3.216 \(\int \frac{\cos ^2(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=201 \[ \frac{b^{3/2} \left (35 a^2+56 a b+24 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^4 f (a+b)^{5/2}}+\frac{b (4 a+3 b) (a+4 b) \tan (e+f x)}{8 a^3 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac{b (2 a+3 b) \tan (e+f x)}{4 a^2 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}+\frac{x (a-6 b)}{2 a^4}+\frac{\sin (e+f x) \cos (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)+b\right )^2} \]

[Out]

((a - 6*b)*x)/(2*a^4) + (b^(3/2)*(35*a^2 + 56*a*b + 24*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(8*a^4
*(a + b)^(5/2)*f) + (Cos[e + f*x]*Sin[e + f*x])/(2*a*f*(a + b + b*Tan[e + f*x]^2)^2) + (b*(2*a + 3*b)*Tan[e +
f*x])/(4*a^2*(a + b)*f*(a + b + b*Tan[e + f*x]^2)^2) + (b*(4*a + 3*b)*(a + 4*b)*Tan[e + f*x])/(8*a^3*(a + b)^2
*f*(a + b + b*Tan[e + f*x]^2))

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Rubi [A]  time = 0.335745, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{b^{3/2} \left (35 a^2+56 a b+24 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^4 f (a+b)^{5/2}}+\frac{b (4 a+3 b) (a+4 b) \tan (e+f x)}{8 a^3 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac{b (2 a+3 b) \tan (e+f x)}{4 a^2 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}+\frac{x (a-6 b)}{2 a^4}+\frac{\sin (e+f x) \cos (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^2/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

((a - 6*b)*x)/(2*a^4) + (b^(3/2)*(35*a^2 + 56*a*b + 24*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(8*a^4
*(a + b)^(5/2)*f) + (Cos[e + f*x]*Sin[e + f*x])/(2*a*f*(a + b + b*Tan[e + f*x]^2)^2) + (b*(2*a + 3*b)*Tan[e +
f*x])/(4*a^2*(a + b)*f*(a + b + b*Tan[e + f*x]^2)^2) + (b*(4*a + 3*b)*(a + 4*b)*Tan[e + f*x])/(8*a^3*(a + b)^2
*f*(a + b + b*Tan[e + f*x]^2))

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]

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\cos ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )^3} \, 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 )^2}-\frac{\operatorname{Subst}\left (\int \frac{-a+b-5 b x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^3} \, 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 )^2}+\frac{b (2 a+3 b) \tan (e+f x)}{4 a^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{-2 \left (2 a^2-4 a b-3 b^2\right )-6 b (2 a+3 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{8 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 )^2}+\frac{b (2 a+3 b) \tan (e+f x)}{4 a^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{b (4 a+3 b) (a+4 b) \tan (e+f x)}{8 a^3 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-2 \left (4 a^3-12 a^2 b-25 a b^2-12 b^3\right )-2 b (4 a+3 b) (a+4 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{16 a^3 (a+b)^2 f}\\ &=\frac{\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{b (2 a+3 b) \tan (e+f x)}{4 a^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{b (4 a+3 b) (a+4 b) \tan (e+f x)}{8 a^3 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{(a-6 b) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{2 a^4 f}+\frac{\left (b^2 \left (35 a^2+56 a b+24 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^4 (a+b)^2 f}\\ &=\frac{(a-6 b) x}{2 a^4}+\frac{b^{3/2} \left (35 a^2+56 a b+24 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{8 a^4 (a+b)^{5/2} f}+\frac{\cos (e+f x) \sin (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{b (2 a+3 b) \tan (e+f x)}{4 a^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac{b (4 a+3 b) (a+4 b) \tan (e+f x)}{8 a^3 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 3.7653, size = 156, normalized size = 0.78 \[ \frac{\frac{b^{3/2} \left (35 a^2+56 a b+24 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{(a+b)^{5/2}}+a \sin (2 (e+f x)) \left (\frac{2 b^3 (5 a \cos (2 (e+f x))+3 a+8 b)}{(a+b)^2 (a \cos (2 (e+f x))+a+2 b)^2}+\frac{13 a b^2}{(a+b)^2 (a \cos (2 (e+f x))+a+2 b)}+2\right )+4 (a-6 b) (e+f x)}{8 a^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^2/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

(4*(a - 6*b)*(e + f*x) + (b^(3/2)*(35*a^2 + 56*a*b + 24*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a +
b)^(5/2) + a*(2 + (13*a*b^2)/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])) + (2*b^3*(3*a + 8*b + 5*a*Cos[2*(e + f
*x)]))/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2))*Sin[2*(e + f*x)])/(8*a^4*f)

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Maple [A]  time = 0.113, size = 366, normalized size = 1.8 \begin{align*}{\frac{\tan \left ( fx+e \right ) }{2\,f{a}^{3} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) }}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{2\,f{a}^{3}}}-3\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) b}{f{a}^{4}}}+{\frac{11\,{b}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f{a}^{2} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+{\frac{{b}^{4} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{f{a}^{3} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }}+{\frac{13\,{b}^{2}\tan \left ( fx+e \right ) }{8\,f{a}^{2} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ( a+b \right ) }}+{\frac{{b}^{3}\tan \left ( fx+e \right ) }{f{a}^{3} \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ( a+b \right ) }}+{\frac{35\,{b}^{2}}{8\,f{a}^{2} \left ({a}^{2}+2\,ab+{b}^{2} \right ) }\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}+7\,{\frac{{b}^{3}}{f{a}^{3} \left ({a}^{2}+2\,ab+{b}^{2} \right ) \sqrt{ \left ( a+b \right ) b}}\arctan \left ({\frac{\tan \left ( fx+e \right ) b}{\sqrt{ \left ( a+b \right ) b}}} \right ) }+3\,{\frac{{b}^{4}}{f{a}^{4} \left ({a}^{2}+2\,ab+{b}^{2} \right ) \sqrt{ \left ( a+b \right ) b}}\arctan \left ({\frac{\tan \left ( fx+e \right ) b}{\sqrt{ \left ( a+b \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x)

[Out]

1/2/f/a^3*tan(f*x+e)/(tan(f*x+e)^2+1)+1/2/f/a^3*arctan(tan(f*x+e))-3/f/a^4*arctan(tan(f*x+e))*b+11/8/f/a^2*b^3
/(a+b+b*tan(f*x+e)^2)^2/(a^2+2*a*b+b^2)*tan(f*x+e)^3+1/f*b^4/a^3/(a+b+b*tan(f*x+e)^2)^2/(a^2+2*a*b+b^2)*tan(f*
x+e)^3+13/8/f/a^2*b^2/(a+b+b*tan(f*x+e)^2)^2/(a+b)*tan(f*x+e)+1/f*b^3/a^3/(a+b+b*tan(f*x+e)^2)^2/(a+b)*tan(f*x
+e)+35/8/f/a^2*b^2/(a^2+2*a*b+b^2)/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))+7/f/a^3*b^3/(a^2+2*a*b
+b^2)/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))+3/f*b^4/a^4/(a^2+2*a*b+b^2)/((a+b)*b)^(1/2)*arctan(
tan(f*x+e)*b/((a+b)*b)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.785094, size = 2188, normalized size = 10.89 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[1/32*(16*(a^5 - 4*a^4*b - 11*a^3*b^2 - 6*a^2*b^3)*f*x*cos(f*x + e)^4 + 32*(a^4*b - 4*a^3*b^2 - 11*a^2*b^3 - 6
*a*b^4)*f*x*cos(f*x + e)^2 + 16*(a^3*b^2 - 4*a^2*b^3 - 11*a*b^4 - 6*b^5)*f*x + (35*a^2*b^3 + 56*a*b^4 + 24*b^5
+ (35*a^4*b + 56*a^3*b^2 + 24*a^2*b^3)*cos(f*x + e)^4 + 2*(35*a^3*b^2 + 56*a^2*b^3 + 24*a*b^4)*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*(4*(a^5 + 2*a^4*b + a^3*b^2)*cos(f*x + e)^5 + (8*a^4*b + 29*a^3*b^2 + 18
*a^2*b^3)*cos(f*x + e)^3 + (4*a^3*b^2 + 19*a^2*b^3 + 12*a*b^4)*cos(f*x + e))*sin(f*x + e))/((a^8 + 2*a^7*b + a
^6*b^2)*f*cos(f*x + e)^4 + 2*(a^7*b + 2*a^6*b^2 + a^5*b^3)*f*cos(f*x + e)^2 + (a^6*b^2 + 2*a^5*b^3 + a^4*b^4)*
f), 1/16*(8*(a^5 - 4*a^4*b - 11*a^3*b^2 - 6*a^2*b^3)*f*x*cos(f*x + e)^4 + 16*(a^4*b - 4*a^3*b^2 - 11*a^2*b^3 -
6*a*b^4)*f*x*cos(f*x + e)^2 + 8*(a^3*b^2 - 4*a^2*b^3 - 11*a*b^4 - 6*b^5)*f*x - (35*a^2*b^3 + 56*a*b^4 + 24*b^
5 + (35*a^4*b + 56*a^3*b^2 + 24*a^2*b^3)*cos(f*x + e)^4 + 2*(35*a^3*b^2 + 56*a^2*b^3 + 24*a*b^4)*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*(4*(a^5 + 2*a^4*b + a^3*b^2)*cos(f*x + e)^5 + (8*a^4*b + 29*a^3*b^2 + 18*a^2*b^3)*cos(f*x + e)^3 + (4*a^3*b^
2 + 19*a^2*b^3 + 12*a*b^4)*cos(f*x + e))*sin(f*x + e))/((a^8 + 2*a^7*b + a^6*b^2)*f*cos(f*x + e)^4 + 2*(a^7*b
+ 2*a^6*b^2 + a^5*b^3)*f*cos(f*x + e)^2 + (a^6*b^2 + 2*a^5*b^3 + a^4*b^4)*f)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**2/(a+b*sec(f*x+e)**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.35934, size = 323, normalized size = 1.61 \begin{align*} \frac{\frac{{\left (35 \, a^{2} b^{2} + 56 \, a b^{3} + 24 \, b^{4}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )}}{{\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} \sqrt{a b + b^{2}}} + \frac{11 \, a b^{3} \tan \left (f x + e\right )^{3} + 8 \, b^{4} \tan \left (f x + e\right )^{3} + 13 \, a^{2} b^{2} \tan \left (f x + e\right ) + 21 \, a b^{3} \tan \left (f x + e\right ) + 8 \, b^{4} \tan \left (f x + e\right )}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )}{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} + \frac{4 \,{\left (f x + e\right )}{\left (a - 6 \, b\right )}}{a^{4}} + \frac{4 \, \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )} a^{3}}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")

[Out]

1/8*((35*a^2*b^2 + 56*a*b^3 + 24*b^4)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b +
b^2)))/((a^6 + 2*a^5*b + a^4*b^2)*sqrt(a*b + b^2)) + (11*a*b^3*tan(f*x + e)^3 + 8*b^4*tan(f*x + e)^3 + 13*a^2*
b^2*tan(f*x + e) + 21*a*b^3*tan(f*x + e) + 8*b^4*tan(f*x + e))/((a^5 + 2*a^4*b + a^3*b^2)*(b*tan(f*x + e)^2 +
a + b)^2) + 4*(f*x + e)*(a - 6*b)/a^4 + 4*tan(f*x + e)/((tan(f*x + e)^2 + 1)*a^3))/f