Integrand size = 27, antiderivative size = 60 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2 a^2 \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (a+a \sin (c+d x))^2}{3 d}-\frac {2 a^2 \tan (c+d x)}{3 d} \]
Time = 1.44 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.20 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=\frac {a^2 \left (3 \cos \left (\frac {1}{2} (c+d x)\right )-2 \cos \left (\frac {3}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \]
(a^2*(3*Cos[(c + d*x)/2] - 2*Cos[(3*(c + d*x))/2] - 3*Sin[(c + d*x)/2]))/( 3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3)
Time = 0.37 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 3334, 3042, 3148, 3042, 4254, 24}
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.
\(\displaystyle \int \tan (c+d x) \sec ^3(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x) (a \sin (c+d x)+a)^2}{\cos (c+d x)^4}dx\) |
\(\Big \downarrow \) 3334 |
\(\displaystyle \frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d}-\frac {2}{3} a \int \sec ^2(c+d x) (\sin (c+d x) a+a)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d}-\frac {2}{3} a \int \frac {\sin (c+d x) a+a}{\cos (c+d x)^2}dx\) |
\(\Big \downarrow \) 3148 |
\(\displaystyle \frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d}-\frac {2}{3} a \left (a \int \sec ^2(c+d x)dx+\frac {a \sec (c+d x)}{d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d}-\frac {2}{3} a \left (a \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {a \sec (c+d x)}{d}\right )\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d}-\frac {2}{3} a \left (\frac {a \sec (c+d x)}{d}-\frac {a \int 1d(-\tan (c+d x))}{d}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d}-\frac {2}{3} a \left (\frac {a \tan (c+d x)}{d}+\frac {a \sec (c+d x)}{d}\right )\) |
(Sec[c + d*x]^3*(a + a*Sin[c + d*x])^2)/(3*d) - (2*a*((a*Sec[c + d*x])/d + (a*Tan[c + d*x])/d))/3
3.9.7.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b* c + a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p + 1))) , x] + Simp[b*((a*d*m + b*c*(m + p + 1))/(a*g^2*(p + 1))) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.58
method | result | size |
parallelrisch | \(\frac {2 a^{2} \left (1-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\) | \(35\) |
risch | \(-\frac {2 a^{2} \left (-3 i {\mathrm e}^{i \left (d x +c \right )}+3 \,{\mathrm e}^{2 i \left (d x +c \right )}-2\right )}{3 d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3}}\) | \(48\) |
derivativedivides | \(\frac {a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {2 a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+\frac {a^{2}}{3 \cos \left (d x +c \right )^{3}}}{d}\) | \(99\) |
default | \(\frac {a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {2 a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}+\frac {a^{2}}{3 \cos \left (d x +c \right )^{3}}}{d}\) | \(99\) |
norman | \(\frac {\frac {2 a^{2}}{3 d}-\frac {16 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {32 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {16 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {28 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\) | \(174\) |
Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.63 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=\frac {2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right ) - a^{2} - {\left (2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]
1/3*(2*a^2*cos(d*x + c)^2 + a^2*cos(d*x + c) - a^2 - (2*a^2*cos(d*x + c) + a^2)*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)
\[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=a^{2} \left (\int \sin {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 \sin ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(sin(c + d*x)*sec(c + d*x)**4, x) + Integral(2*sin(c + d*x)* *2*sec(c + d*x)**4, x) + Integral(sin(c + d*x)**3*sec(c + d*x)**4, x))
Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.93 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=\frac {2 \, a^{2} \tan \left (d x + c\right )^{3} - \frac {{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2}}{\cos \left (d x + c\right )^{3}} + \frac {a^{2}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]
1/3*(2*a^2*tan(d*x + c)^3 - (3*cos(d*x + c)^2 - 1)*a^2/cos(d*x + c)^3 + a^ 2/cos(d*x + c)^3)/d
Time = 0.31 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.63 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}\right )}}{3 \, d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} \]
Time = 10.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.57 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^2 \tan (c+d x) \, dx=-\frac {2\,a^2\,\left (3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3} \]