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

Optimal. Leaf size=142 \[ \frac {b^{3/2} (5 a+4 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^3 f (a+b)^{3/2}}+\frac {x (a-4 b)}{2 a^3}+\frac {b (a+2 b) \tan (e+f x)}{2 a^2 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )}+\frac {\sin (e+f x) \cos (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

[Out]

1/2*(a-4*b)*x/a^3+1/2*b^(3/2)*(5*a+4*b)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/a^3/(a+b)^(3/2)/f+1/2*cos(f*x+e
)*sin(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2)+1/2*b*(a+2*b)*tan(f*x+e)/a^2/(a+b)/f/(a+b+b*tan(f*x+e)^2)

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Rubi [A]  time = 0.19, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4146, 414, 527, 522, 203, 205} \[ \frac {b^{3/2} (5 a+4 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^3 f (a+b)^{3/2}}+\frac {b (a+2 b) \tan (e+f x)}{2 a^2 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )}+\frac {x (a-4 b)}{2 a^3}+\frac {\sin (e+f x) \cos (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - 4*b)*x)/(2*a^3) + (b^(3/2)*(5*a + 4*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*a^3*(a + b)^(3/2)*
f) + (Cos[e + f*x]*Sin[e + f*x])/(2*a*f*(a + b + b*Tan[e + f*x]^2)) + (b*(a + 2*b)*Tan[e + f*x])/(2*a^2*(a + b
)*f*(a + b + b*Tan[e + f*x]^2))

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 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

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)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {-a+b-3 b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{2 a f}\\ &=\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {b (a+2 b) \tan (e+f x)}{2 a^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {-2 \left (a^2-2 a b-2 b^2\right )-2 b (a+2 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{4 a^2 (a+b) f}\\ &=\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {b (a+2 b) \tan (e+f x)}{2 a^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {(a-4 b) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{2 a^3 f}+\frac {\left (b^2 (5 a+4 b)\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^3 (a+b) f}\\ &=\frac {(a-4 b) x}{2 a^3}+\frac {b^{3/2} (5 a+4 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^3 (a+b)^{3/2} f}+\frac {\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {b (a+2 b) \tan (e+f x)}{2 a^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.33, size = 103, normalized size = 0.73 \[ \frac {\frac {2 b^{3/2} (5 a+4 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}+\sin (2 (e+f x)) \left (\frac {2 a b^2}{(a+b) (a \cos (2 (e+f x))+a+2 b)}+a\right )+2 (a-4 b) (e+f x)}{4 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a - 4*b)*(e + f*x) + (2*b^(3/2)*(5*a + 4*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(3/2) + (a
+ (2*a*b^2)/((a + b)*(a + 2*b + a*Cos[2*(e + f*x)])))*Sin[2*(e + f*x)])/(4*a^3*f)

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fricas [A]  time = 0.79, size = 544, normalized size = 3.83 \[ \left [\frac {4 \, {\left (a^{3} - 3 \, a^{2} b - 4 \, a b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 4 \, {\left (a^{2} b - 3 \, a b^{2} - 4 \, b^{3}\right )} f x + {\left (5 \, a b^{2} + 4 \, b^{3} + {\left (5 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) + 4 \, {\left ({\left (a^{3} + a^{2} b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} b + 2 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, {\left ({\left (a^{5} + a^{4} b\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b + a^{3} b^{2}\right )} f\right )}}, \frac {2 \, {\left (a^{3} - 3 \, a^{2} b - 4 \, a b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 2 \, {\left (a^{2} b - 3 \, a b^{2} - 4 \, b^{3}\right )} f x - {\left (5 \, a b^{2} + 4 \, b^{3} + {\left (5 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + 2 \, {\left ({\left (a^{3} + a^{2} b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} b + 2 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{4 \, {\left ({\left (a^{5} + a^{4} b\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b + a^{3} b^{2}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(4*(a^3 - 3*a^2*b - 4*a*b^2)*f*x*cos(f*x + e)^2 + 4*(a^2*b - 3*a*b^2 - 4*b^3)*f*x + (5*a*b^2 + 4*b^3 + (5
*a^2*b + 4*a*b^2)*cos(f*x + e)^2)*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)) + 4*((a^3 + a^2*b)*cos(f*x + e)^3 + (a^2*b +
2*a*b^2)*cos(f*x + e))*sin(f*x + e))/((a^5 + a^4*b)*f*cos(f*x + e)^2 + (a^4*b + a^3*b^2)*f), 1/4*(2*(a^3 - 3*
a^2*b - 4*a*b^2)*f*x*cos(f*x + e)^2 + 2*(a^2*b - 3*a*b^2 - 4*b^3)*f*x - (5*a*b^2 + 4*b^3 + (5*a^2*b + 4*a*b^2)
*cos(f*x + e)^2)*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))) + 2*((a^3 + a^2*b)*cos(f*x + e)^3 + (a^2*b + 2*a*b^2)*cos(f*x + e))*sin(f*x + e))/((a^5 + a^4*b)*f
*cos(f*x + e)^2 + (a^4*b + a^3*b^2)*f)]

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giac [A]  time = 0.25, size = 203, normalized size = 1.43 \[ \frac {\frac {{\left (5 \, a b^{2} + 4 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{4} + a^{3} b\right )} \sqrt {a b + b^{2}}} + \frac {a b \tan \left (f x + e\right )^{3} + 2 \, b^{2} \tan \left (f x + e\right )^{3} + a^{2} \tan \left (f x + e\right ) + 2 \, a b \tan \left (f x + e\right ) + 2 \, b^{2} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{4} + a \tan \left (f x + e\right )^{2} + 2 \, b \tan \left (f x + e\right )^{2} + a + b\right )} {\left (a^{3} + a^{2} b\right )}} + \frac {{\left (f x + e\right )} {\left (a - 4 \, b\right )}}{a^{3}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((5*a*b^2 + 4*b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/((a^4 +
a^3*b)*sqrt(a*b + b^2)) + (a*b*tan(f*x + e)^3 + 2*b^2*tan(f*x + e)^3 + a^2*tan(f*x + e) + 2*a*b*tan(f*x + e) +
2*b^2*tan(f*x + e))/((b*tan(f*x + e)^4 + a*tan(f*x + e)^2 + 2*b*tan(f*x + e)^2 + a + b)*(a^3 + a^2*b)) + (f*x
+ e)*(a - 4*b)/a^3)/f

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maple [A]  time = 1.24, size = 174, normalized size = 1.23 \[ \frac {b^{2} \tan \left (f x +e \right )}{2 f \,a^{2} \left (a +b \right ) \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}+\frac {5 b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 f \,a^{2} \left (a +b \right ) \sqrt {\left (a +b \right ) b}}+\frac {2 b^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \,a^{3} \left (a +b \right ) \sqrt {\left (a +b \right ) b}}+\frac {\tan \left (f x +e \right )}{2 f \,a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{2 f \,a^{2}}-\frac {2 \arctan \left (\tan \left (f x +e \right )\right ) b}{f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*b^2/a^2/(a+b)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)+5/2/f*b^2/a^2/(a+b)/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((
a+b)*b)^(1/2))+2/f*b^3/a^3/(a+b)/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))+1/2/f/a^2*tan(f*x+e)/(ta
n(f*x+e)^2+1)+1/2/f/a^2*arctan(tan(f*x+e))-2/f/a^3*arctan(tan(f*x+e))*b

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maxima [A]  time = 0.45, size = 175, normalized size = 1.23 \[ \frac {\frac {{\left (5 \, a b^{2} + 4 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{4} + a^{3} b\right )} \sqrt {{\left (a + b\right )} b}} + \frac {{\left (a b + 2 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + {\left (a^{2} + 2 \, a b + 2 \, b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{3} b + a^{2} b^{2}\right )} \tan \left (f x + e\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} + {\left (a^{4} + 3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} \tan \left (f x + e\right )^{2}} + \frac {{\left (f x + e\right )} {\left (a - 4 \, b\right )}}{a^{3}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((5*a*b^2 + 4*b^3)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^4 + a^3*b)*sqrt((a + b)*b)) + ((a*b + 2*b^2)
*tan(f*x + e)^3 + (a^2 + 2*a*b + 2*b^2)*tan(f*x + e))/((a^3*b + a^2*b^2)*tan(f*x + e)^4 + a^4 + 2*a^3*b + a^2*
b^2 + (a^4 + 3*a^3*b + 2*a^2*b^2)*tan(f*x + e)^2) + (f*x + e)*(a - 4*b)/a^3)/f

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mupad [B]  time = 7.69, size = 2401, normalized size = 16.91 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((tan(e + f*x)*(2*a*b + a^2 + 2*b^2))/(2*a^2*(a + b)) + (b*tan(e + f*x)^3*(a + 2*b))/(2*a^2*(a + b)))/(f*(a +
b + b*tan(e + f*x)^4 + tan(e + f*x)^2*(a + 2*b))) - (atan(((((tan(e + f*x)*(64*a*b^6 + 32*b^7 + 26*a^2*b^5 - 6
*a^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4*b^2)) - (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^3 - 2*a^9*b^2)/(2*a^7*
b + a^8 + a^6*b^2) - (tan(e + f*x)*(a*1i - b*4i)*(32*a^6*b^5 + 80*a^7*b^4 + 64*a^8*b^3 + 16*a^9*b^2))/(8*a^3*(
2*a^5*b + a^6 + a^4*b^2)))*(a*1i - b*4i))/(4*a^3))*(a*1i - b*4i)*1i)/(4*a^3) + (((tan(e + f*x)*(64*a*b^6 + 32*
b^7 + 26*a^2*b^5 - 6*a^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4*b^2)) + (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^3
- 2*a^9*b^2)/(2*a^7*b + a^8 + a^6*b^2) + (tan(e + f*x)*(a*1i - b*4i)*(32*a^6*b^5 + 80*a^7*b^4 + 64*a^8*b^3 + 1
6*a^9*b^2))/(8*a^3*(2*a^5*b + a^6 + a^4*b^2)))*(a*1i - b*4i))/(4*a^3))*(a*1i - b*4i)*1i)/(4*a^3))/((12*a*b^6 +
8*b^7 + (3*a^2*b^5)/2 - (5*a^3*b^4)/4)/(2*a^7*b + a^8 + a^6*b^2) - (((tan(e + f*x)*(64*a*b^6 + 32*b^7 + 26*a^
2*b^5 - 6*a^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4*b^2)) - (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^3 - 2*a^9*b^2
)/(2*a^7*b + a^8 + a^6*b^2) - (tan(e + f*x)*(a*1i - b*4i)*(32*a^6*b^5 + 80*a^7*b^4 + 64*a^8*b^3 + 16*a^9*b^2))
/(8*a^3*(2*a^5*b + a^6 + a^4*b^2)))*(a*1i - b*4i))/(4*a^3))*(a*1i - b*4i))/(4*a^3) + (((tan(e + f*x)*(64*a*b^6
+ 32*b^7 + 26*a^2*b^5 - 6*a^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4*b^2)) + (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^
8*b^3 - 2*a^9*b^2)/(2*a^7*b + a^8 + a^6*b^2) + (tan(e + f*x)*(a*1i - b*4i)*(32*a^6*b^5 + 80*a^7*b^4 + 64*a^8*b
^3 + 16*a^9*b^2))/(8*a^3*(2*a^5*b + a^6 + a^4*b^2)))*(a*1i - b*4i))/(4*a^3))*(a*1i - b*4i))/(4*a^3)))*(a*1i -
b*4i)*1i)/(2*a^3*f) - (atan(((((5*a)/4 + b)*(-b^3*(a + b)^3)^(1/2)*((tan(e + f*x)*(64*a*b^6 + 32*b^7 + 26*a^2*
b^5 - 6*a^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4*b^2)) + (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^3 - 2*a^9*b^2)/
(2*a^7*b + a^8 + a^6*b^2) + (tan(e + f*x)*((5*a)/4 + b)*(-b^3*(a + b)^3)^(1/2)*(32*a^6*b^5 + 80*a^7*b^4 + 64*a
^8*b^3 + 16*a^9*b^2))/(2*(2*a^5*b + a^6 + a^4*b^2)*(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2)))*((5*a)/4 + b)*(-b^3
*(a + b)^3)^(1/2))/(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2))*1i)/(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2) + (((5*a)/
4 + b)*(-b^3*(a + b)^3)^(1/2)*((tan(e + f*x)*(64*a*b^6 + 32*b^7 + 26*a^2*b^5 - 6*a^3*b^4 + a^4*b^3))/(2*(2*a^5
*b + a^6 + a^4*b^2)) - (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^3 - 2*a^9*b^2)/(2*a^7*b + a^8 + a^6*b^2) - (tan(e +
f*x)*((5*a)/4 + b)*(-b^3*(a + b)^3)^(1/2)*(32*a^6*b^5 + 80*a^7*b^4 + 64*a^8*b^3 + 16*a^9*b^2))/(2*(2*a^5*b + a
^6 + a^4*b^2)*(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2)))*((5*a)/4 + b)*(-b^3*(a + b)^3)^(1/2))/(3*a^5*b + a^6 + a
^3*b^3 + 3*a^4*b^2))*1i)/(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2))/((12*a*b^6 + 8*b^7 + (3*a^2*b^5)/2 - (5*a^3*b^
4)/4)/(2*a^7*b + a^8 + a^6*b^2) + (((5*a)/4 + b)*(-b^3*(a + b)^3)^(1/2)*((tan(e + f*x)*(64*a*b^6 + 32*b^7 + 26
*a^2*b^5 - 6*a^3*b^4 + a^4*b^3))/(2*(2*a^5*b + a^6 + a^4*b^2)) + (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^3 - 2*a^9*
b^2)/(2*a^7*b + a^8 + a^6*b^2) + (tan(e + f*x)*((5*a)/4 + b)*(-b^3*(a + b)^3)^(1/2)*(32*a^6*b^5 + 80*a^7*b^4 +
64*a^8*b^3 + 16*a^9*b^2))/(2*(2*a^5*b + a^6 + a^4*b^2)*(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2)))*((5*a)/4 + b)*
(-b^3*(a + b)^3)^(1/2))/(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2)))/(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2) - (((5*a
)/4 + b)*(-b^3*(a + b)^3)^(1/2)*((tan(e + f*x)*(64*a*b^6 + 32*b^7 + 26*a^2*b^5 - 6*a^3*b^4 + a^4*b^3))/(2*(2*a
^5*b + a^6 + a^4*b^2)) - (((4*a^6*b^5 + 8*a^7*b^4 + 2*a^8*b^3 - 2*a^9*b^2)/(2*a^7*b + a^8 + a^6*b^2) - (tan(e
+ f*x)*((5*a)/4 + b)*(-b^3*(a + b)^3)^(1/2)*(32*a^6*b^5 + 80*a^7*b^4 + 64*a^8*b^3 + 16*a^9*b^2))/(2*(2*a^5*b +
a^6 + a^4*b^2)*(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2)))*((5*a)/4 + b)*(-b^3*(a + b)^3)^(1/2))/(3*a^5*b + a^6 +
a^3*b^3 + 3*a^4*b^2)))/(3*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2)))*((5*a)/4 + b)*(-b^3*(a + b)^3)^(1/2)*2i)/(f*(3
*a^5*b + a^6 + a^3*b^3 + 3*a^4*b^2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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