3.192 \(\int \frac{\cos ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=163 \[ \frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a^4 f \sqrt{a+b}}+\frac{\left (5 a^2-6 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x)}{16 a^3 f}+\frac{x \left (-6 a^2 b+5 a^3+8 a b^2-16 b^3\right )}{16 a^4}+\frac{(5 a-6 b) \sin (e+f x) \cos ^3(e+f x)}{24 a^2 f}+\frac{\sin (e+f x) \cos ^5(e+f x)}{6 a f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.244825, antiderivative size = 163, 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} \[ \frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a^4 f \sqrt{a+b}}+\frac{\left (5 a^2-6 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x)}{16 a^3 f}+\frac{x \left (-6 a^2 b+5 a^3+8 a b^2-16 b^3\right )}{16 a^4}+\frac{(5 a-6 b) \sin (e+f x) \cos ^3(e+f x)}{24 a^2 f}+\frac{\sin (e+f x) \cos ^5(e+f x)}{6 a f} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^6/(a + b*Sec[e + f*x]^2),x]

[Out]

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

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 ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right )^4 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cos ^5(e+f x) \sin (e+f x)}{6 a f}-\frac{\operatorname{Subst}\left (\int \frac{-5 a+b-5 b x^2}{\left (1+x^2\right )^3 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{6 a f}\\ &=\frac{(5 a-6 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f}+\frac{\cos ^5(e+f x) \sin (e+f x)}{6 a f}+\frac{\operatorname{Subst}\left (\int \frac{3 \left (5 a^2-a b+2 b^2\right )+3 (5 a-6 b) b x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{24 a^2 f}\\ &=\frac{\left (5 a^2-6 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 a^3 f}+\frac{(5 a-6 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f}+\frac{\cos ^5(e+f x) \sin (e+f x)}{6 a f}-\frac{\operatorname{Subst}\left (\int \frac{-3 \left (5 a^3-a^2 b+2 a b^2-8 b^3\right )-3 b \left (5 a^2-6 a b+8 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{48 a^3 f}\\ &=\frac{\left (5 a^2-6 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 a^3 f}+\frac{(5 a-6 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f}+\frac{\cos ^5(e+f x) \sin (e+f x)}{6 a f}+\frac{b^4 \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{a^4 f}+\frac{\left (5 a^3-6 a^2 b+8 a b^2-16 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 a^4 f}\\ &=\frac{\left (5 a^3-6 a^2 b+8 a b^2-16 b^3\right ) x}{16 a^4}+\frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a^4 \sqrt{a+b} f}+\frac{\left (5 a^2-6 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 a^3 f}+\frac{(5 a-6 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f}+\frac{\cos ^5(e+f x) \sin (e+f x)}{6 a f}\\ \end{align*}

Mathematica [A]  time = 0.936595, size = 133, normalized size = 0.82 \[ \frac{12 \left (-6 a^2 b+5 a^3+8 a b^2-16 b^3\right ) (e+f x)+3 a \left (15 a^2-16 a b+16 b^2\right ) \sin (2 (e+f x))+3 a^2 (3 a-2 b) \sin (4 (e+f x))+a^3 \sin (6 (e+f x))+\frac{192 b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{\sqrt{a+b}}}{192 a^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^6/(a + b*Sec[e + f*x]^2),x]

[Out]

(12*(5*a^3 - 6*a^2*b + 8*a*b^2 - 16*b^3)*(e + f*x) + (192*b^(7/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/
Sqrt[a + b] + 3*a*(15*a^2 - 16*a*b + 16*b^2)*Sin[2*(e + f*x)] + 3*a^2*(3*a - 2*b)*Sin[4*(e + f*x)] + a^3*Sin[6
*(e + f*x)])/(192*a^4*f)

________________________________________________________________________________________

Maple [B]  time = 0.098, size = 359, normalized size = 2.2 \begin{align*}{\frac{5\, \left ( \tan \left ( fx+e \right ) \right ) ^{5}}{16\,fa \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{3\, \left ( \tan \left ( fx+e \right ) \right ) ^{5}b}{8\,f{a}^{2} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{5}{b}^{2}}{2\,f{a}^{3} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}{b}^{2}}{f{a}^{3} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{5\, \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{6\,fa \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}b}{f{a}^{2} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{5\,\tan \left ( fx+e \right ) b}{8\,f{a}^{2} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{\tan \left ( fx+e \right ){b}^{2}}{2\,f{a}^{3} \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{11\,\tan \left ( fx+e \right ) }{16\,fa \left ( \left ( \tan \left ( fx+e \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{3}}{f{a}^{4}}}+{\frac{5\,\arctan \left ( \tan \left ( fx+e \right ) \right ) }{16\,fa}}-{\frac{3\,\arctan \left ( \tan \left ( fx+e \right ) \right ) b}{8\,f{a}^{2}}}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{2}}{2\,f{a}^{3}}}+{\frac{{b}^{4}}{f{a}^{4}}\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}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^6/(a+b*sec(f*x+e)^2),x)

[Out]

5/16/f/a/(tan(f*x+e)^2+1)^3*tan(f*x+e)^5-3/8/f/a^2/(tan(f*x+e)^2+1)^3*tan(f*x+e)^5*b+1/2/f/a^3/(tan(f*x+e)^2+1
)^3*tan(f*x+e)^5*b^2+1/f/a^3/(tan(f*x+e)^2+1)^3*tan(f*x+e)^3*b^2+5/6/f/a/(tan(f*x+e)^2+1)^3*tan(f*x+e)^3-1/f/a
^2/(tan(f*x+e)^2+1)^3*tan(f*x+e)^3*b-5/8/f/a^2/(tan(f*x+e)^2+1)^3*tan(f*x+e)*b+1/2/f/a^3/(tan(f*x+e)^2+1)^3*ta
n(f*x+e)*b^2+11/16/f/a/(tan(f*x+e)^2+1)^3*tan(f*x+e)-1/f/a^4*arctan(tan(f*x+e))*b^3+5/16/f/a*arctan(tan(f*x+e)
)-3/8/f/a^2*arctan(tan(f*x+e))*b+1/2/f/a^3*arctan(tan(f*x+e))*b^2+1/f*b^4/a^4/((a+b)*b)^(1/2)*arctan(tan(f*x+e
)*b/((a+b)*b)^(1/2))

________________________________________________________________________________________

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)^6/(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 0.611514, size = 996, normalized size = 6.11 \begin{align*} \left [\frac{12 \, b^{3} \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 ) + 3 \,{\left (5 \, a^{3} - 6 \, a^{2} b + 8 \, a b^{2} - 16 \, b^{3}\right )} f x +{\left (8 \, a^{3} \cos \left (f x + e\right )^{5} + 2 \,{\left (5 \, a^{3} - 6 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (5 \, a^{3} - 6 \, a^{2} b + 8 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, a^{4} f}, -\frac{24 \, b^{3} \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 ) - 3 \,{\left (5 \, a^{3} - 6 \, a^{2} b + 8 \, a b^{2} - 16 \, b^{3}\right )} f x -{\left (8 \, a^{3} \cos \left (f x + e\right )^{5} + 2 \,{\left (5 \, a^{3} - 6 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (5 \, a^{3} - 6 \, a^{2} b + 8 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, a^{4} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

[1/48*(12*b^3*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)) + 3*(5*a^3 - 6*a^2*b + 8*a*b^2 - 16*b^3)*f*x + (8*a^3*cos(f*x +
e)^5 + 2*(5*a^3 - 6*a^2*b)*cos(f*x + e)^3 + 3*(5*a^3 - 6*a^2*b + 8*a*b^2)*cos(f*x + e))*sin(f*x + e))/(a^4*f),
-1/48*(24*b^3*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))) - 3*(5*a^3 - 6*a^2*b + 8*a*b^2 - 16*b^3)*f*x - (8*a^3*cos(f*x + e)^5 + 2*(5*a^3 - 6*a^2*b)*cos(f*x +
e)^3 + 3*(5*a^3 - 6*a^2*b + 8*a*b^2)*cos(f*x + e))*sin(f*x + e))/(a^4*f)]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.27734, size = 309, normalized size = 1.9 \begin{align*} \frac{\frac{48 \,{\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 )} b^{4}}{\sqrt{a b + b^{2}} a^{4}} + \frac{3 \,{\left (5 \, a^{3} - 6 \, a^{2} b + 8 \, a b^{2} - 16 \, b^{3}\right )}{\left (f x + e\right )}}{a^{4}} + \frac{15 \, a^{2} \tan \left (f x + e\right )^{5} - 18 \, a b \tan \left (f x + e\right )^{5} + 24 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} - 48 \, a b \tan \left (f x + e\right )^{3} + 48 \, b^{2} \tan \left (f x + e\right )^{3} + 33 \, a^{2} \tan \left (f x + e\right ) - 30 \, a b \tan \left (f x + e\right ) + 24 \, b^{2} \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3} a^{3}}}{48 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

1/48*(48*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*b^4/(sqrt(a*b + b^2)*a
^4) + 3*(5*a^3 - 6*a^2*b + 8*a*b^2 - 16*b^3)*(f*x + e)/a^4 + (15*a^2*tan(f*x + e)^5 - 18*a*b*tan(f*x + e)^5 +
24*b^2*tan(f*x + e)^5 + 40*a^2*tan(f*x + e)^3 - 48*a*b*tan(f*x + e)^3 + 48*b^2*tan(f*x + e)^3 + 33*a^2*tan(f*x
+ e) - 30*a*b*tan(f*x + e) + 24*b^2*tan(f*x + e))/((tan(f*x + e)^2 + 1)^3*a^3))/f