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*Sec[c + d*x]^7)/(7*a^4*d) + (4*Sec[c + d*x]^9)/(3*a^4*d) - (8*Sec[c + d*x]^11)/(11*a^4*d) + Tan[c + d*x]^3
/(3*a^4*d) + (2*Tan[c + d*x]^5)/(a^4*d) + (25*Tan[c + d*x]^7)/(7*a^4*d) + (8*Tan[c + d*x]^9)/(3*a^4*d) + (8*Ta
n[c + d*x]^11)/(11*a^4*d)

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Rubi [A]  time = 0.35645, 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 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 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 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]

Rule 8

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

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*}

Mathematica [A]  time = 0.492419, 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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Maple [A]  time = 0.144, size = 218, normalized size = 1.5 \begin{align*} 8\,{\frac{1}{d{a}^{4}} \left ( -{\frac{1}{384\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{3}}}-{\frac{1}{256\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{2}}}-{\frac{1}{128\,\tan \left ( 1/2\,dx+c/2 \right ) -128}}-2/11\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-11}+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-10}-8/3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-9}+9/2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-8}-{\frac{295}{56\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{7}}}+{\frac{71}{16\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{6}}}-{\frac{43}{16\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{5}}}+{\frac{9}{8\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}-{\frac{109}{384\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}+{\frac{5}{256\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}+{\frac{1}{128\,\tan \left ( 1/2\,dx+c/2 \right ) +128}} \right ) } \end{align*}

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 = 1.25567, size = 713, normalized size = 4.99 \begin{align*} \text{result too large to display} \end{align*}

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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Fricas [A]  time = 1.62218, size = 390, normalized size = 2.73 \begin{align*} \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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.35932, size = 267, normalized size = 1.87 \begin{align*} -\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} \end{align*}

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