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]

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

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Rubi [A]  time = 0.26, antiderivative size = 247, 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} \[ \frac {a^4}{160 d (a \sin (c+d x)+a)^5}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}-\frac {17 a^3}{256 d (a \sin (c+d x)+a)^4}-\frac {a^2}{24 d (a-a \sin (c+d x))^3}+\frac {125 a^2}{384 d (a \sin (c+d x)+a)^3}+\frac {109 a}{512 d (a-a \sin (c+d x))^2}-\frac {515 a}{512 d (a \sin (c+d x)+a)^2}-\frac {203}{256 d (a-a \sin (c+d x))}+\frac {5}{2 d (a \sin (c+d x)+a)}-\frac {\sin (c+d x)}{a d}-\frac {437 \log (1-\sin (c+d x))}{512 a d}+\frac {949 \log (\sin (c+d x)+1)}{512 a d} \]

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 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 ^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*}

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Mathematica [A]  time = 6.19, 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]

-1/7680*(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]))/
(a*d)

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fricas [A]  time = 0.55, size = 207, normalized size = 0.84 \[ \frac {7680 \, \cos \left (d x + c\right )^{10} + 17610 \, \cos \left (d x + c\right )^{8} - 27630 \, \cos \left (d x + c\right )^{6} + 15828 \, \cos \left (d x + c\right )^{4} - 5584 \, \cos \left (d x + c\right )^{2} + 14235 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 6555 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3840 \, \cos \left (d x + c\right )^{8} + 3045 \, \cos \left (d x + c\right )^{6} - 1170 \, \cos \left (d x + c\right )^{4} + 344 \, \cos \left (d x + c\right )^{2} - 48\right )} \sin \left (d x + c\right ) + 864}{7680 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \]

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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giac [A]  time = 0.40, size = 167, normalized size = 0.68 \[ \frac {\frac {56940 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {26220 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} - \frac {30720 \, \sin \left (d x + c\right )}{a} + \frac {5 \, {\left (10925 \, \sin \left (d x + c\right )^{4} - 38828 \, \sin \left (d x + c\right )^{3} + 52242 \, \sin \left (d x + c\right )^{2} - 31444 \, \sin \left (d x + c\right ) + 7129\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac {130013 \, \sin \left (d x + c\right )^{5} + 573265 \, \sin \left (d x + c\right )^{4} + 1023830 \, \sin \left (d x + c\right )^{3} + 922030 \, \sin \left (d x + c\right )^{2} + 417605 \, \sin \left (d x + c\right ) + 75961}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{30720 \, d} \]

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

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maple [A]  time = 0.48, size = 212, normalized size = 0.86 \[ -\frac {\sin \left (d x +c \right )}{a d}+\frac {1}{256 a d \left (\sin \left (d x +c \right )-1\right )^{4}}+\frac {1}{24 a d \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {109}{512 a d \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {203}{256 a d \left (\sin \left (d x +c \right )-1\right )}-\frac {437 \ln \left (\sin \left (d x +c \right )-1\right )}{512 a d}+\frac {1}{160 a d \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {17}{256 a d \left (1+\sin \left (d x +c \right )\right )^{4}}+\frac {125}{384 a d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {515}{512 a d \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {5}{2 a d \left (1+\sin \left (d x +c \right )\right )}+\frac {949 \ln \left (1+\sin \left (d x +c \right )\right )}{512 a d} \]

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)/a/d+1/256/a/d/(sin(d*x+c)-1)^4+1/24/a/d/(sin(d*x+c)-1)^3+109/512/a/d/(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/a/d/(1+sin(d*x+c))^5-17/256/a/d/(1+sin(d*x+c))^4+125/384/a/d
/(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 = 0.38, size = 225, normalized size = 0.91 \[ \frac {\frac {2 \, {\left (12645 \, \sin \left (d x + c\right )^{8} + 3045 \, \sin \left (d x + c\right )^{7} - 36765 \, \sin \left (d x + c\right )^{6} - 7965 \, \sin \left (d x + c\right )^{5} + 42339 \, \sin \left (d x + c\right )^{4} + 7139 \, \sin \left (d x + c\right )^{3} - 22171 \, \sin \left (d x + c\right )^{2} - 2171 \, \sin \left (d x + c\right ) + 4384\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} + \frac {14235 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac {6555 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a} - \frac {7680 \, \sin \left (d x + c\right )}{a}}{7680 \, d} \]

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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mupad [B]  time = 10.85, size = 595, normalized size = 2.41 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^11/(cos(c + d*x)^9*(a + a*sin(c + d*x))),x)

[Out]

(949*log(tan(c/2 + (d*x)/2) + 1))/(256*a*d) - (437*log(tan(c/2 + (d*x)/2) - 1))/(256*a*d) - ((693*tan(c/2 + (d
*x)/2))/128 + (565*tan(c/2 + (d*x)/2)^2)/64 - (4439*tan(c/2 + (d*x)/2)^3)/128 - (963*tan(c/2 + (d*x)/2)^4)/16
+ (7091*tan(c/2 + (d*x)/2)^5)/80 + (40031*tan(c/2 + (d*x)/2)^6)/240 - (12829*tan(c/2 + (d*x)/2)^7)/120 - (1796
9*tan(c/2 + (d*x)/2)^8)/80 + (39491*tan(c/2 + (d*x)/2)^9)/960 + (49513*tan(c/2 + (d*x)/2)^10)/480 + (39491*tan
(c/2 + (d*x)/2)^11)/960 - (17969*tan(c/2 + (d*x)/2)^12)/80 - (12829*tan(c/2 + (d*x)/2)^13)/120 + (40031*tan(c/
2 + (d*x)/2)^14)/240 + (7091*tan(c/2 + (d*x)/2)^15)/80 - (963*tan(c/2 + (d*x)/2)^16)/16 - (4439*tan(c/2 + (d*x
)/2)^17)/128 + (565*tan(c/2 + (d*x)/2)^18)/64 + (693*tan(c/2 + (d*x)/2)^19)/128)/(d*(a + 2*a*tan(c/2 + (d*x)/2
) - 6*a*tan(c/2 + (d*x)/2)^2 - 14*a*tan(c/2 + (d*x)/2)^3 + 13*a*tan(c/2 + (d*x)/2)^4 + 40*a*tan(c/2 + (d*x)/2)
^5 - 8*a*tan(c/2 + (d*x)/2)^6 - 56*a*tan(c/2 + (d*x)/2)^7 - 14*a*tan(c/2 + (d*x)/2)^8 + 28*a*tan(c/2 + (d*x)/2
)^9 + 28*a*tan(c/2 + (d*x)/2)^10 + 28*a*tan(c/2 + (d*x)/2)^11 - 14*a*tan(c/2 + (d*x)/2)^12 - 56*a*tan(c/2 + (d
*x)/2)^13 - 8*a*tan(c/2 + (d*x)/2)^14 + 40*a*tan(c/2 + (d*x)/2)^15 + 13*a*tan(c/2 + (d*x)/2)^16 - 14*a*tan(c/2
+ (d*x)/2)^17 - 6*a*tan(c/2 + (d*x)/2)^18 + 2*a*tan(c/2 + (d*x)/2)^19 + a*tan(c/2 + (d*x)/2)^20)) - 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)**9*sin(d*x+c)**11/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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