\(\int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [789]

Optimal result

Integrand size = 29, antiderivative size = 102 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sec (c+d x)}{a^3 d}-\frac {2 \sec ^3(c+d x)}{a^3 d}+\frac {9 \sec ^5(c+d x)}{5 a^3 d}-\frac {4 \sec ^7(c+d x)}{7 a^3 d}+\frac {\tan ^5(c+d x)}{5 a^3 d}+\frac {4 \tan ^7(c+d x)}{7 a^3 d} \]

[Out]

sec(d*x+c)/a^3/d-2*sec(d*x+c)^3/a^3/d+9/5*sec(d*x+c)^5/a^3/d-4/7*sec(d*x+c)^7/a^3/d+1/5*tan(d*x+c)^5/a^3/d+4/7
*tan(d*x+c)^7/a^3/d

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2954, 2952, 2687, 14, 2686, 276, 30, 200} \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4 \tan ^7(c+d x)}{7 a^3 d}+\frac {\tan ^5(c+d x)}{5 a^3 d}-\frac {4 \sec ^7(c+d x)}{7 a^3 d}+\frac {9 \sec ^5(c+d x)}{5 a^3 d}-\frac {2 \sec ^3(c+d x)}{a^3 d}+\frac {\sec (c+d x)}{a^3 d} \]

[In]

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

[Out]

Sec[c + d*x]/(a^3*d) - (2*Sec[c + d*x]^3)/(a^3*d) + (9*Sec[c + d*x]^5)/(5*a^3*d) - (4*Sec[c + d*x]^7)/(7*a^3*d
) + Tan[c + d*x]^5/(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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 200

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

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 2952

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

Rule 2954

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sec ^4(c+d x) (a-a \sin (c+d x))^3 \tan ^4(c+d x) \, dx}{a^6} \\ & = \frac {\int \left (a^3 \sec ^4(c+d x) \tan ^4(c+d x)-3 a^3 \sec ^3(c+d x) \tan ^5(c+d x)+3 a^3 \sec ^2(c+d x) \tan ^6(c+d x)-a^3 \sec (c+d x) \tan ^7(c+d x)\right ) \, dx}{a^6} \\ & = \frac {\int \sec ^4(c+d x) \tan ^4(c+d x) \, dx}{a^3}-\frac {\int \sec (c+d x) \tan ^7(c+d x) \, dx}{a^3}-\frac {3 \int \sec ^3(c+d x) \tan ^5(c+d x) \, dx}{a^3}+\frac {3 \int \sec ^2(c+d x) \tan ^6(c+d x) \, dx}{a^3} \\ & = -\frac {\text {Subst}\left (\int \left (-1+x^2\right )^3 \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac {\text {Subst}\left (\int x^4 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac {3 \text {Subst}\left (\int x^6 \, dx,x,\tan (c+d x)\right )}{a^3 d}-\frac {3 \text {Subst}\left (\int x^2 \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{a^3 d} \\ & = \frac {3 \tan ^7(c+d x)}{7 a^3 d}-\frac {\text {Subst}\left (\int \left (-1+3 x^2-3 x^4+x^6\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac {\text {Subst}\left (\int \left (x^4+x^6\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d}-\frac {3 \text {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d} \\ & = \frac {\sec (c+d x)}{a^3 d}-\frac {2 \sec ^3(c+d x)}{a^3 d}+\frac {9 \sec ^5(c+d x)}{5 a^3 d}-\frac {4 \sec ^7(c+d x)}{7 a^3 d}+\frac {\tan ^5(c+d x)}{5 a^3 d}+\frac {4 \tan ^7(c+d x)}{7 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sec (c+d x) (840-1946 \cos (c+d x)-224 \cos (2 (c+d x))+834 \cos (3 (c+d x))-104 \cos (4 (c+d x))+1344 \sin (c+d x)-1946 \sin (2 (c+d x))+64 \sin (3 (c+d x))+139 \sin (4 (c+d x)))}{2240 a^3 d (1+\sin (c+d x))^3} \]

[In]

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

[Out]

(Sec[c + d*x]*(840 - 1946*Cos[c + d*x] - 224*Cos[2*(c + d*x)] + 834*Cos[3*(c + d*x)] - 104*Cos[4*(c + d*x)] +
1344*Sin[c + d*x] - 1946*Sin[2*(c + d*x)] + 64*Sin[3*(c + d*x)] + 139*Sin[4*(c + d*x)]))/(2240*a^3*d*(1 + Sin[
c + d*x])^3)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.73

method	result	size
parallelrisch	\(\frac {-\frac {16}{35}-\frac {32 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {32 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {96 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{35}}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}\)	\(74\)
risch	\(\frac {-\frac {2 \,{\mathrm e}^{3 i \left (d x +c \right )}}{5}-\frac {22 i {\mathrm e}^{2 i \left (d x +c \right )}}{5}-6 i {\mathrm e}^{4 i \left (d x +c \right )}-10 \,{\mathrm e}^{5 i \left (d x +c \right )}+\frac {86 \,{\mathrm e}^{i \left (d x +c \right )}}{35}+\frac {26 i}{35}+6 i {\mathrm e}^{6 i \left (d x +c \right )}+2 \,{\mathrm e}^{7 i \left (d x +c \right )}}{\left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{7} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) d \,a^{3}}\)	\(120\)
derivativedivides	\(\frac {-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {22}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {32}{256 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+256}}{d \,a^{3}}\)	\(130\)
default	\(\frac {-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {22}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {32}{256 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+256}}{d \,a^{3}}\)	\(130\)
norman	\(\frac {-\frac {32 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}-\frac {16}{35 a d}-\frac {32 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}-\frac {96 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{35 d a}-\frac {256 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d a}-\frac {416 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d a}-\frac {464 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d a}-\frac {544 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(186\)

[In]

int(sec(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

16/35*(-1-14*tan(1/2*d*x+1/2*c)^3-14*tan(1/2*d*x+1/2*c)^2-6*tan(1/2*d*x+1/2*c))/d/a^3/(tan(1/2*d*x+1/2*c)-1)/(
tan(1/2*d*x+1/2*c)+1)^7

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {13 \, \cos \left (d x + c\right )^{4} - 6 \, \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + 5\right )} \sin \left (d x + c\right ) - 15}{35 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) + {\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/35*(13*cos(d*x + c)^4 - 6*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 + 5)*sin(d*x + c) - 15)/(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]

\[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\sin ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)**4/(a+a*sin(d*x+c))**3,x)

[Out]

Integral(sin(c + d*x)**4*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. 230 vs. \(2 (94) = 188\).

Time = 0.21 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.25 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {16 \, {\left (\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {14 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {14 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 1\right )}}{35 \, {\left (a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

16/35*(6*sin(d*x + c)/(cos(d*x + c) + 1) + 14*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 14*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 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 = 0.35 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.18 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {35}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} - \frac {35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 280 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1015 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1673 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 616 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 93}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{280 \, d} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/280*(35/(a^3*(tan(1/2*d*x + 1/2*c) - 1)) - (35*tan(1/2*d*x + 1/2*c)^6 + 280*tan(1/2*d*x + 1/2*c)^5 + 1015*t
an(1/2*d*x + 1/2*c)^4 + 2240*tan(1/2*d*x + 1/2*c)^3 + 1673*tan(1/2*d*x + 1/2*c)^2 + 616*tan(1/2*d*x + 1/2*c) +
93)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^7))/d

Mupad [B] (verification not implemented)

Time = 11.82 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.32 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{35}+\frac {96\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{5}}{a^3\,d\,\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^7} \]

[In]

int(sin(c + d*x)^4/(cos(c + d*x)^2*(a + a*sin(c + d*x))^3),x)

[Out]

((16*cos(c/2 + (d*x)/2)^8)/35 + (96*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2))/35 + (32*cos(c/2 + (d*x)/2)^5*sin
(c/2 + (d*x)/2)^3)/5 + (32*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^2)/5)/(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)