3.848 \(\int \frac {\tan ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=145 \[ \frac {8 \tan ^{11}(c+d x)}{11 a^4 d}+\frac {16 \tan ^9(c+d x)}{9 a^4 d}+\frac {9 \tan ^7(c+d x)}{7 a^4 d}+\frac {\tan ^5(c+d x)}{5 a^4 d}-\frac {8 \sec ^{11}(c+d x)}{11 a^4 d}+\frac {20 \sec ^9(c+d x)}{9 a^4 d}-\frac {16 \sec ^7(c+d x)}{7 a^4 d}+\frac {4 \sec ^5(c+d x)}{5 a^4 d} \]

[Out]

4/5*sec(d*x+c)^5/a^4/d-16/7*sec(d*x+c)^7/a^4/d+20/9*sec(d*x+c)^9/a^4/d-8/11*sec(d*x+c)^11/a^4/d+1/5*tan(d*x+c)
^5/a^4/d+9/7*tan(d*x+c)^7/a^4/d+16/9*tan(d*x+c)^9/a^4/d+8/11*tan(d*x+c)^11/a^4/d

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Rubi [A]  time = 0.31, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2711, 2607, 270, 2606, 14} \[ \frac {8 \tan ^{11}(c+d x)}{11 a^4 d}+\frac {16 \tan ^9(c+d x)}{9 a^4 d}+\frac {9 \tan ^7(c+d x)}{7 a^4 d}+\frac {\tan ^5(c+d x)}{5 a^4 d}-\frac {8 \sec ^{11}(c+d x)}{11 a^4 d}+\frac {20 \sec ^9(c+d x)}{9 a^4 d}-\frac {16 \sec ^7(c+d x)}{7 a^4 d}+\frac {4 \sec ^5(c+d x)}{5 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sec[c + d*x]^5)/(5*a^4*d) - (16*Sec[c + d*x]^7)/(7*a^4*d) + (20*Sec[c + d*x]^9)/(9*a^4*d) - (8*Sec[c + d*x]
^11)/(11*a^4*d) + Tan[c + d*x]^5/(5*a^4*d) + (9*Tan[c + d*x]^7)/(7*a^4*d) + (16*Tan[c + d*x]^9)/(9*a^4*d) + (8
*Tan[c + d*x]^11)/(11*a^4*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 270

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 2606

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 2607

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 2711

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*} \int \frac {\tan ^4(c+d x)}{(a+a \sin (c+d x))^4} \, dx &=\frac {\int \left (a^4 \sec ^8(c+d x) \tan ^4(c+d x)-4 a^4 \sec ^7(c+d x) \tan ^5(c+d x)+6 a^4 \sec ^6(c+d x) \tan ^6(c+d x)-4 a^4 \sec ^5(c+d x) \tan ^7(c+d x)+a^4 \sec ^4(c+d x) \tan ^8(c+d x)\right ) \, dx}{a^8}\\ &=\frac {\int \sec ^8(c+d x) \tan ^4(c+d x) \, dx}{a^4}+\frac {\int \sec ^4(c+d x) \tan ^8(c+d x) \, dx}{a^4}-\frac {4 \int \sec ^7(c+d x) \tan ^5(c+d x) \, dx}{a^4}-\frac {4 \int \sec ^5(c+d x) \tan ^7(c+d x) \, dx}{a^4}+\frac {6 \int \sec ^6(c+d x) \tan ^6(c+d x) \, dx}{a^4}\\ &=\frac {\operatorname {Subst}\left (\int x^8 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a^4 d}+\frac {\operatorname {Subst}\left (\int x^4 \left (1+x^2\right )^3 \, dx,x,\tan (c+d x)\right )}{a^4 d}-\frac {4 \operatorname {Subst}\left (\int x^6 \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{a^4 d}-\frac {4 \operatorname {Subst}\left (\int x^4 \left (-1+x^2\right )^3 \, dx,x,\sec (c+d x)\right )}{a^4 d}+\frac {6 \operatorname {Subst}\left (\int x^6 \left (1+x^2\right )^2 \, dx,x,\tan (c+d x)\right )}{a^4 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (x^8+x^{10}\right ) \, dx,x,\tan (c+d x)\right )}{a^4 d}+\frac {\operatorname {Subst}\left (\int \left (x^4+3 x^6+3 x^8+x^{10}\right ) \, dx,x,\tan (c+d x)\right )}{a^4 d}-\frac {4 \operatorname {Subst}\left (\int \left (-x^4+3 x^6-3 x^8+x^{10}\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d}-\frac {4 \operatorname {Subst}\left (\int \left (x^6-2 x^8+x^{10}\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d}+\frac {6 \operatorname {Subst}\left (\int \left (x^6+2 x^8+x^{10}\right ) \, dx,x,\tan (c+d x)\right )}{a^4 d}\\ &=\frac {4 \sec ^5(c+d x)}{5 a^4 d}-\frac {16 \sec ^7(c+d x)}{7 a^4 d}+\frac {20 \sec ^9(c+d x)}{9 a^4 d}-\frac {8 \sec ^{11}(c+d x)}{11 a^4 d}+\frac {\tan ^5(c+d x)}{5 a^4 d}+\frac {9 \tan ^7(c+d x)}{7 a^4 d}+\frac {16 \tan ^9(c+d x)}{9 a^4 d}+\frac {8 \tan ^{11}(c+d x)}{11 a^4 d}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 166, normalized size = 1.14 \[ \frac {\sec ^3(c+d x) (501600 \sin (c+d x)-70136 \sin (2 (c+d x))-200288 \sin (3 (c+d x))-25504 \sin (4 (c+d x))+48800 \sin (5 (c+d x))+6376 \sin (6 (c+d x))-1952 \sin (7 (c+d x))-78903 \cos (c+d x)-183040 \cos (2 (c+d x))+8767 \cos (3 (c+d x))+62464 \cos (4 (c+d x))+19925 \cos (5 (c+d x))-15616 \cos (6 (c+d x))-797 \cos (7 (c+d x))+168960)}{3548160 a^4 d (\sin (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*(168960 - 78903*Cos[c + d*x] - 183040*Cos[2*(c + d*x)] + 8767*Cos[3*(c + d*x)] + 62464*Cos[4*(
c + d*x)] + 19925*Cos[5*(c + d*x)] - 15616*Cos[6*(c + d*x)] - 797*Cos[7*(c + d*x)] + 501600*Sin[c + d*x] - 701
36*Sin[2*(c + d*x)] - 200288*Sin[3*(c + d*x)] - 25504*Sin[4*(c + d*x)] + 48800*Sin[5*(c + d*x)] + 6376*Sin[6*(
c + d*x)] - 1952*Sin[7*(c + d*x)]))/(3548160*a^4*d*(1 + Sin[c + d*x])^4)

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fricas [A]  time = 0.44, size = 153, normalized size = 1.06 \[ -\frac {488 \, \cos \left (d x + c\right )^{6} - 1220 \, \cos \left (d x + c\right )^{4} + 1120 \, \cos \left (d x + c\right )^{2} + {\left (122 \, \cos \left (d x + c\right )^{6} - 915 \, \cos \left (d x + c\right )^{4} + 1400 \, \cos \left (d x + c\right )^{2} - 735\right )} \sin \left (d x + c\right ) - 420}{3465 \, {\left (a^{4} d \cos \left (d x + c\right )^{7} - 8 \, a^{4} d \cos \left (d x + c\right )^{5} + 8 \, a^{4} d \cos \left (d x + c\right )^{3} - 4 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} - 2 \, a^{4} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3465*(488*cos(d*x + c)^6 - 1220*cos(d*x + c)^4 + 1120*cos(d*x + c)^2 + (122*cos(d*x + c)^6 - 915*cos(d*x +
c)^4 + 1400*cos(d*x + c)^2 - 735)*sin(d*x + c) - 420)/(a^4*d*cos(d*x + c)^7 - 8*a^4*d*cos(d*x + c)^5 + 8*a^4*d
*cos(d*x + c)^3 - 4*(a^4*d*cos(d*x + c)^5 - 2*a^4*d*cos(d*x + c)^3)*sin(d*x + c))

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giac [A]  time = 0.46, size = 172, normalized size = 1.19 \[ -\frac {\frac {1155 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {3465 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 47355 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 309540 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 588588 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 891198 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 747450 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 481140 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 172700 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35233 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3203}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{11}}}{110880 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/110880*(1155*(3*tan(1/2*d*x + 1/2*c) - 1)/(a^4*(tan(1/2*d*x + 1/2*c) - 1)^3) - (3465*tan(1/2*d*x + 1/2*c)^9
+ 47355*tan(1/2*d*x + 1/2*c)^8 + 309540*tan(1/2*d*x + 1/2*c)^7 + 588588*tan(1/2*d*x + 1/2*c)^6 + 891198*tan(1
/2*d*x + 1/2*c)^5 + 747450*tan(1/2*d*x + 1/2*c)^4 + 481140*tan(1/2*d*x + 1/2*c)^3 + 172700*tan(1/2*d*x + 1/2*c
)^2 + 35233*tan(1/2*d*x + 1/2*c) + 3203)/(a^4*(tan(1/2*d*x + 1/2*c) + 1)^11))/d

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maple [A]  time = 0.72, size = 190, normalized size = 1.31 \[ \frac {-\frac {1}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {16}{11 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{11}}+\frac {8}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{10}}-\frac {176}{9 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}+\frac {28}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {179}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {89}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {49}{10 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {7}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/d/a^4*(-1/1536/(tan(1/2*d*x+1/2*c)-1)^3-1/1024/(tan(1/2*d*x+1/2*c)-1)^2-1/22/(tan(1/2*d*x+1/2*c)+1)^11+1/4/
(tan(1/2*d*x+1/2*c)+1)^10-11/18/(tan(1/2*d*x+1/2*c)+1)^9+7/8/(tan(1/2*d*x+1/2*c)+1)^8-179/224/(tan(1/2*d*x+1/2
*c)+1)^7+89/192/(tan(1/2*d*x+1/2*c)+1)^6-49/320/(tan(1/2*d*x+1/2*c)+1)^5+1/64/(tan(1/2*d*x+1/2*c)+1)^4+7/1536/
(tan(1/2*d*x+1/2*c)+1)^3+1/1024/(tan(1/2*d*x+1/2*c)+1)^2)

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maxima [B]  time = 0.35, size = 488, normalized size = 3.37 \[ \frac {32 \, {\left (\frac {16 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {50 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {64 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {22 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {517 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {726 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {1650 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {924 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {693 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 2\right )}}{3465 \, {\left (a^{4} + \frac {8 \, a^{4} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {25 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {32 \, a^{4} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {11 \, a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {88 \, a^{4} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {99 \, a^{4} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {99 \, a^{4} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {88 \, a^{4} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {11 \, a^{4} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {32 \, a^{4} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {25 \, a^{4} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}} - \frac {8 \, a^{4} \sin \left (d x + c\right )^{13}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{13}} - \frac {a^{4} \sin \left (d x + c\right )^{14}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{14}}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/3465*(16*sin(d*x + c)/(cos(d*x + c) + 1) + 50*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 64*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 - 22*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 517*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 726*sin(d
*x + c)^6/(cos(d*x + c) + 1)^6 + 1650*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 924*sin(d*x + c)^8/(cos(d*x + c) +
1)^8 + 693*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 2)/((a^4 + 8*a^4*sin(d*x + c)/(cos(d*x + c) + 1) + 25*a^4*si
n(d*x + c)^2/(cos(d*x + c) + 1)^2 + 32*a^4*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 11*a^4*sin(d*x + c)^4/(cos(d*
x + c) + 1)^4 - 88*a^4*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 99*a^4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 99*a
^4*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 88*a^4*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 11*a^4*sin(d*x + c)^10/(
cos(d*x + c) + 1)^10 - 32*a^4*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 25*a^4*sin(d*x + c)^12/(cos(d*x + c) + 1
)^12 - 8*a^4*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - a^4*sin(d*x + c)^14/(cos(d*x + c) + 1)^14)*d)

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mupad [B]  time = 16.85, size = 279, normalized size = 1.92 \[ \frac {\frac {64\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{3465}+\frac {512\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3465}+\frac {320\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{693}+\frac {2048\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3465}-\frac {64\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{315}+\frac {1504\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{315}+\frac {704\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{105}+\frac {320\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{21}+\frac {128\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{15}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{5}}{a^4\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((64*cos(c/2 + (d*x)/2)^14)/3465 + (512*cos(c/2 + (d*x)/2)^13*sin(c/2 + (d*x)/2))/3465 + (32*cos(c/2 + (d*x)/2
)^5*sin(c/2 + (d*x)/2)^9)/5 + (128*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^8)/15 + (320*cos(c/2 + (d*x)/2)^7*s
in(c/2 + (d*x)/2)^7)/21 + (704*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^6)/105 + (1504*cos(c/2 + (d*x)/2)^9*sin
(c/2 + (d*x)/2)^5)/315 - (64*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d*x)/2)^4)/315 + (2048*cos(c/2 + (d*x)/2)^11*sin
(c/2 + (d*x)/2)^3)/3465 + (320*cos(c/2 + (d*x)/2)^12*sin(c/2 + (d*x)/2)^2)/693)/(a^4*d*(cos(c/2 + (d*x)/2) - s
in(c/2 + (d*x)/2))^3*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^11)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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