3.350 \(\int \frac {\cos ^2(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\)

Optimal. Leaf size=114 \[ \frac {(a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac {b}{a}\right .\right )}{b f \sqrt {a+b \sin ^2(e+f x)}}-\frac {a \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} E\left (e+f x\left |-\frac {b}{a}\right .\right )}{b f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

-a*(cos(f*x+e)^2)^(1/2)/cos(f*x+e)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*(1+b*sin(f*x+e)^2/a)^(1/2)/b/f/(a+b*sin(
f*x+e)^2)^(1/2)+(a+b)*(cos(f*x+e)^2)^(1/2)/cos(f*x+e)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*(1+b*sin(f*x+e)^2/a)^
(1/2)/b/f/(a+b*sin(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 153, normalized size of antiderivative = 1.34, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3192, 423, 426, 424, 421, 419} \[ \frac {(a+b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{b f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{b f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^2/Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

-((Sqrt[Cos[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(b*f*
Sqrt[1 + (b*Sin[e + f*x]^2)/a])) + ((a + b)*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e
+ f*x]*Sqrt[1 + (b*Sin[e + f*x]^2)/a])/(b*f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 421

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d*x^2)/c]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d*x^2)/c]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 423

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[b/d, Int[Sqrt[c + d*x^2]/Sqrt[a + b
*x^2], x], x] - Dist[(b*c - a*d)/d, Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d}, x]
&& PosQ[d/c] && NegQ[b/a]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 426

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b*x^2)/a]
, Int[Sqrt[1 + (b*x^2)/a]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 3192

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[(ff*Sqrt[Cos[e + f*x]^2])/(f*Cos[e + f*x]), 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/2] &&  !IntegerQ
[p]

Rubi steps

\begin {align*} \int \frac {\cos ^2(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx &=\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1-x^2}}{\sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{b f}+\frac {\left ((a+b) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{b f}\\ &=-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {b x^2}{a}}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{b f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\left ((a+b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {b x^2}{a}}} \, dx,x,\sin (e+f x)\right )}{b f \sqrt {a+b \sin ^2(e+f x)}}\\ &=-\frac {\sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{b f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {(a+b) \sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{b f \sqrt {a+b \sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 83, normalized size = 0.73 \[ \frac {\sqrt {\frac {2 a-b \cos (2 (e+f x))+b}{a}} \left ((a+b) F\left (e+f x\left |-\frac {b}{a}\right .\right )-a E\left (e+f x\left |-\frac {b}{a}\right .\right )\right )}{b f \sqrt {2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^2/Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

(Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*(-(a*EllipticE[e + f*x, -(b/a)]) + (a + b)*EllipticF[e + f*x, -(b/a)])
)/(b*f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \cos \left (f x + e\right )^{2}}{b \cos \left (f x + e\right )^{2} - a - b}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-b*cos(f*x + e)^2 + a + b)*cos(f*x + e)^2/(b*cos(f*x + e)^2 - a - b), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (f x + e\right )^{2}}{\sqrt {b \sin \left (f x + e\right )^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(cos(f*x + e)^2/sqrt(b*sin(f*x + e)^2 + a), x)

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maple [A]  time = 1.14, size = 111, normalized size = 0.97 \[ \frac {\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \left (\EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a +\EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -\EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \right )}{b \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2/(a+b*sin(f*x+e)^2)^(1/2),x)

[Out]

(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*(EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a+EllipticF(sin(f*x+e)
,(-1/a*b)^(1/2))*b-EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a)/b/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (f x + e\right )^{2}}{\sqrt {b \sin \left (f x + e\right )^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(cos(f*x + e)^2/sqrt(b*sin(f*x + e)^2 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (e+f\,x\right )}^2}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(e + f*x)^2/(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

int(cos(e + f*x)^2/(a + b*sin(e + f*x)^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{2}{\left (e + f x \right )}}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**2/(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

Integral(cos(e + f*x)**2/sqrt(a + b*sin(e + f*x)**2), x)

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