3.895 \(\int \frac{\sin ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=264 \[ -\frac{a^4}{160 d (a \sin (c+d x)+a)^5}+\frac{a^3}{256 d (a-a \sin (c+d x))^4}+\frac{19 a^3}{256 d (a \sin (c+d x)+a)^4}-\frac{3 a^2}{64 d (a-a \sin (c+d x))^3}-\frac{53 a^2}{128 d (a \sin (c+d x)+a)^3}-\frac{\sin ^2(c+d x)}{2 a d}+\frac{141 a}{512 d (a-a \sin (c+d x))^2}+\frac{765 a}{512 d (a \sin (c+d x)+a)^2}-\frac{39}{32 d (a-a \sin (c+d x))}-\frac{1155}{256 d (a \sin (c+d x)+a)}+\frac{\sin (c+d x)}{a d}-\frac{843 \log (1-\sin (c+d x))}{512 a d}-\frac{2229 \log (\sin (c+d x)+1)}{512 a d} \]

[Out]

(-843*Log[1 - Sin[c + d*x]])/(512*a*d) - (2229*Log[1 + Sin[c + d*x]])/(512*a*d) + Sin[c + d*x]/(a*d) - Sin[c +
d*x]^2/(2*a*d) + a^3/(256*d*(a - a*Sin[c + d*x])^4) - (3*a^2)/(64*d*(a - a*Sin[c + d*x])^3) + (141*a)/(512*d*
(a - a*Sin[c + d*x])^2) - 39/(32*d*(a - a*Sin[c + d*x])) - a^4/(160*d*(a + a*Sin[c + d*x])^5) + (19*a^3)/(256*
d*(a + a*Sin[c + d*x])^4) - (53*a^2)/(128*d*(a + a*Sin[c + d*x])^3) + (765*a)/(512*d*(a + a*Sin[c + d*x])^2) -
1155/(256*d*(a + a*Sin[c + d*x]))

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Rubi [A]  time = 0.282715, antiderivative size = 264, normalized size of antiderivative = 1., 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} \[ -\frac{a^4}{160 d (a \sin (c+d x)+a)^5}+\frac{a^3}{256 d (a-a \sin (c+d x))^4}+\frac{19 a^3}{256 d (a \sin (c+d x)+a)^4}-\frac{3 a^2}{64 d (a-a \sin (c+d x))^3}-\frac{53 a^2}{128 d (a \sin (c+d x)+a)^3}-\frac{\sin ^2(c+d x)}{2 a d}+\frac{141 a}{512 d (a-a \sin (c+d x))^2}+\frac{765 a}{512 d (a \sin (c+d x)+a)^2}-\frac{39}{32 d (a-a \sin (c+d x))}-\frac{1155}{256 d (a \sin (c+d x)+a)}+\frac{\sin (c+d x)}{a d}-\frac{843 \log (1-\sin (c+d x))}{512 a d}-\frac{2229 \log (\sin (c+d x)+1)}{512 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[c + d*x]^3*Tan[c + d*x]^9)/(a + a*Sin[c + d*x]),x]

[Out]

(-843*Log[1 - Sin[c + d*x]])/(512*a*d) - (2229*Log[1 + Sin[c + d*x]])/(512*a*d) + Sin[c + d*x]/(a*d) - Sin[c +
d*x]^2/(2*a*d) + a^3/(256*d*(a - a*Sin[c + d*x])^4) - (3*a^2)/(64*d*(a - a*Sin[c + d*x])^3) + (141*a)/(512*d*
(a - a*Sin[c + d*x])^2) - 39/(32*d*(a - a*Sin[c + d*x])) - a^4/(160*d*(a + a*Sin[c + d*x])^5) + (19*a^3)/(256*
d*(a + a*Sin[c + d*x])^4) - (53*a^2)/(128*d*(a + a*Sin[c + d*x])^3) + (765*a)/(512*d*(a + a*Sin[c + d*x])^2) -
1155/(256*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 ^3(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{a^9 \operatorname{Subst}\left (\int \frac{x^{12}}{a^{12} (a-x)^5 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^{12}}{(a-x)^5 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a+\frac{a^6}{64 (a-x)^5}-\frac{9 a^5}{64 (a-x)^4}+\frac{141 a^4}{256 (a-x)^3}-\frac{39 a^3}{32 (a-x)^2}+\frac{843 a^2}{512 (a-x)}-x+\frac{a^7}{32 (a+x)^6}-\frac{19 a^6}{64 (a+x)^5}+\frac{159 a^5}{128 (a+x)^4}-\frac{765 a^4}{256 (a+x)^3}+\frac{1155 a^3}{256 (a+x)^2}-\frac{2229 a^2}{512 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=-\frac{843 \log (1-\sin (c+d x))}{512 a d}-\frac{2229 \log (1+\sin (c+d x))}{512 a d}+\frac{\sin (c+d x)}{a d}-\frac{\sin ^2(c+d x)}{2 a d}+\frac{a^3}{256 d (a-a \sin (c+d x))^4}-\frac{3 a^2}{64 d (a-a \sin (c+d x))^3}+\frac{141 a}{512 d (a-a \sin (c+d x))^2}-\frac{39}{32 d (a-a \sin (c+d x))}-\frac{a^4}{160 d (a+a \sin (c+d x))^5}+\frac{19 a^3}{256 d (a+a \sin (c+d x))^4}-\frac{53 a^2}{128 d (a+a \sin (c+d x))^3}+\frac{765 a}{512 d (a+a \sin (c+d x))^2}-\frac{1155}{256 d (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.1625, size = 169, normalized size = 0.64 \[ -\frac{1280 \sin ^2(c+d x)-2560 \sin (c+d x)+\frac{3120}{1-\sin (c+d x)}+\frac{11550}{\sin (c+d x)+1}-\frac{705}{(1-\sin (c+d x))^2}-\frac{3825}{(\sin (c+d x)+1)^2}+\frac{120}{(1-\sin (c+d x))^3}+\frac{1060}{(\sin (c+d x)+1)^3}-\frac{10}{(1-\sin (c+d x))^4}-\frac{190}{(\sin (c+d x)+1)^4}+\frac{16}{(\sin (c+d x)+1)^5}+4215 \log (1-\sin (c+d x))+11145 \log (\sin (c+d x)+1)}{2560 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[c + d*x]^3*Tan[c + d*x]^9)/(a + a*Sin[c + d*x]),x]

[Out]

-(4215*Log[1 - Sin[c + d*x]] + 11145*Log[1 + Sin[c + d*x]] - 10/(1 - Sin[c + d*x])^4 + 120/(1 - Sin[c + d*x])^
3 - 705/(1 - Sin[c + d*x])^2 + 3120/(1 - Sin[c + d*x]) - 2560*Sin[c + d*x] + 1280*Sin[c + d*x]^2 + 16/(1 + Sin
[c + d*x])^5 - 190/(1 + Sin[c + d*x])^4 + 1060/(1 + Sin[c + d*x])^3 - 3825/(1 + Sin[c + d*x])^2 + 11550/(1 + S
in[c + d*x]))/(2560*a*d)

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Maple [A]  time = 0.112, size = 227, normalized size = 0.9 \begin{align*} -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,da}}+{\frac{\sin \left ( dx+c \right ) }{da}}+{\frac{1}{256\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{4}}}+{\frac{3}{64\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3}}}+{\frac{141}{512\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}+{\frac{39}{32\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{843\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{512\,da}}-{\frac{1}{160\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{5}}}+{\frac{19}{256\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{53}{128\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}+{\frac{765}{512\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{1155}{256\,da \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{2229\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{512\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*sin(d*x+c)^12/(a+a*sin(d*x+c)),x)

[Out]

-1/2*sin(d*x+c)^2/d/a+sin(d*x+c)/d/a+1/256/d/a/(sin(d*x+c)-1)^4+3/64/d/a/(sin(d*x+c)-1)^3+141/512/d/a/(sin(d*x
+c)-1)^2+39/32/a/d/(sin(d*x+c)-1)-843/512/a/d*ln(sin(d*x+c)-1)-1/160/d/a/(1+sin(d*x+c))^5+19/256/d/a/(1+sin(d*
x+c))^4-53/128/d/a/(1+sin(d*x+c))^3+765/512/a/d/(1+sin(d*x+c))^2-1155/256/a/d/(1+sin(d*x+c))-2229/512*ln(1+sin
(d*x+c))/a/d

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Maxima [A]  time = 1.03996, size = 319, normalized size = 1.21 \begin{align*} -\frac{\frac{2 \,{\left (4215 \, \sin \left (d x + c\right )^{8} - 5385 \, \sin \left (d x + c\right )^{7} - 18655 \, \sin \left (d x + c\right )^{6} + 13345 \, \sin \left (d x + c\right )^{5} + 30113 \, \sin \left (d x + c\right )^{4} - 11487 \, \sin \left (d x + c\right )^{3} - 21257 \, \sin \left (d x + c\right )^{2} + 3383 \, \sin \left (d x + c\right ) + 5568\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{1280 \,{\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )}}{a} + \frac{11145 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac{4215 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{2560 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^12/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/2560*(2*(4215*sin(d*x + c)^8 - 5385*sin(d*x + c)^7 - 18655*sin(d*x + c)^6 + 13345*sin(d*x + c)^5 + 30113*si
n(d*x + c)^4 - 11487*sin(d*x + c)^3 - 21257*sin(d*x + c)^2 + 3383*sin(d*x + c) + 5568)/(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) + 1280*(sin(d*x + c)^2 - 2*sin(d*x + c))/a + 11145*log(sin
(d*x + c) + 1)/a + 4215*log(sin(d*x + c) - 1)/a)/d

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Fricas [A]  time = 1.97715, size = 626, normalized size = 2.37 \begin{align*} -\frac{1280 \, \cos \left (d x + c\right )^{10} + 6510 \, \cos \left (d x + c\right )^{8} + 3590 \, \cos \left (d x + c\right )^{6} - 1124 \, \cos \left (d x + c\right )^{4} + 272 \, \cos \left (d x + c\right )^{2} + 11145 \,{\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 ) + 4215 \,{\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 (640 \, \cos \left (d x + c\right )^{10} + 960 \, \cos \left (d x + c\right )^{8} - 5385 \, \cos \left (d x + c\right )^{6} + 2810 \, \cos \left (d x + c\right )^{4} - 952 \, \cos \left (d x + c\right )^{2} + 144\right )} \sin \left (d x + c\right ) - 32}{2560 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^12/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/2560*(1280*cos(d*x + c)^10 + 6510*cos(d*x + c)^8 + 3590*cos(d*x + c)^6 - 1124*cos(d*x + c)^4 + 272*cos(d*x
+ c)^2 + 11145*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(sin(d*x + c) + 1) + 4215*(cos(d*x + c)^8*sin
(d*x + c) + cos(d*x + c)^8)*log(-sin(d*x + c) + 1) - 2*(640*cos(d*x + c)^10 + 960*cos(d*x + c)^8 - 5385*cos(d*
x + c)^6 + 2810*cos(d*x + c)^4 - 952*cos(d*x + c)^2 + 144)*sin(d*x + c) - 32)/(a*d*cos(d*x + c)^8*sin(d*x + c)
+ a*d*cos(d*x + c)^8)

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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)**9*sin(d*x+c)**12/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.35359, size = 244, normalized size = 0.92 \begin{align*} -\frac{\frac{44580 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} + \frac{16860 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac{5120 \,{\left (a \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right )\right )}}{a^{2}} - \frac{5 \,{\left (7025 \, \sin \left (d x + c\right )^{4} - 25604 \, \sin \left (d x + c\right )^{3} + 35226 \, \sin \left (d x + c\right )^{2} - 21644 \, \sin \left (d x + c\right ) + 5005\right )}}{a{\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac{101791 \, \sin \left (d x + c\right )^{5} + 462755 \, \sin \left (d x + c\right )^{4} + 848410 \, \sin \left (d x + c\right )^{3} + 782370 \, \sin \left (d x + c\right )^{2} + 362335 \, \sin \left (d x + c\right ) + 67347}{a{\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{10240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^12/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/10240*(44580*log(abs(sin(d*x + c) + 1))/a + 16860*log(abs(sin(d*x + c) - 1))/a + 5120*(a*sin(d*x + c)^2 - 2
*a*sin(d*x + c))/a^2 - 5*(7025*sin(d*x + c)^4 - 25604*sin(d*x + c)^3 + 35226*sin(d*x + c)^2 - 21644*sin(d*x +
c) + 5005)/(a*(sin(d*x + c) - 1)^4) - (101791*sin(d*x + c)^5 + 462755*sin(d*x + c)^4 + 848410*sin(d*x + c)^3 +
782370*sin(d*x + c)^2 + 362335*sin(d*x + c) + 67347)/(a*(sin(d*x + c) + 1)^5))/d