\(\int \sec ^6(c+d x) (a+b \tan (c+d x))^3 \, dx\) [537]

Optimal result

Integrand size = 21, antiderivative size = 138 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^4}{4 b^5 d}-\frac {4 a \left (a^2+b^2\right ) (a+b \tan (c+d x))^5}{5 b^5 d}+\frac {\left (3 a^2+b^2\right ) (a+b \tan (c+d x))^6}{3 b^5 d}-\frac {4 a (a+b \tan (c+d x))^7}{7 b^5 d}+\frac {(a+b \tan (c+d x))^8}{8 b^5 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.83 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\frac {1}{4} \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^4-\frac {4}{5} a \left (a^2+b^2\right ) (a+b \tan (c+d x))^5+\frac {1}{3} \left (3 a^2+b^2\right ) (a+b \tan (c+d x))^6-\frac {4}{7} a (a+b \tan (c+d x))^7+\frac {1}{8} (a+b \tan (c+d x))^8}{b^5 d} \] Input:

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

Output:

(((a^2 + b^2)^2*(a + b*Tan[c + d*x])^4)/4 - (4*a*(a^2 + b^2)*(a + b*Tan[c 
+ d*x])^5)/5 + ((3*a^2 + b^2)*(a + b*Tan[c + d*x])^6)/3 - (4*a*(a + b*Tan[ 
c + d*x])^7)/7 + (a + b*Tan[c + d*x])^8/8)/(b^5*d)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.83, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3987

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 476

\(\displaystyle \frac {\int \left ((a+b \tan (c+d x))^7-4 a (a+b \tan (c+d x))^6+2 \left (3 a^2+b^2\right ) (a+b \tan (c+d x))^5-4 a \left (a^2+b^2\right ) (a+b \tan (c+d x))^4+\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^3\right )d(b \tan (c+d x))}{b^5 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{3} \left (3 a^2+b^2\right ) (a+b \tan (c+d x))^6-\frac {4}{5} a \left (a^2+b^2\right ) (a+b \tan (c+d x))^5+\frac {1}{4} \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^4+\frac {1}{8} (a+b \tan (c+d x))^8-\frac {4}{7} a (a+b \tan (c+d x))^7}{b^5 d}\)

Input:

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

Output:

(((a^2 + b^2)^2*(a + b*Tan[c + d*x])^4)/4 - (4*a*(a^2 + b^2)*(a + b*Tan[c 
+ d*x])^5)/5 + ((3*a^2 + b^2)*(a + b*Tan[c + d*x])^6)/3 - (4*a*(a + b*Tan[ 
c + d*x])^7)/7 + (a + b*Tan[c + d*x])^8/8)/(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 = 68.53 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.25

method	result	size
derivativedivides	\(\frac {b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{24 \cos \left (d x +c \right )^{4}}\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \sin \left (d x +c \right )^{3}}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \sin \left (d x +c \right )^{3}}{105 \cos \left (d x +c \right )^{3}}\right )+\frac {a^{2} b}{2 \cos \left (d x +c \right )^{6}}-a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(173\)
default	\(\frac {b^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin \left (d x +c \right )^{4}}{12 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{24 \cos \left (d x +c \right )^{4}}\right )+3 a \,b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \sin \left (d x +c \right )^{3}}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \sin \left (d x +c \right )^{3}}{105 \cos \left (d x +c \right )^{3}}\right )+\frac {a^{2} b}{2 \cos \left (d x +c \right )^{6}}-a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(173\)
risch	\(-\frac {16 \left (-56 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-196 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-210 a^{2} b \,{\mathrm e}^{10 i \left (d x +c \right )}+70 b^{3} {\mathrm e}^{10 i \left (d x +c \right )}-245 i a^{3} {\mathrm e}^{8 i \left (d x +c \right )}-7 i a^{3}-420 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}-70 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+210 i a \,b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+3 i a \,b^{2}-210 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+70 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-42 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+105 i a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-322 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+24 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-70 i a^{3} {\mathrm e}^{10 i \left (d x +c \right )}+84 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{8}}\)	\(275\)

Input:

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

Output:

1/d*(b^3*(1/8*sin(d*x+c)^4/cos(d*x+c)^8+1/12*sin(d*x+c)^4/cos(d*x+c)^6+1/2 
4*sin(d*x+c)^4/cos(d*x+c)^4)+3*a*b^2*(1/7*sin(d*x+c)^3/cos(d*x+c)^7+4/35*s 
in(d*x+c)^3/cos(d*x+c)^5+8/105*sin(d*x+c)^3/cos(d*x+c)^3)+1/2*a^2*b/cos(d* 
x+c)^6-a^3*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.93 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {105 \, b^{3} + 140 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (8 \, {\left (7 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{7} + 4 \, {\left (7 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 45 \, a b^{2} \cos \left (d x + c\right ) + 3 \, {\left (7 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{840 \, d \cos \left (d x + c\right )^{8}} \] Input:

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

Output:

1/840*(105*b^3 + 140*(3*a^2*b - b^3)*cos(d*x + c)^2 + 8*(8*(7*a^3 - 3*a*b^ 
2)*cos(d*x + c)^7 + 4*(7*a^3 - 3*a*b^2)*cos(d*x + c)^5 + 45*a*b^2*cos(d*x 
+ c) + 3*(7*a^3 - 3*a*b^2)*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^8 
)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.03 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {105 \, b^{3} \tan \left (d x + c\right )^{8} + 360 \, a b^{2} \tan \left (d x + c\right )^{7} + 140 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \tan \left (d x + c\right )^{6} + 168 \, {\left (a^{3} + 6 \, a b^{2}\right )} \tan \left (d x + c\right )^{5} + 1260 \, a^{2} b \tan \left (d x + c\right )^{2} + 210 \, {\left (6 \, a^{2} b + b^{3}\right )} \tan \left (d x + c\right )^{4} + 840 \, a^{3} \tan \left (d x + c\right ) + 280 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{3}}{840 \, d} \] Input:

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

Output:

1/840*(105*b^3*tan(d*x + c)^8 + 360*a*b^2*tan(d*x + c)^7 + 140*(3*a^2*b + 
2*b^3)*tan(d*x + c)^6 + 168*(a^3 + 6*a*b^2)*tan(d*x + c)^5 + 1260*a^2*b*ta 
n(d*x + c)^2 + 210*(6*a^2*b + b^3)*tan(d*x + c)^4 + 840*a^3*tan(d*x + c) + 
 280*(2*a^3 + 3*a*b^2)*tan(d*x + c)^3)/d
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.20 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {105 \, b^{3} \tan \left (d x + c\right )^{8} + 360 \, a b^{2} \tan \left (d x + c\right )^{7} + 420 \, a^{2} b \tan \left (d x + c\right )^{6} + 280 \, b^{3} \tan \left (d x + c\right )^{6} + 168 \, a^{3} \tan \left (d x + c\right )^{5} + 1008 \, a b^{2} \tan \left (d x + c\right )^{5} + 1260 \, a^{2} b \tan \left (d x + c\right )^{4} + 210 \, b^{3} \tan \left (d x + c\right )^{4} + 560 \, a^{3} \tan \left (d x + c\right )^{3} + 840 \, a b^{2} \tan \left (d x + c\right )^{3} + 1260 \, a^{2} b \tan \left (d x + c\right )^{2} + 840 \, a^{3} \tan \left (d x + c\right )}{840 \, d} \] Input:

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

Output:

1/840*(105*b^3*tan(d*x + c)^8 + 360*a*b^2*tan(d*x + c)^7 + 420*a^2*b*tan(d 
*x + c)^6 + 280*b^3*tan(d*x + c)^6 + 168*a^3*tan(d*x + c)^5 + 1008*a*b^2*t 
an(d*x + c)^5 + 1260*a^2*b*tan(d*x + c)^4 + 210*b^3*tan(d*x + c)^4 + 560*a 
^3*tan(d*x + c)^3 + 840*a*b^2*tan(d*x + c)^3 + 1260*a^2*b*tan(d*x + c)^2 + 
 840*a^3*tan(d*x + c))/d
 

Mupad [B] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.01 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {2\,a^3}{3}+a\,b^2\right )+{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (\frac {a^3}{5}+\frac {6\,a\,b^2}{5}\right )+{\mathrm {tan}\left (c+d\,x\right )}^6\,\left (\frac {a^2\,b}{2}+\frac {b^3}{3}\right )+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (\frac {3\,a^2\,b}{2}+\frac {b^3}{4}\right )+a^3\,\mathrm {tan}\left (c+d\,x\right )+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^8}{8}+\frac {3\,a^2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {3\,a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^7}{7}}{d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.99 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\sin \left (d x +c \right ) \left (-448 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} a^{3}+192 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} a \,b^{2}+1568 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{3}-672 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a \,b^{2}-1960 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{3}+840 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a \,b^{2}+840 \cos \left (d x +c \right ) a^{3}-420 \sin \left (d x +c \right )^{7} a^{2} b +35 \sin \left (d x +c \right )^{7} b^{3}+1680 \sin \left (d x +c \right )^{5} a^{2} b -140 \sin \left (d x +c \right )^{5} b^{3}-2520 \sin \left (d x +c \right )^{3} a^{2} b +210 \sin \left (d x +c \right )^{3} b^{3}+1260 \sin \left (d x +c \right ) a^{2} b \right )}{840 d \left (\sin \left (d x +c \right )^{8}-4 \sin \left (d x +c \right )^{6}+6 \sin \left (d x +c \right )^{4}-4 \sin \left (d x +c \right )^{2}+1\right )} \] Input:

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

Output:

(sin(c + d*x)*( - 448*cos(c + d*x)*sin(c + d*x)**6*a**3 + 192*cos(c + d*x) 
*sin(c + d*x)**6*a*b**2 + 1568*cos(c + d*x)*sin(c + d*x)**4*a**3 - 672*cos 
(c + d*x)*sin(c + d*x)**4*a*b**2 - 1960*cos(c + d*x)*sin(c + d*x)**2*a**3 
+ 840*cos(c + d*x)*sin(c + d*x)**2*a*b**2 + 840*cos(c + d*x)*a**3 - 420*si 
n(c + d*x)**7*a**2*b + 35*sin(c + d*x)**7*b**3 + 1680*sin(c + d*x)**5*a**2 
*b - 140*sin(c + d*x)**5*b**3 - 2520*sin(c + d*x)**3*a**2*b + 210*sin(c + 
d*x)**3*b**3 + 1260*sin(c + d*x)*a**2*b))/(840*d*(sin(c + d*x)**8 - 4*sin( 
c + d*x)**6 + 6*sin(c + d*x)**4 - 4*sin(c + d*x)**2 + 1))