Integrand size = 23, antiderivative size = 43 \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=-2 \log (1-\sin (x))+128 \log (1+\sin (x))-49 \sin (x)+63 \sin ^2(x)-\frac {49 \sin ^3(x)}{3}-\frac {49 \sin ^5(x)}{5} \] Output:
-2*ln(1-sin(x))+128*ln(1+sin(x))-49*sin(x)+63*sin(x)^2-49/3*sin(x)^3-49/5* sin(x)^5
Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.98 \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=81 \coth ^{-1}(\sin (x))+49 \text {arctanh}(\sin (x))-63 \cos ^2(x)+126 \log (\cos (x))-49 \sin (x)-\frac {49 \sin ^3(x)}{3}-\frac {49 \sin ^5(x)}{5} \] Input:
Integrate[(Cos[x]*(9 - 7*Sin[x]^3)^2)/(1 - Sin[x]^2),x]
Output:
81*ArcCoth[Sin[x]] + 49*ArcTanh[Sin[x]] - 63*Cos[x]^2 + 126*Log[Cos[x]] - 49*Sin[x] - (49*Sin[x]^3)/3 - (49*Sin[x]^5)/5
Time = 0.36 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3654, 3042, 3702, 2341, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (9-7 \sin ^3(x)\right )^2 \cos (x)}{1-\sin ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (9-7 \sin (x)^3\right )^2 \cos (x)}{1-\sin (x)^2}dx\) |
\(\Big \downarrow \) 3654 |
\(\displaystyle \int \left (9-7 \sin ^3(x)\right )^2 \sec (x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (9-7 \sin (x)^3\right )^2}{\cos (x)}dx\) |
\(\Big \downarrow \) 3702 |
\(\displaystyle \int \frac {\left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)}d\sin (x)\) |
\(\Big \downarrow \) 2341 |
\(\displaystyle \int \left (-49 \sin ^4(x)-49 \sin ^2(x)+\frac {2 (65-63 \sin (x))}{1-\sin ^2(x)}+126 \sin (x)-49\right )d\sin (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 130 \text {arctanh}(\sin (x))-\frac {49}{5} \sin ^5(x)-\frac {49 \sin ^3(x)}{3}+63 \sin ^2(x)-49 \sin (x)+63 \log \left (1-\sin ^2(x)\right )\) |
Input:
Int[(Cos[x]*(9 - 7*Sin[x]^3)^2)/(1 - Sin[x]^2),x]
Output:
130*ArcTanh[Sin[x]] + 63*Log[1 - Sin[x]^2] - 49*Sin[x] + 63*Sin[x]^2 - (49 *Sin[x]^3)/3 - (49*Sin[x]^5)/5
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[ a^p Int[ActivateTrig[u*cos[e + f*x]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x _)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Si mp[ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0] || IGtQ[p, 0] || IntegersQ[m, p])
Time = 2.16 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(-\frac {49 \sin \left (x \right )^{5}}{5}-\frac {49 \sin \left (x \right )^{3}}{3}+63 \sin \left (x \right )^{2}-49 \sin \left (x \right )-2 \ln \left (-1+\sin \left (x \right )\right )+128 \ln \left (1+\sin \left (x \right )\right )\) | \(38\) |
default | \(-\frac {49 \sin \left (x \right )^{5}}{5}-\frac {49 \sin \left (x \right )^{3}}{3}+63 \sin \left (x \right )^{2}-49 \sin \left (x \right )-2 \ln \left (-1+\sin \left (x \right )\right )+128 \ln \left (1+\sin \left (x \right )\right )\) | \(38\) |
parallelrisch | \(-\frac {441}{10}-126 \ln \left (\sec \left (\frac {x}{2}\right )^{2}\right )+256 \ln \left (1+\tan \left (\frac {x}{2}\right )\right )-4 \ln \left (\tan \left (\frac {x}{2}\right )-1\right )+\frac {343 \sin \left (3 x \right )}{48}-\frac {539 \sin \left (x \right )}{8}-\frac {49 \sin \left (5 x \right )}{80}-\frac {63 \cos \left (2 x \right )}{2}\) | \(52\) |
risch | \(-126 i x +\frac {539 i {\mathrm e}^{i x}}{16}-\frac {539 i {\mathrm e}^{-i x}}{16}-4 \ln \left ({\mathrm e}^{i x}-i\right )+256 \ln \left ({\mathrm e}^{i x}+i\right )-\frac {49 \sin \left (5 x \right )}{80}+\frac {343 \sin \left (3 x \right )}{48}-\frac {63 \cos \left (2 x \right )}{2}\) | \(62\) |
norman | \(\frac {-1260 \tan \left (\frac {x}{2}\right )^{6}-1008 \tan \left (\frac {x}{2}\right )^{4}-252 \tan \left (\frac {x}{2}\right )^{2}+252 \tan \left (\frac {x}{2}\right )^{14}+1008 \tan \left (\frac {x}{2}\right )^{12}+1260 \tan \left (\frac {x}{2}\right )^{10}+\frac {1862 \tan \left (\frac {x}{2}\right )^{3}}{3}+\frac {7938 \tan \left (\frac {x}{2}\right )^{5}}{5}+\frac {15974 \tan \left (\frac {x}{2}\right )^{7}}{15}-\frac {15974 \tan \left (\frac {x}{2}\right )^{9}}{15}-\frac {7938 \tan \left (\frac {x}{2}\right )^{11}}{5}-\frac {1862 \tan \left (\frac {x}{2}\right )^{13}}{3}-98 \tan \left (\frac {x}{2}\right )^{15}+98 \tan \left (\frac {x}{2}\right )}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{7} \left (\tan \left (\frac {x}{2}\right )^{2}-1\right )}-4 \ln \left (\tan \left (\frac {x}{2}\right )-1\right )+256 \ln \left (1+\tan \left (\frac {x}{2}\right )\right )-126 \ln \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )\) | \(163\) |
Input:
int(cos(x)*(9-7*sin(x)^3)^2/(1-sin(x)^2),x,method=_RETURNVERBOSE)
Output:
-49/5*sin(x)^5-49/3*sin(x)^3+63*sin(x)^2-49*sin(x)-2*ln(-1+sin(x))+128*ln( 1+sin(x))
Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=-63 \, \cos \left (x\right )^{2} - \frac {49}{15} \, {\left (3 \, \cos \left (x\right )^{4} - 11 \, \cos \left (x\right )^{2} + 23\right )} \sin \left (x\right ) + 128 \, \log \left (\sin \left (x\right ) + 1\right ) - 2 \, \log \left (-\sin \left (x\right ) + 1\right ) \] Input:
integrate(cos(x)*(9-7*sin(x)^3)^2/(1-sin(x)^2),x, algorithm="fricas")
Output:
-63*cos(x)^2 - 49/15*(3*cos(x)^4 - 11*cos(x)^2 + 23)*sin(x) + 128*log(sin( x) + 1) - 2*log(-sin(x) + 1)
Time = 0.40 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02 \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=- 2 \log {\left (\sin {\left (x \right )} - 1 \right )} + 128 \log {\left (\sin {\left (x \right )} + 1 \right )} - \frac {49 \sin ^{5}{\left (x \right )}}{5} - \frac {49 \sin ^{3}{\left (x \right )}}{3} - 49 \sin {\left (x \right )} - 63 \cos ^{2}{\left (x \right )} \] Input:
integrate(cos(x)*(9-7*sin(x)**3)**2/(1-sin(x)**2),x)
Output:
-2*log(sin(x) - 1) + 128*log(sin(x) + 1) - 49*sin(x)**5/5 - 49*sin(x)**3/3 - 49*sin(x) - 63*cos(x)**2
Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=-\frac {49}{5} \, \sin \left (x\right )^{5} - \frac {49}{3} \, \sin \left (x\right )^{3} + 63 \, \sin \left (x\right )^{2} + 128 \, \log \left (\sin \left (x\right ) + 1\right ) - 2 \, \log \left (\sin \left (x\right ) - 1\right ) - 49 \, \sin \left (x\right ) \] Input:
integrate(cos(x)*(9-7*sin(x)^3)^2/(1-sin(x)^2),x, algorithm="maxima")
Output:
-49/5*sin(x)^5 - 49/3*sin(x)^3 + 63*sin(x)^2 + 128*log(sin(x) + 1) - 2*log (sin(x) - 1) - 49*sin(x)
Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=-\frac {49}{5} \, \sin \left (x\right )^{5} - \frac {49}{3} \, \sin \left (x\right )^{3} + 63 \, \sin \left (x\right )^{2} + 128 \, \log \left (\sin \left (x\right ) + 1\right ) - 2 \, \log \left (-\sin \left (x\right ) + 1\right ) - 49 \, \sin \left (x\right ) \] Input:
integrate(cos(x)*(9-7*sin(x)^3)^2/(1-sin(x)^2),x, algorithm="giac")
Output:
-49/5*sin(x)^5 - 49/3*sin(x)^3 + 63*sin(x)^2 + 128*log(sin(x) + 1) - 2*log (-sin(x) + 1) - 49*sin(x)
Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=128\,\ln \left (\sin \left (x\right )+1\right )-2\,\ln \left (\sin \left (x\right )-1\right )-49\,\sin \left (x\right )+63\,{\sin \left (x\right )}^2-\frac {49\,{\sin \left (x\right )}^3}{3}-\frac {49\,{\sin \left (x\right )}^5}{5} \] Input:
int(-(cos(x)*(7*sin(x)^3 - 9)^2)/(sin(x)^2 - 1),x)
Output:
128*log(sin(x) + 1) - 2*log(sin(x) - 1) - 49*sin(x) + 63*sin(x)^2 - (49*si n(x)^3)/3 - (49*sin(x)^5)/5
Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=-63 \cos \left (x \right )^{2}-2 \,\mathrm {log}\left (\sin \left (x \right )-1\right )+128 \,\mathrm {log}\left (\sin \left (x \right )+1\right )-\frac {49 \sin \left (x \right )^{5}}{5}-\frac {49 \sin \left (x \right )^{3}}{3}-49 \sin \left (x \right ) \] Input:
int(cos(x)*(9-7*sin(x)^3)^2/(1-sin(x)^2),x)
Output:
( - 945*cos(x)**2 - 30*log(sin(x) - 1) + 1920*log(sin(x) + 1) - 147*sin(x) **5 - 245*sin(x)**3 - 735*sin(x))/15