\(\int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx\) [772]

Optimal result

Integrand size = 29, antiderivative size = 83 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {x}{a}+\frac {\cos (c+d x)}{a d}+\frac {2 \sec (c+d x)}{a d}-\frac {\sec ^3(c+d x)}{3 a d}-\frac {\tan (c+d x)}{a d}+\frac {\tan ^3(c+d x)}{3 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2918, 3554, 8, 2670, 276} \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\cos (c+d x)}{a d}+\frac {\tan ^3(c+d x)}{3 a d}-\frac {\tan (c+d x)}{a d}-\frac {\sec ^3(c+d x)}{3 a d}+\frac {2 \sec (c+d x)}{a d}+\frac {x}{a} \]

[In]

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

[Out]

x/a + Cos[c + d*x]/(a*d) + (2*Sec[c + d*x])/(a*d) - Sec[c + d*x]^3/(3*a*d) - Tan[c + d*x]/(a*d) + Tan[c + d*x]
^3/(3*a*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2670

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 2918

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \tan ^4(c+d x) \, dx}{a}-\frac {\int \sin (c+d x) \tan ^4(c+d x) \, dx}{a} \\ & = \frac {\tan ^3(c+d x)}{3 a d}-\frac {\int \tan ^2(c+d x) \, dx}{a}+\frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^2}{x^4} \, dx,x,\cos (c+d x)\right )}{a d} \\ & = -\frac {\tan (c+d x)}{a d}+\frac {\tan ^3(c+d x)}{3 a d}+\frac {\int 1 \, dx}{a}+\frac {\text {Subst}\left (\int \left (1+\frac {1}{x^4}-\frac {2}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{a d} \\ & = \frac {x}{a}+\frac {\cos (c+d x)}{a d}+\frac {2 \sec (c+d x)}{a d}-\frac {\sec ^3(c+d x)}{3 a d}-\frac {\tan (c+d x)}{a d}+\frac {\tan ^3(c+d x)}{3 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.78 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {18+2 (-11+6 c+6 d x) \cos (c+d x)+14 \cos (2 (c+d x))+11 \sin (c+d x)-11 \sin (2 (c+d x))+6 c \sin (2 (c+d x))+6 d x \sin (2 (c+d x))+3 \sin (3 (c+d x))}{12 a d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) (1+\sin (c+d x))} \]

[In]

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

[Out]

(18 + 2*(-11 + 6*c + 6*d*x)*Cos[c + d*x] + 14*Cos[2*(c + d*x)] + 11*Sin[c + d*x] - 11*Sin[2*(c + d*x)] + 6*c*S
in[2*(c + d*x)] + 6*d*x*Sin[2*(c + d*x)] + 3*Sin[3*(c + d*x)])/(12*a*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(
Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(1 + Sin[c + d*x]))

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.18

method	result	size
parallelrisch	\(\frac {\left (6 d x +16\right ) \cos \left (3 d x +3 c \right )+18 d x \cos \left (d x +c \right )+48 \cos \left (d x +c \right )+36 \cos \left (2 d x +2 c \right )+3 \cos \left (4 d x +4 c \right )-8 \sin \left (3 d x +3 c \right )+25}{6 a d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(98\)
derivativedivides	\(\frac {-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {32}{16+16 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {5}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d a}\)	\(99\)
default	\(\frac {-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {32}{16+16 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {5}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d a}\)	\(99\)
risch	\(\frac {x}{a}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 a d}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 a d}+\frac {\frac {4 \,{\mathrm e}^{i \left (d x +c \right )}}{3}+4 i {\mathrm e}^{2 i \left (d x +c \right )}+4 \,{\mathrm e}^{3 i \left (d x +c \right )}+\frac {8 i}{3}}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{3} d a}\)	\(112\)
norman	\(\frac {\frac {x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {x}{a}+\frac {8 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {2 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-\frac {2 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {2 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {2 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {2 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {16}{3 a d}+\frac {2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {4 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {10 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {20 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {22 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {26 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\)	\(305\)

[In]

int(sec(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/6*((6*d*x+16)*cos(3*d*x+3*c)+18*d*x*cos(d*x+c)+48*cos(d*x+c)+36*cos(2*d*x+2*c)+3*cos(4*d*x+4*c)-8*sin(3*d*x+
3*c)+25)/a/d/(cos(3*d*x+3*c)+3*cos(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 \, d x \cos \left (d x + c\right ) + 7 \, \cos \left (d x + c\right )^{2} + {\left (3 \, d x \cos \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) + 1}{3 \, {\left (a d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )\right )}} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/3*(3*d*x*cos(d*x + c) + 7*cos(d*x + c)^2 + (3*d*x*cos(d*x + c) + 3*cos(d*x + c)^2 + 2)*sin(d*x + c) + 1)/(a*
d*cos(d*x + c)*sin(d*x + c) + a*d*cos(d*x + c))

Sympy [F]

\[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\sin ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)**4/(a+a*sin(d*x+c)),x)

[Out]

Integral(sin(c + d*x)**4*sec(c + d*x)**2/(sin(c + d*x) + 1), x)/a

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (79) = 158\).

Time = 0.29 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.84 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {2 \, {\left (\frac {\frac {13 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {6 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 8}{a + \frac {2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {2 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}\right )}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

2/3*((13*sin(d*x + c)/(cos(d*x + c) + 1) + 2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 - 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 8)/(a + 2*a*sin(d
*x + c)/(cos(d*x + c) + 1) + a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2
*a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + 3*arctan(sin(d*x + c)/(cos(d
*x + c) + 1))/a)/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.51 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {6 \, {\left (d x + c\right )}}{a} - \frac {3 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )} a} + \frac {15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 17}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{3}}}{6 \, d} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)/a - 3*(tan(1/2*d*x + 1/2*c)^2 - 4*tan(1/2*d*x + 1/2*c) + 5)/((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)*a) + (15*tan(1/2*d*x + 1/2*c)^2 + 36*tan(1/2*d*x + 1/2*c) + 17)/(
a*(tan(1/2*d*x + 1/2*c) + 1)^3))/d

Mupad [B] (verification not implemented)

Time = 14.31 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.55 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {x}{a}-\frac {-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {26\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {16}{3}}{a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^3\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )} \]

[In]

int(sin(c + d*x)^4/(cos(c + d*x)^2*(a + a*sin(c + d*x))),x)

[Out]

x/a - ((26*tan(c/2 + (d*x)/2))/3 + (4*tan(c/2 + (d*x)/2)^2)/3 - (4*tan(c/2 + (d*x)/2)^3)/3 - 4*tan(c/2 + (d*x)
/2)^4 - 2*tan(c/2 + (d*x)/2)^5 + 16/3)/(a*d*(tan(c/2 + (d*x)/2) + 1)^3*(tan(c/2 + (d*x)/2) - tan(c/2 + (d*x)/2
)^2 + tan(c/2 + (d*x)/2)^3 - 1))