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

Optimal. Leaf size=178 \[ \frac {\tan ^{10}(c+d x)}{10 a d}+\frac {\tan ^8(c+d x)}{8 a d}-\frac {7 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac {\tan ^7(c+d x) \sec ^3(c+d x)}{10 a d}+\frac {7 \tan ^5(c+d x) \sec ^3(c+d x)}{80 a d}-\frac {7 \tan ^3(c+d x) \sec ^3(c+d x)}{96 a d}+\frac {7 \tan (c+d x) \sec ^3(c+d x)}{128 a d}-\frac {7 \tan (c+d x) \sec (c+d x)}{256 a d} \]

[Out]

-7/256*arctanh(sin(d*x+c))/a/d-7/256*sec(d*x+c)*tan(d*x+c)/a/d+7/128*sec(d*x+c)^3*tan(d*x+c)/a/d-7/96*sec(d*x+
c)^3*tan(d*x+c)^3/a/d+7/80*sec(d*x+c)^3*tan(d*x+c)^5/a/d-1/10*sec(d*x+c)^3*tan(d*x+c)^7/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 = 0.28, antiderivative size = 178, 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, 2607, 14, 2611, 3768, 3770} \[ \frac {\tan ^{10}(c+d x)}{10 a d}+\frac {\tan ^8(c+d x)}{8 a d}-\frac {7 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac {\tan ^7(c+d x) \sec ^3(c+d x)}{10 a d}+\frac {7 \tan ^5(c+d x) \sec ^3(c+d x)}{80 a d}-\frac {7 \tan ^3(c+d x) \sec ^3(c+d x)}{96 a d}+\frac {7 \tan (c+d x) \sec ^3(c+d x)}{128 a d}-\frac {7 \tan (c+d x) \sec (c+d x)}{256 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*ArcTanh[Sin[c + d*x]])/(256*a*d) - (7*Sec[c + d*x]*Tan[c + d*x])/(256*a*d) + (7*Sec[c + d*x]^3*Tan[c + d*x
])/(128*a*d) - (7*Sec[c + d*x]^3*Tan[c + d*x]^3)/(96*a*d) + (7*Sec[c + d*x]^3*Tan[c + d*x]^5)/(80*a*d) - (Sec[
c + d*x]^3*Tan[c + d*x]^7)/(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 ^2(c+d x) \tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int \sec ^4(c+d x) \tan ^7(c+d x) \, dx}{a}-\frac {\int \sec ^3(c+d x) \tan ^8(c+d x) \, dx}{a}\\ &=-\frac {\sec ^3(c+d x) \tan ^7(c+d x)}{10 a d}+\frac {7 \int \sec ^3(c+d x) \tan ^6(c+d x) \, dx}{10 a}+\frac {\operatorname {Subst}\left (\int x^7 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a d}\\ &=\frac {7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}-\frac {\sec ^3(c+d x) \tan ^7(c+d x)}{10 a d}-\frac {7 \int \sec ^3(c+d x) \tan ^4(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 {7 \sec ^3(c+d x) \tan ^3(c+d x)}{96 a d}+\frac {7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}-\frac {\sec ^3(c+d x) \tan ^7(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 {7 \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx}{32 a}\\ &=\frac {7 \sec ^3(c+d x) \tan (c+d x)}{128 a d}-\frac {7 \sec ^3(c+d x) \tan ^3(c+d x)}{96 a d}+\frac {7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}-\frac {\sec ^3(c+d x) \tan ^7(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 {7 \int \sec ^3(c+d x) \, dx}{128 a}\\ &=-\frac {7 \sec (c+d x) \tan (c+d x)}{256 a d}+\frac {7 \sec ^3(c+d x) \tan (c+d x)}{128 a d}-\frac {7 \sec ^3(c+d x) \tan ^3(c+d x)}{96 a d}+\frac {7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}-\frac {\sec ^3(c+d x) \tan ^7(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 {7 \int \sec (c+d x) \, dx}{256 a}\\ &=-\frac {7 \tanh ^{-1}(\sin (c+d x))}{256 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)}{128 a d}-\frac {7 \sec ^3(c+d x) \tan ^3(c+d x)}{96 a d}+\frac {7 \sec ^3(c+d x) \tan ^5(c+d x)}{80 a d}-\frac {\sec ^3(c+d x) \tan ^7(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 = 1.68, size = 124, normalized size = 0.70 \[ -\frac {\frac {210}{1-\sin (c+d x)}-\frac {315}{(1-\sin (c+d x))^2}+\frac {525}{(\sin (c+d x)+1)^2}+\frac {160}{(1-\sin (c+d x))^3}-\frac {580}{(\sin (c+d x)+1)^3}-\frac {30}{(1-\sin (c+d x))^4}+\frac {270}{(\sin (c+d x)+1)^4}-\frac {48}{(\sin (c+d x)+1)^5}+210 \tanh ^{-1}(\sin (c+d x))}{7680 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/7680*(210*ArcTanh[Sin[c + d*x]] - 30/(1 - Sin[c + d*x])^4 + 160/(1 - Sin[c + d*x])^3 - 315/(1 - Sin[c + d*x
])^2 + 210/(1 - Sin[c + d*x]) - 48/(1 + Sin[c + d*x])^5 + 270/(1 + Sin[c + d*x])^4 - 580/(1 + Sin[c + d*x])^3
+ 525/(1 + Sin[c + d*x])^2)/(a*d)

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fricas [A]  time = 0.52, size = 187, normalized size = 1.05 \[ \frac {210 \, \cos \left (d x + c\right )^{8} - 2630 \, \cos \left (d x + c\right )^{6} + 4708 \, \cos \left (d x + c\right )^{4} - 3344 \, \cos \left (d x + c\right )^{2} - 105 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 105 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (105 \, \cos \left (d x + c\right )^{6} - 250 \, \cos \left (d x + c\right )^{4} + 184 \, \cos \left (d x + c\right )^{2} - 48\right )} \sin \left (d x + c\right ) + 864}{7680 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^7/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/7680*(210*cos(d*x + c)^8 - 2630*cos(d*x + c)^6 + 4708*cos(d*x + c)^4 - 3344*cos(d*x + c)^2 - 105*(cos(d*x +
c)^8*sin(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 - 250*cos(d*x + c)^4 + 184*cos(d*x + c)^2 - 48)*sin(d*x + c) +
864)/(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 = 0.35, size = 156, normalized size = 0.88 \[ -\frac {\frac {420 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {420 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {5 \, {\left (175 \, \sin \left (d x + c\right )^{4} - 868 \, \sin \left (d x + c\right )^{3} + 1302 \, \sin \left (d x + c\right )^{2} - 828 \, \sin \left (d x + c\right ) + 195\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac {959 \, \sin \left (d x + c\right )^{5} + 4795 \, \sin \left (d x + c\right )^{4} + 7490 \, \sin \left (d x + c\right )^{3} + 5610 \, \sin \left (d x + c\right )^{2} + 2055 \, \sin \left (d x + c\right ) + 291}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{30720 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^7/(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 - 868*si
n(d*x + c)^3 + 1302*sin(d*x + c)^2 - 828*sin(d*x + c) + 195)/(a*(sin(d*x + c) - 1)^4) - (959*sin(d*x + c)^5 +
4795*sin(d*x + c)^4 + 7490*sin(d*x + c)^3 + 5610*sin(d*x + c)^2 + 2055*sin(d*x + c) + 291)/(a*(sin(d*x + c) +
1)^5))/d

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maple [A]  time = 0.40, size = 180, normalized size = 1.01 \[ \frac {1}{256 a d \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {1}{48 a d \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {21}{512 a d \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {7}{256 a d \left (\sin \left (d x +c \right )-1\right )}+\frac {7 \ln \left (\sin \left (d x +c \right )-1\right )}{512 a d}+\frac {1}{160 a d \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {9}{256 a d \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {29}{384 a d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {35}{512 a d \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {7 \ln \left (1+\sin \left (d x +c \right )\right )}{512 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*sin(d*x+c)^7/(a+a*sin(d*x+c)),x)

[Out]

1/256/a/d/(sin(d*x+c)-1)^4+1/48/a/d/(sin(d*x+c)-1)^3+21/512/a/d/(sin(d*x+c)-1)^2+7/256/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))^5-9/256/a/d/(1+sin(d*x+c))^4+29/384/a/d/(1+sin(d*x+c))^3-35/51
2/a/d/(1+sin(d*x+c))^2-7/512*ln(1+sin(d*x+c))/a/d

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maxima [A]  time = 0.41, size = 214, normalized size = 1.20 \[ \frac {\frac {2 \, {\left (105 \, \sin \left (d x + c\right )^{8} + 105 \, \sin \left (d x + c\right )^{7} + 895 \, \sin \left (d x + c\right )^{6} - 65 \, \sin \left (d x + c\right )^{5} - 961 \, \sin \left (d x + c\right )^{4} - \sin \left (d x + c\right )^{3} + 489 \, \sin \left (d x + c\right )^{2} + 9 \, \sin \left (d x + c\right ) - 96\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {105 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {105 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{7680 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^7/(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 + 895*sin(d*x + c)^6 - 65*sin(d*x + c)^5 - 961*sin(d*x + c)
^4 - sin(d*x + c)^3 + 489*sin(d*x + c)^2 + 9*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*sin(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 = 16.70, size = 496, normalized size = 2.79 \[ \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 {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{96}-\frac {161\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{192}+\frac {469\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{480}+\frac {2681\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{960}-\frac {593\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{480}+\frac {25667\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{960}+\frac {1447\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {25667\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{960}-\frac {593\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{480}+\frac {2681\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{960}+\frac {469\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{480}-\frac {161\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{192}-\frac {35\,{\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)^7/(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 - (35*tan(c/2 + (d*x)/2)^3)/96 - (161*tan(c/2 + (d*x
)/2)^4)/192 + (469*tan(c/2 + (d*x)/2)^5)/480 + (2681*tan(c/2 + (d*x)/2)^6)/960 - (593*tan(c/2 + (d*x)/2)^7)/48
0 + (25667*tan(c/2 + (d*x)/2)^8)/960 + (1447*tan(c/2 + (d*x)/2)^9)/192 + (25667*tan(c/2 + (d*x)/2)^10)/960 - (
593*tan(c/2 + (d*x)/2)^11)/480 + (2681*tan(c/2 + (d*x)/2)^12)/960 + (469*tan(c/2 + (d*x)/2)^13)/480 - (161*tan
(c/2 + (d*x)/2)^14)/192 - (35*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)**7/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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