2*c)^5 + 80*a*b*tan(1/2*d*x + 1/2*c)^5 - 12*b^2*tan(1/2*d*x 
 + 1/2*c)^5 + 9*a^2*tan(1/2*d*x + 1/2*c)^3 - 80*a*b*tan(1/2*d*x + 1/2*c)^3 
 - 12*b^2*tan(1/2*d*x + 1/2*c)^3 + 15*a^2*tan(1/2*d*x + 1/2*c) + 48*a*b*ta 
n(1/2*d*x + 1/2*c) + 12*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 
- 1)^4)/d
 

Mupad [B] (verification not implemented)

Time = 42.13 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.67 \[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a^2}{4}+b^2\right )}{d}+\frac {\left (\frac {5\,a^2}{4}-4\,a\,b+b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,a^2}{4}+\frac {20\,a\,b}{3}-b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,a^2}{4}-\frac {20\,a\,b}{3}-b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,a^2}{4}+4\,a\,b+b^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:

int((a + b*cos(c + d*x))^2/cos(c + d*x)^5,x)
 

Output:

(atanh(tan(c/2 + (d*x)/2))*((3*a^2)/4 + b^2))/d + (tan(c/2 + (d*x)/2)^5*(( 
20*a*b)/3 + (3*a^2)/4 - b^2) + tan(c/2 + (d*x)/2)*(4*a*b + (5*a^2)/4 + b^2 
) + tan(c/2 + (d*x)/2)^7*((5*a^2)/4 - 4*a*b + b^2) - tan(c/2 + (d*x)/2)^3* 
((20*a*b)/3 - (3*a^2)/4 + b^2))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + ( 
d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 482, normalized size of antiderivative = 4.38 \[ \int (a+b \cos (c+d x))^2 \sec ^5(c+d x) \, dx=\frac {-9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a^{2}-12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b^{2}+18 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2}+24 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{2}-9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}-12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}+9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a^{2}+12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b^{2}-18 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2}-24 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{2}+9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}+12 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}-9 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2}-12 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{2}+15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2}+12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{2}+32 \sin \left (d x +c \right )^{5} a b -80 \sin \left (d x +c \right )^{3} a b +48 \sin \left (d x +c \right ) a b}{24 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:

int((a+b*cos(d*x+c))^2*sec(d*x+c)^5,x)
 

Output:

( - 9*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2 - 12*cos 
(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b**2 + 18*cos(c + d*x) 
*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2 + 24*cos(c + d*x)*log(tan( 
(c + d*x)/2) - 1)*sin(c + d*x)**2*b**2 - 9*cos(c + d*x)*log(tan((c + d*x)/ 
2) - 1)*a**2 - 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b**2 + 9*cos(c + 
d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**2 + 12*cos(c + d*x)*log( 
tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b**2 - 18*cos(c + d*x)*log(tan((c + 
d*x)/2) + 1)*sin(c + d*x)**2*a**2 - 24*cos(c + d*x)*log(tan((c + d*x)/2) + 
 1)*sin(c + d*x)**2*b**2 + 9*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2 + 
 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b**2 - 9*cos(c + d*x)*sin(c + d 
*x)**3*a**2 - 12*cos(c + d*x)*sin(c + d*x)**3*b**2 + 15*cos(c + d*x)*sin(c 
 + d*x)*a**2 + 12*cos(c + d*x)*sin(c + d*x)*b**2 + 32*sin(c + d*x)**5*a*b 
- 80*sin(c + d*x)**3*a*b + 48*sin(c + d*x)*a*b)/(24*cos(c + d*x)*d*(sin(c 
+ d*x)**4 - 2*sin(c + d*x)**2 + 1))