Integrand size = 29, antiderivative size = 63 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=a^2 x-\frac {5 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2} \]
Time = 0.14 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.25 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^2 \arctan (\tan (c+d x))}{d}-\frac {2 a^2 \sec (c+d x)}{d}+\frac {2 a^2 \sec ^3(c+d x)}{3 d}-\frac {a^2 \tan (c+d x)}{d}+\frac {2 a^2 \tan ^3(c+d x)}{3 d} \]
(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 = 0.53 (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 |
\(\displaystyle a^4 \left (\frac {\cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {\frac {5 a \cos (c+d x)}{d (a-a \sin (c+d x))}-3 x}{3 a^2}\right )\) |
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))
3.9.6.3.1 Defintions of rubi rules used
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 = 0.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.97
method | result | size |
risch | \(a^{2} x -\frac {2 \left (-9 i a^{2} {\mathrm e}^{i \left (d x +c \right )}+6 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-5 a^{2}\right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}\) | \(61\) |
parallelrisch | \(\frac {a^{2} \left (3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) d x -9 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) x d +9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d x +6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 d x -18 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8\right )}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\) | \(95\) |
derivativedivides | \(\frac {a^{2} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+2 a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}}{d}\) | \(114\) |
default | \(\frac {a^{2} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+2 a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}}{d}\) | \(114\) |
norman | \(\frac {a^{2} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a^{2} x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a^{2} x +\frac {8 a^{2}}{3 d}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {16 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {44 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {16 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+2 a^{2} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 a^{2} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a^{2} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {40 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\) | \(280\) |
Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (58) = 116\).
Time = 0.29 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.24 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {6 \, a^{2} d x - {\left (3 \, a^{2} d x + 5 \, a^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + {\left (3 \, a^{2} d x - 4 \, a^{2}\right )} \cos \left (d x + c\right ) - {\left (6 \, a^{2} d x - a^{2} + {\left (3 \, a^{2} d x - 5 \, a^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]
-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)
Timed out. \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.13 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^{2} \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} - \frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]
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 = 0.31 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.06 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {3 \, {\left (d x + c\right )} a^{2} + \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{3 \, d} \]
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 = 11.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.62 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx=a^2\,x+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^2\,\left (9\,d\,x-18\right )}{3}-3\,a^2\,d\,x\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2\,\left (9\,d\,x-6\right )}{3}-3\,a^2\,d\,x\right )-\frac {a^2\,\left (3\,d\,x-8\right )}{3}+a^2\,d\,x}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3} \]