Integrand size = 25, antiderivative size = 42 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx=-2 a b x+\frac {2 b^2 \cos (c+d x)}{d}+\frac {\sec (c+d x) (a+b \sin (c+d x))^2}{d} \] Output:
-2*a*b*x+2*b^2*cos(d*x+c)/d+sec(d*x+c)*(a+b*sin(d*x+c))^2/d
Time = 0.78 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.57 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx=\frac {\left (2 a^2+3 b^2+b^2 \cos (2 (c+d x))\right ) \sec (c+d x)-2 \left (a^2+b^2+2 a b (c+d x)-2 a b \tan (c+d x)\right )}{2 d} \] Input:
Integrate[Sec[c + d*x]*(a + b*Sin[c + d*x])^2*Tan[c + d*x],x]
Output:
((2*a^2 + 3*b^2 + b^2*Cos[2*(c + d*x)])*Sec[c + d*x] - 2*(a^2 + b^2 + 2*a* b*(c + d*x) - 2*a*b*Tan[c + d*x]))/(2*d)
Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 3340, 27, 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.
\(\displaystyle \int \tan (c+d x) \sec (c+d x) (a+b \sin (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x) (a+b \sin (c+d x))^2}{\cos (c+d x)^2}dx\) |
\(\Big \downarrow \) 3340 |
\(\displaystyle \frac {\sec (c+d x) (a+b \sin (c+d x))^2}{d}-\int 2 b (a+b \sin (c+d x))dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sec (c+d x) (a+b \sin (c+d x))^2}{d}-2 b \int (a+b \sin (c+d x))dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sec (c+d x) (a+b \sin (c+d x))^2}{d}-2 b \left (a x-\frac {b \cos (c+d x)}{d}\right )\) |
Input:
Int[Sec[c + d*x]*(a + b*Sin[c + d*x])^2*Tan[c + d*x],x]
Output:
-2*b*(a*x - (b*Cos[c + d*x])/d) + (Sec[c + d*x]*(a + b*Sin[c + d*x])^2)/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(g* Cos[e + f*x])^(p + 1))*(a + b*Sin[e + f*x])^m*((d + c*Sin[e + f*x])/(f*g*(p + 1))), x] + Simp[1/(g^2*(p + 1)) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Si n[e + f*x])^(m - 1)*Simp[a*c*(p + 2) + b*d*m + b*c*(m + p + 2)*Sin[e + f*x] , x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ [m, 0] && LtQ[p, -1] && IntegerQ[2*m] && !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x])
Time = 1.94 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.79
method | result | size |
derivativedivides | \(\frac {\frac {a^{2}}{\cos \left (d x +c \right )}+2 a b \left (\tan \left (d x +c \right )-d x -c \right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(75\) |
default | \(\frac {\frac {a^{2}}{\cos \left (d x +c \right )}+2 a b \left (\tan \left (d x +c \right )-d x -c \right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(75\) |
risch | \(-2 a b x +\frac {{\mathrm e}^{i \left (d x +c \right )} b^{2}}{2 d}+\frac {{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{2 d}+\frac {4 i a b +2 a^{2} {\mathrm e}^{i \left (d x +c \right )}+2 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\) | \(91\) |
Input:
int(sec(d*x+c)*(a+b*sin(d*x+c))^2*tan(d*x+c),x,method=_RETURNVERBOSE)
Output:
1/d*(a^2/cos(d*x+c)+2*a*b*(tan(d*x+c)-d*x-c)+b^2*(sin(d*x+c)^4/cos(d*x+c)+ (2+sin(d*x+c)^2)*cos(d*x+c)))
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.40 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2 \, a b d x \cos \left (d x + c\right ) - b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}{d \cos \left (d x + c\right )} \] Input:
integrate(sec(d*x+c)*(a+b*sin(d*x+c))^2*tan(d*x+c),x, algorithm="fricas")
Output:
-(2*a*b*d*x*cos(d*x + c) - b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)/(d*cos(d*x + c))
\[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \tan {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)*(a+b*sin(d*x+c))**2*tan(d*x+c),x)
Output:
Integral((a + b*sin(c + d*x))**2*tan(c + d*x)*sec(c + d*x), x)
Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.33 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a b - b^{2} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - \frac {a^{2}}{\cos \left (d x + c\right )}}{d} \] Input:
integrate(sec(d*x+c)*(a+b*sin(d*x+c))^2*tan(d*x+c),x, algorithm="maxima")
Output:
-(2*(d*x + c - tan(d*x + c))*a*b - b^2*(1/cos(d*x + c) + cos(d*x + c)) - a ^2/cos(d*x + c))/d
Time = 0.30 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.95 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2 \, {\left ({\left (d x + c\right )} a b + \frac {2 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2} + 2 \, b^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}\right )}}{d} \] Input:
integrate(sec(d*x+c)*(a+b*sin(d*x+c))^2*tan(d*x+c),x, algorithm="giac")
Output:
-2*((d*x + c)*a*b + (2*a*b*tan(1/2*d*x + 1/2*c)^3 + a^2*tan(1/2*d*x + 1/2* c)^2 + 2*a*b*tan(1/2*d*x + 1/2*c) + a^2 + 2*b^2)/(tan(1/2*d*x + 1/2*c)^4 - 1))/d
Time = 20.19 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.93 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^2+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+4\,b^2}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-1\right )}-2\,a\,b\,x \] Input:
int((tan(c + d*x)*(a + b*sin(c + d*x))^2)/cos(c + d*x),x)
Output:
- (2*a^2*tan(c/2 + (d*x)/2)^2 + 2*a^2 + 4*b^2 + 4*a*b*tan(c/2 + (d*x)/2)^3 + 4*a*b*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d*x)/2)^4 - 1)) - 2*a*b*x
Time = 0.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.12 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan (c+d x) \, dx=\frac {-\cos \left (d x +c \right ) a^{2}-2 \cos \left (d x +c \right ) a b c -2 \cos \left (d x +c \right ) a b d x -2 \cos \left (d x +c \right ) b^{2}-\sin \left (d x +c \right )^{2} b^{2}+2 \sin \left (d x +c \right ) a b +a^{2}+2 b^{2}}{\cos \left (d x +c \right ) d} \] Input:
int(sec(d*x+c)*(a+b*sin(d*x+c))^2*tan(d*x+c),x)
Output:
( - cos(c + d*x)*a**2 - 2*cos(c + d*x)*a*b*c - 2*cos(c + d*x)*a*b*d*x - 2* cos(c + d*x)*b**2 - sin(c + d*x)**2*b**2 + 2*sin(c + d*x)*a*b + a**2 + 2*b **2)/(cos(c + d*x)*d)