3.844 \(\int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=127 \[ \frac {4 \tan ^9(c+d x)}{9 a^3 d}+\frac {9 \tan ^7(c+d x)}{7 a^3 d}+\frac {6 \tan ^5(c+d x)}{5 a^3 d}+\frac {\tan ^3(c+d x)}{3 a^3 d}-\frac {4 \sec ^9(c+d x)}{9 a^3 d}+\frac {5 \sec ^7(c+d x)}{7 a^3 d}-\frac {\sec ^5(c+d x)}{5 a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2875, 2873, 2607, 270, 2606, 14} \[ \frac {4 \tan ^9(c+d x)}{9 a^3 d}+\frac {9 \tan ^7(c+d x)}{7 a^3 d}+\frac {6 \tan ^5(c+d x)}{5 a^3 d}+\frac {\tan ^3(c+d x)}{3 a^3 d}-\frac {4 \sec ^9(c+d x)}{9 a^3 d}+\frac {5 \sec ^7(c+d x)}{7 a^3 d}-\frac {\sec ^5(c+d x)}{5 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sec[c + d*x]^5/(5*a^3*d) + (5*Sec[c + d*x]^7)/(7*a^3*d) - (4*Sec[c + d*x]^9)/(9*a^3*d) + Tan[c + d*x]^3/(3*a^
3*d) + (6*Tan[c + d*x]^5)/(5*a^3*d) + (9*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 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 2873

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 2875

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

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Mathematica [A]  time = 0.33, size = 185, normalized size = 1.46 \[ \frac {73728 \sin (c+d x)-7263 \sin (2 (c+d x))+512 \sin (3 (c+d x))-3228 \sin (4 (c+d x))-1536 \sin (5 (c+d x))+269 \sin (6 (c+d x))-9684 \cos (c+d x)-6912 \cos (2 (c+d x))-538 \cos (3 (c+d x))-3072 \cos (4 (c+d x))+1614 \cos (5 (c+d x))+256 \cos (6 (c+d x))+32256}{322560 d (a \sin (c+d x)+a)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32256 - 9684*Cos[c + d*x] - 6912*Cos[2*(c + d*x)] - 538*Cos[3*(c + d*x)] - 3072*Cos[4*(c + d*x)] + 1614*Cos[5
*(c + d*x)] + 256*Cos[6*(c + d*x)] + 73728*Sin[c + d*x] - 7263*Sin[2*(c + d*x)] + 512*Sin[3*(c + d*x)] - 3228*
Sin[4*(c + d*x)] - 1536*Sin[5*(c + d*x)] + 269*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)

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fricas [A]  time = 0.45, size = 130, normalized size = 1.02 \[ -\frac {8 \, \cos \left (d x + c\right )^{6} - 36 \, \cos \left (d x + c\right )^{4} + 15 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (12 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} - 35\right )} \sin \left (d x + c\right ) + 35}{315 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3} + {\left (a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/315*(8*cos(d*x + c)^6 - 36*cos(d*x + c)^4 + 15*cos(d*x + c)^2 - 2*(12*cos(d*x + c)^4 - 10*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))

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giac [A]  time = 0.34, size = 172, normalized size = 1.35 \[ -\frac {\frac {105 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {945 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 10080 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 23940 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 42840 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 41958 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 32592 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 14148 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5112 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 673}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{9}}}{10080 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10080*(105*(9*tan(1/2*d*x + 1/2*c)^2 - 12*tan(1/2*d*x + 1/2*c) + 7)/(a^3*(tan(1/2*d*x + 1/2*c) - 1)^3) - (9
45*tan(1/2*d*x + 1/2*c)^8 + 10080*tan(1/2*d*x + 1/2*c)^7 + 23940*tan(1/2*d*x + 1/2*c)^6 + 42840*tan(1/2*d*x +
1/2*c)^5 + 41958*tan(1/2*d*x + 1/2*c)^4 + 32592*tan(1/2*d*x + 1/2*c)^3 + 14148*tan(1/2*d*x + 1/2*c)^2 + 5112*t
an(1/2*d*x + 1/2*c) + 673)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^9))/d

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maple [A]  time = 0.54, size = 190, normalized size = 1.50 \[ \frac {-\frac {1}{24 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {8}{9 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {60}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {34}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {99}{10 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {23}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {3}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/d/a^3*(-1/192/(tan(1/2*d*x+1/2*c)-1)^3-1/128/(tan(1/2*d*x+1/2*c)-1)^2-3/256/(tan(1/2*d*x+1/2*c)-1)-1/9/(tan(
1/2*d*x+1/2*c)+1)^9+1/2/(tan(1/2*d*x+1/2*c)+1)^8-15/14/(tan(1/2*d*x+1/2*c)+1)^7+17/12/(tan(1/2*d*x+1/2*c)+1)^6
-99/80/(tan(1/2*d*x+1/2*c)+1)^5+23/32/(tan(1/2*d*x+1/2*c)+1)^4-1/4/(tan(1/2*d*x+1/2*c)+1)^3+1/32/(tan(1/2*d*x+
1/2*c)+1)^2+3/256/(tan(1/2*d*x+1/2*c)+1))

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maxima [B]  time = 0.34, size = 442, normalized size = 3.48 \[ \frac {4 \, {\left (\frac {66 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {132 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {232 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {18 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {108 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {84 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {504 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {315 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {210 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 11\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} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/315*(66*sin(d*x + c)/(cos(d*x + c) + 1) + 132*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 232*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 + 18*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 108*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 84*sin(d*
x + c)^6/(cos(d*x + c) + 1)^6 + 504*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 315*sin(d*x + c)^8/(cos(d*x + c) + 1
)^8 + 210*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 11)/((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
)

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mupad [B]  time = 14.87, size = 279, normalized size = 2.20 \[ \frac {\frac {44\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{315}+\frac {88\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}+\frac {176\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{105}+\frac {928\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{315}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{35}+\frac {48\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{35}+\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{15}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{5}+4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{3}}{a^3\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((44*cos(c/2 + (d*x)/2)^12)/315 + (88*cos(c/2 + (d*x)/2)^11*sin(c/2 + (d*x)/2))/105 + (8*cos(c/2 + (d*x)/2)^3*
sin(c/2 + (d*x)/2)^9)/3 + 4*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^8 + (32*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*
x)/2)^7)/5 + (16*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^6)/15 + (48*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^5
)/35 + (8*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^4)/35 + (928*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2)^3)/315
+ (176*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*(co
s(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^9)

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

[Out]

Timed out

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