Integrand size = 19, antiderivative size = 58 \[ \int (a+b \sec (c+d x)) \sin ^3(c+d x) \, dx=-\frac {a \cos (c+d x)}{d}+\frac {b \cos ^2(c+d x)}{2 d}+\frac {a \cos ^3(c+d x)}{3 d}-\frac {b \log (\cos (c+d x))}{d} \]
Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.98 \[ \int (a+b \sec (c+d x)) \sin ^3(c+d x) \, dx=-\frac {3 a \cos (c+d x)}{4 d}+\frac {a \cos (3 (c+d x))}{12 d}-\frac {b \left (-\frac {1}{2} \cos ^2(c+d x)+\log (\cos (c+d x))\right )}{d} \]
(-3*a*Cos[c + d*x])/(4*d) + (a*Cos[3*(c + d*x)])/(12*d) - (b*(-1/2*Cos[c + d*x]^2 + Log[Cos[c + d*x]]))/d
Time = 0.33 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {3042, 4360, 25, 25, 3042, 25, 3316, 27, 522, 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.
\(\displaystyle \int \sin ^3(c+d x) (a+b \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^3 \left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -\left (\sin ^2(c+d x) \tan (c+d x) (-a \cos (c+d x)-b)\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\left ((b+a \cos (c+d x)) \sin ^2(c+d x) \tan (c+d x)\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \sin ^2(c+d x) \tan (c+d x) (a \cos (c+d x)+b)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cos \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (b+a \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )}{\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle -\frac {\int (b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right ) \sec (c+d x)d(a \cos (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {(b+a \cos (c+d x)) \left (a^2-a^2 \cos ^2(c+d x)\right ) \sec (c+d x)}{a}d(a \cos (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle -\frac {\int \left (-\cos ^2(c+d x) a^2+a^2-b \cos (c+d x) a+b \sec (c+d x) a\right )d(a \cos (c+d x))}{a^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{3} a^3 \cos ^3(c+d x)+a^3 \cos (c+d x)-\frac {1}{2} a^2 b \cos ^2(c+d x)+a^2 b \log (a \cos (c+d x))}{a^2 d}\) |
-((a^3*Cos[c + d*x] - (a^2*b*Cos[c + d*x]^2)/2 - (a^3*Cos[c + d*x]^3)/3 + a^2*b*Log[a*Cos[c + d*x]])/(a^2*d))
3.2.62.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^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 = 1.64 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.81
method | result | size |
derivativedivides | \(\frac {-\frac {a \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+b \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(47\) |
default | \(\frac {-\frac {a \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+b \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(47\) |
parts | \(-\frac {a \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3 d}+\frac {b \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(49\) |
risch | \(i b x +\frac {b \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i b c}{d}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}-\frac {3 a \cos \left (d x +c \right )}{4 d}+\frac {a \cos \left (3 d x +3 c \right )}{12 d}\) | \(90\) |
parallelrisch | \(\frac {12 b \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-12 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-12 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-9 a \cos \left (d x +c \right )+3 \cos \left (2 d x +2 c \right ) b +a \cos \left (3 d x +3 c \right )-8 a -3 b}{12 d}\) | \(90\) |
norman | \(\frac {-\frac {4 a}{3 d}-\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {\left (4 a +2 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {b \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(120\) |
Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.84 \[ \int (a+b \sec (c+d x)) \sin ^3(c+d x) \, dx=\frac {2 \, a \cos \left (d x + c\right )^{3} + 3 \, b \cos \left (d x + c\right )^{2} - 6 \, a \cos \left (d x + c\right ) - 6 \, b \log \left (-\cos \left (d x + c\right )\right )}{6 \, d} \]
\[ \int (a+b \sec (c+d x)) \sin ^3(c+d x) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \sin ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.20 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.81 \[ \int (a+b \sec (c+d x)) \sin ^3(c+d x) \, dx=\frac {2 \, a \cos \left (d x + c\right )^{3} + 3 \, b \cos \left (d x + c\right )^{2} - 6 \, a \cos \left (d x + c\right ) - 6 \, b \log \left (\cos \left (d x + c\right )\right )}{6 \, d} \]
Time = 0.31 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.14 \[ \int (a+b \sec (c+d x)) \sin ^3(c+d x) \, dx=-\frac {b \log \left (\frac {{\left | \cos \left (d x + c\right ) \right |}}{{\left | d \right |}}\right )}{d} + \frac {2 \, a d^{2} \cos \left (d x + c\right )^{3} + 3 \, b d^{2} \cos \left (d x + c\right )^{2} - 6 \, a d^{2} \cos \left (d x + c\right )}{6 \, d^{3}} \]
-b*log(abs(cos(d*x + c))/abs(d))/d + 1/6*(2*a*d^2*cos(d*x + c)^3 + 3*b*d^2 *cos(d*x + c)^2 - 6*a*d^2*cos(d*x + c))/d^3
Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.78 \[ \int (a+b \sec (c+d x)) \sin ^3(c+d x) \, dx=-\frac {a\,\cos \left (c+d\,x\right )-\frac {a\,{\cos \left (c+d\,x\right )}^3}{3}-\frac {b\,{\cos \left (c+d\,x\right )}^2}{2}+b\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]