Integrand size = 19, antiderivative size = 52 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {4 a^3 \log (1-\sin (c+d x))}{d}-\frac {3 a^3 \sin (c+d x)}{d}-\frac {a^3 \sin ^2(c+d x)}{2 d} \] Output:
-4*a^3*ln(1-sin(d*x+c))/d-3*a^3*sin(d*x+c)/d-1/2*a^3*sin(d*x+c)^2/d
Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.79 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \left (-4 \log (1-\sin (c+d x))-3 \sin (c+d x)-\frac {1}{2} \sin ^2(c+d x)\right )}{d} \] Input:
Integrate[Sec[c + d*x]*(a + a*Sin[c + d*x])^3,x]
Output:
(a^3*(-4*Log[1 - Sin[c + d*x]] - 3*Sin[c + d*x] - Sin[c + d*x]^2/2))/d
Time = 0.23 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3146, 49, 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.
\(\displaystyle \int \sec (c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (c+d x)+a)^3}{\cos (c+d x)}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {a \int \frac {(\sin (c+d x) a+a)^2}{a-a \sin (c+d x)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {a \int \left (\frac {4 a^2}{a-a \sin (c+d x)}-\sin (c+d x) a-3 a\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a \left (-\frac {1}{2} a^2 \sin ^2(c+d x)-3 a^2 \sin (c+d x)-4 a^2 \log (a-a \sin (c+d x))\right )}{d}\) |
Input:
Int[Sec[c + d*x]*(a + a*Sin[c + d*x])^3,x]
Output:
(a*(-4*a^2*Log[a - a*Sin[c + d*x]] - 3*a^2*Sin[c + d*x] - (a^2*Sin[c + d*x ]^2)/2))/d
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
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 + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 0.18 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.12
method | result | size |
parallelrisch | \(-\frac {a^{3} \left (32 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-16 \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+1-\cos \left (2 d x +2 c \right )+12 \sin \left (d x +c \right )\right )}{4 d}\) | \(58\) |
derivativedivides | \(\frac {a^{3} \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-3 a^{3} \ln \left (\cos \left (d x +c \right )\right )+a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(88\) |
default | \(\frac {a^{3} \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-3 a^{3} \ln \left (\cos \left (d x +c \right )\right )+a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(88\) |
risch | \(4 i a^{3} x +\frac {3 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {3 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {8 i a^{3} c}{d}-\frac {8 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a^{3} \cos \left (2 d x +2 c \right )}{4 d}\) | \(93\) |
norman | \(\frac {-\frac {6 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {12 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {6 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-\frac {8 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {4 a^{3} \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}\) | \(154\) |
Input:
int(sec(d*x+c)*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
-1/4*a^3*(32*ln(tan(1/2*d*x+1/2*c)-1)-16*ln(sec(1/2*d*x+1/2*c)^2)+1-cos(2* d*x+2*c)+12*sin(d*x+c))/d
Time = 0.13 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \cos \left (d x + c\right )^{2} - 8 \, a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 6 \, a^{3} \sin \left (d x + c\right )}{2 \, d} \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="fricas")
Output:
1/2*(a^3*cos(d*x + c)^2 - 8*a^3*log(-sin(d*x + c) + 1) - 6*a^3*sin(d*x + c ))/d
\[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \, dx=a^{3} \left (\int 3 \sin {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \sec {\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))**3,x)
Output:
a**3*(Integral(3*sin(c + d*x)*sec(c + d*x), x) + Integral(3*sin(c + d*x)** 2*sec(c + d*x), x) + Integral(sin(c + d*x)**3*sec(c + d*x), x) + Integral( sec(c + d*x), x))
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.83 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^{3} \sin \left (d x + c\right )^{2} + 8 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, a^{3} \sin \left (d x + c\right )}{2 \, d} \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="maxima")
Output:
-1/2*(a^3*sin(d*x + c)^2 + 8*a^3*log(sin(d*x + c) - 1) + 6*a^3*sin(d*x + c ))/d
Time = 0.13 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.96 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {4 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {a^{3} d \sin \left (d x + c\right )^{2} + 6 \, a^{3} d \sin \left (d x + c\right )}{2 \, d^{2}} \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="giac")
Output:
-4*a^3*log(abs(sin(d*x + c) - 1))/d - 1/2*(a^3*d*sin(d*x + c)^2 + 6*a^3*d* sin(d*x + c))/d^2
Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.69 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3\,\left (8\,\ln \left (\sin \left (c+d\,x\right )-1\right )+6\,\sin \left (c+d\,x\right )+{\sin \left (c+d\,x\right )}^2\right )}{2\,d} \] Input:
int((a + a*sin(c + d*x))^3/cos(c + d*x),x)
Output:
-(a^3*(8*log(sin(c + d*x) - 1) + 6*sin(c + d*x) + sin(c + d*x)^2))/(2*d)
Time = 0.17 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.12 \[ \int \sec (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \left (8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )-16 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\sin \left (d x +c \right )^{2}-6 \sin \left (d x +c \right )+2\right )}{2 d} \] Input:
int(sec(d*x+c)*(a+a*sin(d*x+c))^3,x)
Output:
(a**3*(8*log(tan((c + d*x)/2)**2 + 1) - 16*log(tan((c + d*x)/2) - 1) - sin (c + d*x)**2 - 6*sin(c + d*x) + 2))/(2*d)