\(\int \frac {\sin (x)}{(5 \cos ^2(x)-2 \sin ^2(x))^{7/2}} \, dx\) [422]

Optimal result

Integrand size = 20, antiderivative size = 55 \[ \int \frac {\sin (x)}{\left (5 \cos ^2(x)-2 \sin ^2(x)\right )^{7/2}} \, dx=\frac {\cos (x)}{10 \left (-2+7 \cos ^2(x)\right )^{5/2}}-\frac {\cos (x)}{15 \left (-2+7 \cos ^2(x)\right )^{3/2}}+\frac {\cos (x)}{15 \sqrt {-2+7 \cos ^2(x)}} \]

[Out]

1/10*cos(x)/(-2+7*cos(x)^2)^(5/2)-1/15*cos(x)/(-2+7*cos(x)^2)^(3/2)+1/15*cos(x)/(-2+7*cos(x)^2)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4442, 198, 197} \[ \int \frac {\sin (x)}{\left (5 \cos ^2(x)-2 \sin ^2(x)\right )^{7/2}} \, dx=\frac {\cos (x)}{15 \sqrt {7 \cos ^2(x)-2}}-\frac {\cos (x)}{15 \left (7 \cos ^2(x)-2\right )^{3/2}}+\frac {\cos (x)}{10 \left (7 \cos ^2(x)-2\right )^{5/2}} \]

[In]

Int[Sin[x]/(5*Cos[x]^2 - 2*Sin[x]^2)^(7/2),x]

[Out]

Cos[x]/(10*(-2 + 7*Cos[x]^2)^(5/2)) - Cos[x]/(15*(-2 + 7*Cos[x]^2)^(3/2)) + Cos[x]/(15*Sqrt[-2 + 7*Cos[x]^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 4442

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, Dist[-d/(
b*c), Subst[Int[SubstFor[1, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a
+ b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Sin] || EqQ[F, sin])

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {1}{\left (-2+7 x^2\right )^{7/2}} \, dx,x,\cos (x)\right ) \\ & = \frac {\cos (x)}{10 \left (-2+7 \cos ^2(x)\right )^{5/2}}+\frac {2}{5} \text {Subst}\left (\int \frac {1}{\left (-2+7 x^2\right )^{5/2}} \, dx,x,\cos (x)\right ) \\ & = \frac {\cos (x)}{10 \left (-2+7 \cos ^2(x)\right )^{5/2}}-\frac {\cos (x)}{15 \left (-2+7 \cos ^2(x)\right )^{3/2}}-\frac {2}{15} \text {Subst}\left (\int \frac {1}{\left (-2+7 x^2\right )^{3/2}} \, dx,x,\cos (x)\right ) \\ & = \frac {\cos (x)}{10 \left (-2+7 \cos ^2(x)\right )^{5/2}}-\frac {\cos (x)}{15 \left (-2+7 \cos ^2(x)\right )^{3/2}}+\frac {\cos (x)}{15 \sqrt {-2+7 \cos ^2(x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.67 \[ \int \frac {\sin (x)}{\left (5 \cos ^2(x)-2 \sin ^2(x)\right )^{7/2}} \, dx=\frac {\cos (x) (67+56 \cos (2 x)+49 \cos (4 x))}{15 \sqrt {2} (3+7 \cos (2 x))^{5/2}} \]

[In]

Integrate[Sin[x]/(5*Cos[x]^2 - 2*Sin[x]^2)^(7/2),x]

[Out]

(Cos[x]*(67 + 56*Cos[2*x] + 49*Cos[4*x]))/(15*Sqrt[2]*(3 + 7*Cos[2*x])^(5/2))

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {\cos \left (x \right )}{10 {\left (-2+7 \left (\cos ^{2}\left (x \right )\right )\right )}^{\frac {5}{2}}}-\frac {\cos \left (x \right )}{15 {\left (-2+7 \left (\cos ^{2}\left (x \right )\right )\right )}^{\frac {3}{2}}}+\frac {\cos \left (x \right )}{15 \sqrt {-2+7 \left (\cos ^{2}\left (x \right )\right )}}\)	\(44\)
default	\(\frac {\cos \left (x \right )}{10 {\left (-2+7 \left (\cos ^{2}\left (x \right )\right )\right )}^{\frac {5}{2}}}-\frac {\cos \left (x \right )}{15 {\left (-2+7 \left (\cos ^{2}\left (x \right )\right )\right )}^{\frac {3}{2}}}+\frac {\cos \left (x \right )}{15 \sqrt {-2+7 \left (\cos ^{2}\left (x \right )\right )}}\)	\(44\)

[In]

int(sin(x)/(5*cos(x)^2-2*sin(x)^2)^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/10*cos(x)/(-2+7*cos(x)^2)^(5/2)-1/15*cos(x)/(-2+7*cos(x)^2)^(3/2)+1/15*cos(x)/(-2+7*cos(x)^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int \frac {\sin (x)}{\left (5 \cos ^2(x)-2 \sin ^2(x)\right )^{7/2}} \, dx=\frac {{\left (98 \, \cos \left (x\right )^{5} - 70 \, \cos \left (x\right )^{3} + 15 \, \cos \left (x\right )\right )} \sqrt {7 \, \cos \left (x\right )^{2} - 2}}{30 \, {\left (343 \, \cos \left (x\right )^{6} - 294 \, \cos \left (x\right )^{4} + 84 \, \cos \left (x\right )^{2} - 8\right )}} \]

[In]

integrate(sin(x)/(5*cos(x)^2-2*sin(x)^2)^(7/2),x, algorithm="fricas")

[Out]

1/30*(98*cos(x)^5 - 70*cos(x)^3 + 15*cos(x))*sqrt(7*cos(x)^2 - 2)/(343*cos(x)^6 - 294*cos(x)^4 + 84*cos(x)^2 -
8)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin (x)}{\left (5 \cos ^2(x)-2 \sin ^2(x)\right )^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate(sin(x)/(5*cos(x)**2-2*sin(x)**2)**(7/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int \frac {\sin (x)}{\left (5 \cos ^2(x)-2 \sin ^2(x)\right )^{7/2}} \, dx=\frac {\cos \left (x\right )}{15 \, \sqrt {7 \, \cos \left (x\right )^{2} - 2}} - \frac {\cos \left (x\right )}{15 \, {\left (7 \, \cos \left (x\right )^{2} - 2\right )}^{\frac {3}{2}}} + \frac {\cos \left (x\right )}{10 \, {\left (7 \, \cos \left (x\right )^{2} - 2\right )}^{\frac {5}{2}}} \]

[In]

integrate(sin(x)/(5*cos(x)^2-2*sin(x)^2)^(7/2),x, algorithm="maxima")

[Out]

1/15*cos(x)/sqrt(7*cos(x)^2 - 2) - 1/15*cos(x)/(7*cos(x)^2 - 2)^(3/2) + 1/10*cos(x)/(7*cos(x)^2 - 2)^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.55 \[ \int \frac {\sin (x)}{\left (5 \cos ^2(x)-2 \sin ^2(x)\right )^{7/2}} \, dx=\frac {{\left (14 \, {\left (7 \, \cos \left (x\right )^{2} - 5\right )} \cos \left (x\right )^{2} + 15\right )} \cos \left (x\right )}{30 \, {\left (7 \, \cos \left (x\right )^{2} - 2\right )}^{\frac {5}{2}}} \]

[In]

integrate(sin(x)/(5*cos(x)^2-2*sin(x)^2)^(7/2),x, algorithm="giac")

[Out]

1/30*(14*(7*cos(x)^2 - 5)*cos(x)^2 + 15)*cos(x)/(7*cos(x)^2 - 2)^(5/2)

Mupad [B] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.51 \[ \int \frac {\sin (x)}{\left (5 \cos ^2(x)-2 \sin ^2(x)\right )^{7/2}} \, dx=\frac {\cos \left (x\right )\,\left (98\,{\cos \left (x\right )}^4-70\,{\cos \left (x\right )}^2+15\right )}{30\,{\left (7\,{\cos \left (x\right )}^2-2\right )}^{5/2}} \]

[In]

int(sin(x)/(5*cos(x)^2 - 2*sin(x)^2)^(7/2),x)

[Out]

(cos(x)*(98*cos(x)^4 - 70*cos(x)^2 + 15))/(30*(7*cos(x)^2 - 2)^(5/2))