3.897 \(\int \frac {\sin (c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=233 \[ -\frac {a^4}{160 d (a \sin (c+d x)+a)^5}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}+\frac {15 a^3}{256 d (a \sin (c+d x)+a)^4}-\frac {7 a^2}{192 d (a-a \sin (c+d x))^3}-\frac {95 a^2}{384 d (a \sin (c+d x)+a)^3}+\frac {81 a}{512 d (a-a \sin (c+d x))^2}+\frac {325 a}{512 d (a \sin (c+d x)+a)^2}-\frac {61}{128 d (a-a \sin (c+d x))}-\frac {315}{256 d (a \sin (c+d x)+a)}-\frac {193 \log (1-\sin (c+d x))}{512 a d}-\frac {319 \log (\sin (c+d x)+1)}{512 a d} \]

[Out]

-193/512*ln(1-sin(d*x+c))/a/d-319/512*ln(1+sin(d*x+c))/a/d+1/256*a^3/d/(a-a*sin(d*x+c))^4-7/192*a^2/d/(a-a*sin
(d*x+c))^3+81/512*a/d/(a-a*sin(d*x+c))^2-61/128/d/(a-a*sin(d*x+c))-1/160*a^4/d/(a+a*sin(d*x+c))^5+15/256*a^3/d
/(a+a*sin(d*x+c))^4-95/384*a^2/d/(a+a*sin(d*x+c))^3+325/512*a/d/(a+a*sin(d*x+c))^2-315/256/d/(a+a*sin(d*x+c))

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Rubi [A]  time = 0.23, antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2836, 12, 88} \[ -\frac {a^4}{160 d (a \sin (c+d x)+a)^5}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}+\frac {15 a^3}{256 d (a \sin (c+d x)+a)^4}-\frac {7 a^2}{192 d (a-a \sin (c+d x))^3}-\frac {95 a^2}{384 d (a \sin (c+d x)+a)^3}+\frac {81 a}{512 d (a-a \sin (c+d x))^2}+\frac {325 a}{512 d (a \sin (c+d x)+a)^2}-\frac {61}{128 d (a-a \sin (c+d x))}-\frac {315}{256 d (a \sin (c+d x)+a)}-\frac {193 \log (1-\sin (c+d x))}{512 a d}-\frac {319 \log (\sin (c+d x)+1)}{512 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[c + d*x]*Tan[c + d*x]^9)/(a + a*Sin[c + d*x]),x]

[Out]

(-193*Log[1 - Sin[c + d*x]])/(512*a*d) - (319*Log[1 + Sin[c + d*x]])/(512*a*d) + a^3/(256*d*(a - a*Sin[c + d*x
])^4) - (7*a^2)/(192*d*(a - a*Sin[c + d*x])^3) + (81*a)/(512*d*(a - a*Sin[c + d*x])^2) - 61/(128*d*(a - a*Sin[
c + d*x])) - a^4/(160*d*(a + a*Sin[c + d*x])^5) + (15*a^3)/(256*d*(a + a*Sin[c + d*x])^4) - (95*a^2)/(384*d*(a
+ a*Sin[c + d*x])^3) + (325*a)/(512*d*(a + a*Sin[c + d*x])^2) - 315/(256*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 (c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {a^9 \operatorname {Subst}\left (\int \frac {x^{10}}{a^{10} (a-x)^5 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^{10}}{(a-x)^5 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {a^4}{64 (a-x)^5}-\frac {7 a^3}{64 (a-x)^4}+\frac {81 a^2}{256 (a-x)^3}-\frac {61 a}{128 (a-x)^2}+\frac {193}{512 (a-x)}+\frac {a^5}{32 (a+x)^6}-\frac {15 a^4}{64 (a+x)^5}+\frac {95 a^3}{128 (a+x)^4}-\frac {325 a^2}{256 (a+x)^3}+\frac {315 a}{256 (a+x)^2}-\frac {319}{512 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=-\frac {193 \log (1-\sin (c+d x))}{512 a d}-\frac {319 \log (1+\sin (c+d x))}{512 a d}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}-\frac {7 a^2}{192 d (a-a \sin (c+d x))^3}+\frac {81 a}{512 d (a-a \sin (c+d x))^2}-\frac {61}{128 d (a-a \sin (c+d x))}-\frac {a^4}{160 d (a+a \sin (c+d x))^5}+\frac {15 a^3}{256 d (a+a \sin (c+d x))^4}-\frac {95 a^2}{384 d (a+a \sin (c+d x))^3}+\frac {325 a}{512 d (a+a \sin (c+d x))^2}-\frac {315}{256 d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 4.97, size = 137, normalized size = 0.59 \[ -\frac {\frac {2 \left (2895 \sin ^8(c+d x)-6705 \sin ^7(c+d x)-13815 \sin ^6(c+d x)+14985 \sin ^5(c+d x)+23049 \sin ^4(c+d x)-12151 \sin ^3(c+d x)-16561 \sin ^2(c+d x)+3439 \sin (c+d x)+4384\right )}{(\sin (c+d x)-1)^4 (\sin (c+d x)+1)^5}+2895 \log (1-\sin (c+d x))+4785 \log (\sin (c+d x)+1)}{7680 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[c + d*x]*Tan[c + d*x]^9)/(a + a*Sin[c + d*x]),x]

[Out]

-1/7680*(2895*Log[1 - Sin[c + d*x]] + 4785*Log[1 + Sin[c + d*x]] + (2*(4384 + 3439*Sin[c + d*x] - 16561*Sin[c
+ d*x]^2 - 12151*Sin[c + d*x]^3 + 23049*Sin[c + d*x]^4 + 14985*Sin[c + d*x]^5 - 13815*Sin[c + d*x]^6 - 6705*Si
n[c + d*x]^7 + 2895*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 = 0.54, size = 187, normalized size = 0.80 \[ -\frac {5790 \, \cos \left (d x + c\right )^{8} + 4470 \, \cos \left (d x + c\right )^{6} - 2052 \, \cos \left (d x + c\right )^{4} + 656 \, \cos \left (d x + c\right )^{2} + 4785 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2895 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6705 \, \cos \left (d x + c\right )^{6} - 5130 \, \cos \left (d x + c\right )^{4} + 2296 \, \cos \left (d x + c\right )^{2} - 432\right )} \sin \left (d x + c\right ) - 96}{7680 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^10/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/7680*(5790*cos(d*x + c)^8 + 4470*cos(d*x + c)^6 - 2052*cos(d*x + c)^4 + 656*cos(d*x + c)^2 + 4785*(cos(d*x
+ c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(sin(d*x + c) + 1) + 2895*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)
^8)*log(-sin(d*x + c) + 1) + 2*(6705*cos(d*x + c)^6 - 5130*cos(d*x + c)^4 + 2296*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 = 0.37, size = 156, normalized size = 0.67 \[ -\frac {\frac {19140 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} + \frac {11580 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} - \frac {5 \, {\left (4825 \, \sin \left (d x + c\right )^{4} - 16372 \, \sin \left (d x + c\right )^{3} + 21138 \, \sin \left (d x + c\right )^{2} - 12236 \, \sin \left (d x + c\right ) + 2669\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac {43703 \, \sin \left (d x + c\right )^{5} + 180715 \, \sin \left (d x + c\right )^{4} + 305330 \, \sin \left (d x + c\right )^{3} + 261130 \, \sin \left (d x + c\right )^{2} + 112415 \, \sin \left (d x + c\right ) + 19411}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{30720 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^10/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/30720*(19140*log(abs(sin(d*x + c) + 1))/a + 11580*log(abs(sin(d*x + c) - 1))/a - 5*(4825*sin(d*x + c)^4 - 1
6372*sin(d*x + c)^3 + 21138*sin(d*x + c)^2 - 12236*sin(d*x + c) + 2669)/(a*(sin(d*x + c) - 1)^4) - (43703*sin(
d*x + c)^5 + 180715*sin(d*x + c)^4 + 305330*sin(d*x + c)^3 + 261130*sin(d*x + c)^2 + 112415*sin(d*x + c) + 194
11)/(a*(sin(d*x + c) + 1)^5))/d

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maple [A]  time = 0.48, size = 198, normalized size = 0.85 \[ \frac {1}{256 a d \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {7}{192 a d \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {81}{512 a d \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {61}{128 a d \left (\sin \left (d x +c \right )-1\right )}-\frac {193 \ln \left (\sin \left (d x +c \right )-1\right )}{512 a d}-\frac {1}{160 a d \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {15}{256 a d \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {95}{384 a d \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {325}{512 a d \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {315}{256 a d \left (1+\sin \left (d x +c \right )\right )}-\frac {319 \ln \left (1+\sin \left (d x +c \right )\right )}{512 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*sin(d*x+c)^10/(a+a*sin(d*x+c)),x)

[Out]

1/256/a/d/(sin(d*x+c)-1)^4+7/192/a/d/(sin(d*x+c)-1)^3+81/512/a/d/(sin(d*x+c)-1)^2+61/128/a/d/(sin(d*x+c)-1)-19
3/512/a/d*ln(sin(d*x+c)-1)-1/160/a/d/(1+sin(d*x+c))^5+15/256/a/d/(1+sin(d*x+c))^4-95/384/a/d/(1+sin(d*x+c))^3+
325/512/a/d/(1+sin(d*x+c))^2-315/256/a/d/(1+sin(d*x+c))-319/512*ln(1+sin(d*x+c))/a/d

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maxima [A]  time = 0.53, size = 214, normalized size = 0.92 \[ -\frac {\frac {2 \, {\left (2895 \, \sin \left (d x + c\right )^{8} - 6705 \, \sin \left (d x + c\right )^{7} - 13815 \, \sin \left (d x + c\right )^{6} + 14985 \, \sin \left (d x + c\right )^{5} + 23049 \, \sin \left (d x + c\right )^{4} - 12151 \, \sin \left (d x + c\right )^{3} - 16561 \, \sin \left (d x + c\right )^{2} + 3439 \, \sin \left (d x + c\right ) + 4384\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} + \frac {4785 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {2895 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{7680 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^10/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/7680*(2*(2895*sin(d*x + c)^8 - 6705*sin(d*x + c)^7 - 13815*sin(d*x + c)^6 + 14985*sin(d*x + c)^5 + 23049*si
n(d*x + c)^4 - 12151*sin(d*x + c)^3 - 16561*sin(d*x + c)^2 + 3439*sin(d*x + c) + 4384)/(a*sin(d*x + c)^9 + a*s
in(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) + 4785*log(sin(d*x + c) + 1)/a + 2895*log(sin(d*x + c) - 1
)/a)/d

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mupad [B]  time = 9.40, size = 539, normalized size = 2.31 \[ \frac {\frac {63\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{128}-\frac {65\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}}{64}-\frac {233\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{32}+\frac {413\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{64}+\frac {6527\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{160}-\frac {14911\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{960}-\frac {59737\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{480}+\frac {12763\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{960}+\frac {45791\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {12763\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{960}-\frac {59737\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{480}-\frac {14911\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{960}+\frac {6527\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160}+\frac {413\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64}-\frac {233\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{32}-\frac {65\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{64}+\frac {63\,\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 {193\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{256\,a\,d}-\frac {319\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{256\,a\,d}+\frac {\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)^10/(cos(c + d*x)^9*(a + a*sin(c + d*x))),x)

[Out]

((63*tan(c/2 + (d*x)/2))/128 - (65*tan(c/2 + (d*x)/2)^2)/64 - (233*tan(c/2 + (d*x)/2)^3)/32 + (413*tan(c/2 + (
d*x)/2)^4)/64 + (6527*tan(c/2 + (d*x)/2)^5)/160 - (14911*tan(c/2 + (d*x)/2)^6)/960 - (59737*tan(c/2 + (d*x)/2)
^7)/480 + (12763*tan(c/2 + (d*x)/2)^8)/960 + (45791*tan(c/2 + (d*x)/2)^9)/192 + (12763*tan(c/2 + (d*x)/2)^10)/
960 - (59737*tan(c/2 + (d*x)/2)^11)/480 - (14911*tan(c/2 + (d*x)/2)^12)/960 + (6527*tan(c/2 + (d*x)/2)^13)/160
+ (413*tan(c/2 + (d*x)/2)^14)/64 - (233*tan(c/2 + (d*x)/2)^15)/32 - (65*tan(c/2 + (d*x)/2)^16)/64 + (63*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)) - (193*log(t
an(c/2 + (d*x)/2) - 1))/(256*a*d) - (319*log(tan(c/2 + (d*x)/2) + 1))/(256*a*d) + log(tan(c/2 + (d*x)/2)^2 + 1
)/(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)**10/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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