3.878 \(\int \frac {\sin ^4(c+d x) \tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=236 \[ \frac {a^3}{64 d (a \sin (c+d x)+a)^4}+\frac {a^2}{96 d (a-a \sin (c+d x))^3}-\frac {3 a^2}{16 d (a \sin (c+d x)+a)^3}+\frac {\sin ^3(c+d x)}{3 a d}-\frac {\sin ^2(c+d x)}{2 a d}-\frac {17 a}{128 d (a-a \sin (c+d x))^2}+\frac {71 a}{64 d (a \sin (c+d x)+a)^2}+\frac {125}{128 d (a-a \sin (c+d x))}-\frac {5}{d (a \sin (c+d x)+a)}+\frac {5 \sin (c+d x)}{a d}+\frac {515 \log (1-\sin (c+d x))}{256 a d}-\frac {1795 \log (\sin (c+d x)+1)}{256 a d} \]

[Out]

515/256*ln(1-sin(d*x+c))/a/d-1795/256*ln(1+sin(d*x+c))/a/d+5*sin(d*x+c)/a/d-1/2*sin(d*x+c)^2/a/d+1/3*sin(d*x+c
)^3/a/d+1/96*a^2/d/(a-a*sin(d*x+c))^3-17/128*a/d/(a-a*sin(d*x+c))^2+125/128/d/(a-a*sin(d*x+c))+1/64*a^3/d/(a+a
*sin(d*x+c))^4-3/16*a^2/d/(a+a*sin(d*x+c))^3+71/64*a/d/(a+a*sin(d*x+c))^2-5/d/(a+a*sin(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2836, 12, 88} \[ \frac {a^3}{64 d (a \sin (c+d x)+a)^4}+\frac {a^2}{96 d (a-a \sin (c+d x))^3}-\frac {3 a^2}{16 d (a \sin (c+d x)+a)^3}+\frac {\sin ^3(c+d x)}{3 a d}-\frac {\sin ^2(c+d x)}{2 a d}-\frac {17 a}{128 d (a-a \sin (c+d x))^2}+\frac {71 a}{64 d (a \sin (c+d x)+a)^2}+\frac {125}{128 d (a-a \sin (c+d x))}-\frac {5}{d (a \sin (c+d x)+a)}+\frac {5 \sin (c+d x)}{a d}+\frac {515 \log (1-\sin (c+d x))}{256 a d}-\frac {1795 \log (\sin (c+d x)+1)}{256 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[c + d*x]^4*Tan[c + d*x]^7)/(a + a*Sin[c + d*x]),x]

[Out]

(515*Log[1 - Sin[c + d*x]])/(256*a*d) - (1795*Log[1 + Sin[c + d*x]])/(256*a*d) + (5*Sin[c + d*x])/(a*d) - Sin[
c + d*x]^2/(2*a*d) + Sin[c + d*x]^3/(3*a*d) + a^2/(96*d*(a - a*Sin[c + d*x])^3) - (17*a)/(128*d*(a - a*Sin[c +
d*x])^2) + 125/(128*d*(a - a*Sin[c + d*x])) + a^3/(64*d*(a + a*Sin[c + d*x])^4) - (3*a^2)/(16*d*(a + a*Sin[c
+ d*x])^3) + (71*a)/(64*d*(a + a*Sin[c + d*x])^2) - 5/(d*(a + a*Sin[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2836

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

Rubi steps

\begin {align*} \int \frac {\sin ^4(c+d x) \tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {a^7 \operatorname {Subst}\left (\int \frac {x^{11}}{a^{11} (a-x)^4 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^{11}}{(a-x)^4 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{a^4 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (5 a^2+\frac {a^6}{32 (a-x)^4}-\frac {17 a^5}{64 (a-x)^3}+\frac {125 a^4}{128 (a-x)^2}-\frac {515 a^3}{256 (a-x)}-a x+x^2-\frac {a^7}{16 (a+x)^5}+\frac {9 a^6}{16 (a+x)^4}-\frac {71 a^5}{32 (a+x)^3}+\frac {5 a^4}{(a+x)^2}-\frac {1795 a^3}{256 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{a^4 d}\\ &=\frac {515 \log (1-\sin (c+d x))}{256 a d}-\frac {1795 \log (1+\sin (c+d x))}{256 a d}+\frac {5 \sin (c+d x)}{a d}-\frac {\sin ^2(c+d x)}{2 a d}+\frac {\sin ^3(c+d x)}{3 a d}+\frac {a^2}{96 d (a-a \sin (c+d x))^3}-\frac {17 a}{128 d (a-a \sin (c+d x))^2}+\frac {125}{128 d (a-a \sin (c+d x))}+\frac {a^3}{64 d (a+a \sin (c+d x))^4}-\frac {3 a^2}{16 d (a+a \sin (c+d x))^3}+\frac {71 a}{64 d (a+a \sin (c+d x))^2}-\frac {5}{d (a+a \sin (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 6.13, size = 153, normalized size = 0.65 \[ \frac {256 \sin ^3(c+d x)-384 \sin ^2(c+d x)+3840 \sin (c+d x)+\frac {750}{1-\sin (c+d x)}-\frac {3840}{\sin (c+d x)+1}-\frac {102}{(1-\sin (c+d x))^2}+\frac {852}{(\sin (c+d x)+1)^2}+\frac {8}{(1-\sin (c+d x))^3}-\frac {144}{(\sin (c+d x)+1)^3}+\frac {12}{(\sin (c+d x)+1)^4}+1545 \log (1-\sin (c+d x))-5385 \log (\sin (c+d x)+1)}{768 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[c + d*x]^4*Tan[c + d*x]^7)/(a + a*Sin[c + d*x]),x]

[Out]

(1545*Log[1 - Sin[c + d*x]] - 5385*Log[1 + Sin[c + d*x]] + 8/(1 - Sin[c + d*x])^3 - 102/(1 - Sin[c + d*x])^2 +
750/(1 - Sin[c + d*x]) + 3840*Sin[c + d*x] - 384*Sin[c + d*x]^2 + 256*Sin[c + d*x]^3 + 12/(1 + Sin[c + d*x])^
4 - 144/(1 + Sin[c + d*x])^3 + 852/(1 + Sin[c + d*x])^2 - 3840/(1 + Sin[c + d*x]))/(768*a*d)

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 207, normalized size = 0.88 \[ \frac {256 \, \cos \left (d x + c\right )^{10} - 3968 \, \cos \left (d x + c\right )^{8} - 686 \, \cos \left (d x + c\right )^{6} + 2810 \, \cos \left (d x + c\right )^{4} - 796 \, \cos \left (d x + c\right )^{2} - 5385 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 1545 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (64 \, \cos \left (d x + c\right )^{8} + 1952 \, \cos \left (d x + c\right )^{6} + 375 \, \cos \left (d x + c\right )^{4} - 70 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 112}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^11/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/768*(256*cos(d*x + c)^10 - 3968*cos(d*x + c)^8 - 686*cos(d*x + c)^6 + 2810*cos(d*x + c)^4 - 796*cos(d*x + c)
^2 - 5385*(cos(d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(sin(d*x + c) + 1) + 1545*(cos(d*x + c)^6*sin(d*x
+ c) + cos(d*x + c)^6)*log(-sin(d*x + c) + 1) + 2*(64*cos(d*x + c)^8 + 1952*cos(d*x + c)^6 + 375*cos(d*x + c)^
4 - 70*cos(d*x + c)^2 + 8)*sin(d*x + c) + 112)/(a*d*cos(d*x + c)^6*sin(d*x + c) + a*d*cos(d*x + c)^6)

________________________________________________________________________________________

giac [A]  time = 0.36, size = 179, normalized size = 0.76 \[ -\frac {\frac {21540 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {6180 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} - \frac {512 \, {\left (2 \, a^{2} \sin \left (d x + c\right )^{3} - 3 \, a^{2} \sin \left (d x + c\right )^{2} + 30 \, a^{2} \sin \left (d x + c\right )\right )}}{a^{3}} + \frac {2 \, {\left (5665 \, \sin \left (d x + c\right )^{3} - 15495 \, \sin \left (d x + c\right )^{2} + 14199 \, \sin \left (d x + c\right ) - 4353\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac {44875 \, \sin \left (d x + c\right )^{4} + 164140 \, \sin \left (d x + c\right )^{3} + 226578 \, \sin \left (d x + c\right )^{2} + 139660 \, \sin \left (d x + c\right ) + 32395}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{3072 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^11/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/3072*(21540*log(abs(sin(d*x + c) + 1))/a - 6180*log(abs(sin(d*x + c) - 1))/a - 512*(2*a^2*sin(d*x + c)^3 -
3*a^2*sin(d*x + c)^2 + 30*a^2*sin(d*x + c))/a^3 + 2*(5665*sin(d*x + c)^3 - 15495*sin(d*x + c)^2 + 14199*sin(d*
x + c) - 4353)/(a*(sin(d*x + c) - 1)^3) - (44875*sin(d*x + c)^4 + 164140*sin(d*x + c)^3 + 226578*sin(d*x + c)^
2 + 139660*sin(d*x + c) + 32395)/(a*(sin(d*x + c) + 1)^4))/d

________________________________________________________________________________________

maple [A]  time = 0.48, size = 208, normalized size = 0.88 \[ -\frac {1}{96 a d \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {17}{128 a d \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {125}{128 a d \left (\sin \left (d x +c \right )-1\right )}+\frac {515 \ln \left (\sin \left (d x +c \right )-1\right )}{256 a d}+\frac {\sin ^{3}\left (d x +c \right )}{3 d a}-\frac {\sin ^{2}\left (d x +c \right )}{2 a d}+\frac {5 \sin \left (d x +c \right )}{a d}+\frac {1}{64 a d \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {3}{16 a d \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {71}{64 a d \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{a d \left (1+\sin \left (d x +c \right )\right )}-\frac {1795 \ln \left (1+\sin \left (d x +c \right )\right )}{256 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*sin(d*x+c)^11/(a+a*sin(d*x+c)),x)

[Out]

-1/96/a/d/(sin(d*x+c)-1)^3-17/128/a/d/(sin(d*x+c)-1)^2-125/128/a/d/(sin(d*x+c)-1)+515/256/a/d*ln(sin(d*x+c)-1)
+1/3*sin(d*x+c)^3/d/a-1/2*sin(d*x+c)^2/a/d+5*sin(d*x+c)/a/d+1/64/a/d/(1+sin(d*x+c))^4-3/16/a/d/(1+sin(d*x+c))^
3+71/64/a/d/(1+sin(d*x+c))^2-5/a/d/(1+sin(d*x+c))-1795/256*ln(1+sin(d*x+c))/a/d

________________________________________________________________________________________

maxima [A]  time = 0.33, size = 209, normalized size = 0.89 \[ -\frac {\frac {2 \, {\left (2295 \, \sin \left (d x + c\right )^{6} + 375 \, \sin \left (d x + c\right )^{5} - 5480 \, \sin \left (d x + c\right )^{4} - 680 \, \sin \left (d x + c\right )^{3} + 4473 \, \sin \left (d x + c\right )^{2} + 313 \, \sin \left (d x + c\right ) - 1232\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} - \frac {128 \, {\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 30 \, \sin \left (d x + c\right )\right )}}{a} + \frac {5385 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {1545 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{768 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^11/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/768*(2*(2295*sin(d*x + c)^6 + 375*sin(d*x + c)^5 - 5480*sin(d*x + c)^4 - 680*sin(d*x + c)^3 + 4473*sin(d*x
+ c)^2 + 313*sin(d*x + c) - 1232)/(a*sin(d*x + c)^7 + a*sin(d*x + c)^6 - 3*a*sin(d*x + c)^5 - 3*a*sin(d*x + c)
^4 + 3*a*sin(d*x + c)^3 + 3*a*sin(d*x + c)^2 - a*sin(d*x + c) - a) - 128*(2*sin(d*x + c)^3 - 3*sin(d*x + c)^2
+ 30*sin(d*x + c))/a + 5385*log(sin(d*x + c) + 1)/a - 1545*log(sin(d*x + c) - 1)/a)/d

________________________________________________________________________________________

mupad [B]  time = 10.64, size = 567, normalized size = 2.40 \[ \frac {515\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{128\,a\,d}-\frac {1795\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{128\,a\,d}-\frac {-\frac {1155\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{19}}{64}-\frac {835\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}}{32}+\frac {3205\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{64}+\frac {305\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}}{4}+\frac {41\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{16}+\frac {53\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{24}-\frac {5521\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{48}-\frac {2387\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{12}+\frac {6697\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{96}+\frac {6901\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{48}+\frac {6697\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{96}-\frac {2387\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{12}-\frac {5521\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{48}+\frac {53\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{24}+\frac {41\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{16}+\frac {305\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{4}+\frac {3205\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{64}-\frac {835\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{32}-\frac {1155\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{20}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{19}-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}-6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}+8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+16\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+8\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,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 {5\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^11/(cos(c + d*x)^7*(a + a*sin(c + d*x))),x)

[Out]

(515*log(tan(c/2 + (d*x)/2) - 1))/(128*a*d) - (1795*log(tan(c/2 + (d*x)/2) + 1))/(128*a*d) - ((3205*tan(c/2 +
(d*x)/2)^3)/64 - (835*tan(c/2 + (d*x)/2)^2)/32 - (1155*tan(c/2 + (d*x)/2))/64 + (305*tan(c/2 + (d*x)/2)^4)/4 +
(41*tan(c/2 + (d*x)/2)^5)/16 + (53*tan(c/2 + (d*x)/2)^6)/24 - (5521*tan(c/2 + (d*x)/2)^7)/48 - (2387*tan(c/2
+ (d*x)/2)^8)/12 + (6697*tan(c/2 + (d*x)/2)^9)/96 + (6901*tan(c/2 + (d*x)/2)^10)/48 + (6697*tan(c/2 + (d*x)/2)
^11)/96 - (2387*tan(c/2 + (d*x)/2)^12)/12 - (5521*tan(c/2 + (d*x)/2)^13)/48 + (53*tan(c/2 + (d*x)/2)^14)/24 +
(41*tan(c/2 + (d*x)/2)^15)/16 + (305*tan(c/2 + (d*x)/2)^16)/4 + (3205*tan(c/2 + (d*x)/2)^17)/64 - (835*tan(c/2
+ (d*x)/2)^18)/32 - (1155*tan(c/2 + (d*x)/2)^19)/64)/(d*(a + 2*a*tan(c/2 + (d*x)/2) - 2*a*tan(c/2 + (d*x)/2)^
2 - 6*a*tan(c/2 + (d*x)/2)^3 - 3*a*tan(c/2 + (d*x)/2)^4 + 8*a*tan(c/2 + (d*x)/2)^6 + 16*a*tan(c/2 + (d*x)/2)^7
+ 2*a*tan(c/2 + (d*x)/2)^8 - 12*a*tan(c/2 + (d*x)/2)^9 - 12*a*tan(c/2 + (d*x)/2)^10 - 12*a*tan(c/2 + (d*x)/2)
^11 + 2*a*tan(c/2 + (d*x)/2)^12 + 16*a*tan(c/2 + (d*x)/2)^13 + 8*a*tan(c/2 + (d*x)/2)^14 - 3*a*tan(c/2 + (d*x)
/2)^16 - 6*a*tan(c/2 + (d*x)/2)^17 - 2*a*tan(c/2 + (d*x)/2)^18 + 2*a*tan(c/2 + (d*x)/2)^19 + a*tan(c/2 + (d*x)
/2)^20)) + (5*log(tan(c/2 + (d*x)/2)^2 + 1))/(a*d)

________________________________________________________________________________________

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)**7*sin(d*x+c)**11/(a+a*sin(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________