3.4.5 \(\int \frac {\cos ^3(x)}{a+b \sin ^2(x)} \, dx\) [305]

Optimal. Leaf size=36 \[ \frac {(a+b) \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}-\frac {\sin (x)}{b} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3269, 396, 211} \begin {gather*} \frac {(a+b) \text {ArcTan}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}-\frac {\sin (x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[x]^3/(a + b*Sin[x]^2),x]

[Out]

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

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 396

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

Rule 3269

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

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 36, normalized size = 1.00 \begin {gather*} \frac {(a+b) \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}-\frac {\sin (x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^3/(a + b*Sin[x]^2),x]

[Out]

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

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Maple [A]
time = 0.21, size = 31, normalized size = 0.86

method	result	size
default	\(-\frac {\sin \left (x \right )}{b}+\frac {\left (a +b \right ) \arctan \left (\frac {b \sin \left (x \right )}{\sqrt {a b}}\right )}{b \sqrt {a b}}\)	\(31\)
risch	\(\frac {i {\mathrm e}^{i x}}{2 b}-\frac {i {\mathrm e}^{-i x}}{2 b}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right ) a}{2 \sqrt {-a b}\, b}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{2 \sqrt {-a b}}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right ) a}{2 \sqrt {-a b}\, b}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{2 \sqrt {-a b}}\)	\(156\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^3/(a+b*sin(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 30, normalized size = 0.83 \begin {gather*} \frac {{\left (a + b\right )} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{\sqrt {a b} b} - \frac {\sin \left (x\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^3/(a+b*sin(x)^2),x, algorithm="maxima")

[Out]

(a + b)*arctan(b*sin(x)/sqrt(a*b))/(sqrt(a*b)*b) - sin(x)/b

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Fricas [A]
time = 0.43, size = 101, normalized size = 2.81 \begin {gather*} \left [-\frac {2 \, a b \sin \left (x\right ) + \sqrt {-a b} {\left (a + b\right )} \log \left (-\frac {b \cos \left (x\right )^{2} + 2 \, \sqrt {-a b} \sin \left (x\right ) + a - b}{b \cos \left (x\right )^{2} - a - b}\right )}{2 \, a b^{2}}, -\frac {a b \sin \left (x\right ) - \sqrt {a b} {\left (a + b\right )} \arctan \left (\frac {\sqrt {a b} \sin \left (x\right )}{a}\right )}{a b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^3/(a+b*sin(x)^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]
time = 0.44, size = 30, normalized size = 0.83 \begin {gather*} \frac {{\left (a + b\right )} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{\sqrt {a b} b} - \frac {\sin \left (x\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^3/(a+b*sin(x)^2),x, algorithm="giac")

[Out]

(a + b)*arctan(b*sin(x)/sqrt(a*b))/(sqrt(a*b)*b) - sin(x)/b

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Mupad [B]
time = 0.09, size = 28, normalized size = 0.78 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sin \left (x\right )}{\sqrt {a}}\right )\,\left (a+b\right )}{\sqrt {a}\,b^{3/2}}-\frac {\sin \left (x\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^3/(a + b*sin(x)^2),x)

[Out]

(atan((b^(1/2)*sin(x))/a^(1/2))*(a + b))/(a^(1/2)*b^(3/2)) - sin(x)/b

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