3.314 \(\int \frac{\cos ^6(x)}{(a+b \sin ^2(x))^2} \, dx\)

Optimal. Leaf size=113 \[ -\frac{(4 a-b) (a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{2 a^{3/2} b^3}+\frac{x (4 a+5 b)}{2 b^3}+\frac{(2 a+b) (a+b) \tan (x)}{2 a b^2 \left ((a+b) \tan ^2(x)+a\right )}-\frac{\sin (x) \cos (x)}{2 b \left ((a+b) \tan ^2(x)+a\right )} \]

[Out]

((4*a + 5*b)*x)/(2*b^3) - ((4*a - b)*(a + b)^(3/2)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(2*a^(3/2)*b^3) - (Co
s[x]*Sin[x])/(2*b*(a + (a + b)*Tan[x]^2)) + ((a + b)*(2*a + b)*Tan[x])/(2*a*b^2*(a + (a + b)*Tan[x]^2))

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Rubi [A]  time = 0.223141, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3191, 414, 527, 522, 203, 205} \[ -\frac{(4 a-b) (a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{2 a^{3/2} b^3}+\frac{x (4 a+5 b)}{2 b^3}+\frac{(2 a+b) (a+b) \tan (x)}{2 a b^2 \left ((a+b) \tan ^2(x)+a\right )}-\frac{\sin (x) \cos (x)}{2 b \left ((a+b) \tan ^2(x)+a\right )} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^6/(a + b*Sin[x]^2)^2,x]

[Out]

((4*a + 5*b)*x)/(2*b^3) - ((4*a - b)*(a + b)^(3/2)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(2*a^(3/2)*b^3) - (Co
s[x]*Sin[x])/(2*b*(a + (a + b)*Tan[x]^2)) + ((a + b)*(2*a + b)*Tan[x])/(2*a*b^2*(a + (a + b)*Tan[x]^2))

Rule 3191

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\cos ^6(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right )^2 \left (a+(a+b) x^2\right )^2} \, dx,x,\tan (x)\right )\\ &=-\frac{\cos (x) \sin (x)}{2 b \left (a+(a+b) \tan ^2(x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{a+2 b-3 (a+b) x^2}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )^2} \, dx,x,\tan (x)\right )}{2 b}\\ &=-\frac{\cos (x) \sin (x)}{2 b \left (a+(a+b) \tan ^2(x)\right )}+\frac{(a+b) (2 a+b) \tan (x)}{2 a b^2 \left (a+(a+b) \tan ^2(x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{2 \left (2 a^2+2 a b-b^2\right )-2 (a+b) (2 a+b) x^2}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )} \, dx,x,\tan (x)\right )}{4 a b^2}\\ &=-\frac{\cos (x) \sin (x)}{2 b \left (a+(a+b) \tan ^2(x)\right )}+\frac{(a+b) (2 a+b) \tan (x)}{2 a b^2 \left (a+(a+b) \tan ^2(x)\right )}-\frac{\left ((4 a-b) (a+b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (x)\right )}{2 a b^3}+\frac{(4 a+5 b) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (x)\right )}{2 b^3}\\ &=\frac{(4 a+5 b) x}{2 b^3}-\frac{(4 a-b) (a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{2 a^{3/2} b^3}-\frac{\cos (x) \sin (x)}{2 b \left (a+(a+b) \tan ^2(x)\right )}+\frac{(a+b) (2 a+b) \tan (x)}{2 a b^2 \left (a+(a+b) \tan ^2(x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.294315, size = 90, normalized size = 0.8 \[ \frac{-\frac{2 (4 a-b) (a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{a^{3/2}}+2 x (4 a+5 b)+\frac{2 b (a+b)^2 \sin (2 x)}{a (2 a-b \cos (2 x)+b)}+b \sin (2 x)}{4 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^6/(a + b*Sin[x]^2)^2,x]

[Out]

(2*(4*a + 5*b)*x - (2*(4*a - b)*(a + b)^(3/2)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/a^(3/2) + b*Sin[2*x] + (2*
b*(a + b)^2*Sin[2*x])/(a*(2*a + b - b*Cos[2*x])))/(4*b^3)

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Maple [B]  time = 0.091, size = 211, normalized size = 1.9 \begin{align*}{\frac{\tan \left ( x \right ) }{2\,{b}^{2} \left ( \left ( \tan \left ( x \right ) \right ) ^{2}+1 \right ) }}+{\frac{5\,\arctan \left ( \tan \left ( x \right ) \right ) }{2\,{b}^{2}}}+2\,{\frac{\arctan \left ( \tan \left ( x \right ) \right ) a}{{b}^{3}}}+{\frac{a\tan \left ( x \right ) }{2\,{b}^{2} \left ( \left ( \tan \left ( x \right ) \right ) ^{2}a+ \left ( \tan \left ( x \right ) \right ) ^{2}b+a \right ) }}+{\frac{\tan \left ( x \right ) }{b \left ( \left ( \tan \left ( x \right ) \right ) ^{2}a+ \left ( \tan \left ( x \right ) \right ) ^{2}b+a \right ) }}+{\frac{\tan \left ( x \right ) }{2\,a \left ( \left ( \tan \left ( x \right ) \right ) ^{2}a+ \left ( \tan \left ( x \right ) \right ) ^{2}b+a \right ) }}-2\,{\frac{{a}^{2}}{{b}^{3}\sqrt{a \left ( a+b \right ) }}\arctan \left ({\frac{ \left ( a+b \right ) \tan \left ( x \right ) }{\sqrt{a \left ( a+b \right ) }}} \right ) }-{\frac{7\,a}{2\,{b}^{2}}\arctan \left ({ \left ( a+b \right ) \tan \left ( x \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}}-{\frac{1}{b}\arctan \left ({ \left ( a+b \right ) \tan \left ( x \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}}+{\frac{1}{2\,a}\arctan \left ({ \left ( a+b \right ) \tan \left ( x \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^6/(a+b*sin(x)^2)^2,x)

[Out]

1/2/b^2*tan(x)/(tan(x)^2+1)+5/2/b^2*arctan(tan(x))+2/b^3*arctan(tan(x))*a+1/2/b^2*a*tan(x)/(tan(x)^2*a+tan(x)^
2*b+a)+1/b*tan(x)/(tan(x)^2*a+tan(x)^2*b+a)+1/2/a*tan(x)/(tan(x)^2*a+tan(x)^2*b+a)-2/b^3/(a*(a+b))^(1/2)*arcta
n((a+b)*tan(x)/(a*(a+b))^(1/2))*a^2-7/2/b^2/(a*(a+b))^(1/2)*arctan((a+b)*tan(x)/(a*(a+b))^(1/2))*a-1/b/(a*(a+b
))^(1/2)*arctan((a+b)*tan(x)/(a*(a+b))^(1/2))+1/2/a/(a*(a+b))^(1/2)*arctan((a+b)*tan(x)/(a*(a+b))^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^6/(a+b*sin(x)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.40978, size = 1127, normalized size = 9.97 \begin{align*} \left [\frac{4 \,{\left (4 \, a^{2} b + 5 \, a b^{2}\right )} x \cos \left (x\right )^{2} +{\left (4 \, a^{3} + 7 \, a^{2} b + 2 \, a b^{2} - b^{3} -{\left (4 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (x\right )^{2}\right )} \sqrt{-\frac{a + b}{a}} \log \left (\frac{{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (x\right )^{4} - 2 \,{\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (x\right )^{2} - 4 \,{\left ({\left (2 \, a^{2} + a b\right )} \cos \left (x\right )^{3} -{\left (a^{2} + a b\right )} \cos \left (x\right )\right )} \sqrt{-\frac{a + b}{a}} \sin \left (x\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (x\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (x\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - 4 \,{\left (4 \, a^{3} + 9 \, a^{2} b + 5 \, a b^{2}\right )} x + 4 \,{\left (a b^{2} \cos \left (x\right )^{3} -{\left (2 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{8 \,{\left (a b^{4} \cos \left (x\right )^{2} - a^{2} b^{3} - a b^{4}\right )}}, \frac{2 \,{\left (4 \, a^{2} b + 5 \, a b^{2}\right )} x \cos \left (x\right )^{2} -{\left (4 \, a^{3} + 7 \, a^{2} b + 2 \, a b^{2} - b^{3} -{\left (4 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (x\right )^{2}\right )} \sqrt{\frac{a + b}{a}} \arctan \left (\frac{{\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a - b\right )} \sqrt{\frac{a + b}{a}}}{2 \,{\left (a + b\right )} \cos \left (x\right ) \sin \left (x\right )}\right ) - 2 \,{\left (4 \, a^{3} + 9 \, a^{2} b + 5 \, a b^{2}\right )} x + 2 \,{\left (a b^{2} \cos \left (x\right )^{3} -{\left (2 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{4 \,{\left (a b^{4} \cos \left (x\right )^{2} - a^{2} b^{3} - a b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^6/(a+b*sin(x)^2)^2,x, algorithm="fricas")

[Out]

[1/8*(4*(4*a^2*b + 5*a*b^2)*x*cos(x)^2 + (4*a^3 + 7*a^2*b + 2*a*b^2 - b^3 - (4*a^2*b + 3*a*b^2 - b^3)*cos(x)^2
)*sqrt(-(a + b)/a)*log(((8*a^2 + 8*a*b + b^2)*cos(x)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(x)^2 - 4*((2*a^2 + a*b)*c
os(x)^3 - (a^2 + a*b)*cos(x))*sqrt(-(a + b)/a)*sin(x) + a^2 + 2*a*b + b^2)/(b^2*cos(x)^4 - 2*(a*b + b^2)*cos(x
)^2 + a^2 + 2*a*b + b^2)) - 4*(4*a^3 + 9*a^2*b + 5*a*b^2)*x + 4*(a*b^2*cos(x)^3 - (2*a^2*b + 3*a*b^2 + b^3)*co
s(x))*sin(x))/(a*b^4*cos(x)^2 - a^2*b^3 - a*b^4), 1/4*(2*(4*a^2*b + 5*a*b^2)*x*cos(x)^2 - (4*a^3 + 7*a^2*b + 2
*a*b^2 - b^3 - (4*a^2*b + 3*a*b^2 - b^3)*cos(x)^2)*sqrt((a + b)/a)*arctan(1/2*((2*a + b)*cos(x)^2 - a - b)*sqr
t((a + b)/a)/((a + b)*cos(x)*sin(x))) - 2*(4*a^3 + 9*a^2*b + 5*a*b^2)*x + 2*(a*b^2*cos(x)^3 - (2*a^2*b + 3*a*b
^2 + b^3)*cos(x))*sin(x))/(a*b^4*cos(x)^2 - a^2*b^3 - a*b^4)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)**6/(a+b*sin(x)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.12205, size = 236, normalized size = 2.09 \begin{align*} \frac{{\left (4 \, a + 5 \, b\right )} x}{2 \, b^{3}} - \frac{{\left (4 \, a^{3} + 7 \, a^{2} b + 2 \, a b^{2} - b^{3}\right )}{\left (\pi \left \lfloor \frac{x}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac{a \tan \left (x\right ) + b \tan \left (x\right )}{\sqrt{a^{2} + a b}}\right )\right )}}{2 \, \sqrt{a^{2} + a b} a b^{3}} + \frac{2 \, a^{2} \tan \left (x\right )^{3} + 3 \, a b \tan \left (x\right )^{3} + b^{2} \tan \left (x\right )^{3} + 2 \, a^{2} \tan \left (x\right ) + 2 \, a b \tan \left (x\right ) + b^{2} \tan \left (x\right )}{2 \,{\left (a \tan \left (x\right )^{4} + b \tan \left (x\right )^{4} + 2 \, a \tan \left (x\right )^{2} + b \tan \left (x\right )^{2} + a\right )} a b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^6/(a+b*sin(x)^2)^2,x, algorithm="giac")

[Out]

1/2*(4*a + 5*b)*x/b^3 - 1/2*(4*a^3 + 7*a^2*b + 2*a*b^2 - b^3)*(pi*floor(x/pi + 1/2)*sgn(2*a + 2*b) + arctan((a
*tan(x) + b*tan(x))/sqrt(a^2 + a*b)))/(sqrt(a^2 + a*b)*a*b^3) + 1/2*(2*a^2*tan(x)^3 + 3*a*b*tan(x)^3 + b^2*tan
(x)^3 + 2*a^2*tan(x) + 2*a*b*tan(x) + b^2*tan(x))/((a*tan(x)^4 + b*tan(x)^4 + 2*a*tan(x)^2 + b*tan(x)^2 + a)*a
*b^2)