3.905 \(\int \frac {\sec ^7(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=172 \[ -\frac {\sec ^{10}(c+d x)}{10 a d}+\frac {\sec ^8(c+d x)}{8 a d}-\frac {7 \tanh ^{-1}(\sin (c+d x))}{256 a d}+\frac {\tan (c+d x) \sec ^9(c+d x)}{10 a d}-\frac {\tan (c+d x) \sec ^7(c+d x)}{80 a d}-\frac {7 \tan (c+d x) \sec ^5(c+d x)}{480 a d}-\frac {7 \tan (c+d x) \sec ^3(c+d x)}{384 a d}-\frac {7 \tan (c+d x) \sec (c+d x)}{256 a d} \]

[Out]

-7/256*arctanh(sin(d*x+c))/a/d+1/8*sec(d*x+c)^8/a/d-1/10*sec(d*x+c)^10/a/d-7/256*sec(d*x+c)*tan(d*x+c)/a/d-7/3
84*sec(d*x+c)^3*tan(d*x+c)/a/d-7/480*sec(d*x+c)^5*tan(d*x+c)/a/d-1/80*sec(d*x+c)^7*tan(d*x+c)/a/d+1/10*sec(d*x
+c)^9*tan(d*x+c)/a/d

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Rubi [A]  time = 0.22, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2835, 2611, 3768, 3770, 2606, 14} \[ -\frac {\sec ^{10}(c+d x)}{10 a d}+\frac {\sec ^8(c+d x)}{8 a d}-\frac {7 \tanh ^{-1}(\sin (c+d x))}{256 a d}+\frac {\tan (c+d x) \sec ^9(c+d x)}{10 a d}-\frac {\tan (c+d x) \sec ^7(c+d x)}{80 a d}-\frac {7 \tan (c+d x) \sec ^5(c+d x)}{480 a d}-\frac {7 \tan (c+d x) \sec ^3(c+d x)}{384 a d}-\frac {7 \tan (c+d x) \sec (c+d x)}{256 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^7*Tan[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

(-7*ArcTanh[Sin[c + d*x]])/(256*a*d) + Sec[c + d*x]^8/(8*a*d) - Sec[c + d*x]^10/(10*a*d) - (7*Sec[c + d*x]*Tan
[c + d*x])/(256*a*d) - (7*Sec[c + d*x]^3*Tan[c + d*x])/(384*a*d) - (7*Sec[c + d*x]^5*Tan[c + d*x])/(480*a*d) -
(Sec[c + d*x]^7*Tan[c + d*x])/(80*a*d) + (Sec[c + d*x]^9*Tan[c + d*x])/(10*a*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 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 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2835

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\sec ^7(c+d x) \tan ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int \sec ^9(c+d x) \tan ^2(c+d x) \, dx}{a}-\frac {\int \sec ^8(c+d x) \tan ^3(c+d x) \, dx}{a}\\ &=\frac {\sec ^9(c+d x) \tan (c+d x)}{10 a d}-\frac {\int \sec ^9(c+d x) \, dx}{10 a}-\frac {\operatorname {Subst}\left (\int x^7 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{a d}\\ &=-\frac {\sec ^7(c+d x) \tan (c+d x)}{80 a d}+\frac {\sec ^9(c+d x) \tan (c+d x)}{10 a d}-\frac {7 \int \sec ^7(c+d x) \, dx}{80 a}-\frac {\operatorname {Subst}\left (\int \left (-x^7+x^9\right ) \, dx,x,\sec (c+d x)\right )}{a d}\\ &=\frac {\sec ^8(c+d x)}{8 a d}-\frac {\sec ^{10}(c+d x)}{10 a d}-\frac {7 \sec ^5(c+d x) \tan (c+d x)}{480 a d}-\frac {\sec ^7(c+d x) \tan (c+d x)}{80 a d}+\frac {\sec ^9(c+d x) \tan (c+d x)}{10 a d}-\frac {7 \int \sec ^5(c+d x) \, dx}{96 a}\\ &=\frac {\sec ^8(c+d x)}{8 a d}-\frac {\sec ^{10}(c+d x)}{10 a d}-\frac {7 \sec ^3(c+d x) \tan (c+d x)}{384 a d}-\frac {7 \sec ^5(c+d x) \tan (c+d x)}{480 a d}-\frac {\sec ^7(c+d x) \tan (c+d x)}{80 a d}+\frac {\sec ^9(c+d x) \tan (c+d x)}{10 a d}-\frac {7 \int \sec ^3(c+d x) \, dx}{128 a}\\ &=\frac {\sec ^8(c+d x)}{8 a d}-\frac {\sec ^{10}(c+d x)}{10 a d}-\frac {7 \sec (c+d x) \tan (c+d x)}{256 a d}-\frac {7 \sec ^3(c+d x) \tan (c+d x)}{384 a d}-\frac {7 \sec ^5(c+d x) \tan (c+d x)}{480 a d}-\frac {\sec ^7(c+d x) \tan (c+d x)}{80 a d}+\frac {\sec ^9(c+d x) \tan (c+d x)}{10 a d}-\frac {7 \int \sec (c+d x) \, dx}{256 a}\\ &=-\frac {7 \tanh ^{-1}(\sin (c+d x))}{256 a d}+\frac {\sec ^8(c+d x)}{8 a d}-\frac {\sec ^{10}(c+d x)}{10 a d}-\frac {7 \sec (c+d x) \tan (c+d x)}{256 a d}-\frac {7 \sec ^3(c+d x) \tan (c+d x)}{384 a d}-\frac {7 \sec ^5(c+d x) \tan (c+d x)}{480 a d}-\frac {\sec ^7(c+d x) \tan (c+d x)}{80 a d}+\frac {\sec ^9(c+d x) \tan (c+d x)}{10 a d}\\ \end {align*}

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Mathematica [A]  time = 2.62, size = 122, normalized size = 0.71 \[ -\frac {210 \tanh ^{-1}(\sin (c+d x))-\frac {2 \left (105 \sin ^8(c+d x)+105 \sin ^7(c+d x)-385 \sin ^6(c+d x)-385 \sin ^5(c+d x)+511 \sin ^4(c+d x)+511 \sin ^3(c+d x)-279 \sin ^2(c+d x)+201 \sin (c+d x)+96\right )}{(\sin (c+d x)-1)^4 (\sin (c+d x)+1)^5}}{7680 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^7*Tan[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

-1/7680*(210*ArcTanh[Sin[c + d*x]] - (2*(96 + 201*Sin[c + d*x] - 279*Sin[c + d*x]^2 + 511*Sin[c + d*x]^3 + 511
*Sin[c + d*x]^4 - 385*Sin[c + d*x]^5 - 385*Sin[c + d*x]^6 + 105*Sin[c + d*x]^7 + 105*Sin[c + d*x]^8))/((-1 + S
in[c + d*x])^4*(1 + Sin[c + d*x])^5))/(a*d)

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fricas [A]  time = 0.53, size = 187, normalized size = 1.09 \[ \frac {210 \, \cos \left (d x + c\right )^{8} - 70 \, \cos \left (d x + c\right )^{6} - 28 \, \cos \left (d x + c\right )^{4} - 16 \, \cos \left (d x + c\right )^{2} - 105 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (105 \, \cos \left (d x + c\right )^{6} + 70 \, \cos \left (d x + c\right )^{4} + 56 \, \cos \left (d x + c\right )^{2} - 432\right )} \sin \left (d x + c\right ) + 96}{7680 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/7680*(210*cos(d*x + c)^8 - 70*cos(d*x + c)^6 - 28*cos(d*x + c)^4 - 16*cos(d*x + c)^2 - 105*(cos(d*x + c)^8*s
in(d*x + c) + cos(d*x + c)^8)*log(sin(d*x + c) + 1) + 105*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(-
sin(d*x + c) + 1) - 2*(105*cos(d*x + c)^6 + 70*cos(d*x + c)^4 + 56*cos(d*x + c)^2 - 432)*sin(d*x + c) + 96)/(a
*d*cos(d*x + c)^8*sin(d*x + c) + a*d*cos(d*x + c)^8)

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giac [A]  time = 0.29, size = 156, normalized size = 0.91 \[ -\frac {\frac {420 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {420 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {5 \, {\left (175 \, \sin \left (d x + c\right )^{4} - 748 \, \sin \left (d x + c\right )^{3} + 1182 \, \sin \left (d x + c\right )^{2} - 788 \, \sin \left (d x + c\right ) + 155\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac {959 \, \sin \left (d x + c\right )^{5} + 5395 \, \sin \left (d x + c\right )^{4} + 12290 \, \sin \left (d x + c\right )^{3} + 14170 \, \sin \left (d x + c\right )^{2} + 8135 \, \sin \left (d x + c\right ) + 1627}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{30720 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/30720*(420*log(abs(sin(d*x + c) + 1))/a - 420*log(abs(sin(d*x + c) - 1))/a + 5*(175*sin(d*x + c)^4 - 748*si
n(d*x + c)^3 + 1182*sin(d*x + c)^2 - 788*sin(d*x + c) + 155)/(a*(sin(d*x + c) - 1)^4) - (959*sin(d*x + c)^5 +
5395*sin(d*x + c)^4 + 12290*sin(d*x + c)^3 + 14170*sin(d*x + c)^2 + 8135*sin(d*x + c) + 1627)/(a*(sin(d*x + c)
+ 1)^5))/d

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maple [A]  time = 0.35, size = 198, normalized size = 1.15 \[ \frac {1}{256 a d \left (\sin \left (d x +c \right )-1\right )^{4}}-\frac {1}{192 a d \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {1}{512 a d \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {1}{128 a d \left (\sin \left (d x +c \right )-1\right )}+\frac {7 \ln \left (\sin \left (d x +c \right )-1\right )}{512 a d}-\frac {1}{160 a d \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {1}{256 a d \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {1}{384 a d \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {5}{512 a d \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5}{256 a d \left (1+\sin \left (d x +c \right )\right )}-\frac {7 \ln \left (1+\sin \left (d x +c \right )\right )}{512 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*sin(d*x+c)^2/(a+a*sin(d*x+c)),x)

[Out]

1/256/a/d/(sin(d*x+c)-1)^4-1/192/a/d/(sin(d*x+c)-1)^3+1/512/a/d/(sin(d*x+c)-1)^2+1/128/a/d/(sin(d*x+c)-1)+7/51
2/a/d*ln(sin(d*x+c)-1)-1/160/a/d/(1+sin(d*x+c))^5-1/256/a/d/(1+sin(d*x+c))^4+1/384/a/d/(1+sin(d*x+c))^3+5/512/
a/d/(1+sin(d*x+c))^2+5/256/a/d/(1+sin(d*x+c))-7/512*ln(1+sin(d*x+c))/a/d

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maxima [A]  time = 0.49, size = 214, normalized size = 1.24 \[ \frac {\frac {2 \, {\left (105 \, \sin \left (d x + c\right )^{8} + 105 \, \sin \left (d x + c\right )^{7} - 385 \, \sin \left (d x + c\right )^{6} - 385 \, \sin \left (d x + c\right )^{5} + 511 \, \sin \left (d x + c\right )^{4} + 511 \, \sin \left (d x + c\right )^{3} - 279 \, \sin \left (d x + c\right )^{2} + 201 \, \sin \left (d x + c\right ) + 96\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{7680 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/7680*(2*(105*sin(d*x + c)^8 + 105*sin(d*x + c)^7 - 385*sin(d*x + c)^6 - 385*sin(d*x + c)^5 + 511*sin(d*x + c
)^4 + 511*sin(d*x + c)^3 - 279*sin(d*x + c)^2 + 201*sin(d*x + c) + 96)/(a*sin(d*x + c)^9 + a*sin(d*x + c)^8 -
4*a*sin(d*x + c)^7 - 4*a*sin(d*x + c)^6 + 6*a*sin(d*x + c)^5 + 6*a*sin(d*x + c)^4 - 4*a*sin(d*x + c)^3 - 4*a*s
in(d*x + c)^2 + a*sin(d*x + c) + a) - 105*log(sin(d*x + c) + 1)/a + 105*log(sin(d*x + c) - 1)/a)/d

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mupad [B]  time = 16.88, size = 496, normalized size = 2.88 \[ \frac {\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{128}+\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}}{64}+\frac {221\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{96}+\frac {95\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{192}+\frac {2261\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{480}+\frac {889\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{960}+\frac {7343\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{480}+\frac {1603\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{960}+\frac {2471\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {1603\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{960}+\frac {7343\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{480}+\frac {889\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{960}+\frac {2261\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{480}+\frac {95\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{192}+\frac {221\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96}+\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{64}+\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}-7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}-16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}+20\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+56\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-28\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-112\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+14\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+140\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+14\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-112\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-28\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+56\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+20\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}-\frac {7\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{128\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^2/(cos(c + d*x)^9*(a + a*sin(c + d*x))),x)

[Out]

((7*tan(c/2 + (d*x)/2))/128 + (7*tan(c/2 + (d*x)/2)^2)/64 + (221*tan(c/2 + (d*x)/2)^3)/96 + (95*tan(c/2 + (d*x
)/2)^4)/192 + (2261*tan(c/2 + (d*x)/2)^5)/480 + (889*tan(c/2 + (d*x)/2)^6)/960 + (7343*tan(c/2 + (d*x)/2)^7)/4
80 + (1603*tan(c/2 + (d*x)/2)^8)/960 + (2471*tan(c/2 + (d*x)/2)^9)/192 + (1603*tan(c/2 + (d*x)/2)^10)/960 + (7
343*tan(c/2 + (d*x)/2)^11)/480 + (889*tan(c/2 + (d*x)/2)^12)/960 + (2261*tan(c/2 + (d*x)/2)^13)/480 + (95*tan(
c/2 + (d*x)/2)^14)/192 + (221*tan(c/2 + (d*x)/2)^15)/96 + (7*tan(c/2 + (d*x)/2)^16)/64 + (7*tan(c/2 + (d*x)/2)
^17)/128)/(d*(a + 2*a*tan(c/2 + (d*x)/2) - 7*a*tan(c/2 + (d*x)/2)^2 - 16*a*tan(c/2 + (d*x)/2)^3 + 20*a*tan(c/2
+ (d*x)/2)^4 + 56*a*tan(c/2 + (d*x)/2)^5 - 28*a*tan(c/2 + (d*x)/2)^6 - 112*a*tan(c/2 + (d*x)/2)^7 + 14*a*tan(
c/2 + (d*x)/2)^8 + 140*a*tan(c/2 + (d*x)/2)^9 + 14*a*tan(c/2 + (d*x)/2)^10 - 112*a*tan(c/2 + (d*x)/2)^11 - 28*
a*tan(c/2 + (d*x)/2)^12 + 56*a*tan(c/2 + (d*x)/2)^13 + 20*a*tan(c/2 + (d*x)/2)^14 - 16*a*tan(c/2 + (d*x)/2)^15
- 7*a*tan(c/2 + (d*x)/2)^16 + 2*a*tan(c/2 + (d*x)/2)^17 + a*tan(c/2 + (d*x)/2)^18)) - (7*atanh(tan(c/2 + (d*x
)/2)))/(128*a*d)

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

[Out]

Timed out

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