3.204 \(\int \frac{\cos ^4(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=203 \[ -\frac{b^{5/2} (7 a+6 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^4 f (a+b)^{3/2}}+\frac{x \left (3 a^2-8 a b+24 b^2\right )}{8 a^4}+\frac{b (a-3 b) (3 a+4 b) \tan (e+f x)}{8 a^3 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )}+\frac{3 (a-2 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

[Out]

((3*a^2 - 8*a*b + 24*b^2)*x)/(8*a^4) - (b^(5/2)*(7*a + 6*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*a^4
*(a + b)^(3/2)*f) + (3*(a - 2*b)*Cos[e + f*x]*Sin[e + f*x])/(8*a^2*f*(a + b + b*Tan[e + f*x]^2)) + (Cos[e + f*
x]^3*Sin[e + f*x])/(4*a*f*(a + b + b*Tan[e + f*x]^2)) + ((a - 3*b)*b*(3*a + 4*b)*Tan[e + f*x])/(8*a^3*(a + b)*
f*(a + b + b*Tan[e + f*x]^2))

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Rubi [A]  time = 0.289948, antiderivative size = 203, 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} \[ -\frac{b^{5/2} (7 a+6 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^4 f (a+b)^{3/2}}+\frac{x \left (3 a^2-8 a b+24 b^2\right )}{8 a^4}+\frac{b (a-3 b) (3 a+4 b) \tan (e+f x)}{8 a^3 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )}+\frac{3 (a-2 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac{\sin (e+f x) \cos ^3(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^4/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

((3*a^2 - 8*a*b + 24*b^2)*x)/(8*a^4) - (b^(5/2)*(7*a + 6*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*a^4
*(a + b)^(3/2)*f) + (3*(a - 2*b)*Cos[e + f*x]*Sin[e + f*x])/(8*a^2*f*(a + b + b*Tan[e + f*x]^2)) + (Cos[e + f*
x]^3*Sin[e + f*x])/(4*a*f*(a + b + b*Tan[e + f*x]^2)) + ((a - 3*b)*b*(3*a + 4*b)*Tan[e + f*x])/(8*a^3*(a + b)*
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 ^4(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 )^3 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-3 a+b-5 b x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 a f}\\ &=\frac{3 (a-2 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{3 a^2+a b+6 b^2+9 (a-2 b) 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 f}\\ &=\frac{3 (a-2 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{(a-3 b) b (3 a+4 b) \tan (e+f x)}{8 a^3 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{2 \left (3 a^3-2 a^2 b+11 a b^2+12 b^3\right )+2 (a-3 b) b (3 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) f}\\ &=\frac{3 (a-2 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{(a-3 b) b (3 a+4 b) \tan (e+f x)}{8 a^3 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac{\left (b^3 (7 a+6 b)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^4 (a+b) f}+\frac{\left (3 a^2-8 a b+24 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 a^4 f}\\ &=\frac{\left (3 a^2-8 a b+24 b^2\right ) x}{8 a^4}-\frac{b^{5/2} (7 a+6 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^4 (a+b)^{3/2} f}+\frac{3 (a-2 b) \cos (e+f x) \sin (e+f x)}{8 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\cos ^3(e+f x) \sin (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{(a-3 b) b (3 a+4 b) \tan (e+f x)}{8 a^3 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.58873, size = 138, normalized size = 0.68 \[ \frac{4 \left (3 a^2-8 a b+24 b^2\right ) (e+f x)+a^2 \sin (4 (e+f x))-\frac{16 b^{5/2} (7 a+6 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{(a+b)^{3/2}}-\frac{16 a b^3 \sin (2 (e+f x))}{(a+b) (a \cos (2 (e+f x))+a+2 b)}+8 a (a-2 b) \sin (2 (e+f x))}{32 a^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^4/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

(4*(3*a^2 - 8*a*b + 24*b^2)*(e + f*x) - (16*b^(5/2)*(7*a + 6*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a
+ b)^(3/2) + 8*a*(a - 2*b)*Sin[2*(e + f*x)] - (16*a*b^3*Sin[2*(e + f*x)])/((a + b)*(a + 2*b + a*Cos[2*(e + f*
x)])) + a^2*Sin[4*(e + f*x)])/(32*a^4*f)

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Maple [A]  time = 0.106, size = 276, normalized size = 1.4 \begin{align*}{\frac{3\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f{a}^{2} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}b}{f{a}^{3} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{\tan \left ( fx+e \right ) b}{f{a}^{3} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{5\,\tan \left ( fx+e \right ) }{8\,f{a}^{2} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{2}}}+3\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{2}}{f{a}^{4}}}+{\frac{3\,\arctan \left ( \tan \left ( fx+e \right ) \right ) }{8\,f{a}^{2}}}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) b}{f{a}^{3}}}-{\frac{{b}^{3}\tan \left ( fx+e \right ) }{2\,f{a}^{3} \left ( a+b \right ) \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{7\,{b}^{3}}{2\,f{a}^{3} \left ( a+b \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}}}}-3\,{\frac{{b}^{4}}{f{a}^{4} \left ( a+b \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)^4/(a+b*sec(f*x+e)^2)^2,x)

[Out]

3/8/f/a^2/(tan(f*x+e)^2+1)^2*tan(f*x+e)^3-1/f/a^3/(tan(f*x+e)^2+1)^2*tan(f*x+e)^3*b-1/f/a^3/(tan(f*x+e)^2+1)^2
*tan(f*x+e)*b+5/8/f/a^2/(tan(f*x+e)^2+1)^2*tan(f*x+e)+3/f/a^4*arctan(tan(f*x+e))*b^2+3/8/f/a^2*arctan(tan(f*x+
e))-1/f/a^3*arctan(tan(f*x+e))*b-1/2/f*b^3/a^3/(a+b)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)-7/2/f*b^3/a^3/(a+b)/((a+b
)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-3/f*b^4/a^4/(a+b)/((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)^4/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.710846, size = 1492, normalized size = 7.35 \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)^4/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

[1/8*((3*a^4 - 5*a^3*b + 16*a^2*b^2 + 24*a*b^3)*f*x*cos(f*x + e)^2 + (3*a^3*b - 5*a^2*b^2 + 16*a*b^3 + 24*b^4)
*f*x + (7*a*b^3 + 6*b^4 + (7*a^2*b^2 + 6*a*b^3)*cos(f*x + e)^2)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*co
s(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)) + (2*(a^4 + a^
3*b)*cos(f*x + e)^5 + 3*(a^4 - a^3*b - 2*a^2*b^2)*cos(f*x + e)^3 + (3*a^3*b - 5*a^2*b^2 - 12*a*b^3)*cos(f*x +
e))*sin(f*x + e))/((a^6 + a^5*b)*f*cos(f*x + e)^2 + (a^5*b + a^4*b^2)*f), 1/8*((3*a^4 - 5*a^3*b + 16*a^2*b^2 +
24*a*b^3)*f*x*cos(f*x + e)^2 + (3*a^3*b - 5*a^2*b^2 + 16*a*b^3 + 24*b^4)*f*x + 2*(7*a*b^3 + 6*b^4 + (7*a^2*b^
2 + 6*a*b^3)*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^4 + a^3*b)*cos(f*x + e)^5 + 3*(a^4 - a^3*b - 2*a^2*b^2)*cos(f*x + e)^3 + (3*a^
3*b - 5*a^2*b^2 - 12*a*b^3)*cos(f*x + e))*sin(f*x + e))/((a^6 + a^5*b)*f*cos(f*x + e)^2 + (a^5*b + a^4*b^2)*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)**4/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.25679, size = 277, normalized size = 1.36 \begin{align*} -\frac{\frac{4 \, b^{3} \tan \left (f x + e\right )}{{\left (a^{4} + a^{3} b\right )}{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}} + \frac{4 \,{\left (7 \, a b^{3} + 6 \, 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^{5} + a^{4} b\right )} \sqrt{a b + b^{2}}} - \frac{{\left (3 \, a^{2} - 8 \, a b + 24 \, b^{2}\right )}{\left (f x + e\right )}}{a^{4}} - \frac{3 \, a \tan \left (f x + e\right )^{3} - 8 \, b \tan \left (f x + e\right )^{3} + 5 \, a \tan \left (f x + e\right ) - 8 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2} a^{3}}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

-1/8*(4*b^3*tan(f*x + e)/((a^4 + a^3*b)*(b*tan(f*x + e)^2 + a + b)) + 4*(7*a*b^3 + 6*b^4)*(pi*floor((f*x + e)/
pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/((a^5 + a^4*b)*sqrt(a*b + b^2)) - (3*a^2 - 8*a*b +
24*b^2)*(f*x + e)/a^4 - (3*a*tan(f*x + e)^3 - 8*b*tan(f*x + e)^3 + 5*a*tan(f*x + e) - 8*b*tan(f*x + e))/((tan(
f*x + e)^2 + 1)^2*a^3))/f