Integrand size = 21, antiderivative size = 89 \[ \int \frac {\sin ^3(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{a^3 d}-\frac {b \cos ^2(c+d x)}{2 a^2 d}+\frac {\cos ^3(c+d x)}{3 a d}+\frac {b \left (a^2-b^2\right ) \log (b+a \cos (c+d x))}{a^4 d} \]
-(a^2-b^2)*cos(d*x+c)/a^3/d-1/2*b*cos(d*x+c)^2/a^2/d+1/3*cos(d*x+c)^3/a/d+ b*(a^2-b^2)*ln(b+a*cos(d*x+c))/a^4/d
Time = 0.10 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \frac {\sin ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\left (-9 a^3+12 a b^2\right ) \cos (c+d x)-3 a^2 b \cos (2 (c+d x))+a^3 \cos (3 (c+d x))+12 a^2 b \log (b+a \cos (c+d x))-12 b^3 \log (b+a \cos (c+d x))}{12 a^4 d} \]
((-9*a^3 + 12*a*b^2)*Cos[c + d*x] - 3*a^2*b*Cos[2*(c + d*x)] + a^3*Cos[3*( c + d*x)] + 12*a^2*b*Log[b + a*Cos[c + d*x]] - 12*b^3*Log[b + a*Cos[c + d* x]])/(12*a^4*d)
Time = 0.40 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.90, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, 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.
\(\displaystyle \int \frac {\sin ^3(c+d x)}{a+b \sec (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos \left (c+d x-\frac {\pi }{2}\right )^3}{a-b \csc \left (c+d x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -\frac {\sin ^3(c+d x) \cos (c+d x)}{-a \cos (c+d x)-b}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\frac {\cos (c+d x) \sin ^3(c+d x)}{b+a \cos (c+d x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\sin ^3(c+d x) \cos (c+d x)}{a \cos (c+d x)+b}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \cos \left (c+d x+\frac {\pi }{2}\right )^3}{a \sin \left (c+d x+\frac {\pi }{2}\right )+b}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}{b+a \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle -\frac {\int \frac {\cos (c+d x) \left (a^2-a^2 \cos ^2(c+d x)\right )}{b+a \cos (c+d x)}d(a \cos (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {a \cos (c+d x) \left (a^2-a^2 \cos ^2(c+d x)\right )}{b+a \cos (c+d x)}d(a \cos (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle -\frac {\int \left (-\cos ^2(c+d x) a^2+\left (1-\frac {b^2}{a^2}\right ) a^2+b \cos (c+d x) a+\frac {b^3-a^2 b}{b+a \cos (c+d x)}\right )d(a \cos (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{3} a^3 \cos ^3(c+d x)+a \left (a^2-b^2\right ) \cos (c+d x)-b \left (a^2-b^2\right ) \log (a \cos (c+d x)+b)+\frac {1}{2} a^2 b \cos ^2(c+d x)}{a^4 d}\) |
-((a*(a^2 - b^2)*Cos[c + d*x] + (a^2*b*Cos[c + d*x]^2)/2 - (a^3*Cos[c + d* x]^3)/3 - b*(a^2 - b^2)*Log[b + a*Cos[c + d*x]])/(a^4*d))
3.2.98.3.1 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 = 0.78 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {\frac {\frac {\cos \left (d x +c \right )^{3} a^{2}}{3}-\frac {a b \cos \left (d x +c \right )^{2}}{2}-\cos \left (d x +c \right ) a^{2}+\cos \left (d x +c \right ) b^{2}}{a^{3}}+\frac {b \left (a^{2}-b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{a^{4}}}{d}\) | \(82\) |
default | \(\frac {\frac {\frac {\cos \left (d x +c \right )^{3} a^{2}}{3}-\frac {a b \cos \left (d x +c \right )^{2}}{2}-\cos \left (d x +c \right ) a^{2}+\cos \left (d x +c \right ) b^{2}}{a^{3}}+\frac {b \left (a^{2}-b^{2}\right ) \ln \left (b +a \cos \left (d x +c \right )\right )}{a^{4}}}{d}\) | \(82\) |
parallelrisch | \(\frac {-9 a^{3} \cos \left (d x +c \right )+12 \cos \left (d x +c \right ) a \,b^{2}+a^{3} \cos \left (3 d x +3 c \right )-3 a^{2} b \cos \left (2 d x +2 c \right )-12 \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a^{2} b +12 \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) b^{3}+12 \ln \left (-2 a +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a -b \right )\right ) a^{2} b -12 \ln \left (-2 a +\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a -b \right )\right ) b^{3}-8 a^{3}+3 a^{2} b +12 a \,b^{2}}{12 a^{4} d}\) | \(168\) |
norman | \(\frac {\frac {\left (4 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a^{3} d}+\frac {-2 a b +2 b^{2}}{3 d \,a^{3}}+\frac {\left (4 a^{2}-2 a b -4 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d \,a^{3}}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+\frac {\left (a +b \right ) b \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}{a^{4} d}-\frac {\left (a +b \right ) b \left (a -b \right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a^{4} d}\) | \(182\) |
risch | \(-\frac {i x b}{a^{2}}+\frac {i x \,b^{3}}{a^{4}}-\frac {b \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{2} d}-\frac {3 \,{\mathrm e}^{i \left (d x +c \right )}}{8 d a}+\frac {{\mathrm e}^{i \left (d x +c \right )} b^{2}}{2 a^{3} d}-\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a d}+\frac {{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{2 a^{3} d}-\frac {b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{2} d}-\frac {2 i b c}{a^{2} d}+\frac {2 i b^{3} c}{a^{4} d}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{2} d}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{4} d}+\frac {\cos \left (3 d x +3 c \right )}{12 a d}\) | \(244\) |
1/d*(1/a^3*(1/3*cos(d*x+c)^3*a^2-1/2*a*b*cos(d*x+c)^2-cos(d*x+c)*a^2+cos(d *x+c)*b^2)+b*(a^2-b^2)/a^4*ln(b+a*cos(d*x+c)))
Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.88 \[ \int \frac {\sin ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {2 \, a^{3} \cos \left (d x + c\right )^{3} - 3 \, a^{2} b \cos \left (d x + c\right )^{2} - 6 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) + 6 \, {\left (a^{2} b - b^{3}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{6 \, a^{4} d} \]
1/6*(2*a^3*cos(d*x + c)^3 - 3*a^2*b*cos(d*x + c)^2 - 6*(a^3 - a*b^2)*cos(d *x + c) + 6*(a^2*b - b^3)*log(a*cos(d*x + c) + b))/(a^4*d)
Timed out. \[ \int \frac {\sin ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.90 \[ \int \frac {\sin ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {2 \, a^{2} \cos \left (d x + c\right )^{3} - 3 \, a b \cos \left (d x + c\right )^{2} - 6 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )}{a^{3}} + \frac {6 \, {\left (a^{2} b - b^{3}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{4}}}{6 \, d} \]
1/6*((2*a^2*cos(d*x + c)^3 - 3*a*b*cos(d*x + c)^2 - 6*(a^2 - b^2)*cos(d*x + c))/a^3 + 6*(a^2*b - b^3)*log(a*cos(d*x + c) + b)/a^4)/d
Time = 0.33 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.15 \[ \int \frac {\sin ^3(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {{\left (a^{2} b - b^{3}\right )} \log \left ({\left | -a \cos \left (d x + c\right ) - b \right |}\right )}{a^{4} d} + \frac {2 \, a^{2} d^{2} \cos \left (d x + c\right )^{3} - 3 \, a b d^{2} \cos \left (d x + c\right )^{2} - 6 \, a^{2} d^{2} \cos \left (d x + c\right ) + 6 \, b^{2} d^{2} \cos \left (d x + c\right )}{6 \, a^{3} d^{3}} \]
(a^2*b - b^3)*log(abs(-a*cos(d*x + c) - b))/(a^4*d) + 1/6*(2*a^2*d^2*cos(d *x + c)^3 - 3*a*b*d^2*cos(d*x + c)^2 - 6*a^2*d^2*cos(d*x + c) + 6*b^2*d^2* cos(d*x + c))/(a^3*d^3)
Time = 13.51 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89 \[ \int \frac {\sin ^3(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\cos \left (c+d\,x\right )\,\left (\frac {1}{a}-\frac {b^2}{a^3}\right )-\frac {{\cos \left (c+d\,x\right )}^3}{3\,a}+\frac {b\,{\cos \left (c+d\,x\right )}^2}{2\,a^2}-\frac {\ln \left (b+a\,\cos \left (c+d\,x\right )\right )\,\left (a^2\,b-b^3\right )}{a^4}}{d} \]