3.901 \(\int \frac {\sec ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=176 \[ -\frac {\tan ^{10}(c+d x)}{10 a d}-\frac {\tan ^8(c+d x)}{8 a d}-\frac {3 \tanh ^{-1}(\sin (c+d x))}{256 a d}+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 a d}-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{16 a d}+\frac {\tan (c+d x) \sec ^5(c+d x)}{32 a d}-\frac {\tan (c+d x) \sec ^3(c+d x)}{128 a d}-\frac {3 \tan (c+d x) \sec (c+d x)}{256 a d} \]

[Out]

-3/256*arctanh(sin(d*x+c))/a/d-3/256*sec(d*x+c)*tan(d*x+c)/a/d-1/128*sec(d*x+c)^3*tan(d*x+c)/a/d+1/32*sec(d*x+
c)^5*tan(d*x+c)/a/d-1/16*sec(d*x+c)^5*tan(d*x+c)^3/a/d+1/10*sec(d*x+c)^5*tan(d*x+c)^5/a/d-1/8*tan(d*x+c)^8/a/d
-1/10*tan(d*x+c)^10/a/d

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Rubi [A]  time = 0.27, antiderivative size = 176, 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, 2607, 14} \[ -\frac {\tan ^{10}(c+d x)}{10 a d}-\frac {\tan ^8(c+d x)}{8 a d}-\frac {3 \tanh ^{-1}(\sin (c+d x))}{256 a d}+\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 a d}-\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{16 a d}+\frac {\tan (c+d x) \sec ^5(c+d x)}{32 a d}-\frac {\tan (c+d x) \sec ^3(c+d x)}{128 a d}-\frac {3 \tan (c+d x) \sec (c+d x)}{256 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*Tan[c + d*x]^6)/(a + a*Sin[c + d*x]),x]

[Out]

(-3*ArcTanh[Sin[c + d*x]])/(256*a*d) - (3*Sec[c + d*x]*Tan[c + d*x])/(256*a*d) - (Sec[c + d*x]^3*Tan[c + d*x])
/(128*a*d) + (Sec[c + d*x]^5*Tan[c + d*x])/(32*a*d) - (Sec[c + d*x]^5*Tan[c + d*x]^3)/(16*a*d) + (Sec[c + d*x]
^5*Tan[c + d*x]^5)/(10*a*d) - Tan[c + d*x]^8/(8*a*d) - Tan[c + d*x]^10/(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 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 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 ^3(c+d x) \tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int \sec ^5(c+d x) \tan ^6(c+d x) \, dx}{a}-\frac {\int \sec ^4(c+d x) \tan ^7(c+d x) \, dx}{a}\\ &=\frac {\sec ^5(c+d x) \tan ^5(c+d x)}{10 a d}-\frac {\int \sec ^5(c+d x) \tan ^4(c+d x) \, dx}{2 a}-\frac {\operatorname {Subst}\left (\int x^7 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a d}\\ &=-\frac {\sec ^5(c+d x) \tan ^3(c+d x)}{16 a d}+\frac {\sec ^5(c+d x) \tan ^5(c+d x)}{10 a d}+\frac {3 \int \sec ^5(c+d x) \tan ^2(c+d x) \, dx}{16 a}-\frac {\operatorname {Subst}\left (\int \left (x^7+x^9\right ) \, dx,x,\tan (c+d x)\right )}{a d}\\ &=\frac {\sec ^5(c+d x) \tan (c+d x)}{32 a d}-\frac {\sec ^5(c+d x) \tan ^3(c+d x)}{16 a d}+\frac {\sec ^5(c+d x) \tan ^5(c+d x)}{10 a d}-\frac {\tan ^8(c+d x)}{8 a d}-\frac {\tan ^{10}(c+d x)}{10 a d}-\frac {\int \sec ^5(c+d x) \, dx}{32 a}\\ &=-\frac {\sec ^3(c+d x) \tan (c+d x)}{128 a d}+\frac {\sec ^5(c+d x) \tan (c+d x)}{32 a d}-\frac {\sec ^5(c+d x) \tan ^3(c+d x)}{16 a d}+\frac {\sec ^5(c+d x) \tan ^5(c+d x)}{10 a d}-\frac {\tan ^8(c+d x)}{8 a d}-\frac {\tan ^{10}(c+d x)}{10 a d}-\frac {3 \int \sec ^3(c+d x) \, dx}{128 a}\\ &=-\frac {3 \sec (c+d x) \tan (c+d x)}{256 a d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{128 a d}+\frac {\sec ^5(c+d x) \tan (c+d x)}{32 a d}-\frac {\sec ^5(c+d x) \tan ^3(c+d x)}{16 a d}+\frac {\sec ^5(c+d x) \tan ^5(c+d x)}{10 a d}-\frac {\tan ^8(c+d x)}{8 a d}-\frac {\tan ^{10}(c+d x)}{10 a d}-\frac {3 \int \sec (c+d x) \, dx}{256 a}\\ &=-\frac {3 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac {3 \sec (c+d x) \tan (c+d x)}{256 a d}-\frac {\sec ^3(c+d x) \tan (c+d x)}{128 a d}+\frac {\sec ^5(c+d x) \tan (c+d x)}{32 a d}-\frac {\sec ^5(c+d x) \tan ^3(c+d x)}{16 a d}+\frac {\sec ^5(c+d x) \tan ^5(c+d x)}{10 a d}-\frac {\tan ^8(c+d x)}{8 a d}-\frac {\tan ^{10}(c+d x)}{10 a d}\\ \end {align*}

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Mathematica [A]  time = 2.76, size = 122, normalized size = 0.69 \[ -\frac {30 \tanh ^{-1}(\sin (c+d x))-\frac {2 \left (15 \sin ^8(c+d x)+15 \sin ^7(c+d x)-55 \sin ^6(c+d x)+265 \sin ^5(c+d x)+137 \sin ^4(c+d x)-183 \sin ^3(c+d x)-113 \sin ^2(c+d x)+47 \sin (c+d x)+32\right )}{(\sin (c+d x)-1)^4 (\sin (c+d x)+1)^5}}{2560 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^3*Tan[c + d*x]^6)/(a + a*Sin[c + d*x]),x]

[Out]

-1/2560*(30*ArcTanh[Sin[c + d*x]] - (2*(32 + 47*Sin[c + d*x] - 113*Sin[c + d*x]^2 - 183*Sin[c + d*x]^3 + 137*S
in[c + d*x]^4 + 265*Sin[c + d*x]^5 - 55*Sin[c + d*x]^6 + 15*Sin[c + d*x]^7 + 15*Sin[c + d*x]^8))/((-1 + Sin[c
+ d*x])^4*(1 + Sin[c + d*x])^5))/(a*d)

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fricas [A]  time = 0.51, size = 187, normalized size = 1.06 \[ \frac {30 \, \cos \left (d x + c\right )^{8} - 10 \, \cos \left (d x + c\right )^{6} + 124 \, \cos \left (d x + c\right )^{4} - 112 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (15 \, \cos \left (d x + c\right )^{6} - 310 \, \cos \left (d x + c\right )^{4} + 392 \, \cos \left (d x + c\right )^{2} - 144\right )} \sin \left (d x + c\right ) + 32}{2560 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/2560*(30*cos(d*x + c)^8 - 10*cos(d*x + c)^6 + 124*cos(d*x + c)^4 - 112*cos(d*x + c)^2 - 15*(cos(d*x + c)^8*s
in(d*x + c) + cos(d*x + c)^8)*log(sin(d*x + c) + 1) + 15*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(-s
in(d*x + c) + 1) - 2*(15*cos(d*x + c)^6 - 310*cos(d*x + c)^4 + 392*cos(d*x + c)^2 - 144)*sin(d*x + c) + 32)/(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 = 0.36, size = 156, normalized size = 0.89 \[ -\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {5 \, {\left (25 \, \sin \left (d x + c\right )^{4} - 84 \, \sin \left (d x + c\right )^{3} + 66 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right ) - 3\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac {137 \, \sin \left (d x + c\right )^{5} + 885 \, \sin \left (d x + c\right )^{4} + 2270 \, \sin \left (d x + c\right )^{3} + 2470 \, \sin \left (d x + c\right )^{2} + 1265 \, \sin \left (d x + c\right ) + 253}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{10240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/10240*(60*log(abs(sin(d*x + c) + 1))/a - 60*log(abs(sin(d*x + c) - 1))/a + 5*(25*sin(d*x + c)^4 - 84*sin(d*
x + c)^3 + 66*sin(d*x + c)^2 - 12*sin(d*x + c) - 3)/(a*(sin(d*x + c) - 1)^4) - (137*sin(d*x + c)^5 + 885*sin(d
*x + c)^4 + 2270*sin(d*x + c)^3 + 2470*sin(d*x + c)^2 + 1265*sin(d*x + c) + 253)/(a*(sin(d*x + c) + 1)^5))/d

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maple [A]  time = 0.41, size = 198, normalized size = 1.12 \[ \frac {1}{256 a d \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {1}{64 a d \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {9}{512 a d \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {1}{128 a d \left (\sin \left (d x +c \right )-1\right )}+\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{512 a d}-\frac {1}{160 a d \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {7}{256 a d \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{128 a d \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {5}{512 a d \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5}{256 a d \left (1+\sin \left (d x +c \right )\right )}-\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{512 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*sin(d*x+c)^6/(a+a*sin(d*x+c)),x)

[Out]

1/256/a/d/(sin(d*x+c)-1)^4+1/64/a/d/(sin(d*x+c)-1)^3+9/512/a/d/(sin(d*x+c)-1)^2-1/128/a/d/(sin(d*x+c)-1)+3/512
/a/d*ln(sin(d*x+c)-1)-1/160/a/d/(1+sin(d*x+c))^5+7/256/a/d/(1+sin(d*x+c))^4-5/128/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))-3/512*ln(1+sin(d*x+c))/a/d

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maxima [A]  time = 0.33, size = 214, normalized size = 1.22 \[ \frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{8} + 15 \, \sin \left (d x + c\right )^{7} - 55 \, \sin \left (d x + c\right )^{6} + 265 \, \sin \left (d x + c\right )^{5} + 137 \, \sin \left (d x + c\right )^{4} - 183 \, \sin \left (d x + c\right )^{3} - 113 \, \sin \left (d x + c\right )^{2} + 47 \, \sin \left (d x + c\right ) + 32\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{2560 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/2560*(2*(15*sin(d*x + c)^8 + 15*sin(d*x + c)^7 - 55*sin(d*x + c)^6 + 265*sin(d*x + c)^5 + 137*sin(d*x + c)^4
- 183*sin(d*x + c)^3 - 113*sin(d*x + c)^2 + 47*sin(d*x + c) + 32)/(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*sin(d
*x + c)^2 + a*sin(d*x + c) + a) - 15*log(sin(d*x + c) + 1)/a + 15*log(sin(d*x + c) - 1)/a)/d

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mupad [B]  time = 16.64, size = 496, normalized size = 2.82 \[ \frac {\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{128}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}}{64}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{32}-\frac {23\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{64}+\frac {67\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{160}+\frac {383\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{320}+\frac {2841\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{160}+\frac {741\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{320}+\frac {1377\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{64}+\frac {741\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{320}+\frac {2841\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{160}+\frac {383\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{320}+\frac {67\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160}-\frac {23\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{32}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{64}+\frac {3\,\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 {3\,\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)^6/(cos(c + d*x)^9*(a + a*sin(c + d*x))),x)

[Out]

((3*tan(c/2 + (d*x)/2))/128 + (3*tan(c/2 + (d*x)/2)^2)/64 - (5*tan(c/2 + (d*x)/2)^3)/32 - (23*tan(c/2 + (d*x)/
2)^4)/64 + (67*tan(c/2 + (d*x)/2)^5)/160 + (383*tan(c/2 + (d*x)/2)^6)/320 + (2841*tan(c/2 + (d*x)/2)^7)/160 +
(741*tan(c/2 + (d*x)/2)^8)/320 + (1377*tan(c/2 + (d*x)/2)^9)/64 + (741*tan(c/2 + (d*x)/2)^10)/320 + (2841*tan(
c/2 + (d*x)/2)^11)/160 + (383*tan(c/2 + (d*x)/2)^12)/320 + (67*tan(c/2 + (d*x)/2)^13)/160 - (23*tan(c/2 + (d*x
)/2)^14)/64 - (5*tan(c/2 + (d*x)/2)^15)/32 + (3*tan(c/2 + (d*x)/2)^16)/64 + (3*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)) - (3*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)**6/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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