Integrand size = 21, antiderivative size = 98 \[ \int (a+a \sec (c+d x))^3 \sin ^2(c+d x) \, dx=-\frac {5 a^3 x}{2}+\frac {5 a^3 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {3 a^3 \sin (c+d x)}{d}-\frac {a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {3 a^3 \tan (c+d x)}{d}+\frac {a^3 \sec (c+d x) \tan (c+d x)}{2 d} \]
-5/2*a^3*x+5/2*a^3*arctanh(sin(d*x+c))/d-3*a^3*sin(d*x+c)/d-1/2*a^3*cos(d* x+c)*sin(d*x+c)/d+3*a^3*tan(d*x+c)/d+1/2*a^3*sec(d*x+c)*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(300\) vs. \(2(98)=196\).
Time = 2.20 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.06 \[ \int (a+a \sec (c+d x))^3 \sin ^2(c+d x) \, dx=\frac {1}{32} a^3 (1+\cos (c+d x))^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (-10 x-\frac {10 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {10 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {12 \cos (d x) \sin (c)}{d}-\frac {\cos (2 d x) \sin (2 c)}{d}-\frac {12 \cos (c) \sin (d x)}{d}-\frac {\cos (2 c) \sin (2 d x)}{d}+\frac {1}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {12 \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {1}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {12 \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right ) \]
(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(-10*x - (10*Log[Cos[(c + d*x )/2] - Sin[(c + d*x)/2]])/d + (10*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] )/d - (12*Cos[d*x]*Sin[c])/d - (Cos[2*d*x]*Sin[2*c])/d - (12*Cos[c]*Sin[d* x])/d - (Cos[2*c]*Sin[2*d*x])/d + 1/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2 ])^2) + (12*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin [(c + d*x)/2])) - 1/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (12*Sin[ (d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))) )/32
Time = 0.44 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4360, 25, 25, 3042, 3351, 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.
| \(\displaystyle \int \sin ^2(c+d x) (a \sec (c+d x)+a)^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^2 \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^3dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \tan ^2(c+d x) \sec (c+d x) \left (-(a (-\cos (c+d x))-a)^3\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -(\cos (c+d x) a+a)^3 \sec (c+d x) \tan ^2(c+d x)dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \tan ^2(c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3351 |
| \(\displaystyle \frac {\int \left (\sec ^3(c+d x) a^5-\cos ^2(c+d x) a^5+3 \sec ^2(c+d x) a^5-3 \cos (c+d x) a^5+2 \sec (c+d x) a^5-2 a^5\right )dx}{a^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {5 a^5 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {3 a^5 \sin (c+d x)}{d}+\frac {3 a^5 \tan (c+d x)}{d}-\frac {a^5 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a^5 \tan (c+d x) \sec (c+d x)}{2 d}-\frac {5 a^5 x}{2}}{a^2}\) |
((-5*a^5*x)/2 + (5*a^5*ArcTanh[Sin[c + d*x]])/(2*d) - (3*a^5*Sin[c + d*x]) /d - (a^5*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (3*a^5*Tan[c + d*x])/d + (a^5 *Sec[c + d*x]*Tan[c + d*x])/(2*d))/a^2
3.1.51.3.1 Defintions of rubi rules used
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/a^p Int[Expan dTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x])^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && In tegersQ[m, n, p/2] && ((GtQ[m, 0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (G tQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 1.55 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.29
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} \left (\tan \left (d x +c \right )-d x -c \right )+3 a^{3} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(126\) |
| default | \(\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} \left (\tan \left (d x +c \right )-d x -c \right )+3 a^{3} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(126\) |
| parts | \(\frac {a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{3} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {3 a^{3} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}+\frac {3 a^{3} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\) | \(134\) |
| parallelrisch | \(\frac {a^{3} \left (-20 \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-20 d x \cos \left (2 d x +2 c \right )+20 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (2 d x +2 c \right )-20 d x +20 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-4 \sin \left (d x +c \right )+22 \sin \left (2 d x +2 c \right )-12 \sin \left (3 d x +3 c \right )-\sin \left (4 d x +4 c \right )\right )}{8 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(143\) |
| norman | \(\frac {-\frac {5 a^{3} x}{2}+5 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\frac {5 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2}+\frac {18 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {10 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {5 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {5 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(152\) |
| risch | \(-\frac {5 a^{3} x}{2}+\frac {i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {3 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {3 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i a^{3} \left ({\mathrm e}^{3 i \left (d x +c \right )}-6 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-6\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}\) | \(177\) |
1/d*(a^3*(1/2*sin(d*x+c)^3/cos(d*x+c)^2+1/2*sin(d*x+c)-1/2*ln(sec(d*x+c)+t an(d*x+c)))+3*a^3*(tan(d*x+c)-d*x-c)+3*a^3*(-sin(d*x+c)+ln(sec(d*x+c)+tan( d*x+c)))+a^3*(-1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))
Time = 0.30 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.28 \[ \int (a+a \sec (c+d x))^3 \sin ^2(c+d x) \, dx=-\frac {10 \, a^{3} d x \cos \left (d x + c\right )^{2} - 5 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + 5 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{3} \cos \left (d x + c\right )^{3} + 6 \, a^{3} \cos \left (d x + c\right )^{2} - 6 \, a^{3} \cos \left (d x + c\right ) - a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
-1/4*(10*a^3*d*x*cos(d*x + c)^2 - 5*a^3*cos(d*x + c)^2*log(sin(d*x + c) + 1) + 5*a^3*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(a^3*cos(d*x + c)^3 + 6*a^3*cos(d*x + c)^2 - 6*a^3*cos(d*x + c) - a^3)*sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int (a+a \sec (c+d x))^3 \sin ^2(c+d x) \, dx=a^{3} \left (\int 3 \sin ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )}\, dx\right ) \]
a**3*(Integral(3*sin(c + d*x)**2*sec(c + d*x), x) + Integral(3*sin(c + d*x )**2*sec(c + d*x)**2, x) + Integral(sin(c + d*x)**2*sec(c + d*x)**3, x) + Integral(sin(c + d*x)**2, x))
Time = 0.28 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.30 \[ \int (a+a \sec (c+d x))^3 \sin ^2(c+d x) \, dx=\frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 12 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} - a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \]
1/4*((2*d*x + 2*c - sin(2*d*x + 2*c))*a^3 - 12*(d*x + c - tan(d*x + c))*a^ 3 - a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - log (sin(d*x + c) - 1)) + 6*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2*sin(d*x + c)))/d
Time = 0.37 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.04 \[ \int (a+a \sec (c+d x))^3 \sin ^2(c+d x) \, dx=-\frac {5 \, {\left (d x + c\right )} a^{3} - 5 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 5 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {4 \, {\left (5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )}^{2}}}{2 \, d} \]
-1/2*(5*(d*x + c)*a^3 - 5*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 5*a^3*l og(abs(tan(1/2*d*x + 1/2*c) - 1)) + 4*(5*a^3*tan(1/2*d*x + 1/2*c)^7 - 9*a^ 3*tan(1/2*d*x + 1/2*c)^3)/(tan(1/2*d*x + 1/2*c)^4 - 1)^2)/d
Time = 13.40 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.92 \[ \int (a+a \sec (c+d x))^3 \sin ^2(c+d x) \, dx=\frac {5\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {5\,a^3\,x}{2}+\frac {18\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+1\right )} \]