\(\int (a+b \cos (c+d x))^2 \sec ^3(c+d x) \, dx\) [424]

Optimal result

Integrand size = 21, antiderivative size = 59 \[ \int (a+b \cos (c+d x))^2 \sec ^3(c+d x) \, dx=\frac {\left (a^2+2 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a b \tan (c+d x)}{d}+\frac {a^2 \sec (c+d x) \tan (c+d x)}{2 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.14 \[ \int (a+b \cos (c+d x))^2 \sec ^3(c+d x) \, dx=\frac {b^2 \coth ^{-1}(\sin (c+d x))}{d}+\frac {a^2 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a b \tan (c+d x)}{d}+\frac {a^2 \sec (c+d x) \tan (c+d x)}{2 d} \] Input:

Integrate[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3268, 3042, 3491, 3042, 4254, 24, 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.

\(\displaystyle \int \sec ^3(c+d x) (a+b \cos (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3268

\(\displaystyle \int \left (a^2+b^2 \cos ^2(c+d x)\right ) \sec ^3(c+d x)dx+2 a b \int \sec ^2(c+d x)dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3491

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4257

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

Input:

Int[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^3,x]
 

Output:

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

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

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 4257

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

Maple [A] (verified)

Time = 4.46 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a b \tan \left (d x +c \right )+b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(69\)
default	\(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a b \tan \left (d x +c \right )+b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(69\)
parts	\(\frac {a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 a b \tan \left (d x +c \right )}{d}\)	\(74\)
parallelrisch	\(\frac {-\left (a^{2}+2 b^{2}\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a^{2}+2 b^{2}\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 a^{2} \sin \left (d x +c \right )+4 a b \sin \left (2 d x +2 c \right )}{2 d \left (\cos \left (2 d x +2 c \right )+1\right )}\)	\(111\)
risch	\(-\frac {i a \left ({\mathrm e}^{3 i \left (d x +c \right )} a -4 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a \,{\mathrm e}^{i \left (d x +c \right )}-4 b \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{d}\)	\(144\)
norman	\(\frac {\frac {a \left (a -4 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {a \left (3 a -4 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {a \left (3 a +4 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {a \left (4 b +a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {\left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(172\)

Input:

int((a+cos(d*x+c)*b)^2*sec(d*x+c)^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(a^2*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+2*a*b*t 
an(d*x+c)+b^2*ln(sec(d*x+c)+tan(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.58 \[ \int (a+b \cos (c+d x))^2 \sec ^3(c+d x) \, dx=\frac {{\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, a b \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^3,x, algorithm="fricas")
 

Output:

1/4*((a^2 + 2*b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (a^2 + 2*b^2)*co 
s(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(4*a*b*cos(d*x + c) + a^2)*sin(d*x 
 + c))/(d*cos(d*x + c)^2)
 

Sympy [F]

\[ \int (a+b \cos (c+d x))^2 \sec ^3(c+d x) \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \] Input:

integrate((a+b*cos(d*x+c))**2*sec(d*x+c)**3,x)
 

Output:

Integral((a + b*cos(c + d*x))**2*sec(c + d*x)**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.47 \[ \int (a+b \cos (c+d x))^2 \sec ^3(c+d x) \, dx=-\frac {a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 8 \, a b \tan \left (d x + c\right )}{4 \, d} \] Input:

integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^3,x, algorithm="maxima")
 

Output:

-1/4*(a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + l 
og(sin(d*x + c) - 1)) - 2*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 
1)) - 8*a*b*tan(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (55) = 110\).

Time = 0.51 (sec) , antiderivative size = 127, normalized size of antiderivative = 2.15 \[ \int (a+b \cos (c+d x))^2 \sec ^3(c+d x) \, dx=\frac {{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:

integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^3,x, algorithm="giac")
 

Output:

1/2*((a^2 + 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (a^2 + 2*b^2)*log( 
abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(a^2*tan(1/2*d*x + 1/2*c)^3 - 4*a*b*tan 
(1/2*d*x + 1/2*c)^3 + a^2*tan(1/2*d*x + 1/2*c) + 4*a*b*tan(1/2*d*x + 1/2*c 
))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d
 

Mupad [B] (verification not implemented)

Time = 40.64 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.68 \[ \int (a+b \cos (c+d x))^2 \sec ^3(c+d x) \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2+2\,b^2\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (4\,a\,b-a^2\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+4\,b\,a\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:

int((a + b*cos(c + d*x))^2/cos(c + d*x)^3,x)
 

Output:

(atanh(tan(c/2 + (d*x)/2))*(a^2 + 2*b^2))/d - (tan(c/2 + (d*x)/2)^3*(4*a*b 
 - a^2) - tan(c/2 + (d*x)/2)*(4*a*b + a^2))/(d*(tan(c/2 + (d*x)/2)^4 - 2*t 
an(c/2 + (d*x)/2)^2 + 1))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 279, normalized size of antiderivative = 4.73 \[ \int (a+b \cos (c+d x))^2 \sec ^3(c+d x) \, dx=\frac {-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2}-2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{2}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}+2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}+\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2}+2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{2}-\cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}-2 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}-\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2}+4 \sin \left (d x +c \right )^{3} a b -4 \sin \left (d x +c \right ) a b}{2 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int((a+b*cos(d*x+c))^2*sec(d*x+c)^3,x)
 

Output:

( - cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2 - 2*cos(c 
+ d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**2 + cos(c + d*x)*log(t 
an((c + d*x)/2) - 1)*a**2 + 2*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b**2 
+ cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2 + 2*cos(c + 
d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**2 - cos(c + d*x)*log(tan 
((c + d*x)/2) + 1)*a**2 - 2*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b**2 - 
cos(c + d*x)*sin(c + d*x)*a**2 + 4*sin(c + d*x)**3*a*b - 4*sin(c + d*x)*a* 
b)/(2*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))