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^2-1/2*(2*a-b)*arctan((a+b)^(1/2)*tan(x)/a^(1/2))*(a+b)^(1/2)/a^(3/2)/b^2+1/2*(a+b)*tan(x)/a/b/(a+(a+b)*tan
(x)^2)

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Rubi [A]  time = 0.11, antiderivative size = 75, 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 = {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 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]

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 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]

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*}

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Mathematica [A]  time = 0.32, 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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fricas [B]  time = 0.46, size = 367, normalized size = 4.89 \[ \left [\frac {8 \, a b x \cos \relax (x)^{2} - 4 \, {\left (a b + b^{2}\right )} \cos \relax (x) \sin \relax (x) - {\left ({\left (2 \, a b - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a^{2} - a b + b^{2}\right )} \sqrt {-\frac {a + b}{a}} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \relax (x)^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \relax (x)^{2} - 4 \, {\left ({\left (2 \, a^{2} + a b\right )} \cos \relax (x)^{3} - {\left (a^{2} + a b\right )} \cos \relax (x)\right )} \sqrt {-\frac {a + b}{a}} \sin \relax (x) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \relax (x)^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \relax (x)^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - 8 \, {\left (a^{2} + a b\right )} x}{8 \, {\left (a b^{3} \cos \relax (x)^{2} - a^{2} b^{2} - a b^{3}\right )}}, \frac {4 \, a b x \cos \relax (x)^{2} - 2 \, {\left (a b + b^{2}\right )} \cos \relax (x) \sin \relax (x) + {\left ({\left (2 \, a b - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a^{2} - a b + b^{2}\right )} \sqrt {\frac {a + b}{a}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \relax (x)^{2} - a - b\right )} \sqrt {\frac {a + b}{a}}}{2 \, {\left (a + b\right )} \cos \relax (x) \sin \relax (x)}\right ) - 4 \, {\left (a^{2} + a b\right )} x}{4 \, {\left (a b^{3} \cos \relax (x)^{2} - a^{2} b^{2} - a b^{3}\right )}}\right ] \]

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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giac [A]  time = 0.16, size = 109, normalized size = 1.45 \[ \frac {x}{b^{2}} - \frac {{\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \relax (x) + b \tan \relax (x)}{\sqrt {a^{2} + a b}}\right )\right )} {\left (2 \, a^{2} + a b - b^{2}\right )}}{2 \, \sqrt {a^{2} + a b} a b^{2}} + \frac {a \tan \relax (x) + b \tan \relax (x)}{2 \, {\left (a \tan \relax (x)^{2} + b \tan \relax (x)^{2} + a\right )} a b} \]

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)

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maple [B]  time = 0.21, size = 132, normalized size = 1.76 \[ \frac {\tan \relax (x )}{2 b \left (\left (\tan ^{2}\relax (x )\right ) a +\left (\tan ^{2}\relax (x )\right ) b +a \right )}-\frac {\arctan \left (\frac {\left (a +b \right ) \tan \relax (x )}{\sqrt {a \left (a +b \right )}}\right ) a}{b^{2} \sqrt {a \left (a +b \right )}}-\frac {\arctan \left (\frac {\left (a +b \right ) \tan \relax (x )}{\sqrt {a \left (a +b \right )}}\right )}{2 b \sqrt {a \left (a +b \right )}}+\frac {\tan \relax (x )}{2 a \left (\left (\tan ^{2}\relax (x )\right ) a +\left (\tan ^{2}\relax (x )\right ) b +a \right )}+\frac {\arctan \left (\frac {\left (a +b \right ) \tan \relax (x )}{\sqrt {a \left (a +b \right )}}\right )}{2 a \sqrt {a \left (a +b \right )}}+\frac {x}{b^{2}} \]

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 [A]  time = 0.95, size = 80, normalized size = 1.07 \[ \frac {{\left (a + b\right )} \tan \relax (x)}{2 \, {\left (a^{2} b + {\left (a^{2} b + a b^{2}\right )} \tan \relax (x)^{2}\right )}} + \frac {x}{b^{2}} - \frac {{\left (2 \, a^{2} + a b - b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \relax (x)}{\sqrt {{\left (a + b\right )} a}}\right )}{2 \, \sqrt {{\left (a + b\right )} a} a b^{2}} \]

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]

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

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mupad [B]  time = 14.33, size = 533, normalized size = 7.11 \[ \frac {\mathrm {atan}\left (\frac {5\,\mathrm {tan}\relax (x)}{2\,\left (\frac {3\,a}{2\,b}+\frac {b}{2\,a}-\frac {b^2}{2\,a^2}+\frac {5}{2}\right )}+\frac {\mathrm {tan}\relax (x)}{2\,\left (\frac {5\,a}{2\,b}-\frac {b}{2\,a}+\frac {3\,a^2}{2\,b^2}+\frac {1}{2}\right )}+\frac {3\,a\,\mathrm {tan}\relax (x)}{2\,\left (\frac {3\,a}{2}+\frac {5\,b}{2}+\frac {b^2}{2\,a}-\frac {b^3}{2\,a^2}\right )}-\frac {b\,\mathrm {tan}\relax (x)}{2\,\left (\frac {a}{2}-\frac {b}{2}+\frac {5\,a^2}{2\,b}+\frac {3\,a^3}{2\,b^2}\right )}\right )}{b^2}+\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\relax (x)\,\sqrt {-a^4-b\,a^3}}{a^2-\frac {3\,a\,b}{2}-\frac {b^2}{2}+\frac {b^3}{4\,a}+\frac {13\,a^3}{4\,b}+\frac {3\,a^4}{2\,b^2}}+\frac {3\,\mathrm {tan}\relax (x)\,\sqrt {-a^4-b\,a^3}}{2\,\left (\frac {13\,a\,b}{4}+\frac {3\,a^2}{2}+b^2-\frac {3\,b^3}{2\,a}-\frac {b^4}{2\,a^2}+\frac {b^5}{4\,a^3}\right )}+\frac {13\,\mathrm {tan}\relax (x)\,\sqrt {-a^4-b\,a^3}}{4\,\left (a\,b+\frac {13\,a^2}{4}-\frac {3\,b^2}{2}-\frac {b^3}{2\,a}+\frac {3\,a^3}{2\,b}+\frac {b^4}{4\,a^2}\right )}-\frac {3\,b\,\mathrm {tan}\relax (x)\,\sqrt {-a^4-b\,a^3}}{2\,\left (a^3-\frac {3\,a^2\,b}{2}-\frac {a\,b^2}{2}+\frac {b^3}{4}+\frac {13\,a^4}{4\,b}+\frac {3\,a^5}{2\,b^2}\right )}-\frac {b^2\,\mathrm {tan}\relax (x)\,\sqrt {-a^4-b\,a^3}}{2\,\left (\frac {a\,b^3}{4}-\frac {3\,a^3\,b}{2}+a^4-\frac {a^2\,b^2}{2}+\frac {13\,a^5}{4\,b}+\frac {3\,a^6}{2\,b^2}\right )}+\frac {b^3\,\mathrm {tan}\relax (x)\,\sqrt {-a^4-b\,a^3}}{4\,\left (a^5-\frac {3\,a^4\,b}{2}+\frac {a^2\,b^3}{4}-\frac {a^3\,b^2}{2}+\frac {13\,a^6}{4\,b}+\frac {3\,a^7}{2\,b^2}\right )}\right )\,\sqrt {-a^3\,\left (a+b\right )}\,\left (2\,a-b\right )}{2\,a^3\,b^2}+\frac {\mathrm {tan}\relax (x)\,\left (a+b\right )}{2\,a\,b\,\left (\left (a+b\right )\,{\mathrm {tan}\relax (x)}^2+a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^4/(a + b*sin(x)^2)^2,x)

[Out]

atan((5*tan(x))/(2*((3*a)/(2*b) + b/(2*a) - b^2/(2*a^2) + 5/2)) + tan(x)/(2*((5*a)/(2*b) - b/(2*a) + (3*a^2)/(
2*b^2) + 1/2)) + (3*a*tan(x))/(2*((3*a)/2 + (5*b)/2 + b^2/(2*a) - b^3/(2*a^2))) - (b*tan(x))/(2*(a/2 - b/2 + (
5*a^2)/(2*b) + (3*a^3)/(2*b^2))))/b^2 + (atanh((tan(x)*(- a^3*b - a^4)^(1/2))/(a^2 - (3*a*b)/2 - b^2/2 + b^3/(
4*a) + (13*a^3)/(4*b) + (3*a^4)/(2*b^2)) + (3*tan(x)*(- a^3*b - a^4)^(1/2))/(2*((13*a*b)/4 + (3*a^2)/2 + b^2 -
(3*b^3)/(2*a) - b^4/(2*a^2) + b^5/(4*a^3))) + (13*tan(x)*(- a^3*b - a^4)^(1/2))/(4*(a*b + (13*a^2)/4 - (3*b^2
)/2 - b^3/(2*a) + (3*a^3)/(2*b) + b^4/(4*a^2))) - (3*b*tan(x)*(- a^3*b - a^4)^(1/2))/(2*(a^3 - (3*a^2*b)/2 - (
a*b^2)/2 + b^3/4 + (13*a^4)/(4*b) + (3*a^5)/(2*b^2))) - (b^2*tan(x)*(- a^3*b - a^4)^(1/2))/(2*((a*b^3)/4 - (3*
a^3*b)/2 + a^4 - (a^2*b^2)/2 + (13*a^5)/(4*b) + (3*a^6)/(2*b^2))) + (b^3*tan(x)*(- a^3*b - a^4)^(1/2))/(4*(a^5
- (3*a^4*b)/2 + (a^2*b^3)/4 - (a^3*b^2)/2 + (13*a^6)/(4*b) + (3*a^7)/(2*b^2))))*(-a^3*(a + b))^(1/2)*(2*a - b
))/(2*a^3*b^2) + (tan(x)*(a + b))/(2*a*b*(a + tan(x)^2*(a + b)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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