3.902 \(\int \frac {\sec ^4(c+d x) \tan ^5(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=194 \[ \frac {\sec ^{10}(c+d x)}{10 a d}-\frac {\sec ^8(c+d x)}{4 a d}+\frac {\sec ^6(c+d x)}{6 a d}+\frac {3 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 a d}+\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{16 a d}-\frac {\tan (c+d x) \sec ^5(c+d x)}{32 a d}+\frac {\tan (c+d x) \sec ^3(c+d x)}{128 a d}+\frac {3 \tan (c+d x) \sec (c+d x)}{256 a d} \]

[Out]

3/256*arctanh(sin(d*x+c))/a/d+1/6*sec(d*x+c)^6/a/d-1/4*sec(d*x+c)^8/a/d+1/10*sec(d*x+c)^10/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

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Rubi [A]  time = 0.27, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2835, 2606, 266, 43, 2611, 3768, 3770} \[ \frac {\sec ^{10}(c+d x)}{10 a d}-\frac {\sec ^8(c+d x)}{4 a d}+\frac {\sec ^6(c+d x)}{6 a d}+\frac {3 \tanh ^{-1}(\sin (c+d x))}{256 a d}-\frac {\tan ^5(c+d x) \sec ^5(c+d x)}{10 a d}+\frac {\tan ^3(c+d x) \sec ^5(c+d x)}{16 a d}-\frac {\tan (c+d x) \sec ^5(c+d x)}{32 a d}+\frac {\tan (c+d x) \sec ^3(c+d x)}{128 a d}+\frac {3 \tan (c+d x) \sec (c+d x)}{256 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^4*Tan[c + d*x]^5)/(a + a*Sin[c + d*x]),x]

[Out]

(3*ArcTanh[Sin[c + d*x]])/(256*a*d) + Sec[c + d*x]^6/(6*a*d) - Sec[c + d*x]^8/(4*a*d) + Sec[c + d*x]^10/(10*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)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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

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Mathematica [A]  time = 5.49, size = 116, normalized size = 0.60 \[ \frac {-\frac {90}{\sin (c+d x)-1}+\frac {15}{(\sin (c+d x)-1)^2}+\frac {75}{(\sin (c+d x)+1)^2}+\frac {80}{(\sin (c+d x)-1)^3}+\frac {100}{(\sin (c+d x)+1)^3}+\frac {30}{(\sin (c+d x)-1)^4}-\frac {150}{(\sin (c+d x)+1)^4}+\frac {48}{(\sin (c+d x)+1)^5}+90 \tanh ^{-1}(\sin (c+d x))}{7680 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^4*Tan[c + d*x]^5)/(a + a*Sin[c + d*x]),x]

[Out]

(90*ArcTanh[Sin[c + d*x]] + 30/(-1 + Sin[c + d*x])^4 + 80/(-1 + Sin[c + d*x])^3 + 15/(-1 + Sin[c + d*x])^2 - 9
0/(-1 + Sin[c + d*x]) + 48/(1 + Sin[c + d*x])^5 - 150/(1 + Sin[c + d*x])^4 + 100/(1 + Sin[c + d*x])^3 + 75/(1
+ Sin[c + d*x])^2)/(7680*a*d)

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fricas [A]  time = 0.51, size = 187, normalized size = 0.96 \[ -\frac {90 \, \cos \left (d x + c\right )^{8} - 30 \, \cos \left (d x + c\right )^{6} - 1548 \, \cos \left (d x + c\right )^{4} + 2224 \, \cos \left (d x + c\right )^{2} - 45 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 45 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (45 \, \cos \left (d x + c\right )^{6} + 30 \, \cos \left (d x + c\right )^{4} - 104 \, \cos \left (d x + c\right )^{2} + 48\right )} \sin \left (d x + c\right ) - 864}{7680 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/7680*(90*cos(d*x + c)^8 - 30*cos(d*x + c)^6 - 1548*cos(d*x + c)^4 + 2224*cos(d*x + c)^2 - 45*(cos(d*x + c)^
8*sin(d*x + c) + cos(d*x + c)^8)*log(sin(d*x + c) + 1) + 45*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*log
(-sin(d*x + c) + 1) - 2*(45*cos(d*x + c)^6 + 30*cos(d*x + c)^4 - 104*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 = 0.32, size = 156, normalized size = 0.80 \[ \frac {\frac {180 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {180 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {5 \, {\left (75 \, \sin \left (d x + c\right )^{4} - 372 \, \sin \left (d x + c\right )^{3} + 678 \, \sin \left (d x + c\right )^{2} - 476 \, \sin \left (d x + c\right ) + 119\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac {411 \, \sin \left (d x + c\right )^{5} + 2055 \, \sin \left (d x + c\right )^{4} + 3810 \, \sin \left (d x + c\right )^{3} + 2810 \, \sin \left (d x + c\right )^{2} + 955 \, \sin \left (d x + c\right ) + 119}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{30720 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/30720*(180*log(abs(sin(d*x + c) + 1))/a - 180*log(abs(sin(d*x + c) - 1))/a + 5*(75*sin(d*x + c)^4 - 372*sin(
d*x + c)^3 + 678*sin(d*x + c)^2 - 476*sin(d*x + c) + 119)/(a*(sin(d*x + c) - 1)^4) - (411*sin(d*x + c)^5 + 205
5*sin(d*x + c)^4 + 3810*sin(d*x + c)^3 + 2810*sin(d*x + c)^2 + 955*sin(d*x + c) + 119)/(a*(sin(d*x + c) + 1)^5
))/d

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maple [A]  time = 0.40, size = 180, normalized size = 0.93 \[ \frac {1}{256 a d \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {1}{96 a d \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {1}{512 a d \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {3}{256 a d \left (\sin \left (d x +c \right )-1\right )}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{512 a d}+\frac {1}{160 a d \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {5}{256 a d \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {5}{384 a d \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {5}{512 a d \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{512 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*sin(d*x+c)^5/(a+a*sin(d*x+c)),x)

[Out]

1/256/a/d/(sin(d*x+c)-1)^4+1/96/a/d/(sin(d*x+c)-1)^3+1/512/a/d/(sin(d*x+c)-1)^2-3/256/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))^5-5/256/a/d/(1+sin(d*x+c))^4+5/384/a/d/(1+sin(d*x+c))^3+5/512/a
/d/(1+sin(d*x+c))^2+3/512*ln(1+sin(d*x+c))/a/d

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maxima [A]  time = 0.48, size = 214, normalized size = 1.10 \[ -\frac {\frac {2 \, {\left (45 \, \sin \left (d x + c\right )^{8} + 45 \, \sin \left (d x + c\right )^{7} - 165 \, \sin \left (d x + c\right )^{6} - 165 \, \sin \left (d x + c\right )^{5} - 549 \, \sin \left (d x + c\right )^{4} + 91 \, \sin \left (d x + c\right )^{3} + 301 \, \sin \left (d x + c\right )^{2} - 19 \, \sin \left (d x + c\right ) - 64\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {45 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {45 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{7680 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^5/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/7680*(2*(45*sin(d*x + c)^8 + 45*sin(d*x + c)^7 - 165*sin(d*x + c)^6 - 165*sin(d*x + c)^5 - 549*sin(d*x + c)
^4 + 91*sin(d*x + c)^3 + 301*sin(d*x + c)^2 - 19*sin(d*x + c) - 64)/(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) - 45*log(sin(d*x + c) + 1)/a + 45*log(sin(d*x + c) - 1)/a)/d

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mupad [B]  time = 16.62, size = 496, normalized size = 2.56 \[ \frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{128\,a\,d}+\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 {9091\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{960}+\frac {1717\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{480}+\frac {18257\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{960}-\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {18257\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{960}+\frac {1717\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{480}+\frac {9091\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{960}-\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 )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^5/(cos(c + d*x)^9*(a + a*sin(c + d*x))),x)

[Out]

(3*atanh(tan(c/2 + (d*x)/2)))/(128*a*d) + ((5*tan(c/2 + (d*x)/2)^3)/32 - (3*tan(c/2 + (d*x)/2)^2)/64 - (3*tan(
c/2 + (d*x)/2))/128 + (23*tan(c/2 + (d*x)/2)^4)/64 - (67*tan(c/2 + (d*x)/2)^5)/160 + (9091*tan(c/2 + (d*x)/2)^
6)/960 + (1717*tan(c/2 + (d*x)/2)^7)/480 + (18257*tan(c/2 + (d*x)/2)^8)/960 - (35*tan(c/2 + (d*x)/2)^9)/192 +
(18257*tan(c/2 + (d*x)/2)^10)/960 + (1717*tan(c/2 + (d*x)/2)^11)/480 + (9091*tan(c/2 + (d*x)/2)^12)/960 - (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*t
an(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))

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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)**5/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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