\(\int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx\) [753]

Optimal result

Integrand size = 19, antiderivative size = 39 \[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=-a x+\frac {a \cos (c+d x)}{d}+\frac {a \cos (c+d x)}{d (1-\sin (c+d x))} \]

[Out]

-a*x+a*cos(d*x+c)/d+a*cos(d*x+c)/d/(1-sin(d*x+c))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2787, 2825, 12, 2814, 2727} \[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=\frac {a \cos (c+d x)}{d}+\frac {a \cos (c+d x)}{d (1-\sin (c+d x))}-a x \]

[In]

Int[(a + a*Sin[c + d*x])*Tan[c + d*x]^2,x]

[Out]

-(a*x) + (a*Cos[c + d*x])/d + (a*Cos[c + d*x])/(d*(1 - Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2727

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]

Rule 2787

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[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]

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(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]

Rule 2825

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] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = a^2 \int \frac {\sin ^2(c+d x)}{a-a \sin (c+d x)} \, dx \\ & = \frac {a \cos (c+d x)}{d}+a \int \frac {a \sin (c+d x)}{a-a \sin (c+d x)} \, dx \\ & = \frac {a \cos (c+d x)}{d}+a^2 \int \frac {\sin (c+d x)}{a-a \sin (c+d x)} \, dx \\ & = -a x+\frac {a \cos (c+d x)}{d}+a^2 \int \frac {1}{a-a \sin (c+d x)} \, dx \\ & = -a x+\frac {a \cos (c+d x)}{d}+\frac {a^2 \cos (c+d x)}{d (a-a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.21 \[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {a \arctan (\tan (c+d x))}{d}+\frac {a \cos (c+d x)}{d}+\frac {a \sec (c+d x)}{d}+\frac {a \tan (c+d x)}{d} \]

[In]

Integrate[(a + a*Sin[c + d*x])*Tan[c + d*x]^2,x]

[Out]

-((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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.33

method	result	size
parallelrisch	\(\frac {a \left (-2 d x \cos \left (d x +c \right )+\cos \left (2 d x +2 c \right )+2 \sin \left (d x +c \right )+4 \cos \left (d x +c \right )+3\right )}{2 d \cos \left (d x +c \right )}\)	\(52\)
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	\(\frac {a \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+a \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(59\)
default	\(\frac {a \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+a \left (\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(59\)
norman	\(\frac {a x -\frac {4 a}{d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-a x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(89\)

[In]

int(sec(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/2/d*a*(-2*d*x*cos(d*x+c)+cos(2*d*x+2*c)+2*sin(d*x+c)+4*cos(d*x+c)+3)/cos(d*x+c)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (38) = 76\).

Time = 0.27 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.05 \[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {a d x - a \cos \left (d x + c\right )^{2} + {\left (a d x - 2 \, a\right )} \cos \left (d x + c\right ) - {\left (a d x - a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) - a}{d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-(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*co
s(d*x + c) - d*sin(d*x + c) + d)

Sympy [F]

\[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=a \left (\int \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)**2*(a+a*sin(d*x+c)),x)

[Out]

a*(Integral(sin(c + d*x)**2*sec(c + d*x)**2, x) + Integral(sin(c + d*x)**3*sec(c + d*x)**2, x))

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {{\left (d x + c - \tan \left (d x + c\right )\right )} a - a {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{d} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-((d*x + c - tan(d*x + c))*a - a*(1/cos(d*x + c) + cos(d*x + c)))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (38) = 76\).

Time = 0.28 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.08 \[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=-\frac {{\left (d x + c\right )} a + \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}}{d} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^2*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-((d*x + c)*a + 2*(a*tan(1/2*d*x + 1/2*c)^2 - a*tan(1/2*d*x + 1/2*c) + 2*a)/(tan(1/2*d*x + 1/2*c)^3 - tan(1/2*
d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) - 1))/d

Mupad [B] (verification not implemented)

Time = 9.74 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.54 \[ \int (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx=\frac {\left (a\,\left (d\,x-2\right )-a\,d\,x\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (a\,d\,x-a\,\left (d\,x-2\right )\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\,\left (d\,x-4\right )-a\,d\,x}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-a\,x \]

[In]

int((sin(c + d*x)^2*(a + a*sin(c + d*x)))/cos(c + d*x)^2,x)

[Out]

(a*(d*x - 4) - tan(c/2 + (d*x)/2)*(a*(d*x - 2) - a*d*x) + tan(c/2 + (d*x)/2)^2*(a*(d*x - 2) - a*d*x) - a*d*x)/
(d*(tan(c/2 + (d*x)/2) - 1)*(tan(c/2 + (d*x)/2)^2 + 1)) - a*x