Integrand size = 21, antiderivative size = 72 \[ \int (a+a \cos (c+d x))^3 \sec ^4(c+d x) \, dx=\frac {5 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {3 a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^3 \tan ^3(c+d x)}{3 d} \]
5/2*a^3*arctanh(sin(d*x+c))/d+4*a^3*tan(d*x+c)/d+3/2*a^3*sec(d*x+c)*tan(d* x+c)/d+1/3*a^3*tan(d*x+c)^3/d
Time = 0.23 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00 \[ \int (a+a \cos (c+d x))^3 \sec ^4(c+d x) \, dx=\frac {5 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {3 a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^3 \tan ^3(c+d x)}{3 d} \]
(5*a^3*ArcTanh[Sin[c + d*x]])/(2*d) + (4*a^3*Tan[c + d*x])/d + (3*a^3*Sec[ c + d*x]*Tan[c + d*x])/(2*d) + (a^3*Tan[c + d*x]^3)/(3*d)
Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 3236, 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.
\(\displaystyle \int \sec ^4(c+d x) (a \cos (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 3236 |
\(\displaystyle \int \left (a^3 \sec ^4(c+d x)+3 a^3 \sec ^3(c+d x)+3 a^3 \sec ^2(c+d x)+a^3 \sec (c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^3 \tan ^3(c+d x)}{3 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {3 a^3 \tan (c+d x) \sec (c+d x)}{2 d}\) |
(5*a^3*ArcTanh[Sin[c + d*x]])/(2*d) + (4*a^3*Tan[c + d*x])/d + (3*a^3*Sec[ c + d*x]*Tan[c + d*x])/(2*d) + (a^3*Tan[c + d*x]^3)/(3*d)
3.1.30.3.1 Defintions of rubi rules used
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)])^(m_.), x_Symbol] :> Int[ExpandTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IGt Q[m, 0] && RationalQ[n]
Time = 3.45 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.31
method | result | size |
derivativedivides | \(\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} \tan \left (d x +c \right )+3 a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(94\) |
default | \(\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} \tan \left (d x +c \right )+3 a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(94\) |
parts | \(-\frac {a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {3 a^{3} \tan \left (d x +c \right )}{d}\) | \(102\) |
risch | \(-\frac {i a^{3} \left (9 \,{\mathrm e}^{5 i \left (d x +c \right )}-18 \,{\mathrm e}^{4 i \left (d x +c \right )}-48 \,{\mathrm e}^{2 i \left (d x +c \right )}-9 \,{\mathrm e}^{i \left (d x +c \right )}-22\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\) | \(112\) |
parallelrisch | \(-\frac {a^{3} \left (15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (3 d x +3 c \right )-15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (3 d x +3 c \right )+45 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-45 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )-30 \sin \left (d x +c \right )-18 \sin \left (2 d x +2 c \right )-22 \sin \left (3 d x +3 c \right )\right )}{6 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(146\) |
norman | \(\frac {-\frac {11 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {59 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {14 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {5 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {5 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {5 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {5 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(186\) |
1/d*(a^3*ln(sec(d*x+c)+tan(d*x+c))+3*a^3*tan(d*x+c)+3*a^3*(1/2*sec(d*x+c)* tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-a^3*(-2/3-1/3*sec(d*x+c)^2)*tan( d*x+c))
Time = 0.25 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.36 \[ \int (a+a \cos (c+d x))^3 \sec ^4(c+d x) \, dx=\frac {15 \, a^{3} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, a^{3} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (22 \, a^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]
1/12*(15*a^3*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 15*a^3*cos(d*x + c)^3* log(-sin(d*x + c) + 1) + 2*(22*a^3*cos(d*x + c)^2 + 9*a^3*cos(d*x + c) + 2 *a^3)*sin(d*x + c))/(d*cos(d*x + c)^3)
\[ \int (a+a \cos (c+d x))^3 \sec ^4(c+d x) \, dx=a^{3} \left (\int 3 \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int 3 \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \cos ^{3}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]
a**3*(Integral(3*cos(c + d*x)*sec(c + d*x)**4, x) + Integral(3*cos(c + d*x )**2*sec(c + d*x)**4, x) + Integral(cos(c + d*x)**3*sec(c + d*x)**4, x) + Integral(sec(c + d*x)**4, x))
Time = 0.24 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.54 \[ \int (a+a \cos (c+d x))^3 \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} - 9 \, a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, a^{3} \tan \left (d x + c\right )}{12 \, d} \]
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^3 - 9*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)) + 36*a^3*tan(d*x + c))/d
Time = 0.37 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.47 \[ \int (a+a \cos (c+d x))^3 \sec ^4(c+d x) \, dx=\frac {15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]
1/6*(15*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*a^3*log(abs(tan(1/2*d* x + 1/2*c) - 1)) - 2*(15*a^3*tan(1/2*d*x + 1/2*c)^5 - 40*a^3*tan(1/2*d*x + 1/2*c)^3 + 33*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 16.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.56 \[ \int (a+a \cos (c+d x))^3 \sec ^4(c+d x) \, dx=\frac {5\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {5\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {40\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+11\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]