Integrand size = 21, antiderivative size = 50 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-3 a^3 x+\frac {3 a^3 \cos (c+d x)}{d}+\frac {2 a^5 \cos ^3(c+d x)}{d (a-a \sin (c+d x))^2} \]
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.10 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {4 \sqrt {2} a^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2},\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec (c+d x) \sqrt {1+\sin (c+d x)}}{d} \]
(4*Sqrt[2]*a^3*Hypergeometric2F1[-3/2, -1/2, 1/2, (1 - Sin[c + d*x])/2]*Se c[c + d*x]*Sqrt[1 + Sin[c + d*x]])/d
Time = 0.41 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.18, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3149, 3042, 3159, 3042, 3161, 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.
\(\displaystyle \int \sec ^2(c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^3}{\cos (c+d x)^2}dx\) |
\(\Big \downarrow \) 3149 |
\(\displaystyle a^6 \int \frac {\cos ^4(c+d x)}{(a-a \sin (c+d x))^3}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^6 \int \frac {\cos (c+d x)^4}{(a-a \sin (c+d x))^3}dx\) |
\(\Big \downarrow \) 3159 |
\(\displaystyle a^6 \left (\frac {2 \cos ^3(c+d x)}{a d (a-a \sin (c+d x))^2}-\frac {3 \int \frac {\cos ^2(c+d x)}{a-a \sin (c+d x)}dx}{a^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^6 \left (\frac {2 \cos ^3(c+d x)}{a d (a-a \sin (c+d x))^2}-\frac {3 \int \frac {\cos (c+d x)^2}{a-a \sin (c+d x)}dx}{a^2}\right )\) |
\(\Big \downarrow \) 3161 |
\(\displaystyle a^6 \left (\frac {2 \cos ^3(c+d x)}{a d (a-a \sin (c+d x))^2}-\frac {3 \left (\frac {\int 1dx}{a}-\frac {\cos (c+d x)}{a d}\right )}{a^2}\right )\) |
\(\Big \downarrow \) 24 |
\(\displaystyle a^6 \left (\frac {2 \cos ^3(c+d x)}{a d (a-a \sin (c+d x))^2}-\frac {3 \left (\frac {x}{a}-\frac {\cos (c+d x)}{a d}\right )}{a^2}\right )\) |
3.1.34.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(a/g)^(2*m) Int[(g*Cos[e + f*x])^(2*m + p)/( a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2 , 0] && IntegerQ[m] && LtQ[p, -1] && GeQ[2*m + p, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f *x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 ))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & & NeQ[2*m + p + 1, 0] && !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(b*f*(p - 1))), x] + Si mp[g^2/a Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g}, x ] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]
Time = 0.66 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \(\frac {a^{3} \left (-6 \cos \left (d x +c \right ) d x +\cos \left (2 d x +2 c \right )+8 \sin \left (d x +c \right )+10 \cos \left (d x +c \right )+9\right )}{2 d \cos \left (d x +c \right )}\) | \(54\) |
risch | \(-3 a^{3} x +\frac {a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {8 a^{3}}{d \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}\) | \(64\) |
derivativedivides | \(\frac {a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (\tan \left (d x +c \right )-d x -c \right )+\frac {3 a^{3}}{\cos \left (d x +c \right )}+a^{3} \tan \left (d x +c \right )}{d}\) | \(87\) |
default | \(\frac {a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (\tan \left (d x +c \right )-d x -c \right )+\frac {3 a^{3}}{\cos \left (d x +c \right )}+a^{3} \tan \left (d x +c \right )}{d}\) | \(87\) |
norman | \(\frac {3 a^{3} x -\frac {10 a^{3}}{d}-\frac {8 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {24 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {22 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {24 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+6 a^{3} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 a^{3} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{3} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {6 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {26 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\) | \(229\) |
Time = 0.30 (sec) , antiderivative size = 101, normalized size of antiderivative = 2.02 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {3 \, a^{3} d x - a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} + {\left (3 \, a^{3} d x - 5 \, a^{3}\right )} \cos \left (d x + c\right ) - {\left (3 \, a^{3} d x - a^{3} \cos \left (d x + c\right ) + 4 \, a^{3}\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d} \]
-(3*a^3*d*x - a^3*cos(d*x + c)^2 - 4*a^3 + (3*a^3*d*x - 5*a^3)*cos(d*x + c ) - (3*a^3*d*x - a^3*cos(d*x + c) + 4*a^3)*sin(d*x + c))/(d*cos(d*x + c) - d*sin(d*x + c) + d)
\[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=a^{3} \left (\int 3 \sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]
a**3*(Integral(3*sin(c + d*x)*sec(c + d*x)**2, x) + Integral(3*sin(c + d*x )**2*sec(c + d*x)**2, x) + Integral(sin(c + d*x)**3*sec(c + d*x)**2, x) + Integral(sec(c + d*x)**2, x))
Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.36 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {3 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} - a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - a^{3} \tan \left (d x + c\right ) - \frac {3 \, a^{3}}{\cos \left (d x + c\right )}}{d} \]
-(3*(d*x + c - tan(d*x + c))*a^3 - a^3*(1/cos(d*x + c) + cos(d*x + c)) - a ^3*tan(d*x + c) - 3*a^3/cos(d*x + c))/d
Time = 0.34 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.82 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {3 \, {\left (d x + c\right )} a^{3} + \frac {2 \, {\left (4 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}}{d} \]
-(3*(d*x + c)*a^3 + 2*(4*a^3*tan(1/2*d*x + 1/2*c)^2 - a^3*tan(1/2*d*x + 1/ 2*c) + 5*a^3)/(tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d *x + 1/2*c) - 1))/d
Time = 6.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.76 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx=-3\,a^3\,x-\frac {3\,a^3\,\left (c+d\,x\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^3\,\left (c+d\,x\right )-a^3\,\left (3\,c+3\,d\,x-2\right )\right )-a^3\,\left (3\,c+3\,d\,x-10\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (3\,a^3\,\left (c+d\,x\right )-a^3\,\left (3\,c+3\,d\,x-8\right )\right )}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]