\(\int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) [563]

Optimal result

Integrand size = 21, antiderivative size = 116 \[ \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {4 a \left (a^2+b^2\right ) \log (a+b \tan (c+d x))}{b^5 d}+\frac {\left (3 a^2+2 b^2\right ) \tan (c+d x)}{b^4 d}-\frac {a \tan ^2(c+d x)}{b^3 d}+\frac {\tan ^3(c+d x)}{3 b^2 d}-\frac {\left (a^2+b^2\right )^2}{b^5 d (a+b \tan (c+d x))} \] Output:

-4*a*(a^2+b^2)*ln(a+b*tan(d*x+c))/b^5/d+(3*a^2+2*b^2)*tan(d*x+c)/b^4/d-a*t 
an(d*x+c)^2/b^3/d+1/3*tan(d*x+c)^3/b^2/d-(a^2+b^2)^2/b^5/d/(a+b*tan(d*x+c) 
)
 

Mathematica [A] (verified)

Time = 1.99 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.05 \[ \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {4 b \left (2 a^2+b^2\right ) \tan (c+d x)-2 a b^2 \tan ^2(c+d x)+\frac {b^4 \sec ^4(c+d x)-4 \left (a^2+b^2\right ) \left (a^2+b^2+3 a^2 \log (a+b \tan (c+d x))+3 a b \log (a+b \tan (c+d x)) \tan (c+d x)\right )}{a+b \tan (c+d x)}}{3 b^5 d} \] Input:

Integrate[Sec[c + d*x]^6/(a + b*Tan[c + d*x])^2,x]
 

Output:

(4*b*(2*a^2 + b^2)*Tan[c + d*x] - 2*a*b^2*Tan[c + d*x]^2 + (b^4*Sec[c + d* 
x]^4 - 4*(a^2 + b^2)*(a^2 + b^2 + 3*a^2*Log[a + b*Tan[c + d*x]] + 3*a*b*Lo 
g[a + b*Tan[c + d*x]]*Tan[c + d*x]))/(a + b*Tan[c + d*x]))/(3*b^5*d)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.86, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3987, 27, 476, 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.

\(\displaystyle \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^6}{(a+b \tan (c+d x))^2}dx\)

\(\Big \downarrow \) 3987

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 476

\(\displaystyle \frac {\int \left (3 \left (\frac {2 b^2}{3 a^2}+1\right ) a^2-2 b \tan (c+d x) a-\frac {4 \left (a^2+b^2\right ) a}{a+b \tan (c+d x)}+b^2 \tan ^2(c+d x)+\frac {\left (a^2+b^2\right )^2}{(a+b \tan (c+d x))^2}\right )d(b \tan (c+d x))}{b^5 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b \left (3 a^2+2 b^2\right ) \tan (c+d x)-\frac {\left (a^2+b^2\right )^2}{a+b \tan (c+d x)}-4 a \left (a^2+b^2\right ) \log (a+b \tan (c+d x))-a b^2 \tan ^2(c+d x)+\frac {1}{3} b^3 \tan ^3(c+d x)}{b^5 d}\)

Input:

Int[Sec[c + d*x]^6/(a + b*Tan[c + d*x])^2,x]
 

Output:

(-4*a*(a^2 + b^2)*Log[a + b*Tan[c + d*x]] + b*(3*a^2 + 2*b^2)*Tan[c + d*x] 
 - a*b^2*Tan[c + d*x]^2 + (b^3*Tan[c + d*x]^3)/3 - (a^2 + b^2)^2/(a + b*Ta 
n[c + d*x]))/(b^5*d)
 

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 476

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ 
ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, 
 x] && IGtQ[p, 0]
 

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 3987

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[1/(b*f)   Subst[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), 
 x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b^2, 
0] && IntegerQ[m/2]
 

Maple [A] (verified)

Time = 58.23 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(\frac {\frac {\frac {\tan \left (d x +c \right )^{3} b^{2}}{3}-\tan \left (d x +c \right )^{2} a b +3 \tan \left (d x +c \right ) a^{2}+2 \tan \left (d x +c \right ) b^{2}}{b^{4}}-\frac {4 a \left (a^{2}+b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{5}}-\frac {a^{4}+2 b^{2} a^{2}+b^{4}}{b^{5} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(114\)
default	\(\frac {\frac {\frac {\tan \left (d x +c \right )^{3} b^{2}}{3}-\tan \left (d x +c \right )^{2} a b +3 \tan \left (d x +c \right ) a^{2}+2 \tan \left (d x +c \right ) b^{2}}{b^{4}}-\frac {4 a \left (a^{2}+b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{5}}-\frac {a^{4}+2 b^{2} a^{2}+b^{4}}{b^{5} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(114\)
risch	\(-\frac {8 i \left (2 b^{3}-3 i a^{3}-2 i a \,b^{2}+3 a^{2} b -3 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-6 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+4 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-5 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-9 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-9 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right ) b^{4} d}+\frac {4 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{5} d}+\frac {4 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{3} d}-\frac {4 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b^{5} d}-\frac {4 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b^{3} d}\)	\(342\)

Input:

int(sec(d*x+c)^6/(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/b^4*(1/3*tan(d*x+c)^3*b^2-tan(d*x+c)^2*a*b+3*tan(d*x+c)*a^2+2*tan(d 
*x+c)*b^2)-4*a/b^5*(a^2+b^2)*ln(a+b*tan(d*x+c))-1/b^5*(a^4+2*a^2*b^2+b^4)/ 
(a+b*tan(d*x+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (114) = 228\).

Time = 0.11 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.42 \[ \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {4 \, {\left (3 \, a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{4} - b^{4} - 2 \, {\left (3 \, a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left ({\left (a^{4} + a^{2} b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 6 \, {\left ({\left (a^{4} + a^{2} b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) + 2 \, {\left (a b^{3} \cos \left (d x + c\right ) - 2 \, {\left (3 \, a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{3 \, {\left (a b^{5} d \cos \left (d x + c\right )^{4} + b^{6} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right )\right )}} \] Input:

integrate(sec(d*x+c)^6/(a+b*tan(d*x+c))^2,x, algorithm="fricas")
 

Output:

-1/3*(4*(3*a^2*b^2 + 2*b^4)*cos(d*x + c)^4 - b^4 - 2*(3*a^2*b^2 + 2*b^4)*c 
os(d*x + c)^2 + 6*((a^4 + a^2*b^2)*cos(d*x + c)^4 + (a^3*b + a*b^3)*cos(d* 
x + c)^3*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*c 
os(d*x + c)^2 + b^2) - 6*((a^4 + a^2*b^2)*cos(d*x + c)^4 + (a^3*b + a*b^3) 
*cos(d*x + c)^3*sin(d*x + c))*log(cos(d*x + c)^2) + 2*(a*b^3*cos(d*x + c) 
- 2*(3*a^3*b + 2*a*b^3)*cos(d*x + c)^3)*sin(d*x + c))/(a*b^5*d*cos(d*x + c 
)^4 + b^6*d*cos(d*x + c)^3*sin(d*x + c))
 

Sympy [F]

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

integrate(sec(d*x+c)**6/(a+b*tan(d*x+c))**2,x)
 

Output:

Integral(sec(c + d*x)**6/(a + b*tan(c + d*x))**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.99 \[ \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {\frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}}{b^{6} \tan \left (d x + c\right ) + a b^{5}} - \frac {b^{2} \tan \left (d x + c\right )^{3} - 3 \, a b \tan \left (d x + c\right )^{2} + 3 \, {\left (3 \, a^{2} + 2 \, b^{2}\right )} \tan \left (d x + c\right )}{b^{4}} + \frac {12 \, {\left (a^{3} + a b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{5}}}{3 \, d} \] Input:

integrate(sec(d*x+c)^6/(a+b*tan(d*x+c))^2,x, algorithm="maxima")
 

Output:

-1/3*(3*(a^4 + 2*a^2*b^2 + b^4)/(b^6*tan(d*x + c) + a*b^5) - (b^2*tan(d*x 
+ c)^3 - 3*a*b*tan(d*x + c)^2 + 3*(3*a^2 + 2*b^2)*tan(d*x + c))/b^4 + 12*( 
a^3 + a*b^2)*log(b*tan(d*x + c) + a)/b^5)/d
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.18 \[ \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx=-\frac {4 \, {\left (a^{3} + a b^{2}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{5} d} - \frac {a^{4} + 2 \, a^{2} b^{2} + b^{4}}{{\left (b \tan \left (d x + c\right ) + a\right )} b^{5} d} + \frac {b^{4} d^{2} \tan \left (d x + c\right )^{3} - 3 \, a b^{3} d^{2} \tan \left (d x + c\right )^{2} + 9 \, a^{2} b^{2} d^{2} \tan \left (d x + c\right ) + 6 \, b^{4} d^{2} \tan \left (d x + c\right )}{3 \, b^{6} d^{3}} \] Input:

integrate(sec(d*x+c)^6/(a+b*tan(d*x+c))^2,x, algorithm="giac")
 

Output:

-4*(a^3 + a*b^2)*log(abs(b*tan(d*x + c) + a))/(b^5*d) - (a^4 + 2*a^2*b^2 + 
 b^4)/((b*tan(d*x + c) + a)*b^5*d) + 1/3*(b^4*d^2*tan(d*x + c)^3 - 3*a*b^3 
*d^2*tan(d*x + c)^2 + 9*a^2*b^2*d^2*tan(d*x + c) + 6*b^4*d^2*tan(d*x + c)) 
/(b^6*d^3)
 

Mupad [B] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.12 \[ \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,b^2\,d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {2}{b^2}+\frac {3\,a^2}{b^4}\right )}{d}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^2}{b^3\,d}-\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (4\,a^3+4\,a\,b^2\right )}{b^5\,d}-\frac {a^4+2\,a^2\,b^2+b^4}{b\,d\,\left (\mathrm {tan}\left (c+d\,x\right )\,b^5+a\,b^4\right )} \] Input:

int(1/(cos(c + d*x)^6*(a + b*tan(c + d*x))^2),x)
 

Output:

tan(c + d*x)^3/(3*b^2*d) + (tan(c + d*x)*(2/b^2 + (3*a^2)/b^4))/d - (a*tan 
(c + d*x)^2)/(b^3*d) - (log(a + b*tan(c + d*x))*(4*a*b^2 + 4*a^3))/(b^5*d) 
 - (a^4 + b^4 + 2*a^2*b^2)/(b*d*(a*b^4 + b^5*tan(c + d*x)))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 3952, normalized size of antiderivative = 34.07 \[ \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^2} \, dx =\text {Too large to display} \] Input:

int(sec(d*x+c)^6/(a+b*tan(d*x+c))^2,x)
 

Output:

(12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*tan(c + d*x)*a* 
*3*b**2 + 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*tan(c 
+ d*x)*a*b**4 + 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5* 
a**4*b + 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*a**2*b* 
*3 - 24*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*tan(c + d*x 
)*a**3*b**2 - 24*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*ta 
n(c + d*x)*a*b**4 - 24*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x) 
**3*a**4*b - 24*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a** 
2*b**3 + 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*tan(c + d* 
x)*a**3*b**2 + 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*tan( 
c + d*x)*a*b**4 + 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a 
**4*b + 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**2*b**3 + 
 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5*tan(c + d*x)*a* 
*3*b**2 + 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5*tan(c 
+ d*x)*a*b**4 + 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5* 
a**4*b + 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5*a**2*b* 
*3 - 24*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*tan(c + d*x 
)*a**3*b**2 - 24*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*ta 
n(c + d*x)*a*b**4 - 24*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x) 
**3*a**4*b - 24*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*...