\(\int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^4} \, dx\) [74]

Optimal result

Integrand size = 21, antiderivative size = 112 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {11 \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {13 \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {13 \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )} \] Output:

1/7*tan(d*x+c)/d/(a+a*sec(d*x+c))^4-11/35*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^ 
3+13/105*tan(d*x+c)/d/(a^2+a^2*sec(d*x+c))^2+13/105*tan(d*x+c)/d/(a^4+a^4* 
sec(d*x+c))
 

Mathematica [A] (verified)

Time = 1.31 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.78 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (35 \sin \left (\frac {d x}{2}\right )-35 \sin \left (c+\frac {d x}{2}\right )+2 \left (21 \sin \left (c+\frac {3 d x}{2}\right )+7 \sin \left (2 c+\frac {5 d x}{2}\right )+\sin \left (3 c+\frac {7 d x}{2}\right )\right )\right )}{1680 a^4 d} \] Input:

Integrate[Sec[c + d*x]^3/(a + a*Sec[c + d*x])^4,x]
 

Output:

(Sec[c/2]*Sec[(c + d*x)/2]^7*(35*Sin[(d*x)/2] - 35*Sin[c + (d*x)/2] + 2*(2 
1*Sin[c + (3*d*x)/2] + 7*Sin[2*c + (5*d*x)/2] + Sin[3*c + (7*d*x)/2])))/(1 
680*a^4*d)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4286, 25, 3042, 4488, 3042, 4283, 3042, 4281}

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.

\(\displaystyle \int \frac {\sec ^3(c+d x)}{(a \sec (c+d x)+a)^4} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4286

\(\displaystyle \frac {\int -\frac {\sec (c+d x) (4 a-7 a \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}+\frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {\int \frac {\sec (c+d x) (4 a-7 a \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (4 a-7 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}\)

\(\Big \downarrow \) 4488

\(\displaystyle \frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {\frac {11 a \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {13}{5} \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^2}dx}{7 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {\frac {11 a \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {13}{5} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{7 a^2}\)

\(\Big \downarrow \) 4283

\(\displaystyle \frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {\frac {11 a \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {13}{5} \left (\frac {\int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{7 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {\frac {11 a \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {13}{5} \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{7 a^2}\)

\(\Big \downarrow \) 4281

\(\displaystyle \frac {\tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {\frac {11 a \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac {13}{5} \left (\frac {\tan (c+d x)}{3 a d (a \sec (c+d x)+a)}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{7 a^2}\)

Input:

Int[Sec[c + d*x]^3/(a + a*Sec[c + d*x])^4,x]
 

Output:

Tan[c + d*x]/(7*d*(a + a*Sec[c + d*x])^4) - ((11*a*Tan[c + d*x])/(5*d*(a + 
 a*Sec[c + d*x])^3) - (13*(Tan[c + d*x]/(3*d*(a + a*Sec[c + d*x])^2) + Tan 
[c + d*x]/(3*a*d*(a + a*Sec[c + d*x]))))/5)/(7*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

rule 4281

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} 
, x] && EqQ[a^2 - b^2, 0]
 

rule 4283

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ 
Symbol] :> Simp[b*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] 
 + Simp[(m + 1)/(a*(2*m + 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 
1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1) 
] && IntegerQ[2*m]
 

rule 4286

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), 
x_Symbol] :> Simp[b*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), 
x] - Simp[1/(a^2*(2*m + 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1) 
*(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[ 
a^2 - b^2, 0] && LtQ[m, -2^(-1)]
 

rule 4488

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(A*b - a*B)*Cot[e + 
f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[(a*B*m + A*b*(m + 
1))/(a*b*(2*m + 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] 
 /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] 
&& NeQ[a*B*m + A*b*(m + 1), 0] && LtQ[m, -2^(-1)]
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.51

method	result	size
parallelrisch	\(-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}-7\right )}{56 d \,a^{4}}\)	\(57\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(58\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}\)	\(58\)
risch	\(\frac {8 i \left (35 \,{\mathrm e}^{4 i \left (d x +c \right )}+35 \,{\mathrm e}^{3 i \left (d x +c \right )}+42 \,{\mathrm e}^{2 i \left (d x +c \right )}+14 \,{\mathrm e}^{i \left (d x +c \right )}+2\right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(69\)
norman	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}+\frac {31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{420 a d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{280 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{56 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} a^{3}}\)	\(133\)

Input:

int(sec(d*x+c)^3/(a+a*sec(d*x+c))^4,x,method=_RETURNVERBOSE)
 

Output:

-1/56*tan(1/2*d*x+1/2*c)*(tan(1/2*d*x+1/2*c)^6+7/5*tan(1/2*d*x+1/2*c)^4-7/ 
3*tan(1/2*d*x+1/2*c)^2-7)/d/a^4
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.88 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {{\left (8 \, \cos \left (d x + c\right )^{3} + 32 \, \cos \left (d x + c\right )^{2} + 52 \, \cos \left (d x + c\right ) + 13\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \] Input:

integrate(sec(d*x+c)^3/(a+a*sec(d*x+c))^4,x, algorithm="fricas")
 

Output:

1/105*(8*cos(d*x + c)^3 + 32*cos(d*x + c)^2 + 52*cos(d*x + c) + 13)*sin(d* 
x + c)/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + 
c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)
 

Sympy [F]

\[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {\sec ^{3}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:

integrate(sec(d*x+c)**3/(a+a*sec(d*x+c))**4,x)
 

Output:

Integral(sec(c + d*x)**3/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + 
d*x)**2 + 4*sec(c + d*x) + 1), x)/a**4
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.78 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{840 \, a^{4} d} \] Input:

integrate(sec(d*x+c)^3/(a+a*sec(d*x+c))^4,x, algorithm="maxima")
 

Output:

1/840*(105*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x + 
c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/(co 
s(d*x + c) + 1)^7)/(a^4*d)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.53 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^4} \, dx=-\frac {15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \] Input:

integrate(sec(d*x+c)^3/(a+a*sec(d*x+c))^4,x, algorithm="giac")
 

Output:

-1/840*(15*tan(1/2*d*x + 1/2*c)^7 + 21*tan(1/2*d*x + 1/2*c)^5 - 35*tan(1/2 
*d*x + 1/2*c)^3 - 105*tan(1/2*d*x + 1/2*c))/(a^4*d)
 

Mupad [B] (verification not implemented)

Time = 9.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.52 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (-15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+105\right )}{840\,a^4\,d} \] Input:

int(1/(cos(c + d*x)^3*(a + a/cos(c + d*x))^4),x)
 

Output:

(tan(c/2 + (d*x)/2)*(35*tan(c/2 + (d*x)/2)^2 - 21*tan(c/2 + (d*x)/2)^4 - 1 
5*tan(c/2 + (d*x)/2)^6 + 105))/(840*a^4*d)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.52 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+105\right )}{840 a^{4} d} \] Input:

int(sec(d*x+c)^3/(a+a*sec(d*x+c))^4,x)
 

Output:

(tan((c + d*x)/2)*( - 15*tan((c + d*x)/2)**6 - 21*tan((c + d*x)/2)**4 + 35 
*tan((c + d*x)/2)**2 + 105))/(840*a**4*d)