\(\int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) [575]

Optimal result

Integrand size = 21, antiderivative size = 121 \[ \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {2 \left (3 a^2+b^2\right ) \log (a+b \tan (c+d x))}{b^5 d}-\frac {3 a \tan (c+d x)}{b^4 d}+\frac {\tan ^2(c+d x)}{2 b^3 d}-\frac {\left (a^2+b^2\right )^2}{2 b^5 d (a+b \tan (c+d x))^2}+\frac {4 a \left (a^2+b^2\right )}{b^5 d (a+b \tan (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 2.42 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.16 \[ \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {b^4 \sec ^4(c+d x)}{2 (a+b \tan (c+d x))^2}-2 a \left (-2 a \log (a+b \tan (c+d x))+b \tan (c+d x)-\frac {a^2+b^2}{a+b \tan (c+d x)}\right )+2 \left (a^2+b^2\right ) \left (\log (a+b \tan (c+d x))+\frac {3 a^2-b^2+4 a b \tan (c+d x)}{2 (a+b \tan (c+d x))^2}\right )}{b^5 d} \] Input:

Integrate[Sec[c + d*x]^6/(a + b*Tan[c + d*x])^3,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.84, 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.

\(\displaystyle \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (c+d x)^6}{(a+b \tan (c+d x))^3}dx\)

\(\Big \downarrow \) 3987

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 476

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

\(\Big \downarrow \) 2009

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

Input:

Int[Sec[c + d*x]^6/(a + b*Tan[c + d*x])^3,x]
 

Output:

(2*(3*a^2 + b^2)*Log[a + b*Tan[c + d*x]] - 3*a*b*Tan[c + d*x] + (b^2*Tan[c 
 + d*x]^2)/2 - (a^2 + b^2)^2/(2*(a + b*Tan[c + d*x])^2) + (4*a*(a^2 + b^2) 
)/(a + b*Tan[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 = 164.80 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {-\frac {-\frac {b \tan \left (d x +c \right )^{2}}{2}+3 a \tan \left (d x +c \right )}{b^{4}}+\frac {\left (6 a^{2}+2 b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{5}}-\frac {a^{4}+2 b^{2} a^{2}+b^{4}}{2 b^{5} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {4 a \left (a^{2}+b^{2}\right )}{b^{5} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(115\)
default	\(\frac {-\frac {-\frac {b \tan \left (d x +c \right )^{2}}{2}+3 a \tan \left (d x +c \right )}{b^{4}}+\frac {\left (6 a^{2}+2 b^{2}\right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{b^{5}}-\frac {a^{4}+2 b^{2} a^{2}+b^{4}}{2 b^{5} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {4 a \left (a^{2}+b^{2}\right )}{b^{5} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(115\)
risch	\(\frac {12 i a^{3}-12 i a \,b^{2}-24 a^{2} b +4 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-4 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+12 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+12 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+36 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+4 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+36 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-36 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+12 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )^{2} b^{4} d}-\frac {6 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}{b^{5} d}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{3} d}+\frac {6 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) a^{2}}{b^{5} d}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b^{3} d}\)	\(346\)

Input:

int(sec(d*x+c)^6/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (117) = 234\).

Time = 0.12 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.93 \[ \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {24 \, a^{2} b^{2} \cos \left (d x + c\right )^{4} + b^{4} - 2 \, {\left (9 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left ({\left (3 \, a^{4} - 2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + {\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 2 \, {\left ({\left (3 \, a^{4} - 2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + {\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) - 4 \, {\left (a b^{3} \cos \left (d x + c\right ) + 3 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (2 \, a b^{6} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + b^{7} d \cos \left (d x + c\right )^{2} + {\left (a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{4}\right )}} \] Input:

integrate(sec(d*x+c)^6/(a+b*tan(d*x+c))^3,x, algorithm="fricas")
 

Output:

1/2*(24*a^2*b^2*cos(d*x + c)^4 + b^4 - 2*(9*a^2*b^2 + b^4)*cos(d*x + c)^2 
+ 2*((3*a^4 - 2*a^2*b^2 - b^4)*cos(d*x + c)^4 + 2*(3*a^3*b + a*b^3)*cos(d* 
x + c)^3*sin(d*x + c) + (3*a^2*b^2 + b^4)*cos(d*x + c)^2)*log(2*a*b*cos(d* 
x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - 2*((3*a^4 - 2*a^ 
2*b^2 - b^4)*cos(d*x + c)^4 + 2*(3*a^3*b + a*b^3)*cos(d*x + c)^3*sin(d*x + 
 c) + (3*a^2*b^2 + b^4)*cos(d*x + c)^2)*log(cos(d*x + c)^2) - 4*(a*b^3*cos 
(d*x + c) + 3*(a^3*b - a*b^3)*cos(d*x + c)^3)*sin(d*x + c))/(2*a*b^6*d*cos 
(d*x + c)^3*sin(d*x + c) + b^7*d*cos(d*x + c)^2 + (a^2*b^5 - b^7)*d*cos(d* 
x + c)^4)
 

Sympy [F]

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

integrate(sec(d*x+c)**6/(a+b*tan(d*x+c))**3,x)
 

Output:

Integral(sec(c + d*x)**6/(a + b*tan(c + d*x))**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {7 \, a^{4} + 6 \, a^{2} b^{2} - b^{4} + 8 \, {\left (a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )}{b^{7} \tan \left (d x + c\right )^{2} + 2 \, a b^{6} \tan \left (d x + c\right ) + a^{2} b^{5}} + \frac {b \tan \left (d x + c\right )^{2} - 6 \, a \tan \left (d x + c\right )}{b^{4}} + \frac {4 \, {\left (3 \, a^{2} + b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{5}}}{2 \, d} \] Input:

integrate(sec(d*x+c)^6/(a+b*tan(d*x+c))^3,x, algorithm="maxima")
 

Output:

1/2*((7*a^4 + 6*a^2*b^2 - b^4 + 8*(a^3*b + a*b^3)*tan(d*x + c))/(b^7*tan(d 
*x + c)^2 + 2*a*b^6*tan(d*x + c) + a^2*b^5) + (b*tan(d*x + c)^2 - 6*a*tan( 
d*x + c))/b^4 + 4*(3*a^2 + b^2)*log(b*tan(d*x + c) + a)/b^5)/d
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {2 \, {\left (3 \, a^{2} + b^{2}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{5} d} + \frac {b^{3} d \tan \left (d x + c\right )^{2} - 6 \, a b^{2} d \tan \left (d x + c\right )}{2 \, b^{6} d^{2}} + \frac {7 \, a^{4} + 6 \, a^{2} b^{2} - b^{4} + 8 \, {\left (a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )}{2 \, {\left (b \tan \left (d x + c\right ) + a\right )}^{2} b^{5} d} \] Input:

integrate(sec(d*x+c)^6/(a+b*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

2*(3*a^2 + b^2)*log(abs(b*tan(d*x + c) + a))/(b^5*d) + 1/2*(b^3*d*tan(d*x 
+ c)^2 - 6*a*b^2*d*tan(d*x + c))/(b^6*d^2) + 1/2*(7*a^4 + 6*a^2*b^2 - b^4 
+ 8*(a^3*b + a*b^3)*tan(d*x + c))/((b*tan(d*x + c) + a)^2*b^5*d)
 

Mupad [B] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.18 \[ \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {7\,a^4+6\,a^2\,b^2-b^4}{2\,b}+\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a^3+4\,a\,b^2\right )}{d\,\left (a^2\,b^4+2\,a\,b^5\,\mathrm {tan}\left (c+d\,x\right )+b^6\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,b^3\,d}-\frac {3\,a\,\mathrm {tan}\left (c+d\,x\right )}{b^4\,d}+\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (6\,a^2+2\,b^2\right )}{b^5\,d} \] Input:

int(1/(cos(c + d*x)^6*(a + b*tan(c + d*x))^3),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 7160, normalized size of antiderivative = 59.17 \[ \int \frac {\sec ^6(c+d x)}{(a+b \tan (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(sec(d*x+c)^6/(a+b*tan(d*x+c))^3,x)
 

Output:

( - 24*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*tan(c + d*x) 
**2*a**3*b**3 - 8*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5*t 
an(c + d*x)**2*a*b**5 - 48*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + 
d*x)**5*tan(c + d*x)*a**4*b**2 - 16*cos(c + d*x)*log(tan((c + d*x)/2) - 1) 
*sin(c + d*x)**5*tan(c + d*x)*a**2*b**4 - 24*cos(c + d*x)*log(tan((c + d*x 
)/2) - 1)*sin(c + d*x)**5*a**5*b - 8*cos(c + d*x)*log(tan((c + d*x)/2) - 1 
)*sin(c + d*x)**5*a**3*b**3 + 48*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*si 
n(c + d*x)**3*tan(c + d*x)**2*a**3*b**3 + 16*cos(c + d*x)*log(tan((c + d*x 
)/2) - 1)*sin(c + d*x)**3*tan(c + d*x)**2*a*b**5 + 96*cos(c + d*x)*log(tan 
((c + d*x)/2) - 1)*sin(c + d*x)**3*tan(c + d*x)*a**4*b**2 + 32*cos(c + d*x 
)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*tan(c + d*x)*a**2*b**4 + 48*co 
s(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a**5*b + 16*cos(c + d 
*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a**3*b**3 - 24*cos(c + d*x)* 
log(tan((c + d*x)/2) - 1)*sin(c + d*x)*tan(c + d*x)**2*a**3*b**3 - 8*cos(c 
 + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*tan(c + d*x)**2*a*b**5 - 48 
*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*tan(c + d*x)*a**4*b** 
2 - 16*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*tan(c + d*x)*a* 
*2*b**4 - 24*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**5*b - 
8*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*a**3*b**3 - 24*cos(c 
 + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5*tan(c + d*x)**2*a**3*...