Integrand size = 29, antiderivative size = 63 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=a^2 x-\frac {5 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2} \] Output:
a^2*x-5/3*a^2*cos(d*x+c)/d/(1-sin(d*x+c))+1/3*a^4*cos(d*x+c)/d/(a-a*sin(d* x+c))^2
Time = 0.23 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.25 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^2 \arctan (\tan (c+d x))}{d}-\frac {2 a^2 \sec (c+d x)}{d}+\frac {2 a^2 \sec ^3(c+d x)}{3 d}-\frac {a^2 \tan (c+d x)}{d}+\frac {2 a^2 \tan ^3(c+d x)}{3 d} \] Input:
Integrate[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^2*Tan[c + d*x]^2,x]
Output:
(a^2*ArcTan[Tan[c + d*x]])/d - (2*a^2*Sec[c + d*x])/d + (2*a^2*Sec[c + d*x ]^3)/(3*d) - (a^2*Tan[c + d*x])/d + (2*a^2*Tan[c + d*x]^3)/(3*d)
Time = 0.58 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 3348, 3042, 3237, 25, 3042, 3214, 3042, 3127}
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.
\(\displaystyle \int \tan ^2(c+d x) \sec ^2(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2 (a \sin (c+d x)+a)^2}{\cos (c+d x)^4}dx\) |
\(\Big \downarrow \) 3348 |
\(\displaystyle a^4 \int \frac {\sin ^2(c+d x)}{(a-a \sin (c+d x))^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^4 \int \frac {\sin (c+d x)^2}{(a-a \sin (c+d x))^2}dx\) |
\(\Big \downarrow \) 3237 |
\(\displaystyle a^4 \left (\frac {\int -\frac {3 \sin (c+d x) a+2 a}{a-a \sin (c+d x)}dx}{3 a^2}+\frac {\cos (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle a^4 \left (\frac {\cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {\int \frac {3 \sin (c+d x) a+2 a}{a-a \sin (c+d x)}dx}{3 a^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^4 \left (\frac {\cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {\int \frac {3 \sin (c+d x) a+2 a}{a-a \sin (c+d x)}dx}{3 a^2}\right )\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle a^4 \left (\frac {\cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {5 a \int \frac {1}{a-a \sin (c+d x)}dx-3 x}{3 a^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^4 \left (\frac {\cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {5 a \int \frac {1}{a-a \sin (c+d x)}dx-3 x}{3 a^2}\right )\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle a^4 \left (\frac {\cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {\frac {5 a \cos (c+d x)}{d (a-a \sin (c+d x))}-3 x}{3 a^2}\right )\) |
Input:
Int[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^2*Tan[c + d*x]^2,x]
Output:
a^4*(Cos[c + d*x]/(3*d*(a - a*Sin[c + d*x])^2) - (-3*x + (5*a*Cos[c + d*x] )/(d*(a - a*Sin[c + d*x])))/(3*a^2))
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b ^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(a*m - b*(2* m + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[a^(2*m) Int[(d* Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]
Result contains complex when optimal does not.
Time = 0.62 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97
method | result | size |
risch | \(a^{2} x -\frac {2 \left (-9 i a^{2} {\mathrm e}^{i \left (d x +c \right )}+6 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-5 a^{2}\right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}\) | \(61\) |
derivativedivides | \(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+d x +c \right )+2 a^{2} \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+\frac {a^{2} \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}}{d}\) | \(114\) |
default | \(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+d x +c \right )+2 a^{2} \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )+\frac {a^{2} \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}}{d}\) | \(114\) |
Input:
int(sec(d*x+c)^2*(a+a*sin(d*x+c))^2*tan(d*x+c)^2,x,method=_RETURNVERBOSE)
Output:
a^2*x-2/3*(-9*I*a^2*exp(I*(d*x+c))+6*a^2*exp(2*I*(d*x+c))-5*a^2)/(exp(I*(d *x+c))-I)^3/d
Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (58) = 116\).
Time = 0.09 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.24 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {6 \, a^{2} d x - {\left (3 \, a^{2} d x + 5 \, a^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + {\left (3 \, a^{2} d x - 4 \, a^{2}\right )} \cos \left (d x + c\right ) - {\left (6 \, a^{2} d x - a^{2} + {\left (3 \, a^{2} d x - 5 \, a^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \] Input:
integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^2*tan(d*x+c)^2,x, algorithm="frica s")
Output:
-1/3*(6*a^2*d*x - (3*a^2*d*x + 5*a^2)*cos(d*x + c)^2 + a^2 + (3*a^2*d*x - 4*a^2)*cos(d*x + c) - (6*a^2*d*x - a^2 + (3*a^2*d*x - 5*a^2)*cos(d*x + c)) *sin(d*x + c))/(d*cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 2*d) *sin(d*x + c) - 2*d)
\[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=a^{2} \left (\int \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 \sin {\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)**2*(a+a*sin(d*x+c))**2*tan(d*x+c)**2,x)
Output:
a**2*(Integral(tan(c + d*x)**2*sec(c + d*x)**2, x) + Integral(2*sin(c + d* x)*tan(c + d*x)**2*sec(c + d*x)**2, x) + Integral(sin(c + d*x)**2*tan(c + d*x)**2*sec(c + d*x)**2, x))
Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.13 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^{2} \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} - \frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \] Input:
integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^2*tan(d*x+c)^2,x, algorithm="maxim a")
Output:
1/3*(a^2*tan(d*x + c)^3 + (tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))* a^2 - 2*(3*cos(d*x + c)^2 - 1)*a^2/cos(d*x + c)^3)/d
Time = 0.36 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.06 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {3 \, {\left (d x + c\right )} a^{2} + \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{3 \, d} \] Input:
integrate(sec(d*x+c)^2*(a+a*sin(d*x+c))^2*tan(d*x+c)^2,x, algorithm="giac" )
Output:
1/3*(3*(d*x + c)*a^2 + 2*(3*a^2*tan(1/2*d*x + 1/2*c)^2 - 9*a^2*tan(1/2*d*x + 1/2*c) + 4*a^2)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d
Time = 32.90 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.90 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=a^2\,x+\frac {a^2\,\left (c+d\,x\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (9\,c+9\,d\,x-18\right )}{3}\right )-\frac {a^2\,\left (3\,c+3\,d\,x-8\right )}{3}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (3\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (9\,c+9\,d\,x-6\right )}{3}\right )}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3} \] Input:
int((tan(c + d*x)^2*(a + a*sin(c + d*x))^2)/cos(c + d*x)^2,x)
Output:
a^2*x + (a^2*(c + d*x) - tan(c/2 + (d*x)/2)*(3*a^2*(c + d*x) - (a^2*(9*c + 9*d*x - 18))/3) - (a^2*(3*c + 3*d*x - 8))/3 + tan(c/2 + (d*x)/2)^2*(3*a^2 *(c + d*x) - (a^2*(9*c + 9*d*x - 6))/3))/(d*(tan(c/2 + (d*x)/2) - 1)^3)
Time = 0.22 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.90 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^{2} \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} d x +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} d x +9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d x -12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-3 d x +6\right )}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )} \] Input:
int(sec(d*x+c)^2*(a+a*sin(d*x+c))^2*tan(d*x+c)^2,x)
Output:
(a**2*(3*tan((c + d*x)/2)**3*d*x + 2*tan((c + d*x)/2)**3 - 9*tan((c + d*x) /2)**2*d*x + 9*tan((c + d*x)/2)*d*x - 12*tan((c + d*x)/2) - 3*d*x + 6))/(3 *d*(tan((c + d*x)/2)**3 - 3*tan((c + d*x)/2)**2 + 3*tan((c + d*x)/2) - 1))