\(\int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx\) [22]

Optimal result

Integrand size = 19, antiderivative size = 72 \[ \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {5 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {3 a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^3 \tan ^3(c+d x)}{3 d} \] Output:

5/2*a^3*arctanh(sin(d*x+c))/d+4*a^3*tan(d*x+c)/d+3/2*a^3*sec(d*x+c)*tan(d* 
x+c)/d+1/3*a^3*tan(d*x+c)^3/d
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.19 \[ \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {a^3 \coth ^{-1}(\sin (c+d x))}{d}+\frac {3 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {3 a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^3 \tan ^3(c+d x)}{3 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 4278, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4278

\(\displaystyle \int \left (a^3 \sec ^4(c+d x)+3 a^3 \sec ^3(c+d x)+3 a^3 \sec ^2(c+d x)+a^3 \sec (c+d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {5 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^3 \tan ^3(c+d x)}{3 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {3 a^3 \tan (c+d x) \sec (c+d x)}{2 d}\)

Input:

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

Output:

(5*a^3*ArcTanh[Sin[c + d*x]])/(2*d) + (4*a^3*Tan[c + d*x])/d + (3*a^3*Sec[ 
c + d*x]*Tan[c + d*x])/(2*d) + (a^3*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 4278

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Int[ExpandTrig[(a + b*csc[e + f*x])^m*(d*csc[e + f 
*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && I 
GtQ[m, 0] && RationalQ[n]
 

Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.31

method	result	size
derivativedivides	\(\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} \tan \left (d x +c \right )+3 a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(94\)
default	\(\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} \tan \left (d x +c \right )+3 a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(94\)
parts	\(\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {3 a^{3} \tan \left (d x +c \right )}{d}+\frac {3 a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(102\)
risch	\(-\frac {i a^{3} \left (9 \,{\mathrm e}^{5 i \left (d x +c \right )}-18 \,{\mathrm e}^{4 i \left (d x +c \right )}-48 \,{\mathrm e}^{2 i \left (d x +c \right )}-9 \,{\mathrm e}^{i \left (d x +c \right )}-22\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\)	\(112\)
norman	\(\frac {-\frac {11 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {40 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {5 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\frac {5 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {5 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(114\)
parallelrisch	\(\frac {a^{3} \left (15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (3 d x +3 c \right )-15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (3 d x +3 c \right )+45 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )-45 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+22 \sin \left (3 d x +3 c \right )+30 \sin \left (d x +c \right )+18 \sin \left (2 d x +2 c \right )\right )}{6 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(146\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.36 \[ \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {15 \, a^{3} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, a^{3} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (22 \, a^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:

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

Output:

1/12*(15*a^3*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 15*a^3*cos(d*x + c)^3* 
log(-sin(d*x + c) + 1) + 2*(22*a^3*cos(d*x + c)^2 + 9*a^3*cos(d*x + c) + 2 
*a^3)*sin(d*x + c))/(d*cos(d*x + c)^3)
 

Sympy [F]

\[ \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx=a^{3} \left (\int \sec {\left (c + d x \right )}\, dx + \int 3 \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 \sec ^{3}{\left (c + d x \right )}\, dx + \int \sec ^{4}{\left (c + d x \right )}\, dx\right ) \] Input:

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

Output:

a**3*(Integral(sec(c + d*x), x) + Integral(3*sec(c + d*x)**2, x) + Integra 
l(3*sec(c + d*x)**3, x) + Integral(sec(c + d*x)**4, x))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.44 \[ \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} - 9 \, a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 36 \, a^{3} \tan \left (d x + c\right )}{12 \, d} \] Input:

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

Output:

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^3 - 9*a^3*(2*sin(d*x + c)/(sin 
(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 12*a^3 
*log(sec(d*x + c) + tan(d*x + c)) + 36*a^3*tan(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.47 \[ \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:

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

Output:

1/6*(15*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*a^3*log(abs(tan(1/2*d* 
x + 1/2*c) - 1)) - 2*(15*a^3*tan(1/2*d*x + 1/2*c)^5 - 40*a^3*tan(1/2*d*x + 
 1/2*c)^3 + 33*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
 

Mupad [B] (verification not implemented)

Time = 11.72 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.56 \[ \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {5\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {5\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {40\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+11\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:

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

Output:

(5*a^3*atanh(tan(c/2 + (d*x)/2)))/d - (5*a^3*tan(c/2 + (d*x)/2)^5 - (40*a^ 
3*tan(c/2 + (d*x)/2)^3)/3 + 11*a^3*tan(c/2 + (d*x)/2))/(d*(3*tan(c/2 + (d* 
x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.18 \[ \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {a^{3} \left (-15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}-15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-9 \cos \left (d x +c \right ) \sin \left (d x +c \right )+22 \sin \left (d x +c \right )^{3}-24 \sin \left (d x +c \right )\right )}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

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

Output:

(a**3*( - 15*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + 15*c 
os(c + d*x)*log(tan((c + d*x)/2) - 1) + 15*cos(c + d*x)*log(tan((c + d*x)/ 
2) + 1)*sin(c + d*x)**2 - 15*cos(c + d*x)*log(tan((c + d*x)/2) + 1) - 9*co 
s(c + d*x)*sin(c + d*x) + 22*sin(c + d*x)**3 - 24*sin(c + d*x)))/(6*cos(c 
+ d*x)*d*(sin(c + d*x)**2 - 1))