Integrand size = 21, antiderivative size = 97 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a b \sec ^6(c+d x)}{3 d}+\frac {a^2 \tan (c+d x)}{d}+\frac {\left (2 a^2+b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {\left (a^2+2 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {b^2 \tan ^7(c+d x)}{7 d} \] Output:
1/3*a*b*sec(d*x+c)^6/d+a^2*tan(d*x+c)/d+1/3*(2*a^2+b^2)*tan(d*x+c)^3/d+1/5 *(a^2+2*b^2)*tan(d*x+c)^5/d+1/7*b^2*tan(d*x+c)^7/d
Time = 0.48 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.07 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\tan (c+d x) \left (105 a^2+105 a b \tan (c+d x)+35 \left (2 a^2+b^2\right ) \tan ^2(c+d x)+105 a b \tan ^3(c+d x)+21 \left (a^2+2 b^2\right ) \tan ^4(c+d x)+35 a b \tan ^5(c+d x)+15 b^2 \tan ^6(c+d x)\right )}{105 d} \] Input:
Integrate[Sec[c + d*x]^6*(a + b*Tan[c + d*x])^2,x]
Output:
(Tan[c + d*x]*(105*a^2 + 105*a*b*Tan[c + d*x] + 35*(2*a^2 + b^2)*Tan[c + d *x]^2 + 105*a*b*Tan[c + d*x]^3 + 21*(a^2 + 2*b^2)*Tan[c + d*x]^4 + 35*a*b* Tan[c + d*x]^5 + 15*b^2*Tan[c + d*x]^6))/(105*d)
Time = 0.30 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.10, 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, 475, 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.
\(\displaystyle \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^6 (a+b \tan (c+d x))^2dx\) |
\(\Big \downarrow \) 3987 |
\(\displaystyle \frac {\int \frac {(a+b \tan (c+d x))^2 \left (\tan ^2(c+d x) b^2+b^2\right )^2}{b^4}d(b \tan (c+d x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (a+b \tan (c+d x))^2 \left (\tan ^2(c+d x) b^2+b^2\right )^2d(b \tan (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 475 |
\(\displaystyle \frac {\int \left (b^6 \tan ^6(c+d x)+b^4 \left (a^2+2 b^2\right ) \tan ^4(c+d x)+b^4 \left (2 a^2+b^2\right ) \tan ^2(c+d x)+a^2 b^4\right )d(b \tan (c+d x))+\frac {1}{3} a \left (b^2 \tan ^2(c+d x)+b^2\right )^3}{b^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 b^5 \tan (c+d x)+\frac {1}{5} b^5 \left (a^2+2 b^2\right ) \tan ^5(c+d x)+\frac {1}{3} b^5 \left (2 a^2+b^2\right ) \tan ^3(c+d x)+\frac {1}{3} a \left (b^2 \tan ^2(c+d x)+b^2\right )^3+\frac {1}{7} b^7 \tan ^7(c+d x)}{b^5 d}\) |
Input:
Int[Sec[c + d*x]^6*(a + b*Tan[c + d*x])^2,x]
Output:
(a^2*b^5*Tan[c + d*x] + (b^5*(2*a^2 + b^2)*Tan[c + d*x]^3)/3 + (b^5*(a^2 + 2*b^2)*Tan[c + d*x]^5)/5 + (b^7*Tan[c + d*x]^7)/7 + (a*(b^2 + b^2*Tan[c + d*x]^2)^3)/3)/(b^5*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp [d*n*c^(n - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Int[ExpandIntegran d[((c + d*x)^n - d*n*c^(n - 1)*x)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0] && IGtQ[n, 0] && LeQ[n, p]
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]
Time = 30.10 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.13
method | result | size |
derivativedivides | \(\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \sin \left (d x +c \right )^{3}}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \sin \left (d x +c \right )^{3}}{105 \cos \left (d x +c \right )^{3}}\right )+\frac {a b}{3 \cos \left (d x +c \right )^{6}}-a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(110\) |
default | \(\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \sin \left (d x +c \right )^{3}}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \sin \left (d x +c \right )^{3}}{105 \cos \left (d x +c \right )^{3}}\right )+\frac {a b}{3 \cos \left (d x +c \right )^{6}}-a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(110\) |
risch | \(\frac {16 i \left (-140 i a b \,{\mathrm e}^{8 i \left (d x +c \right )}+70 a^{2} {\mathrm e}^{8 i \left (d x +c \right )}-70 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-140 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}+175 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+35 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+147 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-21 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+49 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-7 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+7 a^{2}-b^{2}\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}\) | \(171\) |
Input:
int(sec(d*x+c)^6*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(b^2*(1/7*sin(d*x+c)^3/cos(d*x+c)^7+4/35*sin(d*x+c)^3/cos(d*x+c)^5+8/1 05*sin(d*x+c)^3/cos(d*x+c)^3)+1/3*a*b/cos(d*x+c)^6-a^2*(-8/15-1/5*sec(d*x+ c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c))
Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {35 \, a b \cos \left (d x + c\right ) + {\left (8 \, {\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (7 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{7}} \] Input:
integrate(sec(d*x+c)^6*(a+b*tan(d*x+c))^2,x, algorithm="fricas")
Output:
1/105*(35*a*b*cos(d*x + c) + (8*(7*a^2 - b^2)*cos(d*x + c)^6 + 4*(7*a^2 - b^2)*cos(d*x + c)^4 + 3*(7*a^2 - b^2)*cos(d*x + c)^2 + 15*b^2)*sin(d*x + c ))/(d*cos(d*x + c)^7)
\[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \sec ^{6}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**6*(a+b*tan(d*x+c))**2,x)
Output:
Integral((a + b*tan(c + d*x))**2*sec(c + d*x)**6, x)
Time = 0.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.07 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {15 \, b^{2} \tan \left (d x + c\right )^{7} + 35 \, a b \tan \left (d x + c\right )^{6} + 105 \, a b \tan \left (d x + c\right )^{4} + 21 \, {\left (a^{2} + 2 \, b^{2}\right )} \tan \left (d x + c\right )^{5} + 105 \, a b \tan \left (d x + c\right )^{2} + 35 \, {\left (2 \, a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{3} + 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \] Input:
integrate(sec(d*x+c)^6*(a+b*tan(d*x+c))^2,x, algorithm="maxima")
Output:
1/105*(15*b^2*tan(d*x + c)^7 + 35*a*b*tan(d*x + c)^6 + 105*a*b*tan(d*x + c )^4 + 21*(a^2 + 2*b^2)*tan(d*x + c)^5 + 105*a*b*tan(d*x + c)^2 + 35*(2*a^2 + b^2)*tan(d*x + c)^3 + 105*a^2*tan(d*x + c))/d
Time = 0.21 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.22 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {15 \, b^{2} \tan \left (d x + c\right )^{7} + 35 \, a b \tan \left (d x + c\right )^{6} + 21 \, a^{2} \tan \left (d x + c\right )^{5} + 42 \, b^{2} \tan \left (d x + c\right )^{5} + 105 \, a b \tan \left (d x + c\right )^{4} + 70 \, a^{2} \tan \left (d x + c\right )^{3} + 35 \, b^{2} \tan \left (d x + c\right )^{3} + 105 \, a b \tan \left (d x + c\right )^{2} + 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \] Input:
integrate(sec(d*x+c)^6*(a+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
1/105*(15*b^2*tan(d*x + c)^7 + 35*a*b*tan(d*x + c)^6 + 21*a^2*tan(d*x + c) ^5 + 42*b^2*tan(d*x + c)^5 + 105*a*b*tan(d*x + c)^4 + 70*a^2*tan(d*x + c)^ 3 + 35*b^2*tan(d*x + c)^3 + 105*a*b*tan(d*x + c)^2 + 105*a^2*tan(d*x + c)) /d
Time = 0.69 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.05 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a^2\,\mathrm {tan}\left (c+d\,x\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {2\,a^2}{3}+\frac {b^2}{3}\right )+{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (\frac {a^2}{5}+\frac {2\,b^2}{5}\right )+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^7}{7}+a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^4+\frac {a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^6}{3}}{d} \] Input:
int((a + b*tan(c + d*x))^2/cos(c + d*x)^6,x)
Output:
(a^2*tan(c + d*x) + tan(c + d*x)^3*((2*a^2)/3 + b^2/3) + tan(c + d*x)^5*(a ^2/5 + (2*b^2)/5) + (b^2*tan(c + d*x)^7)/7 + a*b*tan(c + d*x)^2 + a*b*tan( c + d*x)^4 + (a*b*tan(c + d*x)^6)/3)/d
Time = 0.21 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.93 \[ \int \sec ^6(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\sin \left (d x +c \right ) \left (-35 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} a b +105 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a b -105 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b +56 \sin \left (d x +c \right )^{6} a^{2}-8 \sin \left (d x +c \right )^{6} b^{2}-196 \sin \left (d x +c \right )^{4} a^{2}+28 \sin \left (d x +c \right )^{4} b^{2}+245 \sin \left (d x +c \right )^{2} a^{2}-35 \sin \left (d x +c \right )^{2} b^{2}-105 a^{2}\right )}{105 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{6}-3 \sin \left (d x +c \right )^{4}+3 \sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^6*(a+b*tan(d*x+c))^2,x)
Output:
(sin(c + d*x)*( - 35*cos(c + d*x)*sin(c + d*x)**5*a*b + 105*cos(c + d*x)*s in(c + d*x)**3*a*b - 105*cos(c + d*x)*sin(c + d*x)*a*b + 56*sin(c + d*x)** 6*a**2 - 8*sin(c + d*x)**6*b**2 - 196*sin(c + d*x)**4*a**2 + 28*sin(c + d* x)**4*b**2 + 245*sin(c + d*x)**2*a**2 - 35*sin(c + d*x)**2*b**2 - 105*a**2 ))/(105*cos(c + d*x)*d*(sin(c + d*x)**6 - 3*sin(c + d*x)**4 + 3*sin(c + d* x)**2 - 1))