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

Optimal. Leaf size=43 \[ -\frac {49}{5} \sin ^5(x)-\frac {49 \sin ^3(x)}{3}+63 \sin ^2(x)-49 \sin (x)-2 \log (1-\sin (x))+128 \log (\sin (x)+1) \]

[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

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Rubi [A]  time = 0.12, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3175, 3223, 1810, 633, 31} \[ -\frac {49}{5} \sin ^5(x)-\frac {49 \sin ^3(x)}{3}+63 \sin ^2(x)-49 \sin (x)-2 \log (1-\sin (x))+128 \log (\sin (x)+1) \]

Antiderivative was successfully verified.

[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 633

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 1810

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 3175

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 3223

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*} \int \frac {\cos (x) \left (9-7 \sin ^3(x)\right )^2}{1-\sin ^2(x)} \, dx &=\int \sec (x) \left (9-7 \sin ^3(x)\right )^2 \, dx\\ &=\operatorname {Subst}\left (\int \frac {\left (9-7 x^3\right )^2}{1-x^2} \, dx,x,\sin (x)\right )\\ &=\operatorname {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 \operatorname {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 \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sin (x)\right )-128 \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 71, normalized size = 1.65 \[ -\frac {49}{5} \sin ^5(x)-\frac {49 \sin ^3(x)}{3}-49 \sin (x)-63 \cos ^2(x)+49 \tanh ^{-1}(\sin (x))+126 \log (\cos (x))-81 \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+81 \log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

49*ArcTanh[Sin[x]] - 63*Cos[x]^2 + 126*Log[Cos[x]] - 81*Log[Cos[x/2] - Sin[x/2]] + 81*Log[Cos[x/2] + Sin[x/2]]
- 49*Sin[x] - (49*Sin[x]^3)/3 - (49*Sin[x]^5)/5

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fricas [A]  time = 1.95, size = 41, normalized size = 0.95 \[ -63 \, \cos \relax (x)^{2} - \frac {49}{15} \, {\left (3 \, \cos \relax (x)^{4} - 11 \, \cos \relax (x)^{2} + 23\right )} \sin \relax (x) + 128 \, \log \left (\sin \relax (x) + 1\right ) - 2 \, \log \left (-\sin \relax (x) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 6.00, size = 39, normalized size = 0.91 \[ -\frac {49}{5} \, \sin \relax (x)^{5} - \frac {49}{3} \, \sin \relax (x)^{3} + 63 \, \sin \relax (x)^{2} + 128 \, \log \left (\sin \relax (x) + 1\right ) - 2 \, \log \left (-\sin \relax (x) + 1\right ) - 49 \, \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.06, size = 38, normalized size = 0.88 \[ -\frac {49 \left (\sin ^{5}\relax (x )\right )}{5}-\frac {49 \left (\sin ^{3}\relax (x )\right )}{3}+63 \left (\sin ^{2}\relax (x )\right )-49 \sin \relax (x )-2 \ln \left (\sin \relax (x )-1\right )+128 \ln \left (1+\sin \relax (x )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))

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maxima [A]  time = 0.32, size = 37, normalized size = 0.86 \[ -\frac {49}{5} \, \sin \relax (x)^{5} - \frac {49}{3} \, \sin \relax (x)^{3} + 63 \, \sin \relax (x)^{2} + 128 \, \log \left (\sin \relax (x) + 1\right ) - 2 \, \log \left (\sin \relax (x) - 1\right ) - 49 \, \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.08, size = 37, normalized size = 0.86 \[ 128\,\ln \left (\sin \relax (x)+1\right )-2\,\ln \left (\sin \relax (x)-1\right )-49\,\sin \relax (x)+63\,{\sin \relax (x)}^2-\frac {49\,{\sin \relax (x)}^3}{3}-\frac {49\,{\sin \relax (x)}^5}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 2.95, size = 44, normalized size = 1.02 \[ - 2 \log {\left (\sin {\relax (x )} - 1 \right )} + 128 \log {\left (\sin {\relax (x )} + 1 \right )} - \frac {49 \sin ^{5}{\relax (x )}}{5} - \frac {49 \sin ^{3}{\relax (x )}}{3} + 63 \sin ^{2}{\relax (x )} - 49 \sin {\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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