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

Optimal. Leaf size=75 \[ \frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^2 f \sqrt {a+b}}+\frac {x (a-2 b)}{2 a^2}+\frac {\sin (e+f x) \cos (e+f x)}{2 a f} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 75, normalized size of antiderivative = 1.00, 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 = {4146, 414, 522, 203, 205} \[ \frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^2 f \sqrt {a+b}}+\frac {x (a-2 b)}{2 a^2}+\frac {\sin (e+f x) \cos (e+f x)}{2 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4146

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/
2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 67, normalized size = 0.89 \[ \frac {\frac {4 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+2 (a-2 b) (e+f x)+a \sin (2 (e+f x))}{4 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a - 2*b)*(e + f*x) + (4*b^(3/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/Sqrt[a + b] + a*Sin[2*(e + f*x
)])/(4*a^2*f)

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fricas [A]  time = 0.62, size = 272, normalized size = 3.63 \[ \left [\frac {2 \, {\left (a - 2 \, b\right )} f x + 2 \, a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{4 \, a^{2} f}, \frac {{\left (a - 2 \, b\right )} f x + a \cos \left (f x + e\right ) \sin \left (f x + e\right ) - b \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{2 \, a^{2} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*(a - 2*b)*f*x + 2*a*cos(f*x + e)*sin(f*x + e) + b*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x
+ e)^4 - 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 - 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x + e)
)*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)))/(a^2*f), 1/2*((a -
2*b)*f*x + a*cos(f*x + e)*sin(f*x + e) - b*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a
+ b))/(b*cos(f*x + e)*sin(f*x + e))))/(a^2*f)]

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giac [A]  time = 0.23, size = 99, normalized size = 1.32 \[ \frac {\frac {2 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b^{2}}{\sqrt {a b + b^{2}} a^{2}} + \frac {{\left (f x + e\right )} {\left (a - 2 \, b\right )}}{a^{2}} + \frac {\tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )} a}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*b^2/(sqrt(a*b + b^2)*a^2
) + (f*x + e)*(a - 2*b)/a^2 + tan(f*x + e)/((tan(f*x + e)^2 + 1)*a))/f

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maple [A]  time = 1.56, size = 92, normalized size = 1.23 \[ \frac {b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \,a^{2} \sqrt {\left (a +b \right ) b}}+\frac {\tan \left (f x +e \right )}{2 f a \left (1+\tan ^{2}\left (f x +e \right )\right )}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{2 f a}-\frac {\arctan \left (\tan \left (f x +e \right )\right ) b}{f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*b^2/a^2/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))+1/2/f/a*tan(f*x+e)/(tan(f*x+e)^2+1)+1/2/f/a*a
rctan(tan(f*x+e))-1/f/a^2*arctan(tan(f*x+e))*b

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maxima [A]  time = 0.44, size = 72, normalized size = 0.96 \[ \frac {\frac {2 \, b^{2} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{2}} + \frac {{\left (f x + e\right )} {\left (a - 2 \, b\right )}}{a^{2}} + \frac {\tan \left (f x + e\right )}{a \tan \left (f x + e\right )^{2} + a}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.24, size = 373, normalized size = 4.97 \[ -\frac {2\,b^2\,\mathrm {atan}\left (\frac {\sin \left (e+f\,x\right )}{\cos \left (e+f\,x\right )}\right )-a\,\left (\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{2}-b\,\mathrm {atan}\left (\frac {\sin \left (e+f\,x\right )}{\cos \left (e+f\,x\right )}\right )\right )-a^2\,\left (\frac {\sin \left (2\,e+2\,f\,x\right )}{2}+\mathrm {atan}\left (\frac {\sin \left (e+f\,x\right )}{\cos \left (e+f\,x\right )}\right )\right )+\mathrm {atan}\left (\frac {a\,\sin \left (e+f\,x\right )\,{\left (-b^4-a\,b^3\right )}^{3/2}\,4{}\mathrm {i}+b\,\sin \left (e+f\,x\right )\,{\left (-b^4-a\,b^3\right )}^{3/2}\,8{}\mathrm {i}+b^5\,\sin \left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}\,8{}\mathrm {i}+a\,b^4\,\sin \left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}\,12{}\mathrm {i}+a^4\,b\,\sin \left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}\,1{}\mathrm {i}+a^2\,b^3\,\sin \left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}\,1{}\mathrm {i}-a^3\,b^2\,\sin \left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}\,2{}\mathrm {i}}{-\cos \left (e+f\,x\right )\,a^5\,b^2+\cos \left (e+f\,x\right )\,a^4\,b^3+5\,\cos \left (e+f\,x\right )\,a^3\,b^4+3\,\cos \left (e+f\,x\right )\,a^2\,b^5}\right )\,\sqrt {-b^4-a\,b^3}\,2{}\mathrm {i}}{f\,\left (2\,a^3+2\,b\,a^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(atan((a*sin(e + f*x)*(- a*b^3 - b^4)^(3/2)*4i + b*sin(e + f*x)*(- a*b^3 - b^4)^(3/2)*8i + b^5*sin(e + f*x)*(
- a*b^3 - b^4)^(1/2)*8i + a*b^4*sin(e + f*x)*(- a*b^3 - b^4)^(1/2)*12i + a^4*b*sin(e + f*x)*(- a*b^3 - b^4)^(1
/2)*1i + a^2*b^3*sin(e + f*x)*(- a*b^3 - b^4)^(1/2)*1i - a^3*b^2*sin(e + f*x)*(- a*b^3 - b^4)^(1/2)*2i)/(3*a^2
*b^5*cos(e + f*x) + 5*a^3*b^4*cos(e + f*x) + a^4*b^3*cos(e + f*x) - a^5*b^2*cos(e + f*x)))*(- a*b^3 - b^4)^(1/
2)*2i + 2*b^2*atan(sin(e + f*x)/cos(e + f*x)) - a*((b*sin(2*e + 2*f*x))/2 - b*atan(sin(e + f*x)/cos(e + f*x)))
- a^2*(sin(2*e + 2*f*x)/2 + atan(sin(e + f*x)/cos(e + f*x))))/(f*(2*a^2*b + 2*a^3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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