3.4.15 \(\int \frac {\cos ^5(x)}{(a+b \sin ^2(x))^2} \, dx\) [315]

Optimal. Leaf size=72 \[ -\frac {(3 a-b) (a+b) \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{5/2}}+\frac {\sin (x)}{b^2}+\frac {(a+b)^2 \sin (x)}{2 a b^2 \left (a+b \sin ^2(x)\right )} \]

[Out]

-1/2*(3*a-b)*(a+b)*arctan(sin(x)*b^(1/2)/a^(1/2))/a^(3/2)/b^(5/2)+sin(x)/b^2+1/2*(a+b)^2*sin(x)/a/b^2/(a+b*sin
(x)^2)

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Rubi [A]
time = 0.08, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3269, 398, 393, 211} \begin {gather*} -\frac {(3 a-b) (a+b) \text {ArcTan}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{2 a^{3/2} b^{5/2}}+\frac {(a+b)^2 \sin (x)}{2 a b^2 \left (a+b \sin ^2(x)\right )}+\frac {\sin (x)}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[x]^5/(a + b*Sin[x]^2)^2,x]

[Out]

-1/2*((3*a - b)*(a + b)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(a^(3/2)*b^(5/2)) + Sin[x]/b^2 + ((a + b)^2*Sin[x])/
(2*a*b^2*(a + b*Sin[x]^2))

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 393

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

Rule 398

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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

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Mathematica [A]
time = 0.26, size = 118, normalized size = 1.64 \begin {gather*} \frac {\frac {\left (3 a^2+2 a b-b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a} \csc (x)}{\sqrt {b}}\right )}{a^{3/2}}+\frac {\left (-3 a^2-2 a b+b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{a^{3/2}}+4 \sqrt {b} \sin (x)+\frac {4 \sqrt {b} (a+b)^2 \sin (x)}{a (2 a+b-b \cos (2 x))}}{4 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^5/(a + b*Sin[x]^2)^2,x]

[Out]

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

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Maple [A]
time = 0.32, size = 77, normalized size = 1.07

method	result	size
default	\(\frac {\sin \left (x \right )}{b^{2}}-\frac {-\frac {\left (a^{2}+2 a b +b^{2}\right ) \sin \left (x \right )}{2 a \left (a +b \left (\sin ^{2}\left (x \right )\right )\right )}+\frac {\left (3 a^{2}+2 a b -b^{2}\right ) \arctan \left (\frac {b \sin \left (x \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}}{b^{2}}\)	\(77\)
risch	\(-\frac {i {\mathrm e}^{i x}}{2 b^{2}}+\frac {i {\mathrm e}^{-i x}}{2 b^{2}}-\frac {i \left (a^{2}+2 a b +b^{2}\right ) \left ({\mathrm e}^{3 i x}-{\mathrm e}^{i x}\right )}{a \,b^{2} \left (-b \,{\mathrm e}^{4 i x}+4 a \,{\mathrm e}^{2 i x}+2 b \,{\mathrm e}^{2 i x}-b \right )}-\frac {3 a \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{4 \sqrt {-a b}\, b^{2}}-\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{2 \sqrt {-a b}\, b}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{4 \sqrt {-a b}\, a}+\frac {3 a \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{4 \sqrt {-a b}\, b^{2}}+\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{2 \sqrt {-a b}\, b}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{4 \sqrt {-a b}\, a}\)	\(293\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^5/(a+b*sin(x)^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.48, size = 79, normalized size = 1.10 \begin {gather*} \frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \left (x\right )}{2 \, {\left (a b^{3} \sin \left (x\right )^{2} + a^{2} b^{2}\right )}} + \frac {\sin \left (x\right )}{b^{2}} - \frac {{\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^5/(a+b*sin(x)^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (60) = 120\).
time = 0.42, size = 296, normalized size = 4.11 \begin {gather*} \left [-\frac {{\left (3 \, a^{3} + 5 \, a^{2} b + a b^{2} - b^{3} - {\left (3 \, a^{2} b + 2 \, a b^{2} - b^{3}\right )} \cos \left (x\right )^{2}\right )} \sqrt {-a b} \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 \, {\left (2 \, a^{2} b^{2} \cos \left (x\right )^{2} - 3 \, a^{3} b - 4 \, a^{2} b^{2} - a b^{3}\right )} \sin \left (x\right )}{4 \, {\left (a^{2} b^{4} \cos \left (x\right )^{2} - a^{3} b^{3} - a^{2} b^{4}\right )}}, \frac {{\left (3 \, a^{3} + 5 \, a^{2} b + a b^{2} - b^{3} - {\left (3 \, a^{2} b + 2 \, a b^{2} - b^{3}\right )} \cos \left (x\right )^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} \sin \left (x\right )}{a}\right ) + {\left (2 \, a^{2} b^{2} \cos \left (x\right )^{2} - 3 \, a^{3} b - 4 \, a^{2} b^{2} - a b^{3}\right )} \sin \left (x\right )}{2 \, {\left (a^{2} b^{4} \cos \left (x\right )^{2} - a^{3} b^{3} - a^{2} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^5/(a+b*sin(x)^2)^2,x, algorithm="fricas")

[Out]

[-1/4*((3*a^3 + 5*a^2*b + a*b^2 - b^3 - (3*a^2*b + 2*a*b^2 - b^3)*cos(x)^2)*sqrt(-a*b)*log(-(b*cos(x)^2 + 2*sq
rt(-a*b)*sin(x) + a - b)/(b*cos(x)^2 - a - b)) - 2*(2*a^2*b^2*cos(x)^2 - 3*a^3*b - 4*a^2*b^2 - a*b^3)*sin(x))/
(a^2*b^4*cos(x)^2 - a^3*b^3 - a^2*b^4), 1/2*((3*a^3 + 5*a^2*b + a*b^2 - b^3 - (3*a^2*b + 2*a*b^2 - b^3)*cos(x)
^2)*sqrt(a*b)*arctan(sqrt(a*b)*sin(x)/a) + (2*a^2*b^2*cos(x)^2 - 3*a^3*b - 4*a^2*b^2 - a*b^3)*sin(x))/(a^2*b^4
*cos(x)^2 - a^3*b^3 - a^2*b^4)]

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

[Out]

Timed out

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Giac [A]
time = 0.48, size = 82, normalized size = 1.14 \begin {gather*} \frac {\sin \left (x\right )}{b^{2}} - \frac {{\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a b^{2}} + \frac {a^{2} \sin \left (x\right ) + 2 \, a b \sin \left (x\right ) + b^{2} \sin \left (x\right )}{2 \, {\left (b \sin \left (x\right )^{2} + a\right )} a b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^5/(a+b*sin(x)^2)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 14.40, size = 96, normalized size = 1.33 \begin {gather*} \frac {\sin \left (x\right )}{b^2}+\frac {\sin \left (x\right )\,\left (a^2+2\,a\,b+b^2\right )}{2\,a\,\left (b^3\,{\sin \left (x\right )}^2+a\,b^2\right )}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sin \left (x\right )\,\left (a+b\right )\,\left (3\,a-b\right )}{\sqrt {a}\,\left (3\,a^2+2\,a\,b-b^2\right )}\right )\,\left (a+b\right )\,\left (3\,a-b\right )}{2\,a^{3/2}\,b^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^5/(a + b*sin(x)^2)^2,x)

[Out]

sin(x)/b^2 + (sin(x)*(2*a*b + a^2 + b^2))/(2*a*(b^3*sin(x)^2 + a*b^2)) - (atan((b^(1/2)*sin(x)*(a + b)*(3*a -
b))/(a^(1/2)*(2*a*b + 3*a^2 - b^2)))*(a + b)*(3*a - b))/(2*a^(3/2)*b^(5/2))

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