Integrand size = 27, antiderivative size = 55 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sec ^5(c+d x)}{5 a d}-\frac {\tan ^3(c+d x)}{3 a d}-\frac {\tan ^5(c+d x)}{5 a d} \]
Time = 0.59 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.93 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sec ^3(c+d x) (-240+54 \cos (c+d x)+32 \cos (2 (c+d x))+18 \cos (3 (c+d x))+16 \cos (4 (c+d x))-96 \sin (c+d x)+18 \sin (2 (c+d x))-32 \sin (3 (c+d x))+9 \sin (4 (c+d x)))}{960 a d (1+\sin (c+d x))} \]
-1/960*(Sec[c + d*x]^3*(-240 + 54*Cos[c + d*x] + 32*Cos[2*(c + d*x)] + 18* Cos[3*(c + d*x)] + 16*Cos[4*(c + d*x)] - 96*Sin[c + d*x] + 18*Sin[2*(c + d *x)] - 32*Sin[3*(c + d*x)] + 9*Sin[4*(c + d*x)]))/(a*d*(1 + Sin[c + d*x]))
Time = 0.37 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 3318, 3042, 3086, 15, 3087, 244, 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.
\(\displaystyle \int \frac {\tan (c+d x) \sec ^3(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)}{\cos (c+d x)^4 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3318 |
\(\displaystyle \frac {\int \sec ^5(c+d x) \tan (c+d x)dx}{a}-\frac {\int \sec ^4(c+d x) \tan ^2(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sec (c+d x)^5 \tan (c+d x)dx}{a}-\frac {\int \sec (c+d x)^4 \tan (c+d x)^2dx}{a}\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {\int \sec ^4(c+d x)d\sec (c+d x)}{a d}-\frac {\int \sec (c+d x)^4 \tan (c+d x)^2dx}{a}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\sec ^5(c+d x)}{5 a d}-\frac {\int \sec (c+d x)^4 \tan (c+d x)^2dx}{a}\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle \frac {\sec ^5(c+d x)}{5 a d}-\frac {\int \tan ^2(c+d x) \left (\tan ^2(c+d x)+1\right )d\tan (c+d x)}{a d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sec ^5(c+d x)}{5 a d}-\frac {\int \left (\tan ^4(c+d x)+\tan ^2(c+d x)\right )d\tan (c+d x)}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sec ^5(c+d x)}{5 a d}-\frac {\frac {1}{5} \tan ^5(c+d x)+\frac {1}{3} \tan ^3(c+d x)}{a d}\) |
3.9.26.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d) Int [(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0]
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.56
method | result | size |
risch | \(\frac {4 i \left (-2 \,{\mathrm e}^{2 i \left (d x +c \right )}+2 i {\mathrm e}^{i \left (d x +c \right )}-1+6 i {\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{4 i \left (d x +c \right )}\right )}{15 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} d a}\) | \(86\) |
parallelrisch | \(\frac {-\frac {2}{5}-2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {2 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {16 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15}-\frac {6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}}{d a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(113\) |
derivativedivides | \(\frac {-\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {4}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d a}\) | \(130\) |
default | \(\frac {-\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {4}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d a}\) | \(130\) |
norman | \(\frac {-\frac {2 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {2}{5 a d}-\frac {2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {6 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5 d a}-\frac {4 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {16 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\) | \(149\) |
4/15*I*(-2*exp(2*I*(d*x+c))+2*I*exp(I*(d*x+c))-1+6*I*exp(3*I*(d*x+c))+15*e xp(4*I*(d*x+c)))/(exp(I*(d*x+c))-I)^3/(exp(I*(d*x+c))+I)^5/d/a
Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2 \, \cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{2} - {\left (2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) - 4}{15 \, {\left (a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{3}\right )}} \]
-1/15*(2*cos(d*x + c)^4 - cos(d*x + c)^2 - (2*cos(d*x + c)^2 + 1)*sin(d*x + c) - 4)/(a*d*cos(d*x + c)^3*sin(d*x + c) + a*d*cos(d*x + c)^3)
\[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\sin {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]
Leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (49) = 98\).
Time = 0.23 (sec) , antiderivative size = 274, normalized size of antiderivative = 4.98 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {2 \, {\left (\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {9 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {8 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {10 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 3\right )}}{15 \, {\left (a + \frac {2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {6 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {2 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {2 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d} \]
2/15*(6*sin(d*x + c)/(cos(d*x + c) + 1) + 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 8*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 5*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*sin(d*x + c)^6/( cos(d*x + c) + 1)^6 + 3)/((a + 2*a*sin(d*x + c)/(cos(d*x + c) + 1) - 2*a*s in(d*x + c)^2/(cos(d*x + c) + 1)^2 - 6*a*sin(d*x + c)^3/(cos(d*x + c) + 1) ^3 + 6*a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 2*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 2*a*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - a*sin(d*x + c)^8/ (cos(d*x + c) + 1)^8)*d)
Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (49) = 98\).
Time = 0.32 (sec) , antiderivative size = 120, normalized size of antiderivative = 2.18 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {5 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 60 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 70 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 13}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{120 \, d} \]
-1/120*(5*(9*tan(1/2*d*x + 1/2*c)^2 - 12*tan(1/2*d*x + 1/2*c) + 7)/(a*(tan (1/2*d*x + 1/2*c) - 1)^3) - (45*tan(1/2*d*x + 1/2*c)^4 + 60*tan(1/2*d*x + 1/2*c)^3 + 70*tan(1/2*d*x + 1/2*c)^2 + 20*tan(1/2*d*x + 1/2*c) + 13)/(a*(t an(1/2*d*x + 1/2*c) + 1)^5))/d
Time = 11.81 (sec) , antiderivative size = 112, normalized size of antiderivative = 2.04 \[ \int \frac {\sec ^3(c+d x) \tan (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2\,\left (15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+3\right )}{15\,a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5} \]