\(\int \frac {\tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [791]

Optimal result

Integrand size = 21, antiderivative size = 103 \[ \int \frac {\tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\sec ^3(c+d x)}{3 a^3 d}+\frac {\sec ^5(c+d x)}{a^3 d}-\frac {4 \sec ^7(c+d x)}{7 a^3 d}+\frac {\tan ^3(c+d x)}{3 a^3 d}+\frac {\tan ^5(c+d x)}{a^3 d}+\frac {4 \tan ^7(c+d x)}{7 a^3 d} \]

[Out]

-1/3*sec(d*x+c)^3/a^3/d+sec(d*x+c)^5/a^3/d-4/7*sec(d*x+c)^7/a^3/d+1/3*tan(d*x+c)^3/a^3/d+tan(d*x+c)^5/a^3/d+4/
7*tan(d*x+c)^7/a^3/d

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2790, 2687, 276, 2686, 14} \[ \int \frac {\tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4 \tan ^7(c+d x)}{7 a^3 d}+\frac {\tan ^5(c+d x)}{a^3 d}+\frac {\tan ^3(c+d x)}{3 a^3 d}-\frac {4 \sec ^7(c+d x)}{7 a^3 d}+\frac {\sec ^5(c+d x)}{a^3 d}-\frac {\sec ^3(c+d x)}{3 a^3 d} \]

[In]

Int[Tan[c + d*x]^2/(a + a*Sin[c + d*x])^3,x]

[Out]

-1/3*Sec[c + d*x]^3/(a^3*d) + Sec[c + d*x]^5/(a^3*d) - (4*Sec[c + d*x]^7)/(7*a^3*d) + Tan[c + d*x]^3/(3*a^3*d)
+ Tan[c + d*x]^5/(a^3*d) + (4*Tan[c + d*x]^7)/(7*a^3*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 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 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 2790

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[a^(2*
m), Int[ExpandIntegrand[(g*Tan[e + f*x])^p/Sec[e + f*x]^m, (a*Sec[e + f*x] - b*Tan[e + f*x])^(-m), x], x], x]
/; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (a^3 \sec ^6(c+d x) \tan ^2(c+d x)-3 a^3 \sec ^5(c+d x) \tan ^3(c+d x)+3 a^3 \sec ^4(c+d x) \tan ^4(c+d x)-a^3 \sec ^3(c+d x) \tan ^5(c+d x)\right ) \, dx}{a^6} \\ & = \frac {\int \sec ^6(c+d x) \tan ^2(c+d x) \, dx}{a^3}-\frac {\int \sec ^3(c+d x) \tan ^5(c+d x) \, dx}{a^3}-\frac {3 \int \sec ^5(c+d x) \tan ^3(c+d x) \, dx}{a^3}+\frac {3 \int \sec ^4(c+d x) \tan ^4(c+d x) \, dx}{a^3} \\ & = -\frac {\text {Subst}\left (\int x^2 \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac {\text {Subst}\left (\int x^2 \left (1+x^2\right )^2 \, dx,x,\tan (c+d x)\right )}{a^3 d}-\frac {3 \text {Subst}\left (\int x^4 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac {3 \text {Subst}\left (\int x^4 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d} \\ & = -\frac {\text {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac {\text {Subst}\left (\int \left (x^2+2 x^4+x^6\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d}-\frac {3 \text {Subst}\left (\int \left (-x^4+x^6\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac {3 \text {Subst}\left (\int \left (x^4+x^6\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d} \\ & = -\frac {\sec ^3(c+d x)}{3 a^3 d}+\frac {\sec ^5(c+d x)}{a^3 d}-\frac {4 \sec ^7(c+d x)}{7 a^3 d}+\frac {\tan ^3(c+d x)}{3 a^3 d}+\frac {\tan ^5(c+d x)}{a^3 d}+\frac {4 \tan ^7(c+d x)}{7 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int \frac {\tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sec (c+d x) (336-70 \cos (c+d x)-224 \cos (2 (c+d x))+30 \cos (3 (c+d x))+16 \cos (4 (c+d x))+672 \sin (c+d x)-70 \sin (2 (c+d x))-96 \sin (3 (c+d x))+5 \sin (4 (c+d x)))}{1344 a^3 d (1+\sin (c+d x))^3} \]

[In]

Integrate[Tan[c + d*x]^2/(a + a*Sin[c + d*x])^3,x]

[Out]

(Sec[c + d*x]*(336 - 70*Cos[c + d*x] - 224*Cos[2*(c + d*x)] + 30*Cos[3*(c + d*x)] + 16*Cos[4*(c + d*x)] + 672*
Sin[c + d*x] - 70*Sin[2*(c + d*x)] - 96*Sin[3*(c + d*x)] + 5*Sin[4*(c + d*x)]))/(1344*a^3*d*(1 + Sin[c + d*x])
^3)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.43 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.94

method	result	size
risch	\(-\frac {4 \left (-14 i {\mathrm e}^{2 i \left (d x +c \right )}-28 \,{\mathrm e}^{3 i \left (d x +c \right )}+6 \,{\mathrm e}^{i \left (d x +c \right )}+i+21 i {\mathrm e}^{4 i \left (d x +c \right )}+14 \,{\mathrm e}^{5 i \left (d x +c \right )}\right )}{21 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{7} d \,a^{3}}\)	\(97\)
parallelrisch	\(\frac {-\frac {4}{21}-\frac {8 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {16 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{7}}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}\)	\(100\)
derivativedivides	\(\frac {-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {6}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {5}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {13}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+64}}{d \,a^{3}}\)	\(130\)
default	\(\frac {-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {6}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {5}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {13}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+64}}{d \,a^{3}}\)	\(130\)
norman	\(\frac {-\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {4}{21 a d}-\frac {8 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{7 d a}-\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {16 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}\)	\(133\)

[In]

int(sec(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-4/21*(-14*I*exp(2*I*(d*x+c))-28*exp(3*I*(d*x+c))+6*exp(I*(d*x+c))+I+21*I*exp(4*I*(d*x+c))+14*exp(5*I*(d*x+c))
)/(exp(I*(d*x+c))-I)/(exp(I*(d*x+c))+I)^7/d/a^3

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int \frac {\tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, \cos \left (d x + c\right )^{4} - 9 \, \cos \left (d x + c\right )^{2} - 6 \, {\left (\cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 9}{21 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) + {\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/21*(2*cos(d*x + c)^4 - 9*cos(d*x + c)^2 - 6*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 9)/(3*a^3*d*cos(d*x + c)^3
- 4*a^3*d*cos(d*x + c) + (a^3*d*cos(d*x + c)^3 - 4*a^3*d*cos(d*x + c))*sin(d*x + c))

Sympy [F]

\[ \int \frac {\tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

[In]

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

[Out]

Integral(sin(c + d*x)**2*sec(c + d*x)**2/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x)/a**3

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (95) = 190\).

Time = 0.22 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.62 \[ \int \frac {\tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4 \, {\left (\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {14 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {28 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {14 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 1\right )}}{21 \, {\left (a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

4/21*(6*sin(d*x + c)/(cos(d*x + c) + 1) + 14*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 28*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 21*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 14*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 1)/((a^3 + 6*a
^3*sin(d*x + c)/(cos(d*x + c) + 1) + 14*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 14*a^3*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 - 14*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 14*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 6*a
^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)*d)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.17 \[ \int \frac {\tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {21}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} - \frac {21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 168 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 161 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 224 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 63 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 56 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{168 \, d} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/168*(21/(a^3*(tan(1/2*d*x + 1/2*c) - 1)) - (21*tan(1/2*d*x + 1/2*c)^6 + 168*tan(1/2*d*x + 1/2*c)^5 + 161*ta
n(1/2*d*x + 1/2*c)^4 + 224*tan(1/2*d*x + 1/2*c)^3 + 63*tan(1/2*d*x + 1/2*c)^2 + 56*tan(1/2*d*x + 1/2*c) + 11)/
(a^3*(tan(1/2*d*x + 1/2*c) + 1)^7))/d

Mupad [B] (verification not implemented)

Time = 12.50 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.78 \[ \int \frac {\tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{21}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{7}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}}{a^3\,d\,\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^7} \]

[In]

int(sin(c + d*x)^2/(cos(c + d*x)^2*(a + a*sin(c + d*x))^3),x)

[Out]

((4*cos(c/2 + (d*x)/2)^8)/21 + (8*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2))/7 + (8*cos(c/2 + (d*x)/2)^3*sin(c/2
+ (d*x)/2)^5)/3 + 4*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^4 + (16*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^3
)/3 + (8*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^2)/3)/(a^3*d*(cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2))*(cos(c
/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^7)