Integrand size = 19, antiderivative size = 70 \[ \int (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=-\frac {4 a^3 \log (1-\sin (c+d x))}{d}-\frac {4 a^3 \sin (c+d x)}{d}-\frac {3 a^3 \sin ^2(c+d x)}{2 d}-\frac {a^3 \sin ^3(c+d x)}{3 d} \] Output:
-4*a^3*ln(1-sin(d*x+c))/d-4*a^3*sin(d*x+c)/d-3/2*a^3*sin(d*x+c)^2/d-1/3*a^ 3*sin(d*x+c)^3/d
Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.74 \[ \int (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=-\frac {a^3 \left (24 \log (1-\sin (c+d x))+24 \sin (c+d x)+9 \sin ^2(c+d x)+2 \sin ^3(c+d x)\right )}{6 d} \] Input:
Integrate[(a + a*Sin[c + d*x])^3*Tan[c + d*x],x]
Output:
-1/6*(a^3*(24*Log[1 - Sin[c + d*x]] + 24*Sin[c + d*x] + 9*Sin[c + d*x]^2 + 2*Sin[c + d*x]^3))/d
Time = 0.24 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3186, 86, 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.
\(\displaystyle \int \tan (c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x) (a \sin (c+d x)+a)^3dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {a \sin (c+d x) (\sin (c+d x) a+a)^2}{a-a \sin (c+d x)}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (\frac {4 a^3}{a-a \sin (c+d x)}-\sin ^2(c+d x) a^2-3 \sin (c+d x) a^2-4 a^2\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{3} a^3 \sin ^3(c+d x)-\frac {3}{2} a^3 \sin ^2(c+d x)-4 a^3 \sin (c+d x)-4 a^3 \log (a-a \sin (c+d x))}{d}\) |
Input:
Int[(a + a*Sin[c + d*x])^3*Tan[c + d*x],x]
Output:
(-4*a^3*Log[a - a*Sin[c + d*x]] - 4*a^3*Sin[c + d*x] - (3*a^3*Sin[c + d*x] ^2)/2 - (a^3*Sin[c + d*x]^3)/3)/d
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[x^p*((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] && E qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
Time = 3.31 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.54
method | result | size |
derivativedivides | \(\frac {a^{3} \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+3 a^{3} \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-a^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}\) | \(108\) |
default | \(\frac {a^{3} \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+3 a^{3} \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-a^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}\) | \(108\) |
risch | \(4 i a^{3} x +\frac {17 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {17 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {8 i a^{3} c}{d}-\frac {8 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a^{3} \sin \left (3 d x +3 c \right )}{12 d}+\frac {3 a^{3} \cos \left (2 d x +2 c \right )}{4 d}\) | \(110\) |
parts | \(\frac {a^{3} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}+\frac {a^{3} \left (-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}+\frac {3 a^{3} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}+\frac {3 a^{3} \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(120\) |
Input:
int((a+a*sin(d*x+c))^3*tan(d*x+c),x,method=_RETURNVERBOSE)
Output:
1/d*(a^3*(-1/3*sin(d*x+c)^3-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))+3*a^3*(- 1/2*sin(d*x+c)^2-ln(cos(d*x+c)))+3*a^3*(-sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+ c)))-a^3*ln(cos(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.87 \[ \int (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=\frac {9 \, a^{3} \cos \left (d x + c\right )^{2} - 24 \, a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - 13 \, a^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate((a+a*sin(d*x+c))^3*tan(d*x+c),x, algorithm="fricas")
Output:
1/6*(9*a^3*cos(d*x + c)^2 - 24*a^3*log(-sin(d*x + c) + 1) + 2*(a^3*cos(d*x + c)^2 - 13*a^3)*sin(d*x + c))/d
\[ \int (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=a^{3} \left (\int 3 \sin {\left (c + d x \right )} \tan {\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \tan {\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \tan {\left (c + d x \right )}\, dx + \int \tan {\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*sin(d*x+c))**3*tan(d*x+c),x)
Output:
a**3*(Integral(3*sin(c + d*x)*tan(c + d*x), x) + Integral(3*sin(c + d*x)** 2*tan(c + d*x), x) + Integral(sin(c + d*x)**3*tan(c + d*x), x) + Integral( tan(c + d*x), x))
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.81 \[ \int (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=-\frac {2 \, a^{3} \sin \left (d x + c\right )^{3} + 9 \, a^{3} \sin \left (d x + c\right )^{2} + 24 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 24 \, a^{3} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate((a+a*sin(d*x+c))^3*tan(d*x+c),x, algorithm="maxima")
Output:
-1/6*(2*a^3*sin(d*x + c)^3 + 9*a^3*sin(d*x + c)^2 + 24*a^3*log(sin(d*x + c ) - 1) + 24*a^3*sin(d*x + c))/d
Time = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.01 \[ \int (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=-\frac {4 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {2 \, a^{3} d^{2} \sin \left (d x + c\right )^{3} + 9 \, a^{3} d^{2} \sin \left (d x + c\right )^{2} + 24 \, a^{3} d^{2} \sin \left (d x + c\right )}{6 \, d^{3}} \] Input:
integrate((a+a*sin(d*x+c))^3*tan(d*x+c),x, algorithm="giac")
Output:
-4*a^3*log(abs(sin(d*x + c) - 1))/d - 1/6*(2*a^3*d^2*sin(d*x + c)^3 + 9*a^ 3*d^2*sin(d*x + c)^2 + 24*a^3*d^2*sin(d*x + c))/d^3
Time = 18.02 (sec) , antiderivative size = 281, normalized size of antiderivative = 4.01 \[ \int (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=-\frac {\frac {56\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^3\,\left (12\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )-6\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\right )-\frac {2\,a^3\,\left (36\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )-18\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )+9\right )}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^3\,\left (12\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )-6\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\right )-\frac {2\,a^3\,\left (36\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )-18\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )+9\right )}{3}\right )+8\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3}-\frac {2\,a^3\,\left (12\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )-6\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\right )}{3\,d} \] Input:
int(tan(c + d*x)*(a + a*sin(c + d*x))^3,x)
Output:
- ((56*a^3*tan(c/2 + (d*x)/2)^3)/3 + 8*a^3*tan(c/2 + (d*x)/2)^5 - tan(c/2 + (d*x)/2)^2*(2*a^3*(12*log(tan(c/2 + (d*x)/2) - 1) - 6*log(tan(c/2 + (d*x )/2)^2 + 1)) - (2*a^3*(36*log(tan(c/2 + (d*x)/2) - 1) - 18*log(tan(c/2 + ( d*x)/2)^2 + 1) + 9))/3) - tan(c/2 + (d*x)/2)^4*(2*a^3*(12*log(tan(c/2 + (d *x)/2) - 1) - 6*log(tan(c/2 + (d*x)/2)^2 + 1)) - (2*a^3*(36*log(tan(c/2 + (d*x)/2) - 1) - 18*log(tan(c/2 + (d*x)/2)^2 + 1) + 9))/3) + 8*a^3*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d*x)/2)^2 + 1)^3) - (2*a^3*(12*log(tan(c/2 + (d *x)/2) - 1) - 6*log(tan(c/2 + (d*x)/2)^2 + 1)))/(3*d)
Time = 0.19 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.34 \[ \int (a+a \sin (c+d x))^3 \tan (c+d x) \, dx=\frac {a^{3} \left (3 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )+18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )-42 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 \sin \left (d x +c \right )^{3}-9 \sin \left (d x +c \right )^{2}-24 \sin \left (d x +c \right )\right )}{6 d} \] Input:
int((a+a*sin(d*x+c))^3*tan(d*x+c),x)
Output:
(a**3*(3*log(tan(c + d*x)**2 + 1) + 18*log(tan((c + d*x)/2)**2 + 1) - 42*l og(tan((c + d*x)/2) - 1) + 6*log(tan((c + d*x)/2) + 1) - 2*sin(c + d*x)**3 - 9*sin(c + d*x)**2 - 24*sin(c + d*x)))/(6*d)