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]

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

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Rubi [A]  time = 0.18, antiderivative size = 188, 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} \[ -\frac {a^3}{64 d (a \sin (c+d x)+a)^4}+\frac {a^2}{96 d (a-a \sin (c+d x))^3}+\frac {a^2}{8 d (a \sin (c+d x)+a)^3}-\frac {11 a}{128 d (a-a \sin (c+d x))^2}-\frac {29 a}{64 d (a \sin (c+d x)+a)^2}+\frac {47}{128 d (a-a \sin (c+d x))}+\frac {35}{32 d (a \sin (c+d x)+a)}+\frac {93 \log (1-\sin (c+d x))}{256 a d}+\frac {163 \log (\sin (c+d x)+1)}{256 a d} \]

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 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 ^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*}

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Mathematica [A]  time = 3.78, 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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fricas [A]  time = 0.53, size = 167, normalized size = 0.89 \[ \frac {558 \, \cos \left (d x + c\right )^{6} + 326 \, \cos \left (d x + c\right )^{4} - 100 \, \cos \left (d x + c\right )^{2} + 489 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 279 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (489 \, \cos \left (d x + c\right )^{4} - 250 \, \cos \left (d x + c\right )^{2} + 56\right )} \sin \left (d x + c\right ) + 16}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \]

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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giac [A]  time = 0.36, size = 136, normalized size = 0.72 \[ \frac {\frac {1956 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} + \frac {1116 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} - \frac {2 \, {\left (1023 \, \sin \left (d x + c\right )^{3} - 2505 \, \sin \left (d x + c\right )^{2} + 2073 \, \sin \left (d x + c\right ) - 575\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac {4075 \, \sin \left (d x + c\right )^{4} + 12940 \, \sin \left (d x + c\right )^{3} + 15762 \, \sin \left (d x + c\right )^{2} + 8620 \, \sin \left (d x + c\right ) + 1771}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{3072 \, d} \]

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

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maple [A]  time = 0.45, size = 162, normalized size = 0.86 \[ -\frac {1}{96 a d \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {11}{128 a d \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {47}{128 a d \left (\sin \left (d x +c \right )-1\right )}+\frac {93 \ln \left (\sin \left (d x +c \right )-1\right )}{256 a d}-\frac {1}{64 a d \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {1}{8 a d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {29}{64 a d \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {35}{32 a d \left (1+\sin \left (d x +c \right )\right )}+\frac {163 \ln \left (1+\sin \left (d x +c \right )\right )}{256 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96/a/d/(sin(d*x+c)-1)^3-11/128/a/d/(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/a/d/(1+sin(d*x+c))^4+1/8/a/d/(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 = 0.32, size = 175, normalized size = 0.93 \[ \frac {\frac {2 \, {\left (279 \, \sin \left (d x + c\right )^{6} - 489 \, \sin \left (d x + c\right )^{5} - 1000 \, \sin \left (d x + c\right )^{4} + 728 \, \sin \left (d x + c\right )^{3} + 1113 \, \sin \left (d x + c\right )^{2} - 295 \, \sin \left (d x + c\right ) - 400\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} + \frac {489 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {279 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{768 \, d} \]

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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mupad [B]  time = 9.36, size = 432, normalized size = 2.30 \[ \frac {-\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}+\frac {29\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{32}+\frac {629\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{96}-\frac {365\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{96}-\frac {5399\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {203\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{48}+\frac {3019\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{48}+\frac {203\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{48}-\frac {5399\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{192}-\frac {365\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{96}+\frac {629\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96}+\frac {29\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{32}-\frac {35\,\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 )}^{14}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-5\,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}+9\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+30\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-40\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+30\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+9\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-12\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-5\,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 {93\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{128\,a\,d}+\frac {163\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{128\,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)^8/(cos(c + d*x)^7*(a + a*sin(c + d*x))),x)

[Out]

((29*tan(c/2 + (d*x)/2)^2)/32 - (35*tan(c/2 + (d*x)/2))/64 + (629*tan(c/2 + (d*x)/2)^3)/96 - (365*tan(c/2 + (d
*x)/2)^4)/96 - (5399*tan(c/2 + (d*x)/2)^5)/192 + (203*tan(c/2 + (d*x)/2)^6)/48 + (3019*tan(c/2 + (d*x)/2)^7)/4
8 + (203*tan(c/2 + (d*x)/2)^8)/48 - (5399*tan(c/2 + (d*x)/2)^9)/192 - (365*tan(c/2 + (d*x)/2)^10)/96 + (629*ta
n(c/2 + (d*x)/2)^11)/96 + (29*tan(c/2 + (d*x)/2)^12)/32 - (35*tan(c/2 + (d*x)/2)^13)/64)/(d*(a + 2*a*tan(c/2 +
(d*x)/2) - 5*a*tan(c/2 + (d*x)/2)^2 - 12*a*tan(c/2 + (d*x)/2)^3 + 9*a*tan(c/2 + (d*x)/2)^4 + 30*a*tan(c/2 + (
d*x)/2)^5 - 5*a*tan(c/2 + (d*x)/2)^6 - 40*a*tan(c/2 + (d*x)/2)^7 - 5*a*tan(c/2 + (d*x)/2)^8 + 30*a*tan(c/2 + (
d*x)/2)^9 + 9*a*tan(c/2 + (d*x)/2)^10 - 12*a*tan(c/2 + (d*x)/2)^11 - 5*a*tan(c/2 + (d*x)/2)^12 + 2*a*tan(c/2 +
(d*x)/2)^13 + a*tan(c/2 + (d*x)/2)^14)) + (93*log(tan(c/2 + (d*x)/2) - 1))/(128*a*d) + (163*log(tan(c/2 + (d*
x)/2) + 1))/(128*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)**7*sin(d*x+c)**8/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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