\(\int \frac {\tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [842]

Optimal result

Integrand size = 21, antiderivative size = 127 \[ \int \frac {\tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sec ^3(c+d x)}{3 a^3 d}-\frac {6 \sec ^5(c+d x)}{5 a^3 d}+\frac {9 \sec ^7(c+d x)}{7 a^3 d}-\frac {4 \sec ^9(c+d x)}{9 a^3 d}+\frac {\tan ^5(c+d x)}{5 a^3 d}+\frac {5 \tan ^7(c+d x)}{7 a^3 d}+\frac {4 \tan ^9(c+d x)}{9 a^3 d} \]

[Out]

1/3*sec(d*x+c)^3/a^3/d-6/5*sec(d*x+c)^5/a^3/d+9/7*sec(d*x+c)^7/a^3/d-4/9*sec(d*x+c)^9/a^3/d+1/5*tan(d*x+c)^5/a
^3/d+5/7*tan(d*x+c)^7/a^3/d+4/9*tan(d*x+c)^9/a^3/d

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 127, 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} \[ \int \frac {\tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4 \tan ^9(c+d x)}{9 a^3 d}+\frac {5 \tan ^7(c+d x)}{7 a^3 d}+\frac {\tan ^5(c+d x)}{5 a^3 d}-\frac {4 \sec ^9(c+d x)}{9 a^3 d}+\frac {9 \sec ^7(c+d x)}{7 a^3 d}-\frac {6 \sec ^5(c+d x)}{5 a^3 d}+\frac {\sec ^3(c+d x)}{3 a^3 d} \]

[In]

Int[Tan[c + d*x]^4/(a + a*Sin[c + d*x])^3,x]

[Out]

Sec[c + d*x]^3/(3*a^3*d) - (6*Sec[c + d*x]^5)/(5*a^3*d) + (9*Sec[c + d*x]^7)/(7*a^3*d) - (4*Sec[c + d*x]^9)/(9
*a^3*d) + Tan[c + d*x]^5/(5*a^3*d) + (5*Tan[c + d*x]^7)/(7*a^3*d) + (4*Tan[c + d*x]^9)/(9*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 ^4(c+d x)-3 a^3 \sec ^5(c+d x) \tan ^5(c+d x)+3 a^3 \sec ^4(c+d x) \tan ^6(c+d x)-a^3 \sec ^3(c+d x) \tan ^7(c+d x)\right ) \, dx}{a^6} \\ & = \frac {\int \sec ^6(c+d x) \tan ^4(c+d x) \, dx}{a^3}-\frac {\int \sec ^3(c+d x) \tan ^7(c+d x) \, dx}{a^3}-\frac {3 \int \sec ^5(c+d x) \tan ^5(c+d x) \, dx}{a^3}+\frac {3 \int \sec ^4(c+d x) \tan ^6(c+d x) \, dx}{a^3} \\ & = -\frac {\text {Subst}\left (\int x^2 \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 )^2 \, dx,x,\tan (c+d x)\right )}{a^3 d}-\frac {3 \text {Subst}\left (\int x^4 \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac {3 \text {Subst}\left (\int x^6 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d} \\ & = -\frac {\text {Subst}\left (\int \left (-x^2+3 x^4-3 x^6+x^8\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac {\text {Subst}\left (\int \left (x^4+2 x^6+x^8\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d}-\frac {3 \text {Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\sec (c+d x)\right )}{a^3 d}+\frac {3 \text {Subst}\left (\int \left (x^6+x^8\right ) \, dx,x,\tan (c+d x)\right )}{a^3 d} \\ & = \frac {\sec ^3(c+d x)}{3 a^3 d}-\frac {6 \sec ^5(c+d x)}{5 a^3 d}+\frac {9 \sec ^7(c+d x)}{7 a^3 d}-\frac {4 \sec ^9(c+d x)}{9 a^3 d}+\frac {\tan ^5(c+d x)}{5 a^3 d}+\frac {5 \tan ^7(c+d x)}{7 a^3 d}+\frac {4 \tan ^9(c+d x)}{9 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.46 \[ \int \frac {\tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {5376+1116 \cos (c+d x)-21312 \cos (2 (c+d x))+62 \cos (3 (c+d x))+8448 \cos (4 (c+d x))-186 \cos (5 (c+d x))-704 \cos (6 (c+d x))+39168 \sin (c+d x)+837 \sin (2 (c+d x))-28288 \sin (3 (c+d x))+372 \sin (4 (c+d x))+4224 \sin (5 (c+d x))-31 \sin (6 (c+d x))}{322560 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (a+a \sin (c+d x))^3} \]

[In]

Integrate[Tan[c + d*x]^4/(a + a*Sin[c + d*x])^3,x]

[Out]

(5376 + 1116*Cos[c + d*x] - 21312*Cos[2*(c + d*x)] + 62*Cos[3*(c + d*x)] + 8448*Cos[4*(c + d*x)] - 186*Cos[5*(
c + d*x)] - 704*Cos[6*(c + d*x)] + 39168*Sin[c + d*x] + 837*Sin[2*(c + d*x)] - 28288*Sin[3*(c + d*x)] + 372*Si
n[4*(c + d*x)] + 4224*Sin[5*(c + d*x)] - 31*Sin[6*(c + d*x)])/(322560*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^
3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(a + a*Sin[c + d*x])^3)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.99

method	result	size
parallelrisch	\(\frac {\frac {16}{315}-\frac {32 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {32 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{315}-\frac {48 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35}-\frac {288 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35}+\frac {64 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105}+\frac {32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{105}-\frac {32 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}\)	\(126\)
risch	\(\frac {\frac {48 \,{\mathrm e}^{5 i \left (d x +c \right )}}{35}+\frac {44 i}{315}+\frac {8 \,{\mathrm e}^{9 i \left (d x +c \right )}}{3}+\frac {88 \,{\mathrm e}^{i \left (d x +c \right )}}{105}-\frac {928 \,{\mathrm e}^{3 i \left (d x +c \right )}}{315}-\frac {32 \,{\mathrm e}^{7 i \left (d x +c \right )}}{5}-\frac {176 i {\mathrm e}^{2 i \left (d x +c \right )}}{105}+4 i {\mathrm e}^{8 i \left (d x +c \right )}-\frac {16 i {\mathrm e}^{6 i \left (d x +c \right )}}{15}+\frac {8 i {\mathrm e}^{4 i \left (d x +c \right )}}{35}}{\left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{9} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d \,a^{3}}\)	\(143\)
norman	\(\frac {-\frac {32 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {16}{315 a d}-\frac {288 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d a}-\frac {32 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{105 d a}+\frac {64 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d a}+\frac {32 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{315 d a}-\frac {48 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d a}}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}\)	\(171\)
derivativedivides	\(\frac {-\frac {1}{24 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {32}{1024 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1024}-\frac {8}{9 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {52}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {22}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {39}{10 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {3}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d \,a^{3}}\)	\(175\)
default	\(\frac {-\frac {1}{24 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {32}{1024 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1024}-\frac {8}{9 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {52}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {22}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {39}{10 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {3}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d \,a^{3}}\)	\(175\)

[In]

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

[Out]

16/315*(1-126*tan(1/2*d*x+1/2*c)^6+2*tan(1/2*d*x+1/2*c)^3-27*tan(1/2*d*x+1/2*c)^4-162*tan(1/2*d*x+1/2*c)^5+12*
tan(1/2*d*x+1/2*c)^2+6*tan(1/2*d*x+1/2*c)-126*tan(1/2*d*x+1/2*c)^7)/d/a^3/(tan(1/2*d*x+1/2*c)-1)^3/(tan(1/2*d*
x+1/2*c)+1)^9

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.02 \[ \int \frac {\tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {22 \, \cos \left (d x + c\right )^{6} - 99 \, \cos \left (d x + c\right )^{4} + 120 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (33 \, \cos \left (d x + c\right )^{4} - 80 \, \cos \left (d x + c\right )^{2} + 35\right )} \sin \left (d x + c\right ) - 35}{315 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3} + {\left (a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \]

[In]

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

[Out]

1/315*(22*cos(d*x + c)^6 - 99*cos(d*x + c)^4 + 120*cos(d*x + c)^2 - 2*(33*cos(d*x + c)^4 - 80*cos(d*x + c)^2 +
35)*sin(d*x + c) - 35)/(3*a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)^3 + (a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos
(d*x + c)^3)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**4*sin(d*x+c)**4/(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. 402 vs. \(2 (113) = 226\).

Time = 0.22 (sec) , antiderivative size = 402, normalized size of antiderivative = 3.17 \[ \int \frac {\tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {16 \, {\left (\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {12 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {27 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {162 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {126 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {126 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 1\right )}}{315 \, {\left (a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {12 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {27 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {36 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {36 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {27 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {12 \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {a^{3} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}\right )} d} \]

[In]

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

[Out]

-16/315*(6*sin(d*x + c)/(cos(d*x + c) + 1) + 12*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 2*sin(d*x + c)^3/(cos(d*
x + c) + 1)^3 - 27*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 162*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 126*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6 - 126*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 1)/((a^3 + 6*a^3*sin(d*x + c)/(cos(d*
x + c) + 1) + 12*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 2*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 27*a^3*
sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 36*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 36*a^3*sin(d*x + c)^7/(cos(
d*x + c) + 1)^7 + 27*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 2*a^3*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 12*
a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 6*a^3*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - a^3*sin(d*x + c)^12/
(cos(d*x + c) + 1)^12)*d)

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.25 \[ \int \frac {\tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {105 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {315 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 2520 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 7140 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 1638 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8232 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2988 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 432 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 13}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{9}}}{10080 \, d} \]

[In]

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

[Out]

1/10080*(105*(3*tan(1/2*d*x + 1/2*c)^2 - 12*tan(1/2*d*x + 1/2*c) + 5)/(a^3*(tan(1/2*d*x + 1/2*c) - 1)^3) - (31
5*tan(1/2*d*x + 1/2*c)^8 + 2520*tan(1/2*d*x + 1/2*c)^7 + 7140*tan(1/2*d*x + 1/2*c)^6 - 1638*tan(1/2*d*x + 1/2*
c)^4 - 8232*tan(1/2*d*x + 1/2*c)^3 - 2988*tan(1/2*d*x + 1/2*c)^2 - 432*tan(1/2*d*x + 1/2*c) - 13)/(a^3*(tan(1/
2*d*x + 1/2*c) + 1)^9))/d

Mupad [B] (verification not implemented)

Time = 16.41 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.83 \[ \int \frac {\tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{315}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}+\frac {64\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{105}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{315}-\frac {48\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{35}-\frac {288\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{35}-\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{5}-\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{5}}{a^3\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^9} \]

[In]

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

[Out]

-((16*cos(c/2 + (d*x)/2)^12)/315 + (32*cos(c/2 + (d*x)/2)^11*sin(c/2 + (d*x)/2))/105 - (32*cos(c/2 + (d*x)/2)^
5*sin(c/2 + (d*x)/2)^7)/5 - (32*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^6)/5 - (288*cos(c/2 + (d*x)/2)^7*sin(c
/2 + (d*x)/2)^5)/35 - (48*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^4)/35 + (32*cos(c/2 + (d*x)/2)^9*sin(c/2 + (
d*x)/2)^3)/315 + (64*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d*x)/2)^2)/105)/(a^3*d*(cos(c/2 + (d*x)/2) - sin(c/2 + (
d*x)/2))^3*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^9)