3.880 \(\int \frac {\sin ^2(c+d x) \tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=199 \[ \frac {a^3}{64 d (a \sin (c+d x)+a)^4}+\frac {a^2}{96 d (a-a \sin (c+d x))^3}-\frac {7 a^2}{48 d (a \sin (c+d x)+a)^3}-\frac {13 a}{128 d (a-a \sin (c+d x))^2}+\frac {41 a}{64 d (a \sin (c+d x)+a)^2}+\frac {69}{128 d (a-a \sin (c+d x))}-\frac {2}{d (a \sin (c+d x)+a)}+\frac {\sin (c+d x)}{a d}+\frac {187 \log (1-\sin (c+d x))}{256 a d}-\frac {443 \log (\sin (c+d x)+1)}{256 a d} \]

[Out]

187/256*ln(1-sin(d*x+c))/a/d-443/256*ln(1+sin(d*x+c))/a/d+sin(d*x+c)/a/d+1/96*a^2/d/(a-a*sin(d*x+c))^3-13/128*
a/d/(a-a*sin(d*x+c))^2+69/128/d/(a-a*sin(d*x+c))+1/64*a^3/d/(a+a*sin(d*x+c))^4-7/48*a^2/d/(a+a*sin(d*x+c))^3+4
1/64*a/d/(a+a*sin(d*x+c))^2-2/d/(a+a*sin(d*x+c))

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Rubi [A]  time = 0.21, antiderivative size = 199, 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} \[ \frac {a^3}{64 d (a \sin (c+d x)+a)^4}+\frac {a^2}{96 d (a-a \sin (c+d x))^3}-\frac {7 a^2}{48 d (a \sin (c+d x)+a)^3}-\frac {13 a}{128 d (a-a \sin (c+d x))^2}+\frac {41 a}{64 d (a \sin (c+d x)+a)^2}+\frac {69}{128 d (a-a \sin (c+d x))}-\frac {2}{d (a \sin (c+d x)+a)}+\frac {\sin (c+d x)}{a d}+\frac {187 \log (1-\sin (c+d x))}{256 a d}-\frac {443 \log (\sin (c+d x)+1)}{256 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[c + d*x]^2*Tan[c + d*x]^7)/(a + a*Sin[c + d*x]),x]

[Out]

(187*Log[1 - Sin[c + d*x]])/(256*a*d) - (443*Log[1 + Sin[c + d*x]])/(256*a*d) + Sin[c + d*x]/(a*d) + a^2/(96*d
*(a - a*Sin[c + d*x])^3) - (13*a)/(128*d*(a - a*Sin[c + d*x])^2) + 69/(128*d*(a - a*Sin[c + d*x])) + a^3/(64*d
*(a + a*Sin[c + d*x])^4) - (7*a^2)/(48*d*(a + a*Sin[c + d*x])^3) + (41*a)/(64*d*(a + a*Sin[c + d*x])^2) - 2/(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 ^2(c+d x) \tan ^7(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {a^7 \operatorname {Subst}\left (\int \frac {x^9}{a^9 (a-x)^4 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^9}{(a-x)^4 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{a^2 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (1+\frac {a^4}{32 (a-x)^4}-\frac {13 a^3}{64 (a-x)^3}+\frac {69 a^2}{128 (a-x)^2}-\frac {187 a}{256 (a-x)}-\frac {a^5}{16 (a+x)^5}+\frac {7 a^4}{16 (a+x)^4}-\frac {41 a^3}{32 (a+x)^3}+\frac {2 a^2}{(a+x)^2}-\frac {443 a}{256 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{a^2 d}\\ &=\frac {187 \log (1-\sin (c+d x))}{256 a d}-\frac {443 \log (1+\sin (c+d x))}{256 a d}+\frac {\sin (c+d x)}{a d}+\frac {a^2}{96 d (a-a \sin (c+d x))^3}-\frac {13 a}{128 d (a-a \sin (c+d x))^2}+\frac {69}{128 d (a-a \sin (c+d x))}+\frac {a^3}{64 d (a+a \sin (c+d x))^4}-\frac {7 a^2}{48 d (a+a \sin (c+d x))^3}+\frac {41 a}{64 d (a+a \sin (c+d x))^2}-\frac {2}{d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 6.14, size = 133, normalized size = 0.67 \[ \frac {768 \sin (c+d x)+\frac {414}{1-\sin (c+d x)}-\frac {1536}{\sin (c+d x)+1}-\frac {78}{(1-\sin (c+d x))^2}+\frac {492}{(\sin (c+d x)+1)^2}+\frac {8}{(1-\sin (c+d x))^3}-\frac {112}{(\sin (c+d x)+1)^3}+\frac {12}{(\sin (c+d x)+1)^4}+561 \log (1-\sin (c+d x))-1329 \log (\sin (c+d x)+1)}{768 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[c + d*x]^2*Tan[c + d*x]^7)/(a + a*Sin[c + d*x]),x]

[Out]

(561*Log[1 - Sin[c + d*x]] - 1329*Log[1 + Sin[c + d*x]] + 8/(1 - Sin[c + d*x])^3 - 78/(1 - Sin[c + d*x])^2 + 4
14/(1 - Sin[c + d*x]) + 768*Sin[c + d*x] + 12/(1 + Sin[c + d*x])^4 - 112/(1 + Sin[c + d*x])^3 + 492/(1 + Sin[c
+ d*x])^2 - 1536/(1 + Sin[c + d*x]))/(768*a*d)

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fricas [A]  time = 0.52, size = 187, normalized size = 0.94 \[ -\frac {768 \, \cos \left (d x + c\right )^{8} + 1182 \, \cos \left (d x + c\right )^{6} - 1674 \, \cos \left (d x + c\right )^{4} + 636 \, \cos \left (d x + c\right )^{2} + 1329 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 561 \, {\left (\cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{6}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (384 \, \cos \left (d x + c\right )^{6} + 207 \, \cos \left (d x + c\right )^{4} - 54 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) - 112}{768 \, {\left (a d \cos \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/768*(768*cos(d*x + c)^8 + 1182*cos(d*x + c)^6 - 1674*cos(d*x + c)^4 + 636*cos(d*x + c)^2 + 1329*(cos(d*x +
c)^6*sin(d*x + c) + cos(d*x + c)^6)*log(sin(d*x + c) + 1) - 561*(cos(d*x + c)^6*sin(d*x + c) + cos(d*x + c)^6)
*log(-sin(d*x + c) + 1) - 2*(384*cos(d*x + c)^6 + 207*cos(d*x + c)^4 - 54*cos(d*x + c)^2 + 8)*sin(d*x + c) - 1
12)/(a*d*cos(d*x + c)^6*sin(d*x + c) + a*d*cos(d*x + c)^6)

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giac [A]  time = 0.41, size = 147, normalized size = 0.74 \[ -\frac {\frac {5316 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {2244 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} - \frac {3072 \, \sin \left (d x + c\right )}{a} + \frac {2 \, {\left (2057 \, \sin \left (d x + c\right )^{3} - 5343 \, \sin \left (d x + c\right )^{2} + 4671 \, \sin \left (d x + c\right ) - 1369\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{3}} - \frac {11075 \, \sin \left (d x + c\right )^{4} + 38156 \, \sin \left (d x + c\right )^{3} + 49986 \, \sin \left (d x + c\right )^{2} + 29356 \, \sin \left (d x + c\right ) + 6499}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{3072 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/3072*(5316*log(abs(sin(d*x + c) + 1))/a - 2244*log(abs(sin(d*x + c) - 1))/a - 3072*sin(d*x + c)/a + 2*(2057
*sin(d*x + c)^3 - 5343*sin(d*x + c)^2 + 4671*sin(d*x + c) - 1369)/(a*(sin(d*x + c) - 1)^3) - (11075*sin(d*x +
c)^4 + 38156*sin(d*x + c)^3 + 49986*sin(d*x + c)^2 + 29356*sin(d*x + c) + 6499)/(a*(sin(d*x + c) + 1)^4))/d

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maple [A]  time = 0.46, size = 175, normalized size = 0.88 \[ \frac {\sin \left (d x +c \right )}{a d}-\frac {1}{96 a d \left (\sin \left (d x +c \right )-1\right )^{3}}-\frac {13}{128 a d \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {69}{128 a d \left (\sin \left (d x +c \right )-1\right )}+\frac {187 \ln \left (\sin \left (d x +c \right )-1\right )}{256 a d}+\frac {1}{64 a d \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {7}{48 a d \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {41}{64 a d \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {2}{a d \left (1+\sin \left (d x +c \right )\right )}-\frac {443 \ln \left (1+\sin \left (d x +c \right )\right )}{256 a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*sin(d*x+c)^9/(a+a*sin(d*x+c)),x)

[Out]

sin(d*x+c)/a/d-1/96/a/d/(sin(d*x+c)-1)^3-13/128/a/d/(sin(d*x+c)-1)^2-69/128/a/d/(sin(d*x+c)-1)+187/256/a/d*ln(
sin(d*x+c)-1)+1/64/a/d/(1+sin(d*x+c))^4-7/48/a/d/(1+sin(d*x+c))^3+41/64/a/d/(1+sin(d*x+c))^2-2/a/d/(1+sin(d*x+
c))-443/256*ln(1+sin(d*x+c))/a/d

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maxima [A]  time = 0.33, size = 186, normalized size = 0.93 \[ -\frac {\frac {2 \, {\left (975 \, \sin \left (d x + c\right )^{6} + 207 \, \sin \left (d x + c\right )^{5} - 2088 \, \sin \left (d x + c\right )^{4} - 360 \, \sin \left (d x + c\right )^{3} + 1569 \, \sin \left (d x + c\right )^{2} + 161 \, \sin \left (d x + c\right ) - 400\right )}}{a \sin \left (d x + c\right )^{7} + a \sin \left (d x + c\right )^{6} - 3 \, a \sin \left (d x + c\right )^{5} - 3 \, a \sin \left (d x + c\right )^{4} + 3 \, a \sin \left (d x + c\right )^{3} + 3 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} + \frac {1329 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {561 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac {768 \, \sin \left (d x + c\right )}{a}}{768 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*sin(d*x+c)^9/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/768*(2*(975*sin(d*x + c)^6 + 207*sin(d*x + c)^5 - 2088*sin(d*x + c)^4 - 360*sin(d*x + c)^3 + 1569*sin(d*x +
c)^2 + 161*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) + 1329*log(sin(d*x + c) + 1)/a - 561*log(sin(
d*x + c) - 1)/a - 768*sin(d*x + c)/a)/d

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mupad [B]  time = 10.14, size = 485, normalized size = 2.44 \[ \frac {\frac {315\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{64}+\frac {251\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{32}-\frac {1411\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{64}-\frac {607\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{16}+\frac {2183\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{64}+\frac {6287\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{96}-\frac {2749\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}-\frac {803\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{24}-\frac {2749\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{192}+\frac {6287\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{96}+\frac {2183\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{64}-\frac {607\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{16}-\frac {1411\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{64}+\frac {251\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{32}+\frac {315\,\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 )}^{16}+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}-4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+18\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+18\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-10\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-4\,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 {187\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{128\,a\,d}-\frac {443\,\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)^9/(cos(c + d*x)^7*(a + a*sin(c + d*x))),x)

[Out]

((315*tan(c/2 + (d*x)/2))/64 + (251*tan(c/2 + (d*x)/2)^2)/32 - (1411*tan(c/2 + (d*x)/2)^3)/64 - (607*tan(c/2 +
(d*x)/2)^4)/16 + (2183*tan(c/2 + (d*x)/2)^5)/64 + (6287*tan(c/2 + (d*x)/2)^6)/96 - (2749*tan(c/2 + (d*x)/2)^7
)/192 - (803*tan(c/2 + (d*x)/2)^8)/24 - (2749*tan(c/2 + (d*x)/2)^9)/192 + (6287*tan(c/2 + (d*x)/2)^10)/96 + (2
183*tan(c/2 + (d*x)/2)^11)/64 - (607*tan(c/2 + (d*x)/2)^12)/16 - (1411*tan(c/2 + (d*x)/2)^13)/64 + (251*tan(c/
2 + (d*x)/2)^14)/32 + (315*tan(c/2 + (d*x)/2)^15)/64)/(d*(a + 2*a*tan(c/2 + (d*x)/2) - 4*a*tan(c/2 + (d*x)/2)^
2 - 10*a*tan(c/2 + (d*x)/2)^3 + 4*a*tan(c/2 + (d*x)/2)^4 + 18*a*tan(c/2 + (d*x)/2)^5 + 4*a*tan(c/2 + (d*x)/2)^
6 - 10*a*tan(c/2 + (d*x)/2)^7 - 10*a*tan(c/2 + (d*x)/2)^8 - 10*a*tan(c/2 + (d*x)/2)^9 + 4*a*tan(c/2 + (d*x)/2)
^10 + 18*a*tan(c/2 + (d*x)/2)^11 + 4*a*tan(c/2 + (d*x)/2)^12 - 10*a*tan(c/2 + (d*x)/2)^13 - 4*a*tan(c/2 + (d*x
)/2)^14 + 2*a*tan(c/2 + (d*x)/2)^15 + a*tan(c/2 + (d*x)/2)^16)) + (187*log(tan(c/2 + (d*x)/2) - 1))/(128*a*d)
- (443*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)**9/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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