Integrand size = 19, antiderivative size = 39 \[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=-a x+\frac {a \cos (c+d x)}{d}+\frac {a \cos (c+d x)}{d (1-\sin (c+d x))} \] Output:
-a*x+a*cos(d*x+c)/d+a*cos(d*x+c)/d/(1-sin(d*x+c))
Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.21 \[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {a \arctan (\tan (c+d x))}{d}+\frac {a \cos (c+d x)}{d}+\frac {a \sec (c+d x)}{d}+\frac {a \tan (c+d x)}{d} \] Input:
Integrate[(a + a*Sin[c + d*x])*Tan[c + d*x]^2,x]
Output:
-((a*ArcTan[Tan[c + d*x]])/d) + (a*Cos[c + d*x])/d + (a*Sec[c + d*x])/d + (a*Tan[c + d*x])/d
Time = 0.44 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.23, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3042, 3187, 3042, 3225, 25, 3042, 3214, 3042, 3127}
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.
\(\displaystyle \int \tan ^2(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^2 (a \sin (c+d x)+a)dx\) |
\(\Big \downarrow \) 3187 |
\(\displaystyle a^2 \int \frac {\sin ^2(c+d x)}{a-a \sin (c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^2 \int \frac {\sin (c+d x)^2}{a-a \sin (c+d x)}dx\) |
\(\Big \downarrow \) 3225 |
\(\displaystyle a^2 \left (\frac {\cos (c+d x)}{a d}-\frac {\int -\frac {\sin (c+d x)}{1-\sin (c+d x)}dx}{a}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle a^2 \left (\frac {\int \frac {\sin (c+d x)}{1-\sin (c+d x)}dx}{a}+\frac {\cos (c+d x)}{a d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^2 \left (\frac {\int \frac {\sin (c+d x)}{1-\sin (c+d x)}dx}{a}+\frac {\cos (c+d x)}{a d}\right )\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle a^2 \left (\frac {\int \frac {1}{1-\sin (c+d x)}dx-x}{a}+\frac {\cos (c+d x)}{a d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^2 \left (\frac {\int \frac {1}{1-\sin (c+d x)}dx-x}{a}+\frac {\cos (c+d x)}{a d}\right )\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle a^2 \left (\frac {\cos (c+d x)}{a d}+\frac {\frac {\cos (c+d x)}{d (1-\sin (c+d x))}-x}{a}\right )\) |
Input:
Int[(a + a*Sin[c + d*x])*Tan[c + d*x]^2,x]
Output:
a^2*(Cos[c + d*x]/(a*d) + (-x + Cos[c + d*x]/(d*(1 - Sin[c + d*x])))/a)
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b ^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _), x_Symbol] :> Simp[a^p Int[Sin[e + f*x]^p/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ [p, 2*m]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2/((c_.) + (d_.)*sin[(e_.) + (f _.)*(x_)]), x_Symbol] :> Simp[(-b^2)*(Cos[e + f*x]/(d*f)), x] + Simp[1/d Int[Simp[a^2*d - b*(b*c - 2*a*d)*Sin[e + f*x], x]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Result contains complex when optimal does not.
Time = 0.57 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.44
method | result | size |
risch | \(-a x +\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 a}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\) | \(56\) |
derivativedivides | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )+a \left (\tan \left (d x +c \right )-d x -c \right )}{d}\) | \(59\) |
default | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )+a \left (\tan \left (d x +c \right )-d x -c \right )}{d}\) | \(59\) |
parts | \(\frac {a \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {a \left (\frac {\sin \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}+\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )\right )}{d}\) | \(63\) |
Input:
int((a+a*sin(d*x+c))*tan(d*x+c)^2,x,method=_RETURNVERBOSE)
Output:
-a*x+1/2*a/d*exp(I*(d*x+c))+1/2*a/d*exp(-I*(d*x+c))+2*a/d/(exp(I*(d*x+c))- I)
Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (38) = 76\).
Time = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.05 \[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {a d x - a \cos \left (d x + c\right )^{2} + {\left (a d x - 2 \, a\right )} \cos \left (d x + c\right ) - {\left (a d x - a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) - a}{d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d} \] Input:
integrate((a+a*sin(d*x+c))*tan(d*x+c)^2,x, algorithm="fricas")
Output:
-(a*d*x - a*cos(d*x + c)^2 + (a*d*x - 2*a)*cos(d*x + c) - (a*d*x - a*cos(d *x + c) + a)*sin(d*x + c) - a)/(d*cos(d*x + c) - d*sin(d*x + c) + d)
\[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=a \left (\int \sin {\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int \tan ^{2}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*sin(d*x+c))*tan(d*x+c)**2,x)
Output:
a*(Integral(sin(c + d*x)*tan(c + d*x)**2, x) + Integral(tan(c + d*x)**2, x ))
Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {{\left (d x + c - \tan \left (d x + c\right )\right )} a - a {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{d} \] Input:
integrate((a+a*sin(d*x+c))*tan(d*x+c)^2,x, algorithm="maxima")
Output:
-((d*x + c - tan(d*x + c))*a - a*(1/cos(d*x + c) + cos(d*x + c)))/d
Leaf count of result is larger than twice the leaf count of optimal. 1008 vs. \(2 (38) = 76\).
Time = 0.78 (sec) , antiderivative size = 1008, normalized size of antiderivative = 25.85 \[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(d*x+c))*tan(d*x+c)^2,x, algorithm="giac")
Output:
-(a*d*x*tan(d*x)*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c) - a*d*x*tan(1/2*d*x)^4 *tan(1/2*c)^4 - 4*a*d*x*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c) - 2*a* tan(d*x)*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c) + a*tan(d*x)*tan(1/2*d*x)^4*ta n(1/2*c)^4 + a*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c) + 4*a*d*x*tan(1/2*d*x)^3 *tan(1/2*c)^3 + 2*a*tan(1/2*d*x)^4*tan(1/2*c)^4 - a*d*x*tan(d*x)*tan(1/2*d *x)^4*tan(c) - 4*a*d*x*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)*tan(c) - 4*a*d*x *tan(d*x)*tan(1/2*d*x)*tan(1/2*c)^3*tan(c) + 8*a*tan(d*x)*tan(1/2*d*x)^3*t an(1/2*c)^3*tan(c) - a*d*x*tan(d*x)*tan(1/2*c)^4*tan(c) - 4*a*tan(d*x)*tan (1/2*d*x)^3*tan(1/2*c)^3 - 4*a*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c) + a*d*x* tan(1/2*d*x)^4 + 4*a*d*x*tan(1/2*d*x)^3*tan(1/2*c) + 4*a*d*x*tan(1/2*d*x)* tan(1/2*c)^3 - 8*a*tan(1/2*d*x)^3*tan(1/2*c)^3 + a*d*x*tan(1/2*c)^4 - 2*a* tan(d*x)*tan(1/2*d*x)^4*tan(c) - 4*a*d*x*tan(d*x)*tan(1/2*d*x)*tan(1/2*c)* tan(c) - 8*a*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)*tan(c) - 24*a*tan(d*x)*tan (1/2*d*x)^2*tan(1/2*c)^2*tan(c) - 8*a*tan(d*x)*tan(1/2*d*x)*tan(1/2*c)^3*t an(c) - 2*a*tan(d*x)*tan(1/2*c)^4*tan(c) - a*tan(d*x)*tan(1/2*d*x)^4 - 4*a *tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c) - 4*a*tan(d*x)*tan(1/2*d*x)*tan(1/2*c) ^3 - a*tan(d*x)*tan(1/2*c)^4 - a*tan(1/2*d*x)^4*tan(c) - 4*a*tan(1/2*d*x)^ 3*tan(1/2*c)*tan(c) - 4*a*tan(1/2*d*x)*tan(1/2*c)^3*tan(c) - a*tan(1/2*c)^ 4*tan(c) + 2*a*tan(1/2*d*x)^4 + 4*a*d*x*tan(1/2*d*x)*tan(1/2*c) + 8*a*tan( 1/2*d*x)^3*tan(1/2*c) + 24*a*tan(1/2*d*x)^2*tan(1/2*c)^2 + 8*a*tan(1/2*...
Time = 32.16 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.85 \[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=\frac {\left (a\,\left (c+d\,x-2\right )-a\,\left (c+d\,x\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (a\,\left (c+d\,x\right )-a\,\left (c+d\,x-2\right )\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,\left (c+d\,x\right )+a\,\left (c+d\,x-4\right )}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-a\,x \] Input:
int(tan(c + d*x)^2*(a + a*sin(c + d*x)),x)
Output:
(tan(c/2 + (d*x)/2)*(a*(c + d*x) - a*(c + d*x - 2)) - tan(c/2 + (d*x)/2)^2 *(a*(c + d*x) - a*(c + d*x - 2)) - a*(c + d*x) + a*(c + d*x - 4))/(d*(tan( c/2 + (d*x)/2) - 1)*(tan(c/2 + (d*x)/2)^2 + 1)) - a*x
Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.00 \[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=\frac {a \left (\tan \left (d x +c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\tan \left (d x +c \right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} d x -4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+d x \right )}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-1\right )} \] Input:
int((a+a*sin(d*x+c))*tan(d*x+c)^2,x)
Output:
(a*(tan(c + d*x)*tan((c + d*x)/2)**4 - tan(c + d*x) - tan((c + d*x)/2)**4* d*x - 4*tan((c + d*x)/2)**4 + d*x))/(d*(tan((c + d*x)/2)**4 - 1))