∫ cos(e+fx)___∘ a+b sec2(e+fx)dx

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

Optimal. Leaf size=105 \[ \frac{\sqrt{a+b} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )|\frac{a+b}{a}\right )}{\sqrt{a} f \sqrt{\cos ^2(e+f x)} \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}} \]

[Out]

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

b)])/(Sqrt[a]*f*Sqrt[Cos[e + f*x]^2]*Sqrt[Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)])

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Rubi [A] time = 0.166239, antiderivative size = 128, normalized size of antiderivative = 1.22, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4148, 6722, 1974, 427, 424} \[ \frac{\sqrt{a+b} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{a \cos ^2(e+f x)+b} E\left (\sin ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )|\frac{a+b}{a}\right )}{\sqrt{a} f \sqrt{\cos ^2(e+f x)} \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

[e + f*x]^2])

Rule 4148

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fr

eeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + b/(1 - ff^2*x^2)^(n/2))^p/(1 - ff^2*x^2)^((m + 1)/2), x

], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] && !IntegerQ

[p]

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/

v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] && !IntegerQ[p] && ILtQ[n, 0] &

& BinomialQ[v, x] && !LinearQ[v, x]

Rule 1974

Int[(u_)^(p_.)*(v_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{p, q}, x] &&

BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] && !BinomialMatchQ[{u, v}, x]

Rule 427

Int[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[Sqrt[a + b*x^2]/Sqrt[1 + (d*x^2)/c], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && !GtQ[c, 0]

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]

Rubi steps

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

Mathematica [C] time = 5.27637, size = 279, normalized size = 2.66 \[ \frac{\sin (e+f x) \csc (2 (e+f x)) \left (a^2 \sqrt{-\frac{1}{a+b}} \sqrt{\frac{a \cos (2 (e+f x))+a+2 b}{a+b}} \text{EllipticF}\left (e+f x,\frac{a}{a+b}\right )-2 i \csc (2 (e+f x)) \sqrt{\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{-\frac{a \cos ^2(e+f x)}{b}} \sqrt{a \cos (2 (e+f x))+a+2 b} \left (a \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{1}{a+b}} \sqrt{a \cos (2 (e+f x))+a+2 b}}{\sqrt{2}}\right ),\frac{a+b}{b}\right )+2 b E\left (i \sinh ^{-1}\left (\frac{\sqrt{-\frac{1}{a+b}} \sqrt{\cos (2 (e+f x)) a+a+2 b}}{\sqrt{2}}\right )|\frac{a+b}{b}\right )\right )\right )}{\sqrt{2} a^2 f \sqrt{-\frac{1}{a+b}} \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

[e + f*x, a/(a + b)] - (2*I)*Sqrt[-((a*Cos[e + f*x]^2)/b)]*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*Csc[2*(e + f*x)]

*(2*b*EllipticE[I*ArcSinh[(Sqrt[-(a + b)^(-1)]*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]])/Sqrt[2]], (a + b)/b] + a*El

lipticF[I*ArcSinh[(Sqrt[-(a + b)^(-1)]*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]])/Sqrt[2]], (a + b)/b])*Sqrt[(a*Sin[e

+ f*x]^2)/(a + b)]))/(Sqrt[2]*a^2*Sqrt[-(a + b)^(-1)]*f*Sqrt[a + b*Sec[e + f*x]^2])

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Maple [C] time = 0.424, size = 2985, normalized size = 28.4 \begin{align*} \text{output too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

*b^(3/2)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2

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

1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^

2+6*a*b-b^2)/(a+b)^2)^(1/2))+2*I*a^(3/2)*b^(1/2)*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1

/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+

e)-b)/(1+cos(f*x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*

I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*sin(f*x+e)-2*I*a^(1/2)*b^(3/2)*2^(1/2)*(1

/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(

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

*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^

2)^(1/2))*sin(f*x+e)+4*I*cos(f*x+e)*a^(1/2)*b^(3/2)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)-4*I*a^(1/2)*b^(3/2

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

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

b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a^(3/2)*b^(1/2)

*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)-cos(f*

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

)*EllipticE((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1

/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*

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

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

(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*2^(1/2)*(1/(a+b)*(I

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

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

cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+

6*a*b-b^2)/(a+b)^2)^(1/2))*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1

+cos(f*x+e)))^(1/2)*b^2+4*cos(f*x+e)*sin(f*x+e)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*co

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

,(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1

/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*a*b+4*I*cos(f*x+e)^3*a^(3/2)*b^(1/2)*((2*I

*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)-2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+

e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f

*x+e)))^(1/2)*EllipticE((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1

/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a^2*sin(f*x+e)-2*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)

*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(

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

^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b*sin(f*x+e)-2^(

1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*

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

)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)

/(a+b)^2)^(1/2))*b^2*sin(f*x+e)+4*2^(1/2)*(1/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e

)+b)/(1+cos(f*x+e)))^(1/2)*(-2/(a+b)*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*

x+e)))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/

2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b*sin(f*x+e)-4*I*cos(f*x+e)^2*a^(3/2)*b^(1/2)*((2*I*a^

(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)-2*cos(f*x+e)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2+2*cos(f*x+e)^3*((2*

I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b+2*cos(f*x+e)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2-2*cos(f*x+e

)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b-2*cos(f*x+e)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b+2*cos

(f*x+e)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^2+2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b-2*((2*I*a^(1

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

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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (f x + e\right )}{\sqrt{b \sec \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\cos \left (f x + e\right )}{\sqrt{b \sec \left (f x + e\right )^{2} + a}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos{\left (e + f x \right )}}{\sqrt{a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (f x + e\right )}{\sqrt{b \sec \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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