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

Optimal. Leaf size=188 \[ -\frac{a^3}{64 d (a \sin (c+d x)+a)^4}+\frac{a^2}{96 d (a-a \sin (c+d x))^3}+\frac{a^2}{8 d (a \sin (c+d x)+a)^3}-\frac{11 a}{128 d (a-a \sin (c+d x))^2}-\frac{29 a}{64 d (a \sin (c+d x)+a)^2}+\frac{47}{128 d (a-a \sin (c+d x))}+\frac{35}{32 d (a \sin (c+d x)+a)}+\frac{93 \log (1-\sin (c+d x))}{256 a d}+\frac{163 \log (\sin (c+d x)+1)}{256 a d} \]

[Out]

(93*Log[1 - Sin[c + d*x]])/(256*a*d) + (163*Log[1 + Sin[c + d*x]])/(256*a*d) + a^2/(96*d*(a - a*Sin[c + d*x])^
3) - (11*a)/(128*d*(a - a*Sin[c + d*x])^2) + 47/(128*d*(a - a*Sin[c + d*x])) - a^3/(64*d*(a + a*Sin[c + d*x])^
4) + a^2/(8*d*(a + a*Sin[c + d*x])^3) - (29*a)/(64*d*(a + a*Sin[c + d*x])^2) + 35/(32*d*(a + a*Sin[c + d*x]))

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Rubi [A]  time = 0.183577, antiderivative size = 188, normalized size of antiderivative = 1., 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} \[ -\frac{a^3}{64 d (a \sin (c+d x)+a)^4}+\frac{a^2}{96 d (a-a \sin (c+d x))^3}+\frac{a^2}{8 d (a \sin (c+d x)+a)^3}-\frac{11 a}{128 d (a-a \sin (c+d x))^2}-\frac{29 a}{64 d (a \sin (c+d x)+a)^2}+\frac{47}{128 d (a-a \sin (c+d x))}+\frac{35}{32 d (a \sin (c+d x)+a)}+\frac{93 \log (1-\sin (c+d x))}{256 a d}+\frac{163 \log (\sin (c+d x)+1)}{256 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(93*Log[1 - Sin[c + d*x]])/(256*a*d) + (163*Log[1 + Sin[c + d*x]])/(256*a*d) + a^2/(96*d*(a - a*Sin[c + d*x])^
3) - (11*a)/(128*d*(a - a*Sin[c + d*x])^2) + 47/(128*d*(a - a*Sin[c + d*x])) - a^3/(64*d*(a + a*Sin[c + d*x])^
4) + a^2/(8*d*(a + a*Sin[c + d*x])^3) - (29*a)/(64*d*(a + a*Sin[c + d*x])^2) + 35/(32*d*(a + a*Sin[c + d*x]))

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]

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]))

Rubi steps

\begin{align*} \int \frac{\sin (c+d x) \tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{a^7 \operatorname{Subst}\left (\int \frac{x^8}{a^8 (a-x)^4 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^8}{(a-x)^4 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a^3}{32 (a-x)^4}-\frac{11 a^2}{64 (a-x)^3}+\frac{47 a}{128 (a-x)^2}-\frac{93}{256 (a-x)}+\frac{a^4}{16 (a+x)^5}-\frac{3 a^3}{8 (a+x)^4}+\frac{29 a^2}{32 (a+x)^3}-\frac{35 a}{32 (a+x)^2}+\frac{163}{256 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac{93 \log (1-\sin (c+d x))}{256 a d}+\frac{163 \log (1+\sin (c+d x))}{256 a d}+\frac{a^2}{96 d (a-a \sin (c+d x))^3}-\frac{11 a}{128 d (a-a \sin (c+d x))^2}+\frac{47}{128 d (a-a \sin (c+d x))}-\frac{a^3}{64 d (a+a \sin (c+d x))^4}+\frac{a^2}{8 d (a+a \sin (c+d x))^3}-\frac{29 a}{64 d (a+a \sin (c+d x))^2}+\frac{35}{32 d (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 3.95042, size = 117, normalized size = 0.62 \[ \frac{\frac{2 \left (279 \sin ^6(c+d x)-489 \sin ^5(c+d x)-1000 \sin ^4(c+d x)+728 \sin ^3(c+d x)+1113 \sin ^2(c+d x)-295 \sin (c+d x)-400\right )}{(\sin (c+d x)-1)^3 (\sin (c+d x)+1)^4}+279 \log (1-\sin (c+d x))+489 \log (\sin (c+d x)+1)}{768 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(279*Log[1 - Sin[c + d*x]] + 489*Log[1 + Sin[c + d*x]] + (2*(-400 - 295*Sin[c + d*x] + 1113*Sin[c + d*x]^2 + 7
28*Sin[c + d*x]^3 - 1000*Sin[c + d*x]^4 - 489*Sin[c + d*x]^5 + 279*Sin[c + d*x]^6))/((-1 + Sin[c + d*x])^3*(1
+ Sin[c + d*x])^4))/(768*a*d)

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Maple [A]  time = 0.104, size = 162, normalized size = 0.9 \begin{align*} -{\frac{1}{96\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3}}}-{\frac{11}{128\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}-{\frac{47}{128\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{93\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{256\,da}}-{\frac{1}{64\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{1}{8\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{29}{64\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{35}{32\,da \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{163\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{256\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*sin(d*x+c)^8/(a+a*sin(d*x+c)),x)

[Out]

-1/96/d/a/(sin(d*x+c)-1)^3-11/128/d/a/(sin(d*x+c)-1)^2-47/128/a/d/(sin(d*x+c)-1)+93/256/a/d*ln(sin(d*x+c)-1)-1
/64/d/a/(1+sin(d*x+c))^4+1/8/d/a/(1+sin(d*x+c))^3-29/64/a/d/(1+sin(d*x+c))^2+35/32/a/d/(1+sin(d*x+c))+163/256*
ln(1+sin(d*x+c))/a/d

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Maxima [A]  time = 1.13474, size = 236, normalized size = 1.26 \begin{align*} \frac{\frac{2 \,{\left (279 \, \sin \left (d x + c\right )^{6} - 489 \, \sin \left (d x + c\right )^{5} - 1000 \, \sin \left (d x + c\right )^{4} + 728 \, \sin \left (d x + c\right )^{3} + 1113 \, \sin \left (d x + c\right )^{2} - 295 \, \sin \left (d x + c\right ) - 400\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{489 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac{279 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{768 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^8/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/768*(2*(279*sin(d*x + c)^6 - 489*sin(d*x + c)^5 - 1000*sin(d*x + c)^4 + 728*sin(d*x + c)^3 + 1113*sin(d*x +
c)^2 - 295*sin(d*x + c) - 400)/(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) + 489*log(sin(d*x + c) + 1)/a + 279*log(sin(d*
x + c) - 1)/a)/d

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Fricas [A]  time = 1.59043, size = 466, normalized size = 2.48 \begin{align*} \frac{558 \, \cos \left (d x + c\right )^{6} + 326 \, \cos \left (d x + c\right )^{4} - 100 \, \cos \left (d x + c\right )^{2} + 489 \,{\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 ) + 279 \,{\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 (489 \, \cos \left (d x + c\right )^{4} - 250 \, \cos \left (d x + c\right )^{2} + 56\right )} \sin \left (d x + c\right ) + 16}{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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^8/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/768*(558*cos(d*x + c)^6 + 326*cos(d*x + c)^4 - 100*cos(d*x + c)^2 + 489*(cos(d*x + c)^6*sin(d*x + c) + cos(d
*x + c)^6)*log(sin(d*x + c) + 1) + 279*(cos(d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(-sin(d*x + c) + 1) +
2*(489*cos(d*x + c)^4 - 250*cos(d*x + c)^2 + 56)*sin(d*x + c) + 16)/(a*d*cos(d*x + c)^6*sin(d*x + c) + a*d*co
s(d*x + c)^6)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**7*sin(d*x+c)**8/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.36621, size = 184, normalized size = 0.98 \begin{align*} \frac{\frac{1956 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} + \frac{1116 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} - \frac{2 \,{\left (1023 \, \sin \left (d x + c\right )^{3} - 2505 \, \sin \left (d x + c\right )^{2} + 2073 \, \sin \left (d x + c\right ) - 575\right )}}{a{\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac{4075 \, \sin \left (d x + c\right )^{4} + 12940 \, \sin \left (d x + c\right )^{3} + 15762 \, \sin \left (d x + c\right )^{2} + 8620 \, \sin \left (d x + c\right ) + 1771}{a{\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{3072 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^8/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/3072*(1956*log(abs(sin(d*x + c) + 1))/a + 1116*log(abs(sin(d*x + c) - 1))/a - 2*(1023*sin(d*x + c)^3 - 2505*
sin(d*x + c)^2 + 2073*sin(d*x + c) - 575)/(a*(sin(d*x + c) - 1)^3) - (4075*sin(d*x + c)^4 + 12940*sin(d*x + c)
^3 + 15762*sin(d*x + c)^2 + 8620*sin(d*x + c) + 1771)/(a*(sin(d*x + c) + 1)^4))/d