Integrand size = 21, antiderivative size = 80 \[ \int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx=-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{d}+\frac {a b \cos ^2(c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {b^2 \sec (c+d x)}{d} \]
-(a^2-b^2)*cos(d*x+c)/d+a*b*cos(d*x+c)^2/d+1/3*a^2*cos(d*x+c)^3/d-2*a*b*ln (cos(d*x+c))/d+b^2*sec(d*x+c)/d
Time = 0.68 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90 \[ \int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx=\frac {\left (-9 a^2+12 b^2\right ) \cos (c+d x)+6 a b \cos (2 (c+d x))+a^2 \cos (3 (c+d x))-24 a b \log (\cos (c+d x))+12 b^2 \sec (c+d x)}{12 d} \]
((-9*a^2 + 12*b^2)*Cos[c + d*x] + 6*a*b*Cos[2*(c + d*x)] + a^2*Cos[3*(c + d*x)] - 24*a*b*Log[Cos[c + d*x]] + 12*b^2*Sec[c + d*x])/(12*d)
Time = 0.41 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4360, 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.
\(\displaystyle \int \sin ^3(c+d x) (a+b \sec (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^3 \left (a-b \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \sin (c+d x) \tan ^2(c+d x) (-a \cos (c+d x)-b)^2dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cos \left (c+d x+\frac {\pi }{2}\right )^3 \left (-a \sin \left (c+d x+\frac {\pi }{2}\right )-b\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^3 \left (b+a \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^2}{\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle -\frac {\int (b+a \cos (c+d x))^2 \left (a^2-a^2 \cos ^2(c+d x)\right ) \sec ^2(c+d x)d(a \cos (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {(b+a \cos (c+d x))^2 \left (a^2-a^2 \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a^2}d(a \cos (c+d x))}{a d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle -\frac {\int \left (-\cos ^2(c+d x) a^2+\left (1-\frac {b^2}{a^2}\right ) a^2-2 b \cos (c+d x) a+2 b \sec (c+d x) a+b^2 \sec ^2(c+d x)\right )d(a \cos (c+d x))}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{3} a^3 \cos ^3(c+d x)+a \left (a^2-b^2\right ) \cos (c+d x)-a^2 b \cos ^2(c+d x)+2 a^2 b \log (a \cos (c+d x))-a b^2 \sec (c+d x)}{a d}\) |
-((a*(a^2 - b^2)*Cos[c + d*x] - a^2*b*Cos[c + d*x]^2 - (a^3*Cos[c + d*x]^3 )/3 + 2*a^2*b*Log[a*Cos[c + d*x]] - a*b^2*Sec[c + d*x])/(a*d))
3.2.75.3.1 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 = 1.80 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.12
method | result | size |
derivativedivides | \(\frac {-\frac {a^{2} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+2 a b \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(90\) |
default | \(\frac {-\frac {a^{2} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}+2 a b \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(90\) |
parts | \(-\frac {a^{2} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3 d}+\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )}{d}+\frac {2 a b \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(95\) |
parallelrisch | \(\frac {48 a b \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \cos \left (d x +c \right )-48 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-48 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+\left (-8 a^{2}+12 b^{2}\right ) \cos \left (2 d x +2 c \right )+6 a b \cos \left (3 d x +3 c \right )+\cos \left (4 d x +4 c \right ) a^{2}+\left (-16 a^{2}-6 a b +48 b^{2}\right ) \cos \left (d x +c \right )-9 a^{2}+36 b^{2}}{24 d \cos \left (d x +c \right )}\) | \(160\) |
norman | \(\frac {\frac {4 a^{2}-12 b^{2}}{3 d}-\frac {4 \left (a^{2}+b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}+\frac {2 \left (4 a^{2}+6 a b -12 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {2 a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}+\frac {2 a b \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}\) | \(181\) |
risch | \(2 i a b x +\frac {a b \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 d}-\frac {3 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {{\mathrm e}^{i \left (d x +c \right )} b^{2}}{2 d}-\frac {3 \,{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{8 d}+\frac {{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{2 d}+\frac {a b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 d}+\frac {4 i a b c}{d}+\frac {2 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}+\frac {\cos \left (3 d x +3 c \right ) a^{2}}{12 d}\) | \(183\) |
1/d*(-1/3*a^2*(2+sin(d*x+c)^2)*cos(d*x+c)+2*a*b*(-1/2*sin(d*x+c)^2-ln(cos( d*x+c)))+b^2*(sin(d*x+c)^4/cos(d*x+c)+(2+sin(d*x+c)^2)*cos(d*x+c)))
Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.15 \[ \int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx=\frac {2 \, a^{2} \cos \left (d x + c\right )^{4} + 6 \, a b \cos \left (d x + c\right )^{3} - 12 \, a b \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - 3 \, a b \cos \left (d x + c\right ) - 6 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \, b^{2}}{6 \, d \cos \left (d x + c\right )} \]
1/6*(2*a^2*cos(d*x + c)^4 + 6*a*b*cos(d*x + c)^3 - 12*a*b*cos(d*x + c)*log (-cos(d*x + c)) - 3*a*b*cos(d*x + c) - 6*(a^2 - b^2)*cos(d*x + c)^2 + 6*b^ 2)/(d*cos(d*x + c))
\[ \int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sin ^{3}{\left (c + d x \right )}\, dx \]
Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx=\frac {a^{2} \cos \left (d x + c\right )^{3} + 3 \, a b \cos \left (d x + c\right )^{2} - 6 \, a b \log \left (\cos \left (d x + c\right )\right ) - 3 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right ) + \frac {3 \, b^{2}}{\cos \left (d x + c\right )}}{3 \, d} \]
1/3*(a^2*cos(d*x + c)^3 + 3*a*b*cos(d*x + c)^2 - 6*a*b*log(cos(d*x + c)) - 3*(a^2 - b^2)*cos(d*x + c) + 3*b^2/cos(d*x + c))/d
Time = 0.36 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.25 \[ \int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx=-\frac {2 \, a b \log \left (\frac {{\left | \cos \left (d x + c\right ) \right |}}{{\left | d \right |}}\right )}{d} + \frac {b^{2}}{d \cos \left (d x + c\right )} + \frac {a^{2} d^{5} \cos \left (d x + c\right )^{3} + 3 \, a b d^{5} \cos \left (d x + c\right )^{2} - 3 \, a^{2} d^{5} \cos \left (d x + c\right ) + 3 \, b^{2} d^{5} \cos \left (d x + c\right )}{3 \, d^{6}} \]
-2*a*b*log(abs(cos(d*x + c))/abs(d))/d + b^2/(d*cos(d*x + c)) + 1/3*(a^2*d ^5*cos(d*x + c)^3 + 3*a*b*d^5*cos(d*x + c)^2 - 3*a^2*d^5*cos(d*x + c) + 3* b^2*d^5*cos(d*x + c))/d^6
Time = 14.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.86 \[ \int (a+b \sec (c+d x))^2 \sin ^3(c+d x) \, dx=\frac {\frac {a^2\,{\cos \left (c+d\,x\right )}^3}{3}-\cos \left (c+d\,x\right )\,\left (a^2-b^2\right )+\frac {b^2}{\cos \left (c+d\,x\right )}+a\,b\,{\cos \left (c+d\,x\right )}^2-2\,a\,b\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]