\(\int \frac {\sec ^6(c+d x)}{a+b \tan (c+d x)} \, dx\) [552]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.82 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {12 \left (a^2+b^2\right )^2 \log (a+b \tan (c+d x))+3 b^4 \sec ^4(c+d x)-12 a b \left (a^2+2 b^2\right ) \tan (c+d x)+6 b^2 \left (a^2+b^2\right ) \tan ^2(c+d x)-4 a b^3 \tan ^3(c+d x)}{12 b^5 d} \] Input:

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

Output:

(12*(a^2 + b^2)^2*Log[a + b*Tan[c + d*x]] + 3*b^4*Sec[c + d*x]^4 - 12*a*b* 
(a^2 + 2*b^2)*Tan[c + d*x] + 6*b^2*(a^2 + b^2)*Tan[c + d*x]^2 - 4*a*b^3*Ta 
n[c + d*x]^3)/(12*b^5*d)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.89, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3987

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 476

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((a^2 + b^2)^2*Log[a + b*Tan[c + d*x]] - a*b*(a^2 + 2*b^2)*Tan[c + d*x] + 
(b^2*(a^2 + 2*b^2)*Tan[c + d*x]^2)/2 - (a*b^3*Tan[c + d*x]^3)/3 + (b^4*Tan 
[c + d*x]^4)/4)/(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 = 29.98 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {-\frac {-\frac {\tan \left (d x +c \right )^{4} b^{3}}{4}+\frac {\tan \left (d x +c \right )^{3} a \,b^{2}}{3}-\frac {\left (a^{2}+2 b^{2}\right ) \tan \left (d x +c \right )^{2} b}{2}+a \left (a^{2}+2 b^{2}\right ) \tan \left (d x +c \right )}{b^{4}}+\frac {\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{5}}}{d}\)	\(106\)
default	\(\frac {-\frac {-\frac {\tan \left (d x +c \right )^{4} b^{3}}{4}+\frac {\tan \left (d x +c \right )^{3} a \,b^{2}}{3}-\frac {\left (a^{2}+2 b^{2}\right ) \tan \left (d x +c \right )^{2} b}{2}+a \left (a^{2}+2 b^{2}\right ) \tan \left (d x +c \right )}{b^{4}}+\frac {\left (a^{4}+2 b^{2} a^{2}+b^{4}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{5}}}{d}\)	\(106\)
risch	\(\frac {-2 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-2 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+2 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-6 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-10 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+4 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+8 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-6 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-\frac {34 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{3}+2 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-2 i a^{3}-\frac {10 i a \,b^{2}}{3}}{b^{4} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{4}}{b^{5} d}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}{b^{3} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) a^{4}}{b^{5} d}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) a^{2}}{b^{3} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b d}\)	\(398\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.58 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {6 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 6 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\cos \left (d x + c\right )^{2}\right ) + 3 \, b^{4} + 6 \, {\left (a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left (a b^{3} \cos \left (d x + c\right ) + {\left (3 \, a^{3} b + 5 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{12 \, b^{5} d \cos \left (d x + c\right )^{4}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.93 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {3 \, b^{3} \tan \left (d x + c\right )^{4} - 4 \, a b^{2} \tan \left (d x + c\right )^{3} + 6 \, {\left (a^{2} b + 2 \, b^{3}\right )} \tan \left (d x + c\right )^{2} - 12 \, {\left (a^{3} + 2 \, a b^{2}\right )} \tan \left (d x + c\right )}{b^{4}} + \frac {12 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{5}}}{12 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.20 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{5} d} + \frac {3 \, b^{3} d^{3} \tan \left (d x + c\right )^{4} - 4 \, a b^{2} d^{3} \tan \left (d x + c\right )^{3} + 6 \, a^{2} b d^{3} \tan \left (d x + c\right )^{2} + 12 \, b^{3} d^{3} \tan \left (d x + c\right )^{2} - 12 \, a^{3} d^{3} \tan \left (d x + c\right ) - 24 \, a b^{2} d^{3} \tan \left (d x + c\right )}{12 \, b^{4} d^{4}} \] Input:

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

Output:

(a^4 + 2*a^2*b^2 + b^4)*log(abs(b*tan(d*x + c) + a))/(b^5*d) + 1/12*(3*b^3 
*d^3*tan(d*x + c)^4 - 4*a*b^2*d^3*tan(d*x + c)^3 + 6*a^2*b*d^3*tan(d*x + c 
)^2 + 12*b^3*d^3*tan(d*x + c)^2 - 12*a^3*d^3*tan(d*x + c) - 24*a*b^2*d^3*t 
an(d*x + c))/(b^4*d^4)
 

Mupad [B] (verification not implemented)

Time = 0.92 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.03 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^4}{4\,b\,d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {1}{b}+\frac {a^2}{2\,b^3}\right )}{d}+\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{b^5\,d}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,b^2\,d}-\frac {a\,\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {2}{b}+\frac {a^2}{b^3}\right )}{b\,d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 954, normalized size of antiderivative = 8.22 \[ \int \frac {\sec ^6(c+d x)}{a+b \tan (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(12*cos(c + d*x)*sin(c + d*x)**3*a**3*b + 20*cos(c + d*x)*sin(c + d*x)**3* 
a*b**3 - 12*cos(c + d*x)*sin(c + d*x)*a**3*b - 24*cos(c + d*x)*sin(c + d*x 
)*a*b**3 - 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**4 - 24*log(tan( 
(c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b**2 - 12*log(tan((c + d*x)/2) - 1) 
*sin(c + d*x)**4*b**4 + 24*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4 
+ 48*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**2 + 24*log(tan((c + 
 d*x)/2) - 1)*sin(c + d*x)**2*b**4 - 12*log(tan((c + d*x)/2) - 1)*a**4 - 2 
4*log(tan((c + d*x)/2) - 1)*a**2*b**2 - 12*log(tan((c + d*x)/2) - 1)*b**4 
- 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**4 - 24*log(tan((c + d*x) 
/2) + 1)*sin(c + d*x)**4*a**2*b**2 - 12*log(tan((c + d*x)/2) + 1)*sin(c + 
d*x)**4*b**4 + 24*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4 + 48*log( 
tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**2 + 24*log(tan((c + d*x)/2) 
+ 1)*sin(c + d*x)**2*b**4 - 12*log(tan((c + d*x)/2) + 1)*a**4 - 24*log(tan 
((c + d*x)/2) + 1)*a**2*b**2 - 12*log(tan((c + d*x)/2) + 1)*b**4 + 12*log( 
tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)**4*a**4 + 2 
4*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + d*x)**4*a* 
*2*b**2 + 12*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*sin(c + 
 d*x)**4*b**4 - 24*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b - a)*s 
in(c + d*x)**2*a**4 - 48*log(tan((c + d*x)/2)**2*a - 2*tan((c + d*x)/2)*b 
- a)*sin(c + d*x)**2*a**2*b**2 - 24*log(tan((c + d*x)/2)**2*a - 2*tan((...