\(\int \frac {\cos (x) (9-7 \sin ^3(x))^2}{1-\sin ^2(x)} \, dx\) [901]

Optimal result

Integrand size = 23, antiderivative size = 43 \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=-2 \log (1-\sin (x))+128 \log (1+\sin (x))-49 \sin (x)+63 \sin ^2(x)-\frac {49 \sin ^3(x)}{3}-\frac {49 \sin ^5(x)}{5} \]

[Out]

-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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3254, 3302, 1824, 647, 31} \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=-\frac {49}{5} \sin ^5(x)-\frac {49 \sin ^3(x)}{3}+63 \sin ^2(x)-49 \sin (x)-2 \log (1-\sin (x))+128 \log (\sin (x)+1) \]

[In]

Int[(Cos[x]*(9 - 7*Sin[x]^3)^2)/(1 - Sin[x]^2),x]

[Out]

-2*Log[1 - Sin[x]] + 128*Log[1 + Sin[x]] - 49*Sin[x] + 63*Sin[x]^2 - (49*Sin[x]^3)/3 - (49*Sin[x]^5)/5

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 1824

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]

Rule 3254

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[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]

Rule 3302

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]}, Dist[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])

Rubi steps \begin{align*} \text {integral}& = \int \sec (x) \left (9-7 \sin ^3(x)\right )^2 \, dx \\ & = \text {Subst}\left (\int \frac {\left (9-7 x^3\right )^2}{1-x^2} \, dx,x,\sin (x)\right ) \\ & = \text {Subst}\left (\int \left (-49+126 x-49 x^2-49 x^4+\frac {2 (65-63 x)}{1-x^2}\right ) \, dx,x,\sin (x)\right ) \\ & = -49 \sin (x)+63 \sin ^2(x)-\frac {49 \sin ^3(x)}{3}-\frac {49 \sin ^5(x)}{5}+2 \text {Subst}\left (\int \frac {65-63 x}{1-x^2} \, dx,x,\sin (x)\right ) \\ & = -49 \sin (x)+63 \sin ^2(x)-\frac {49 \sin ^3(x)}{3}-\frac {49 \sin ^5(x)}{5}+2 \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sin (x)\right )-128 \text {Subst}\left (\int \frac {1}{-1-x} \, dx,x,\sin (x)\right ) \\ & = -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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=130 \text {arctanh}(\sin (x))-63 \cos ^2(x)+126 \log (\cos (x))-49 \sin (x)-\frac {49 \sin ^3(x)}{3}-\frac {49 \sin ^5(x)}{5} \]

[In]

Integrate[(Cos[x]*(9 - 7*Sin[x]^3)^2)/(1 - Sin[x]^2),x]

[Out]

130*ArcTanh[Sin[x]] - 63*Cos[x]^2 + 126*Log[Cos[x]] - 49*Sin[x] - (49*Sin[x]^3)/3 - (49*Sin[x]^5)/5

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(-\frac {49 \sin \left (x \right )^{5}}{5}-\frac {49 \sin \left (x \right )^{3}}{3}+63 \sin \left (x \right )^{2}-49 \sin \left (x \right )-2 \ln \left (\sin \left (x \right )-1\right )+128 \ln \left (1+\sin \left (x \right )\right )\)	\(38\)
default	\(-\frac {49 \sin \left (x \right )^{5}}{5}-\frac {49 \sin \left (x \right )^{3}}{3}+63 \sin \left (x \right )^{2}-49 \sin \left (x \right )-2 \ln \left (\sin \left (x \right )-1\right )+128 \ln \left (1+\sin \left (x \right )\right )\)	\(38\)
parallelrisch	\(-\frac {441}{10}-126 \ln \left (2\right )-126 \ln \left (\frac {1}{\cos \left (x \right )+1}\right )+256 \ln \left (\csc \left (x \right )-\cot \left (x \right )+1\right )-4 \ln \left (-\cot \left (x \right )+\csc \left (x \right )-1\right )+\frac {343 \sin \left (3 x \right )}{48}-\frac {539 \sin \left (x \right )}{8}-\frac {49 \sin \left (5 x \right )}{80}-\frac {63 \cos \left (2 x \right )}{2}\)	\(60\)
risch	\(-126 i x +\frac {539 i {\mathrm e}^{i x}}{16}-\frac {539 i {\mathrm e}^{-i x}}{16}-4 \ln \left ({\mathrm e}^{i x}-i\right )+256 \ln \left (i+{\mathrm e}^{i x}\right )-\frac {49 \sin \left (5 x \right )}{80}+\frac {343 \sin \left (3 x \right )}{48}-\frac {63 \cos \left (2 x \right )}{2}\)	\(62\)
norman	\(\frac {-1260 \tan \left (\frac {x}{2}\right )^{6}-1008 \tan \left (\frac {x}{2}\right )^{4}-252 \tan \left (\frac {x}{2}\right )^{2}+252 \tan \left (\frac {x}{2}\right )^{14}+1008 \tan \left (\frac {x}{2}\right )^{12}+1260 \tan \left (\frac {x}{2}\right )^{10}+\frac {1862 \tan \left (\frac {x}{2}\right )^{3}}{3}+\frac {7938 \tan \left (\frac {x}{2}\right )^{5}}{5}+\frac {15974 \tan \left (\frac {x}{2}\right )^{7}}{15}-\frac {15974 \tan \left (\frac {x}{2}\right )^{9}}{15}-\frac {7938 \tan \left (\frac {x}{2}\right )^{11}}{5}-\frac {1862 \tan \left (\frac {x}{2}\right )^{13}}{3}-98 \tan \left (\frac {x}{2}\right )^{15}+98 \tan \left (\frac {x}{2}\right )}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{7} \left (\tan \left (\frac {x}{2}\right )^{2}-1\right )}-4 \ln \left (\tan \left (\frac {x}{2}\right )-1\right )+256 \ln \left (\tan \left (\frac {x}{2}\right )+1\right )-126 \ln \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )\)	\(163\)

[In]

int(cos(x)*(9-7*sin(x)^3)^2/(1-sin(x)^2),x,method=_RETURNVERBOSE)

[Out]

-49/5*sin(x)^5-49/3*sin(x)^3+63*sin(x)^2-49*sin(x)-2*ln(sin(x)-1)+128*ln(1+sin(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=-63 \, \cos \left (x\right )^{2} - \frac {49}{15} \, {\left (3 \, \cos \left (x\right )^{4} - 11 \, \cos \left (x\right )^{2} + 23\right )} \sin \left (x\right ) + 128 \, \log \left (\sin \left (x\right ) + 1\right ) - 2 \, \log \left (-\sin \left (x\right ) + 1\right ) \]

[In]

integrate(cos(x)*(9-7*sin(x)^3)^2/(1-sin(x)^2),x, algorithm="fricas")

[Out]

-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)

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02 \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=- 2 \log {\left (\sin {\left (x \right )} - 1 \right )} + 128 \log {\left (\sin {\left (x \right )} + 1 \right )} - \frac {49 \sin ^{5}{\left (x \right )}}{5} - \frac {49 \sin ^{3}{\left (x \right )}}{3} + 63 \sin ^{2}{\left (x \right )} - 49 \sin {\left (x \right )} \]

[In]

integrate(cos(x)*(9-7*sin(x)**3)**2/(1-sin(x)**2),x)

[Out]

-2*log(sin(x) - 1) + 128*log(sin(x) + 1) - 49*sin(x)**5/5 - 49*sin(x)**3/3 + 63*sin(x)**2 - 49*sin(x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=-\frac {49}{5} \, \sin \left (x\right )^{5} - \frac {49}{3} \, \sin \left (x\right )^{3} + 63 \, \sin \left (x\right )^{2} + 128 \, \log \left (\sin \left (x\right ) + 1\right ) - 2 \, \log \left (\sin \left (x\right ) - 1\right ) - 49 \, \sin \left (x\right ) \]

[In]

integrate(cos(x)*(9-7*sin(x)^3)^2/(1-sin(x)^2),x, algorithm="maxima")

[Out]

-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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=-\frac {49}{5} \, \sin \left (x\right )^{5} - \frac {49}{3} \, \sin \left (x\right )^{3} + 63 \, \sin \left (x\right )^{2} + 128 \, \log \left (\sin \left (x\right ) + 1\right ) - 2 \, \log \left (-\sin \left (x\right ) + 1\right ) - 49 \, \sin \left (x\right ) \]

[In]

integrate(cos(x)*(9-7*sin(x)^3)^2/(1-sin(x)^2),x, algorithm="giac")

[Out]

-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)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx=128\,\ln \left (\sin \left (x\right )+1\right )-2\,\ln \left (\sin \left (x\right )-1\right )-49\,\sin \left (x\right )+63\,{\sin \left (x\right )}^2-\frac {49\,{\sin \left (x\right )}^3}{3}-\frac {49\,{\sin \left (x\right )}^5}{5} \]

[In]

int(-(cos(x)*(7*sin(x)^3 - 9)^2)/(sin(x)^2 - 1),x)

[Out]

128*log(sin(x) + 1) - 2*log(sin(x) - 1) - 49*sin(x) + 63*sin(x)^2 - (49*sin(x)^3)/3 - (49*sin(x)^5)/5