\(\int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [788]

Optimal result

Integrand size = 29, antiderivative size = 142 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {x}{a^3}-\frac {3 \sec (c+d x)}{a^3 d}+\frac {10 \sec ^3(c+d x)}{3 a^3 d}-\frac {11 \sec ^5(c+d x)}{5 a^3 d}+\frac {4 \sec ^7(c+d x)}{7 a^3 d}+\frac {\tan (c+d x)}{a^3 d}-\frac {\tan ^3(c+d x)}{3 a^3 d}+\frac {\tan ^5(c+d x)}{5 a^3 d}-\frac {4 \tan ^7(c+d x)}{7 a^3 d} \]

[Out]

-x/a^3-3*sec(d*x+c)/a^3/d+10/3*sec(d*x+c)^3/a^3/d-11/5*sec(d*x+c)^5/a^3/d+4/7*sec(d*x+c)^7/a^3/d+tan(d*x+c)/a^
3/d-1/3*tan(d*x+c)^3/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 = 0.24 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2954, 2952, 2686, 276, 2687, 30, 200, 3554, 8} \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {4 \tan ^7(c+d x)}{7 a^3 d}+\frac {\tan ^5(c+d x)}{5 a^3 d}-\frac {\tan ^3(c+d x)}{3 a^3 d}+\frac {\tan (c+d x)}{a^3 d}+\frac {4 \sec ^7(c+d x)}{7 a^3 d}-\frac {11 \sec ^5(c+d x)}{5 a^3 d}+\frac {10 \sec ^3(c+d x)}{3 a^3 d}-\frac {3 \sec (c+d x)}{a^3 d}-\frac {x}{a^3} \]

[In]

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

[Out]

-(x/a^3) - (3*Sec[c + d*x])/(a^3*d) + (10*Sec[c + d*x]^3)/(3*a^3*d) - (11*Sec[c + d*x]^5)/(5*a^3*d) + (4*Sec[c
+ d*x]^7)/(7*a^3*d) + Tan[c + d*x]/(a^3*d) - Tan[c + d*x]^3/(3*a^3*d) + Tan[c + d*x]^5/(5*a^3*d) - (4*Tan[c +
d*x]^7)/(7*a^3*d)

Rule 8

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

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]

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 \sec ^3(c+d x) (a-a \sin (c+d x))^3 \tan ^5(c+d x) \, dx}{a^6} \\ & = \frac {\int \left (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)+3 a^3 \sec (c+d x) \tan ^7(c+d x)-a^3 \tan ^8(c+d x)\right ) \, dx}{a^6} \\ & = \frac {\int \sec ^3(c+d x) \tan ^5(c+d x) \, dx}{a^3}-\frac {\int \tan ^8(c+d x) \, dx}{a^3}-\frac {3 \int \sec ^2(c+d x) \tan ^6(c+d x) \, dx}{a^3}+\frac {3 \int \sec (c+d x) \tan ^7(c+d x) \, dx}{a^3} \\ & = -\frac {\tan ^7(c+d x)}{7 a^3 d}+\frac {\int \tan ^6(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 {3 \text {Subst}\left (\int x^6 \, dx,x,\tan (c+d x)\right )}{a^3 d}+\frac {3 \text {Subst}\left (\int \left (-1+x^2\right )^3 \, dx,x,\sec (c+d x)\right )}{a^3 d} \\ & = \frac {\tan ^5(c+d x)}{5 a^3 d}-\frac {4 \tan ^7(c+d x)}{7 a^3 d}-\frac {\int \tan ^4(c+d x) \, dx}{a^3}+\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 {3 \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 {3 \sec (c+d x)}{a^3 d}+\frac {10 \sec ^3(c+d x)}{3 a^3 d}-\frac {11 \sec ^5(c+d x)}{5 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)}{5 a^3 d}-\frac {4 \tan ^7(c+d x)}{7 a^3 d}+\frac {\int \tan ^2(c+d x) \, dx}{a^3} \\ & = -\frac {3 \sec (c+d x)}{a^3 d}+\frac {10 \sec ^3(c+d x)}{3 a^3 d}-\frac {11 \sec ^5(c+d x)}{5 a^3 d}+\frac {4 \sec ^7(c+d x)}{7 a^3 d}+\frac {\tan (c+d x)}{a^3 d}-\frac {\tan ^3(c+d x)}{3 a^3 d}+\frac {\tan ^5(c+d x)}{5 a^3 d}-\frac {4 \tan ^7(c+d x)}{7 a^3 d}-\frac {\int 1 \, dx}{a^3} \\ & = -\frac {x}{a^3}-\frac {3 \sec (c+d x)}{a^3 d}+\frac {10 \sec ^3(c+d x)}{3 a^3 d}-\frac {11 \sec ^5(c+d x)}{5 a^3 d}+\frac {4 \sec ^7(c+d x)}{7 a^3 d}+\frac {\tan (c+d x)}{a^3 d}-\frac {\tan ^3(c+d x)}{3 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 = 1.15 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.51 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {4200+14 (-1663+840 c+840 d x) \cos (c+d x)+6272 \cos (2 (c+d x))+9978 \cos (3 (c+d x))-5040 c \cos (3 (c+d x))-5040 d x \cos (3 (c+d x))-1768 \cos (4 (c+d x))+2688 \sin (c+d x)-23282 \sin (2 (c+d x))+11760 c \sin (2 (c+d x))+11760 d x \sin (2 (c+d x))+5568 \sin (3 (c+d x))+1663 \sin (4 (c+d x))-840 c \sin (4 (c+d x))-840 d x \sin (4 (c+d x))}{6720 a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^7} \]

[In]

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

[Out]

-1/6720*(4200 + 14*(-1663 + 840*c + 840*d*x)*Cos[c + d*x] + 6272*Cos[2*(c + d*x)] + 9978*Cos[3*(c + d*x)] - 50
40*c*Cos[3*(c + d*x)] - 5040*d*x*Cos[3*(c + d*x)] - 1768*Cos[4*(c + d*x)] + 2688*Sin[c + d*x] - 23282*Sin[2*(c
+ d*x)] + 11760*c*Sin[2*(c + d*x)] + 11760*d*x*Sin[2*(c + d*x)] + 5568*Sin[3*(c + d*x)] + 1663*Sin[4*(c + d*x
)] - 840*c*Sin[4*(c + d*x)] - 840*d*x*Sin[4*(c + d*x)])/(a^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c +
d*x)/2] + Sin[(c + d*x)/2])^7)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.67 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.89

method	result	size
risch	\(-\frac {x}{a^{3}}-\frac {2 \left (1155 i {\mathrm e}^{6 i \left (d x +c \right )}+315 \,{\mathrm e}^{7 i \left (d x +c \right )}-525 i {\mathrm e}^{4 i \left (d x +c \right )}-1715 \,{\mathrm e}^{5 i \left (d x +c \right )}-1939 i {\mathrm e}^{2 i \left (d x +c \right )}-1379 \,{\mathrm e}^{3 i \left (d x +c \right )}+221 i+1011 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{105 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{7} d \,a^{3}}\)	\(127\)
derivativedivides	\(\frac {-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {18}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {5}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {7}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {15}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d \,a^{3}}\)	\(142\)
default	\(\frac {-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {18}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {5}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {7}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {15}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d \,a^{3}}\)	\(142\)
parallelrisch	\(\frac {-105 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) x d +\left (-630 d x -210\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-1470 d x -1260\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-1470 d x -2870\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2520 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (1470 d x +938\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (1470 d x +2548\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (630 d x +1422\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+105 d x +272}{105 d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}\)	\(175\)
norman	\(\frac {\frac {46 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {32 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {17 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {29 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {45 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {29 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {46 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {45 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {32 x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {272}{105 a d}-\frac {17 x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {6 x \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {x}{a}-\frac {100 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {3364 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d a}-\frac {x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {474 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{35 d a}-\frac {1528 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d a}+\frac {6 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-\frac {1186 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}+\frac {5204 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d a}-\frac {1256 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}+\frac {842 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21 d a}-\frac {904 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d a}+\frac {396 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7 d a}-\frac {60 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {12 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{a^{2} \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 )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}\)	\(524\)

[In]

int(sec(d*x+c)^2*sin(d*x+c)^5/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-x/a^3-2/105*(1155*I*exp(6*I*(d*x+c))+315*exp(7*I*(d*x+c))-525*I*exp(4*I*(d*x+c))-1715*exp(5*I*(d*x+c))-1939*I
*exp(2*I*(d*x+c))-1379*exp(3*I*(d*x+c))+221*I+1011*exp(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 = 0.29 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.06 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {315 \, d x \cos \left (d x + c\right )^{3} + 221 \, \cos \left (d x + c\right )^{4} - 420 \, d x \cos \left (d x + c\right ) - 417 \, \cos \left (d x + c\right )^{2} + 3 \, {\left (35 \, d x \cos \left (d x + c\right )^{3} - 140 \, d x \cos \left (d x + c\right ) - 116 \, \cos \left (d x + c\right )^{2} + 15\right )} \sin \left (d x + c\right ) + 60}{105 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) + {\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/105*(315*d*x*cos(d*x + c)^3 + 221*cos(d*x + c)^4 - 420*d*x*cos(d*x + c) - 417*cos(d*x + c)^2 + 3*(35*d*x*co
s(d*x + c)^3 - 140*d*x*cos(d*x + c) - 116*cos(d*x + c)^2 + 15)*sin(d*x + c) + 60)/(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(-1)]

Timed out. \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)**5/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (130) = 260\).

Time = 0.30 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.36 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, {\left (\frac {\frac {711 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {1274 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {469 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {1260 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {1435 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {630 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {105 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 136}{a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac {105 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{105 \, d} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-2/105*((711*sin(d*x + c)/(cos(d*x + c) + 1) + 1274*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 469*sin(d*x + c)^3/(
cos(d*x + c) + 1)^3 - 1260*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 1435*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 63
0*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 105*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 136)/(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) + 105*arctan(sin(d*x + c)/(cos(d*x + c) +
1))/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.67 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.91 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {840 \, {\left (d x + c\right )}}{a^{3}} + \frac {105}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} + \frac {1575 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 10920 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 31675 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 48160 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36981 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 14392 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2281}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, d} \]

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^5/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/840*(840*(d*x + c)/a^3 + 105/(a^3*(tan(1/2*d*x + 1/2*c) - 1)) + (1575*tan(1/2*d*x + 1/2*c)^6 + 10920*tan(1/
2*d*x + 1/2*c)^5 + 31675*tan(1/2*d*x + 1/2*c)^4 + 48160*tan(1/2*d*x + 1/2*c)^3 + 36981*tan(1/2*d*x + 1/2*c)^2
+ 14392*tan(1/2*d*x + 1/2*c) + 2281)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^7))/d

Mupad [B] (verification not implemented)

Time = 20.17 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.92 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {82\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {134\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{15}+\frac {364\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+\frac {474\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}+\frac {272}{105}}{a^3\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^7}-\frac {x}{a^3} \]

[In]

int(sin(c + d*x)^5/(cos(c + d*x)^2*(a + a*sin(c + d*x))^3),x)

[Out]

((474*tan(c/2 + (d*x)/2))/35 + (364*tan(c/2 + (d*x)/2)^2)/15 + (134*tan(c/2 + (d*x)/2)^3)/15 - 24*tan(c/2 + (d
*x)/2)^4 - (82*tan(c/2 + (d*x)/2)^5)/3 - 12*tan(c/2 + (d*x)/2)^6 - 2*tan(c/2 + (d*x)/2)^7 + 272/105)/(a^3*d*(t
an(c/2 + (d*x)/2) - 1)*(tan(c/2 + (d*x)/2) + 1)^7) - x/a^3