3.316 \(\int \frac{\cos ^4(x)}{(a+b \sin ^2(x))^2} \, dx\)

Optimal. Leaf size=75 \[ -\frac{(2 a-b) \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{2 a^{3/2} b^2}+\frac{(a+b) \tan (x)}{2 a b \left ((a+b) \tan ^2(x)+a\right )}+\frac{x}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.105546, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3191, 414, 522, 203, 205} \[ -\frac{(2 a-b) \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{2 a^{3/2} b^2}+\frac{(a+b) \tan (x)}{2 a b \left ((a+b) \tan ^2(x)+a\right )}+\frac{x}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^4/(a + b*Sin[x]^2)^2,x]

[Out]

x/b^2 - ((2*a - b)*Sqrt[a + b]*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(2*a^(3/2)*b^2) + ((a + b)*Tan[x])/(2*a*b
*(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 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 ^4(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )^2} \, dx,x,\tan (x)\right )\\ &=\frac{(a+b) \tan (x)}{2 a b \left (a+(a+b) \tan ^2(x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{a-b+(-a-b) x^2}{\left (1+x^2\right ) \left (a+(a+b) x^2\right )} \, dx,x,\tan (x)\right )}{2 a b}\\ &=\frac{(a+b) \tan (x)}{2 a b \left (a+(a+b) \tan ^2(x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (x)\right )}{b^2}-\frac{((2 a-b) (a+b)) \operatorname{Subst}\left (\int \frac{1}{a+(a+b) x^2} \, dx,x,\tan (x)\right )}{2 a b^2}\\ &=\frac{x}{b^2}-\frac{(2 a-b) \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{2 a^{3/2} b^2}+\frac{(a+b) \tan (x)}{2 a b \left (a+(a+b) \tan ^2(x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.311862, size = 79, normalized size = 1.05 \[ \frac{\frac{\left (-2 a^2-a b+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{a^{3/2} \sqrt{a+b}}+\frac{b (a+b) \sin (2 x)}{a (2 a-b \cos (2 x)+b)}+2 x}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^4/(a + b*Sin[x]^2)^2,x]

[Out]

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

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Maple [B]  time = 0.089, size = 132, normalized size = 1.8 \begin{align*}{\frac{\tan \left ( x \right ) }{2\,b \left ( \left ( \tan \left ( x \right ) \right ) ^{2}a+ \left ( \tan \left ( x \right ) \right ) ^{2}b+a \right ) }}-{\frac{a}{{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}{2\,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{\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 ) }}+{\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 ) }}}}+{\frac{x}{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^4/(a+b*sin(x)^2)^2,x)

[Out]

1/2/b*tan(x)/(tan(x)^2*a+tan(x)^2*b+a)-1/b^2/(a*(a+b))^(1/2)*arctan((a+b)*tan(x)/(a*(a+b))^(1/2))*a-1/2/b/(a*(
a+b))^(1/2)*arctan((a+b)*tan(x)/(a*(a+b))^(1/2))+1/2/a*tan(x)/(tan(x)^2*a+tan(x)^2*b+a)+1/2/a/(a*(a+b))^(1/2)*
arctan((a+b)*tan(x)/(a*(a+b))^(1/2))+x/b^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)^4/(a+b*sin(x)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.1054, size = 882, normalized size = 11.76 \begin{align*} \left [\frac{8 \, a b x \cos \left (x\right )^{2} - 4 \,{\left (a b + b^{2}\right )} \cos \left (x\right ) \sin \left (x\right ) -{\left ({\left (2 \, a b - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a^{2} - a b + b^{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 ) - 8 \,{\left (a^{2} + a b\right )} x}{8 \,{\left (a b^{3} \cos \left (x\right )^{2} - a^{2} b^{2} - a b^{3}\right )}}, \frac{4 \, a b x \cos \left (x\right )^{2} - 2 \,{\left (a b + b^{2}\right )} \cos \left (x\right ) \sin \left (x\right ) +{\left ({\left (2 \, a b - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a^{2} - a b + b^{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 ) - 4 \,{\left (a^{2} + a b\right )} x}{4 \,{\left (a b^{3} \cos \left (x\right )^{2} - a^{2} b^{2} - a b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^4/(a+b*sin(x)^2)^2,x, algorithm="fricas")

[Out]

[1/8*(8*a*b*x*cos(x)^2 - 4*(a*b + b^2)*cos(x)*sin(x) - ((2*a*b - b^2)*cos(x)^2 - 2*a^2 - a*b + b^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)*cos(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)) - 8*(a^2 + a*b)*x)/(a*b^3*cos(x)^2 - a^2*b^2 - a*b^3), 1/4*(4*a*b*x*cos(x)^2 - 2*(a*b + b^2)*cos
(x)*sin(x) + ((2*a*b - b^2)*cos(x)^2 - 2*a^2 - a*b + b^2)*sqrt((a + b)/a)*arctan(1/2*((2*a + b)*cos(x)^2 - a -
b)*sqrt((a + b)/a)/((a + b)*cos(x)*sin(x))) - 4*(a^2 + a*b)*x)/(a*b^3*cos(x)^2 - a^2*b^2 - a*b^3)]

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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)**4/(a+b*sin(x)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.1135, size = 147, normalized size = 1.96 \begin{align*} \frac{x}{b^{2}} - \frac{{\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 )}{\left (2 \, a^{2} + a b - b^{2}\right )}}{2 \, \sqrt{a^{2} + a b} a b^{2}} + \frac{a \tan \left (x\right ) + b \tan \left (x\right )}{2 \,{\left (a \tan \left (x\right )^{2} + b \tan \left (x\right )^{2} + a\right )} a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^4/(a+b*sin(x)^2)^2,x, algorithm="giac")

[Out]

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