Integrand size = 19, antiderivative size = 80 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {(a+b)^3 \log (1-\sin (c+d x))}{2 d}+\frac {(a-b)^3 \log (1+\sin (c+d x))}{2 d}-\frac {3 a b^2 \sin (c+d x)}{d}-\frac {b^3 \sin ^2(c+d x)}{2 d} \]
-1/2*(a+b)^3*ln(1-sin(d*x+c))/d+1/2*(a-b)^3*ln(1+sin(d*x+c))/d-3*a*b^2*sin (d*x+c)/d-1/2*b^3*sin(d*x+c)^2/d
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.84 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {(a+b)^3 \log (1-\sin (c+d x))-(a-b)^3 \log (1+\sin (c+d x))+6 a b^2 \sin (c+d x)+b^3 \sin ^2(c+d x)}{2 d} \]
-1/2*((a + b)^3*Log[1 - Sin[c + d*x]] - (a - b)^3*Log[1 + Sin[c + d*x]] + 6*a*b^2*Sin[c + d*x] + b^3*Sin[c + d*x]^2)/d
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3147, 477, 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.
\(\displaystyle \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \sin (c+d x))^3}{\cos (c+d x)}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {b \int \frac {(a+b \sin (c+d x))^3}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 477 |
\(\displaystyle \frac {\int \left (\frac {b (a-b)^3}{2 (\sin (c+d x) b+b)}-3 a b^2-b^3 \sin (c+d x)+\frac {b (a+b)^3}{2 (b-b \sin (c+d x))}\right )d(b \sin (c+d x))}{b d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-3 a b^3 \sin (c+d x)-\frac {1}{2} b (a+b)^3 \log (b-b \sin (c+d x))+\frac {1}{2} b (a-b)^3 \log (b \sin (c+d x)+b)-\frac {1}{2} b^4 \sin ^2(c+d x)}{b d}\) |
(-1/2*(b*(a + b)^3*Log[b - b*Sin[c + d*x]]) + ((a - b)^3*b*Log[b + b*Sin[c + d*x]])/2 - 3*a*b^3*Sin[c + d*x] - (b^4*Sin[c + d*x]^2)/2)/(b*d)
3.5.3.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ a^p Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 ]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & & NiceSqrtQ[-b/a] && !FractionalPowerFactorQ[Rt[-b/a, 2]]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.76 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.12
method | result | size |
derivativedivides | \(\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-3 a^{2} b \ln \left (\cos \left (d x +c \right )\right )+3 a \,b^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(90\) |
default | \(\frac {a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-3 a^{2} b \ln \left (\cos \left (d x +c \right )\right )+3 a \,b^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(90\) |
parallelrisch | \(\frac {4 \left (3 a^{2} b +b^{3}\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (a +b \right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 \left (a -b \right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-12 a \,b^{2} \sin \left (d x +c \right )+\cos \left (2 d x +2 c \right ) b^{3}-b^{3}}{4 d}\) | \(101\) |
norman | \(\frac {-\frac {2 b^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {12 a \,b^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a \,b^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}+\frac {b \left (3 a^{2}+b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(216\) |
risch | \(\frac {3 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {2 i b^{3} c}{d}+\frac {6 i a^{2} b c}{d}+3 i a^{2} b x -\frac {3 i a \,b^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+i b^{3} x +\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} b}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{3}}{d}-\frac {a^{3} \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{d}-\frac {3 \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) a^{2} b}{d}-\frac {3 \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) a \,b^{2}}{d}-\frac {\ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) b^{3}}{d}+\frac {b^{3} \cos \left (2 d x +2 c \right )}{4 d}\) | \(264\) |
1/d*(a^3*ln(sec(d*x+c)+tan(d*x+c))-3*a^2*b*ln(cos(d*x+c))+3*a*b^2*(-sin(d* x+c)+ln(sec(d*x+c)+tan(d*x+c)))+b^3*(-1/2*sin(d*x+c)^2-ln(cos(d*x+c))))
Time = 0.30 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.16 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {b^{3} \cos \left (d x + c\right )^{2} - 6 \, a b^{2} \sin \left (d x + c\right ) + {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, d} \]
1/2*(b^3*cos(d*x + c)^2 - 6*a*b^2*sin(d*x + c) + (a^3 - 3*a^2*b + 3*a*b^2 - b^3)*log(sin(d*x + c) + 1) - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*log(-sin(d* x + c) + 1))/d
\[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \sec {\left (c + d x \right )}\, dx \]
Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.14 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {b^{3} \sin \left (d x + c\right )^{2} + 6 \, a b^{2} \sin \left (d x + c\right ) - {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{2 \, d} \]
-1/2*(b^3*sin(d*x + c)^2 + 6*a*b^2*sin(d*x + c) - (a^3 - 3*a^2*b + 3*a*b^2 - b^3)*log(sin(d*x + c) + 1) + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*log(sin(d* x + c) - 1))/d
Time = 0.32 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.16 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {b^{3} \sin \left (d x + c\right )^{2} + 6 \, a b^{2} \sin \left (d x + c\right ) - {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, d} \]
-1/2*(b^3*sin(d*x + c)^2 + 6*a*b^2*sin(d*x + c) - (a^3 - 3*a^2*b + 3*a*b^2 - b^3)*log(abs(sin(d*x + c) + 1)) + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*log(a bs(sin(d*x + c) - 1)))/d
Time = 4.99 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.81 \[ \int \sec (c+d x) (a+b \sin (c+d x))^3 \, dx=-\frac {\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,{\left (a+b\right )}^3}{2}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,{\left (a-b\right )}^3}{2}+\frac {b^3\,{\sin \left (c+d\,x\right )}^2}{2}+3\,a\,b^2\,\sin \left (c+d\,x\right )}{d} \]