3.1.36 \(\int \frac {\cos ^4(x)}{a+b \cos ^2(x)} \, dx\) [36]

Optimal. Leaf size=60 \[ -\frac {(2 a-b) x}{2 b^2}-\frac {a^{3/2} \text {ArcTan}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{b^2 \sqrt {a+b}}+\frac {\cos (x) \sin (x)}{2 b} \]

[Out]

-1/2*(2*a-b)*x/b^2+1/2*cos(x)*sin(x)/b-a^(3/2)*arctan(cot(x)*(a+b)^(1/2)/a^(1/2))/b^2/(a+b)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 60, normalized size of antiderivative = 1.00, 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 = {3266, 481, 536, 209, 211} \begin {gather*} -\frac {a^{3/2} \text {ArcTan}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{b^2 \sqrt {a+b}}-\frac {x (2 a-b)}{2 b^2}+\frac {\sin (x) \cos (x)}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[x]^4/(a + b*Cos[x]^2),x]

[Out]

-1/2*((2*a - b)*x)/b^2 - (a^(3/2)*ArcTan[(Sqrt[a + b]*Cot[x])/Sqrt[a]])/(b^2*Sqrt[a + b]) + (Cos[x]*Sin[x])/(2
*b)

Rule 209

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

Rule 211

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

Rule 481

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-a)*e^(
2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Dist[e^
(2*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1)
+ (a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 536

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 3266

Int[sin[(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^(m + 1)/f, Subst[Int[x^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p +
1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]
time = 0.14, size = 52, normalized size = 0.87 \begin {gather*} \frac {2 (-2 a+b) x+\frac {4 a^{3/2} \text {ArcTan}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+b \sin (2 x)}{4 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.13, size = 59, normalized size = 0.98

method	result	size
default	\(-\frac {-\frac {b \tan \left (x \right )}{2 \left (1+\tan ^{2}\left (x \right )\right )}+\frac {\left (2 a -b \right ) \arctan \left (\tan \left (x \right )\right )}{2}}{b^{2}}+\frac {a^{2} \arctan \left (\frac {a \tan \left (x \right )}{\sqrt {\left (a +b \right ) a}}\right )}{b^{2} \sqrt {\left (a +b \right ) a}}\)	\(59\)
risch	\(-\frac {a x}{b^{2}}+\frac {x}{2 b}-\frac {i {\mathrm e}^{2 i x}}{8 b}+\frac {i {\mathrm e}^{-2 i x}}{8 b}+\frac {\sqrt {-\left (a +b \right ) a}\, a \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-\left (a +b \right ) a}-2 a -b}{b}\right )}{2 \left (a +b \right ) b^{2}}-\frac {\sqrt {-\left (a +b \right ) a}\, a \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-\left (a +b \right ) a}+2 a +b}{b}\right )}{2 \left (a +b \right ) b^{2}}\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^4/(a+b*cos(x)^2),x,method=_RETURNVERBOSE)

[Out]

-1/b^2*(-1/2*b*tan(x)/(tan(x)^2+1)+1/2*(2*a-b)*arctan(tan(x)))+1/b^2*a^2/((a+b)*a)^(1/2)*arctan(a*tan(x)/((a+b
)*a)^(1/2))

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Maxima [A]
time = 0.47, size = 54, normalized size = 0.90 \begin {gather*} \frac {a^{2} \arctan \left (\frac {a \tan \left (x\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} b^{2}} - \frac {{\left (2 \, a - b\right )} x}{2 \, b^{2}} + \frac {\tan \left (x\right )}{2 \, {\left (b \tan \left (x\right )^{2} + b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.46, size = 213, normalized size = 3.55 \begin {gather*} \left [\frac {2 \, b \cos \left (x\right ) \sin \left (x\right ) + a \sqrt {-\frac {a}{a + b}} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (x\right )^{4} - 2 \, {\left (4 \, a^{2} + 3 \, a b\right )} \cos \left (x\right )^{2} - 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (x\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (x\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (x\right ) + a^{2}}{b^{2} \cos \left (x\right )^{4} + 2 \, a b \cos \left (x\right )^{2} + a^{2}}\right ) - 2 \, {\left (2 \, a - b\right )} x}{4 \, b^{2}}, \frac {b \cos \left (x\right ) \sin \left (x\right ) - a \sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (x\right ) \sin \left (x\right )}\right ) - {\left (2 \, a - b\right )} x}{2 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)**4/(a+b*cos(x)**2),x)

[Out]

Timed out

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Giac [A]
time = 0.41, size = 72, normalized size = 1.20 \begin {gather*} \frac {{\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (x\right )}{\sqrt {a^{2} + a b}}\right )\right )} a^{2}}{\sqrt {a^{2} + a b} b^{2}} - \frac {{\left (2 \, a - b\right )} x}{2 \, b^{2}} + \frac {\tan \left (x\right )}{2 \, {\left (\tan \left (x\right )^{2} + 1\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.61, size = 291, normalized size = 4.85 \begin {gather*} -\frac {2\,a^2\,\mathrm {atan}\left (\frac {\sin \left (x\right )}{\cos \left (x\right )}\right )-b^2\,\mathrm {atan}\left (\frac {\sin \left (x\right )}{\cos \left (x\right )}\right )-\frac {b^2\,\sin \left (2\,x\right )}{2}+a\,b\,\mathrm {atan}\left (\frac {\sin \left (x\right )}{\cos \left (x\right )}\right )-\frac {a\,b\,\sin \left (2\,x\right )}{2}+\mathrm {atan}\left (\frac {a\,\sin \left (x\right )\,{\left (-a^4-b\,a^3\right )}^{3/2}\,8{}\mathrm {i}+b\,\sin \left (x\right )\,{\left (-a^4-b\,a^3\right )}^{3/2}\,4{}\mathrm {i}+a^5\,\sin \left (x\right )\,\sqrt {-a^4-b\,a^3}\,8{}\mathrm {i}-a^2\,b^3\,\sin \left (x\right )\,\sqrt {-a^4-b\,a^3}\,2{}\mathrm {i}+a^3\,b^2\,\sin \left (x\right )\,\sqrt {-a^4-b\,a^3}\,1{}\mathrm {i}+a\,b^4\,\sin \left (x\right )\,\sqrt {-a^4-b\,a^3}\,1{}\mathrm {i}+a^4\,b\,\sin \left (x\right )\,\sqrt {-a^4-b\,a^3}\,12{}\mathrm {i}}{3\,\cos \left (x\right )\,a^5\,b^2+5\,\cos \left (x\right )\,a^4\,b^3+\cos \left (x\right )\,a^3\,b^4-\cos \left (x\right )\,a^2\,b^5}\right )\,\sqrt {-a^4-b\,a^3}\,2{}\mathrm {i}}{2\,b^3+2\,a\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^4/(a + b*cos(x)^2),x)

[Out]

-(2*a^2*atan(sin(x)/cos(x)) - b^2*atan(sin(x)/cos(x)) + atan((a*sin(x)*(- a^3*b - a^4)^(3/2)*8i + b*sin(x)*(-
a^3*b - a^4)^(3/2)*4i + a^5*sin(x)*(- a^3*b - a^4)^(1/2)*8i - a^2*b^3*sin(x)*(- a^3*b - a^4)^(1/2)*2i + a^3*b^
2*sin(x)*(- a^3*b - a^4)^(1/2)*1i + a*b^4*sin(x)*(- a^3*b - a^4)^(1/2)*1i + a^4*b*sin(x)*(- a^3*b - a^4)^(1/2)
*12i)/(a^3*b^4*cos(x) - a^2*b^5*cos(x) + 5*a^4*b^3*cos(x) + 3*a^5*b^2*cos(x)))*(- a^3*b - a^4)^(1/2)*2i - (b^2
*sin(2*x))/2 + a*b*atan(sin(x)/cos(x)) - (a*b*sin(2*x))/2)/(2*a*b^2 + 2*b^3)

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