3.896 \(\int \frac{\sin ^2(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=247 \[ \frac{a^4}{160 d (a \sin (c+d x)+a)^5}+\frac{a^3}{256 d (a-a \sin (c+d x))^4}-\frac{17 a^3}{256 d (a \sin (c+d x)+a)^4}-\frac{a^2}{24 d (a-a \sin (c+d x))^3}+\frac{125 a^2}{384 d (a \sin (c+d x)+a)^3}+\frac{109 a}{512 d (a-a \sin (c+d x))^2}-\frac{515 a}{512 d (a \sin (c+d x)+a)^2}-\frac{203}{256 d (a-a \sin (c+d x))}+\frac{5}{2 d (a \sin (c+d x)+a)}-\frac{\sin (c+d x)}{a d}-\frac{437 \log (1-\sin (c+d x))}{512 a d}+\frac{949 \log (\sin (c+d x)+1)}{512 a d} \]

[Out]

(-437*Log[1 - Sin[c + d*x]])/(512*a*d) + (949*Log[1 + Sin[c + d*x]])/(512*a*d) - Sin[c + d*x]/(a*d) + a^3/(256
*d*(a - a*Sin[c + d*x])^4) - a^2/(24*d*(a - a*Sin[c + d*x])^3) + (109*a)/(512*d*(a - a*Sin[c + d*x])^2) - 203/
(256*d*(a - a*Sin[c + d*x])) + a^4/(160*d*(a + a*Sin[c + d*x])^5) - (17*a^3)/(256*d*(a + a*Sin[c + d*x])^4) +
(125*a^2)/(384*d*(a + a*Sin[c + d*x])^3) - (515*a)/(512*d*(a + a*Sin[c + d*x])^2) + 5/(2*d*(a + a*Sin[c + d*x]
))

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Rubi [A]  time = 0.26132, antiderivative size = 247, 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} \[ \frac{a^4}{160 d (a \sin (c+d x)+a)^5}+\frac{a^3}{256 d (a-a \sin (c+d x))^4}-\frac{17 a^3}{256 d (a \sin (c+d x)+a)^4}-\frac{a^2}{24 d (a-a \sin (c+d x))^3}+\frac{125 a^2}{384 d (a \sin (c+d x)+a)^3}+\frac{109 a}{512 d (a-a \sin (c+d x))^2}-\frac{515 a}{512 d (a \sin (c+d x)+a)^2}-\frac{203}{256 d (a-a \sin (c+d x))}+\frac{5}{2 d (a \sin (c+d x)+a)}-\frac{\sin (c+d x)}{a d}-\frac{437 \log (1-\sin (c+d x))}{512 a d}+\frac{949 \log (\sin (c+d x)+1)}{512 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[c + d*x]^2*Tan[c + d*x]^9)/(a + a*Sin[c + d*x]),x]

[Out]

(-437*Log[1 - Sin[c + d*x]])/(512*a*d) + (949*Log[1 + Sin[c + d*x]])/(512*a*d) - Sin[c + d*x]/(a*d) + a^3/(256
*d*(a - a*Sin[c + d*x])^4) - a^2/(24*d*(a - a*Sin[c + d*x])^3) + (109*a)/(512*d*(a - a*Sin[c + d*x])^2) - 203/
(256*d*(a - a*Sin[c + d*x])) + a^4/(160*d*(a + a*Sin[c + d*x])^5) - (17*a^3)/(256*d*(a + a*Sin[c + d*x])^4) +
(125*a^2)/(384*d*(a + a*Sin[c + d*x])^3) - (515*a)/(512*d*(a + a*Sin[c + d*x])^2) + 5/(2*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 ^2(c+d x) \tan ^9(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{a^9 \operatorname{Subst}\left (\int \frac{x^{11}}{a^{11} (a-x)^5 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^{11}}{(a-x)^5 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{a^2 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-1+\frac{a^5}{64 (a-x)^5}-\frac{a^4}{8 (a-x)^4}+\frac{109 a^3}{256 (a-x)^3}-\frac{203 a^2}{256 (a-x)^2}+\frac{437 a}{512 (a-x)}-\frac{a^6}{32 (a+x)^6}+\frac{17 a^5}{64 (a+x)^5}-\frac{125 a^4}{128 (a+x)^4}+\frac{515 a^3}{256 (a+x)^3}-\frac{5 a^2}{2 (a+x)^2}+\frac{949 a}{512 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{a^2 d}\\ &=-\frac{437 \log (1-\sin (c+d x))}{512 a d}+\frac{949 \log (1+\sin (c+d x))}{512 a d}-\frac{\sin (c+d x)}{a d}+\frac{a^3}{256 d (a-a \sin (c+d x))^4}-\frac{a^2}{24 d (a-a \sin (c+d x))^3}+\frac{109 a}{512 d (a-a \sin (c+d x))^2}-\frac{203}{256 d (a-a \sin (c+d x))}+\frac{a^4}{160 d (a+a \sin (c+d x))^5}-\frac{17 a^3}{256 d (a+a \sin (c+d x))^4}+\frac{125 a^2}{384 d (a+a \sin (c+d x))^3}-\frac{515 a}{512 d (a+a \sin (c+d x))^2}+\frac{5}{2 d (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.17697, size = 159, normalized size = 0.64 \[ -\frac{7680 \sin (c+d x)+\frac{6090}{1-\sin (c+d x)}-\frac{19200}{\sin (c+d x)+1}-\frac{1635}{(1-\sin (c+d x))^2}+\frac{7725}{(\sin (c+d x)+1)^2}+\frac{320}{(1-\sin (c+d x))^3}-\frac{2500}{(\sin (c+d x)+1)^3}-\frac{30}{(1-\sin (c+d x))^4}+\frac{510}{(\sin (c+d x)+1)^4}-\frac{48}{(\sin (c+d x)+1)^5}+6555 \log (1-\sin (c+d x))-14235 \log (\sin (c+d x)+1)}{7680 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[c + d*x]^2*Tan[c + d*x]^9)/(a + a*Sin[c + d*x]),x]

[Out]

-(6555*Log[1 - Sin[c + d*x]] - 14235*Log[1 + Sin[c + d*x]] - 30/(1 - Sin[c + d*x])^4 + 320/(1 - Sin[c + d*x])^
3 - 1635/(1 - Sin[c + d*x])^2 + 6090/(1 - Sin[c + d*x]) + 7680*Sin[c + d*x] - 48/(1 + Sin[c + d*x])^5 + 510/(1
+ Sin[c + d*x])^4 - 2500/(1 + Sin[c + d*x])^3 + 7725/(1 + Sin[c + d*x])^2 - 19200/(1 + Sin[c + d*x]))/(7680*a
*d)

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Maple [A]  time = 0.113, size = 212, normalized size = 0.9 \begin{align*} -{\frac{\sin \left ( dx+c \right ) }{da}}+{\frac{1}{256\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{4}}}+{\frac{1}{24\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{3}}}+{\frac{109}{512\,da \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}}}+{\frac{203}{256\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{437\,\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{17}{256\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{125}{384\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{3}}}-{\frac{515}{512\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{5}{2\,da \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{949\,\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)^11/(a+a*sin(d*x+c)),x)

[Out]

-sin(d*x+c)/d/a+1/256/d/a/(sin(d*x+c)-1)^4+1/24/d/a/(sin(d*x+c)-1)^3+109/512/d/a/(sin(d*x+c)-1)^2+203/256/a/d/
(sin(d*x+c)-1)-437/512/a/d*ln(sin(d*x+c)-1)+1/160/d/a/(1+sin(d*x+c))^5-17/256/d/a/(1+sin(d*x+c))^4+125/384/d/a
/(1+sin(d*x+c))^3-515/512/a/d/(1+sin(d*x+c))^2+5/2/a/d/(1+sin(d*x+c))+949/512*ln(1+sin(d*x+c))/a/d

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Maxima [A]  time = 1.05169, size = 304, normalized size = 1.23 \begin{align*} \frac{\frac{2 \,{\left (12645 \, \sin \left (d x + c\right )^{8} + 3045 \, \sin \left (d x + c\right )^{7} - 36765 \, \sin \left (d x + c\right )^{6} - 7965 \, \sin \left (d x + c\right )^{5} + 42339 \, \sin \left (d x + c\right )^{4} + 7139 \, \sin \left (d x + c\right )^{3} - 22171 \, \sin \left (d x + c\right )^{2} - 2171 \, \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{14235 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac{6555 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac{7680 \, \sin \left (d x + c\right )}{a}}{7680 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^11/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/7680*(2*(12645*sin(d*x + c)^8 + 3045*sin(d*x + c)^7 - 36765*sin(d*x + c)^6 - 7965*sin(d*x + c)^5 + 42339*sin
(d*x + c)^4 + 7139*sin(d*x + c)^3 - 22171*sin(d*x + c)^2 - 2171*sin(d*x + c) + 4384)/(a*sin(d*x + c)^9 + a*sin
(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) + 14235*log(sin(d*x + c) + 1)/a - 6555*log(sin(d*x + c) - 1)
/a - 7680*sin(d*x + c)/a)/d

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Fricas [A]  time = 1.89572, size = 602, normalized size = 2.44 \begin{align*} \frac{7680 \, \cos \left (d x + c\right )^{10} + 17610 \, \cos \left (d x + c\right )^{8} - 27630 \, \cos \left (d x + c\right )^{6} + 15828 \, \cos \left (d x + c\right )^{4} - 5584 \, \cos \left (d x + c\right )^{2} + 14235 \,{\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 ) - 6555 \,{\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 (3840 \, \cos \left (d x + c\right )^{8} + 3045 \, \cos \left (d x + c\right )^{6} - 1170 \, \cos \left (d x + c\right )^{4} + 344 \, \cos \left (d x + c\right )^{2} - 48\right )} \sin \left (d x + c\right ) + 864}{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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^11/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/7680*(7680*cos(d*x + c)^10 + 17610*cos(d*x + c)^8 - 27630*cos(d*x + c)^6 + 15828*cos(d*x + c)^4 - 5584*cos(d
*x + c)^2 + 14235*(cos(d*x + c)^8*sin(d*x + c) + cos(d*x + c)^8)*log(sin(d*x + c) + 1) - 6555*(cos(d*x + c)^8*
sin(d*x + c) + cos(d*x + c)^8)*log(-sin(d*x + c) + 1) - 2*(3840*cos(d*x + c)^8 + 3045*cos(d*x + c)^6 - 1170*co
s(d*x + c)^4 + 344*cos(d*x + c)^2 - 48)*sin(d*x + c) + 864)/(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)**11/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.32017, size = 225, normalized size = 0.91 \begin{align*} \frac{\frac{56940 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac{26220 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} - \frac{30720 \, \sin \left (d x + c\right )}{a} + \frac{5 \,{\left (10925 \, \sin \left (d x + c\right )^{4} - 38828 \, \sin \left (d x + c\right )^{3} + 52242 \, \sin \left (d x + c\right )^{2} - 31444 \, \sin \left (d x + c\right ) + 7129\right )}}{a{\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac{130013 \, \sin \left (d x + c\right )^{5} + 573265 \, \sin \left (d x + c\right )^{4} + 1023830 \, \sin \left (d x + c\right )^{3} + 922030 \, \sin \left (d x + c\right )^{2} + 417605 \, \sin \left (d x + c\right ) + 75961}{a{\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{30720 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*sin(d*x+c)^11/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/30720*(56940*log(abs(sin(d*x + c) + 1))/a - 26220*log(abs(sin(d*x + c) - 1))/a - 30720*sin(d*x + c)/a + 5*(1
0925*sin(d*x + c)^4 - 38828*sin(d*x + c)^3 + 52242*sin(d*x + c)^2 - 31444*sin(d*x + c) + 7129)/(a*(sin(d*x + c
) - 1)^4) - (130013*sin(d*x + c)^5 + 573265*sin(d*x + c)^4 + 1023830*sin(d*x + c)^3 + 922030*sin(d*x + c)^2 +
417605*sin(d*x + c) + 75961)/(a*(sin(d*x + c) + 1)^5))/d