\(\int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx\) [825]

Optimal result

Integrand size = 29, antiderivative size = 73 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sec ^3(c+d x)}{3 a d}-\frac {\sec ^5(c+d x)}{5 a d}+\frac {\tan ^3(c+d x)}{3 a d}+\frac {\tan ^5(c+d x)}{5 a d} \]

[Out]

1/3*sec(d*x+c)^3/a/d-1/5*sec(d*x+c)^5/a/d+1/3*tan(d*x+c)^3/a/d+1/5*tan(d*x+c)^5/a/d

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2918, 2687, 14, 2686} \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\tan ^5(c+d x)}{5 a d}+\frac {\tan ^3(c+d x)}{3 a d}-\frac {\sec ^5(c+d x)}{5 a d}+\frac {\sec ^3(c+d x)}{3 a d} \]

[In]

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

[Out]

Sec[c + d*x]^3/(3*a*d) - Sec[c + d*x]^5/(5*a*d) + Tan[c + d*x]^3/(3*a*d) + Tan[c + d*x]^5/(5*a*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sec ^4(c+d x) \tan ^2(c+d x) \, dx}{a}-\frac {\int \sec ^3(c+d x) \tan ^3(c+d x) \, dx}{a} \\ & = -\frac {\text {Subst}\left (\int x^2 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{a d}+\frac {\text {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a d} \\ & = -\frac {\text {Subst}\left (\int \left (-x^2+x^4\right ) \, dx,x,\sec (c+d x)\right )}{a d}+\frac {\text {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,\tan (c+d x)\right )}{a d} \\ & = \frac {\sec ^3(c+d x)}{3 a d}-\frac {\sec ^5(c+d x)}{5 a d}+\frac {\tan ^3(c+d x)}{3 a d}+\frac {\tan ^5(c+d x)}{5 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.45 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sec ^3(c+d x) (-80+66 \cos (c+d x)-32 \cos (2 (c+d x))+22 \cos (3 (c+d x))-16 \cos (4 (c+d x))-224 \sin (c+d x)+22 \sin (2 (c+d x))+32 \sin (3 (c+d x))+11 \sin (4 (c+d x)))}{960 a d (1+\sin (c+d x))} \]

[In]

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

[Out]

-1/960*(Sec[c + d*x]^3*(-80 + 66*Cos[c + d*x] - 32*Cos[2*(c + d*x)] + 22*Cos[3*(c + d*x)] - 16*Cos[4*(c + d*x)
] - 224*Sin[c + d*x] + 22*Sin[2*(c + d*x)] + 32*Sin[3*(c + d*x)] + 11*Sin[4*(c + d*x)]))/(a*d*(1 + Sin[c + d*x
]))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.33

method	result	size
risch	\(\frac {-\frac {16 \,{\mathrm e}^{3 i \left (d x +c \right )}}{15}+\frac {4 i {\mathrm e}^{4 i \left (d x +c \right )}}{3}+\frac {8 i {\mathrm e}^{2 i \left (d x +c \right )}}{15}+\frac {8 \,{\mathrm e}^{5 i \left (d x +c \right )}}{3}+\frac {8 \,{\mathrm e}^{i \left (d x +c \right )}}{15}+\frac {4 i}{15}}{\left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d a}\)	\(97\)
parallelrisch	\(\frac {-\frac {4}{15}-\frac {8 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {16 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15}+\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{15}}{d a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(100\)
derivativedivides	\(\frac {-\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+64}}{d a}\)	\(130\)
default	\(\frac {-\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+64}}{d a}\)	\(130\)
norman	\(\frac {-\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {4}{15 a d}-\frac {8 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{15 d a}+\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}-\frac {16 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(130\)

[In]

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

[Out]

4/15*(-4*exp(3*I*(d*x+c))+5*I*exp(4*I*(d*x+c))+2*I*exp(2*I*(d*x+c))+10*exp(5*I*(d*x+c))+2*exp(I*(d*x+c))+I)/(e
xp(I*(d*x+c))+I)^5/(exp(I*(d*x+c))-I)^3/d/a

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {2 \, \cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 1}{15 \, {\left (a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{3}\right )}} \]

[In]

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

[Out]

1/15*(2*cos(d*x + c)^4 - cos(d*x + c)^2 - 2*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 1)/(a*d*cos(d*x + c)^3*sin(d*x
+ c) + a*d*cos(d*x + c)^3)

Sympy [F]

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

[In]

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

[Out]

Integral(sin(c + d*x)**2*sec(c + d*x)**4/(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. 254 vs. \(2 (65) = 130\).

Time = 0.24 (sec) , antiderivative size = 254, normalized size of antiderivative = 3.48 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {4 \, {\left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {10 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 1\right )}}{15 \, {\left (a + \frac {2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {6 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {2 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {2 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d} \]

[In]

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

[Out]

4/15*(2*sin(d*x + c)/(cos(d*x + c) + 1) - 2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 5*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 1)/((a + 2*a*sin(
d*x + c)/(cos(d*x + c) + 1) - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 6*a*sin(d*x + c)^3/(cos(d*x + c) + 1)^
3 + 6*a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 2*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 2*a*sin(d*x + c)^7/(co
s(d*x + c) + 1)^7 - a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)*d)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.49 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {5 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {3 \, {\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 50 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{120 \, d} \]

[In]

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

[Out]

-1/120*(5*(3*tan(1/2*d*x + 1/2*c)^2 + 1)/(a*(tan(1/2*d*x + 1/2*c) - 1)^3) - 3*(5*tan(1/2*d*x + 1/2*c)^4 + 40*t
an(1/2*d*x + 1/2*c)^3 + 50*tan(1/2*d*x + 1/2*c)^2 + 40*tan(1/2*d*x + 1/2*c) + 9)/(a*(tan(1/2*d*x + 1/2*c) + 1)
^5))/d

Mupad [B] (verification not implemented)

Time = 11.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {4\,\left (10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{15\,a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5} \]

[In]

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

[Out]

-(4*(2*tan(c/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^2 + 4*tan(c/2 + (d*x)/2)^3 + 5*tan(c/2 + (d*x)/2)^4 + 10*tan(
c/2 + (d*x)/2)^5 + 1))/(15*a*d*(tan(c/2 + (d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2) + 1)^5)