3.850 \(\int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=143 \[ \frac {8 \tan ^{11}(c+d x)}{11 a^4 d}+\frac {8 \tan ^9(c+d x)}{3 a^4 d}+\frac {25 \tan ^7(c+d x)}{7 a^4 d}+\frac {2 \tan ^5(c+d x)}{a^4 d}+\frac {\tan ^3(c+d x)}{3 a^4 d}-\frac {8 \sec ^{11}(c+d x)}{11 a^4 d}+\frac {4 \sec ^9(c+d x)}{3 a^4 d}-\frac {4 \sec ^7(c+d x)}{7 a^4 d} \]

[Out]

-4/7*sec(d*x+c)^7/a^4/d+4/3*sec(d*x+c)^9/a^4/d-8/11*sec(d*x+c)^11/a^4/d+1/3*tan(d*x+c)^3/a^4/d+2*tan(d*x+c)^5/
a^4/d+25/7*tan(d*x+c)^7/a^4/d+8/3*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 = 0.36, antiderivative size = 184, normalized size of antiderivative = 1.29, number of steps used = 8, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2870, 2672, 3767, 8} \[ \frac {8 \tan (c+d x)}{231 a^4 d}-\frac {4 \sec (c+d x)}{231 d \left (a^4 \sin (c+d x)+a^4\right )}-\frac {4 \sec (c+d x)}{231 d \left (a^2 \sin (c+d x)+a^2\right )^2}+\frac {\sec ^3(c+d x)}{6 a d (a \sin (c+d x)+a)^3}-\frac {5 \sec (c+d x)}{231 a d (a \sin (c+d x)+a)^3}-\frac {\sec (c+d x)}{33 d (a \sin (c+d x)+a)^4}-\frac {a \sec (c+d x)}{22 d (a \sin (c+d x)+a)^5} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^2*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^4,x]

[Out]

-(a*Sec[c + d*x])/(22*d*(a + a*Sin[c + d*x])^5) - Sec[c + d*x]/(33*d*(a + a*Sin[c + d*x])^4) - (5*Sec[c + d*x]
)/(231*a*d*(a + a*Sin[c + d*x])^3) + Sec[c + d*x]^3/(6*a*d*(a + a*Sin[c + d*x])^3) - (4*Sec[c + d*x])/(231*d*(
a^2 + a^2*Sin[c + d*x])^2) - (4*Sec[c + d*x])/(231*d*(a^4 + a^4*Sin[c + d*x])) + (8*Tan[c + d*x])/(231*a^4*d)

Rule 8

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

Rule 2672

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 2870

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> -Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(2*b*f*g*(m + 1)), x] + Dist[a/(2
*g^2), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] && EqQ[m - p, 0]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx &=\frac {\sec ^3(c+d x)}{6 a d (a+a \sin (c+d x))^3}+\frac {1}{2} a \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^5} \, dx\\ &=-\frac {a \sec (c+d x)}{22 d (a+a \sin (c+d x))^5}+\frac {\sec ^3(c+d x)}{6 a d (a+a \sin (c+d x))^3}+\frac {3}{11} \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx\\ &=-\frac {a \sec (c+d x)}{22 d (a+a \sin (c+d x))^5}-\frac {\sec (c+d x)}{33 d (a+a \sin (c+d x))^4}+\frac {\sec ^3(c+d x)}{6 a d (a+a \sin (c+d x))^3}+\frac {5 \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx}{33 a}\\ &=-\frac {a \sec (c+d x)}{22 d (a+a \sin (c+d x))^5}-\frac {\sec (c+d x)}{33 d (a+a \sin (c+d x))^4}-\frac {5 \sec (c+d x)}{231 a d (a+a \sin (c+d x))^3}+\frac {\sec ^3(c+d x)}{6 a d (a+a \sin (c+d x))^3}+\frac {20 \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx}{231 a^2}\\ &=-\frac {a \sec (c+d x)}{22 d (a+a \sin (c+d x))^5}-\frac {\sec (c+d x)}{33 d (a+a \sin (c+d x))^4}-\frac {5 \sec (c+d x)}{231 a d (a+a \sin (c+d x))^3}+\frac {\sec ^3(c+d x)}{6 a d (a+a \sin (c+d x))^3}-\frac {4 \sec (c+d x)}{231 d \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {4 \int \frac {\sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx}{77 a^3}\\ &=-\frac {a \sec (c+d x)}{22 d (a+a \sin (c+d x))^5}-\frac {\sec (c+d x)}{33 d (a+a \sin (c+d x))^4}-\frac {5 \sec (c+d x)}{231 a d (a+a \sin (c+d x))^3}+\frac {\sec ^3(c+d x)}{6 a d (a+a \sin (c+d x))^3}-\frac {4 \sec (c+d x)}{231 d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {4 \sec (c+d x)}{231 d \left (a^4+a^4 \sin (c+d x)\right )}+\frac {8 \int \sec ^2(c+d x) \, dx}{231 a^4}\\ &=-\frac {a \sec (c+d x)}{22 d (a+a \sin (c+d x))^5}-\frac {\sec (c+d x)}{33 d (a+a \sin (c+d x))^4}-\frac {5 \sec (c+d x)}{231 a d (a+a \sin (c+d x))^3}+\frac {\sec ^3(c+d x)}{6 a d (a+a \sin (c+d x))^3}-\frac {4 \sec (c+d x)}{231 d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {4 \sec (c+d x)}{231 d \left (a^4+a^4 \sin (c+d x)\right )}-\frac {8 \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{231 a^4 d}\\ &=-\frac {a \sec (c+d x)}{22 d (a+a \sin (c+d x))^5}-\frac {\sec (c+d x)}{33 d (a+a \sin (c+d x))^4}-\frac {5 \sec (c+d x)}{231 a d (a+a \sin (c+d x))^3}+\frac {\sec ^3(c+d x)}{6 a d (a+a \sin (c+d x))^3}-\frac {4 \sec (c+d x)}{231 d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {4 \sec (c+d x)}{231 d \left (a^4+a^4 \sin (c+d x)\right )}+\frac {8 \tan (c+d x)}{231 a^4 d}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 166, normalized size = 1.16 \[ \frac {\sec ^3(c+d x) (26048 \sin (c+d x)-1144 \sin (2 (c+d x))-704 \sin (3 (c+d x))-416 \sin (4 (c+d x))-1600 \sin (5 (c+d x))+104 \sin (6 (c+d x))+64 \sin (7 (c+d x))-1287 \cos (c+d x)-5632 \cos (2 (c+d x))+143 \cos (3 (c+d x))-2048 \cos (4 (c+d x))+325 \cos (5 (c+d x))+512 \cos (6 (c+d x))-13 \cos (7 (c+d x))+11264)}{118272 a^4 d (\sin (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^2*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^4,x]

[Out]

(Sec[c + d*x]^3*(11264 - 1287*Cos[c + d*x] - 5632*Cos[2*(c + d*x)] + 143*Cos[3*(c + d*x)] - 2048*Cos[4*(c + d*
x)] + 325*Cos[5*(c + d*x)] + 512*Cos[6*(c + d*x)] - 13*Cos[7*(c + d*x)] + 26048*Sin[c + d*x] - 1144*Sin[2*(c +
d*x)] - 704*Sin[3*(c + d*x)] - 416*Sin[4*(c + d*x)] - 1600*Sin[5*(c + d*x)] + 104*Sin[6*(c + d*x)] + 64*Sin[7
*(c + d*x)]))/(118272*a^4*d*(1 + Sin[c + d*x])^4)

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fricas [A]  time = 0.45, size = 153, normalized size = 1.07 \[ \frac {32 \, \cos \left (d x + c\right )^{6} - 80 \, \cos \left (d x + c\right )^{4} + 28 \, \cos \left (d x + c\right )^{2} + {\left (8 \, \cos \left (d x + c\right )^{6} - 60 \, \cos \left (d x + c\right )^{4} + 35 \, \cos \left (d x + c\right )^{2} + 49\right )} \sin \left (d x + c\right ) + 28}{231 \, {\left (a^{4} d \cos \left (d x + c\right )^{7} - 8 \, a^{4} d \cos \left (d x + c\right )^{5} + 8 \, a^{4} d \cos \left (d x + c\right )^{3} - 4 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} - 2 \, a^{4} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/231*(32*cos(d*x + c)^6 - 80*cos(d*x + c)^4 + 28*cos(d*x + c)^2 + (8*cos(d*x + c)^6 - 60*cos(d*x + c)^4 + 35*
cos(d*x + c)^2 + 49)*sin(d*x + c) + 28)/(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 = 0.40, size = 198, normalized size = 1.38 \[ -\frac {\frac {77 \, {\left (6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {462 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 5775 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 14399 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 29260 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30800 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 27874 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12650 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6556 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1210 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 935 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 127}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{11}}}{7392 \, d} \]

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))^4,x, algorithm="giac")

[Out]

-1/7392*(77*(6*tan(1/2*d*x + 1/2*c)^2 - 9*tan(1/2*d*x + 1/2*c) + 5)/(a^4*(tan(1/2*d*x + 1/2*c) - 1)^3) - (462*
tan(1/2*d*x + 1/2*c)^10 + 5775*tan(1/2*d*x + 1/2*c)^9 + 14399*tan(1/2*d*x + 1/2*c)^8 + 29260*tan(1/2*d*x + 1/2
*c)^7 + 30800*tan(1/2*d*x + 1/2*c)^6 + 27874*tan(1/2*d*x + 1/2*c)^5 + 12650*tan(1/2*d*x + 1/2*c)^4 + 6556*tan(
1/2*d*x + 1/2*c)^3 + 1210*tan(1/2*d*x + 1/2*c)^2 + 935*tan(1/2*d*x + 1/2*c) + 127)/(a^4*(tan(1/2*d*x + 1/2*c)
+ 1)^11))/d

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maple [A]  time = 0.73, size = 218, normalized size = 1.52 \[ \frac {-\frac {1}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {16}{11 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{11}}+\frac {8}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{10}}-\frac {64}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}+\frac {36}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {295}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {71}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {43}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {9}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {109}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {5}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+128}}{a^{4} d} \]

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))^4,x)

[Out]

8/d/a^4*(-1/384/(tan(1/2*d*x+1/2*c)-1)^3-1/256/(tan(1/2*d*x+1/2*c)-1)^2-1/128/(tan(1/2*d*x+1/2*c)-1)-2/11/(tan
(1/2*d*x+1/2*c)+1)^11+1/(tan(1/2*d*x+1/2*c)+1)^10-8/3/(tan(1/2*d*x+1/2*c)+1)^9+9/2/(tan(1/2*d*x+1/2*c)+1)^8-29
5/56/(tan(1/2*d*x+1/2*c)+1)^7+71/16/(tan(1/2*d*x+1/2*c)+1)^6-43/16/(tan(1/2*d*x+1/2*c)+1)^5+9/8/(tan(1/2*d*x+1
/2*c)+1)^4-109/384/(tan(1/2*d*x+1/2*c)+1)^3+5/256/(tan(1/2*d*x+1/2*c)+1)^2+1/128/(tan(1/2*d*x+1/2*c)+1))

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maxima [B]  time = 0.35, size = 528, normalized size = 3.69 \[ \frac {8 \, {\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 {141 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {132 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {132 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {44 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {110 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {154 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {308 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {154 \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {77 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} + 2\right )}}{231 \, {\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)^2/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

8/231*(16*sin(d*x + c)/(cos(d*x + c) + 1) + 50*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 141*sin(d*x + c)^3/(cos(d
*x + c) + 1)^3 + 132*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 132*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 44*sin(d*
x + c)^6/(cos(d*x + c) + 1)^6 + 110*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 154*sin(d*x + c)^8/(cos(d*x + c) + 1
)^8 + 308*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 154*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 77*sin(d*x + c)^11
/(cos(d*x + c) + 1)^11 + 2)/((a^4 + 8*a^4*sin(d*x + c)/(cos(d*x + c) + 1) + 25*a^4*sin(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*si
n(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 = 16.00, size = 327, normalized size = 2.29 \[ \frac {\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{231}+\frac {128\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{231}+\frac {400\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{231}+\frac {376\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{77}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{7}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{7}-\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{21}+\frac {80\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{21}+\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{3}+\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{3}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{3}}{a^4\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((16*cos(c/2 + (d*x)/2)^14)/231 + (128*cos(c/2 + (d*x)/2)^13*sin(c/2 + (d*x)/2))/231 + (8*cos(c/2 + (d*x)/2)^3
*sin(c/2 + (d*x)/2)^11)/3 + (16*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^10)/3 + (32*cos(c/2 + (d*x)/2)^5*sin(c
/2 + (d*x)/2)^9)/3 + (16*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^8)/3 + (80*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*
x)/2)^7)/21 - (32*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^6)/21 + (32*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2)^
5)/7 + (32*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d*x)/2)^4)/7 + (376*cos(c/2 + (d*x)/2)^11*sin(c/2 + (d*x)/2)^3)/77
+ (400*cos(c/2 + (d*x)/2)^12*sin(c/2 + (d*x)/2)^2)/231)/(a^4*d*(cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2))^3*(c
os(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)**2/(a+a*sin(d*x+c))**4,x)

[Out]

Timed out

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