3.198 \(\int \frac{\cos ^5(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=157 \[ \frac{b^4 \sin (e+f x)}{2 a^4 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{\left (a^2-2 a b+3 b^2\right ) \sin (e+f x)}{a^4 f}-\frac{b^3 (8 a+7 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{2 a^{9/2} f (a+b)^{3/2}}-\frac{2 (a-b) \sin ^3(e+f x)}{3 a^3 f}+\frac{\sin ^5(e+f x)}{5 a^2 f} \]

[Out]

-(b^3*(8*a + 7*b)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(2*a^(9/2)*(a + b)^(3/2)*f) + ((a^2 - 2*a*b + 3
*b^2)*Sin[e + f*x])/(a^4*f) - (2*(a - b)*Sin[e + f*x]^3)/(3*a^3*f) + Sin[e + f*x]^5/(5*a^2*f) + (b^4*Sin[e + f
*x])/(2*a^4*(a + b)*f*(a + b - a*Sin[e + f*x]^2))

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Rubi [A]  time = 0.168473, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4147, 390, 385, 208} \[ \frac{b^4 \sin (e+f x)}{2 a^4 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{\left (a^2-2 a b+3 b^2\right ) \sin (e+f x)}{a^4 f}-\frac{b^3 (8 a+7 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{2 a^{9/2} f (a+b)^{3/2}}-\frac{2 (a-b) \sin ^3(e+f x)}{3 a^3 f}+\frac{\sin ^5(e+f x)}{5 a^2 f} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^5/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

-(b^3*(8*a + 7*b)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(2*a^(9/2)*(a + b)^(3/2)*f) + ((a^2 - 2*a*b + 3
*b^2)*Sin[e + f*x])/(a^4*f) - (2*(a - b)*Sin[e + f*x]^3)/(3*a^3*f) + Sin[e + f*x]^5/(5*a^2*f) + (b^4*Sin[e + f
*x])/(2*a^4*(a + b)*f*(a + b - a*Sin[e + f*x]^2))

Rule 4147

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

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\cos ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^4}{\left (a+b-a x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a^2-2 a b+3 b^2}{a^4}-\frac{2 (a-b) x^2}{a^3}+\frac{x^4}{a^2}-\frac{b^3 (4 a+3 b)-4 a b^3 x^2}{a^4 \left (a+b-a x^2\right )^2}\right ) \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\left (a^2-2 a b+3 b^2\right ) \sin (e+f x)}{a^4 f}-\frac{2 (a-b) \sin ^3(e+f x)}{3 a^3 f}+\frac{\sin ^5(e+f x)}{5 a^2 f}-\frac{\operatorname{Subst}\left (\int \frac{b^3 (4 a+3 b)-4 a b^3 x^2}{\left (a+b-a x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{a^4 f}\\ &=\frac{\left (a^2-2 a b+3 b^2\right ) \sin (e+f x)}{a^4 f}-\frac{2 (a-b) \sin ^3(e+f x)}{3 a^3 f}+\frac{\sin ^5(e+f x)}{5 a^2 f}+\frac{b^4 \sin (e+f x)}{2 a^4 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )}-\frac{\left (b^3 (8 a+7 b)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b-a x^2} \, dx,x,\sin (e+f x)\right )}{2 a^4 (a+b) f}\\ &=-\frac{b^3 (8 a+7 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{2 a^{9/2} (a+b)^{3/2} f}+\frac{\left (a^2-2 a b+3 b^2\right ) \sin (e+f x)}{a^4 f}-\frac{2 (a-b) \sin ^3(e+f x)}{3 a^3 f}+\frac{\sin ^5(e+f x)}{5 a^2 f}+\frac{b^4 \sin (e+f x)}{2 a^4 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 2.14099, size = 171, normalized size = 1.09 \[ \frac{30 \sqrt{a} \sin (e+f x) \left (5 a^2+8 b^2 \left (\frac{b^2}{(a+b) (a \cos (2 (e+f x))+a+2 b)}+3\right )-12 a b\right )+5 a^{3/2} (5 a-8 b) \sin (3 (e+f x))+3 a^{5/2} \sin (5 (e+f x))+\frac{60 b^3 (8 a+7 b) \left (\log \left (\sqrt{a+b}-\sqrt{a} \sin (e+f x)\right )-\log \left (\sqrt{a+b}+\sqrt{a} \sin (e+f x)\right )\right )}{(a+b)^{3/2}}}{240 a^{9/2} f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^5/(a + b*Sec[e + f*x]^2)^2,x]

[Out]

((60*b^3*(8*a + 7*b)*(Log[Sqrt[a + b] - Sqrt[a]*Sin[e + f*x]] - Log[Sqrt[a + b] + Sqrt[a]*Sin[e + f*x]]))/(a +
b)^(3/2) + 30*Sqrt[a]*(5*a^2 - 12*a*b + 8*b^2*(3 + b^2/((a + b)*(a + 2*b + a*Cos[2*(e + f*x)]))))*Sin[e + f*x
] + 5*a^(3/2)*(5*a - 8*b)*Sin[3*(e + f*x)] + 3*a^(5/2)*Sin[5*(e + f*x)])/(240*a^(9/2)*f)

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Maple [A]  time = 0.105, size = 158, normalized size = 1. \begin{align*}{\frac{1}{f} \left ({\frac{1}{{a}^{4}} \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{5}{a}^{2}}{5}}-{\frac{2\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}{a}^{2}}{3}}+{\frac{2\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}ab}{3}}+{a}^{2}\sin \left ( fx+e \right ) -2\,ab\sin \left ( fx+e \right ) +3\,{b}^{2}\sin \left ( fx+e \right ) \right ) }+{\frac{{b}^{3}}{{a}^{4}} \left ( -{\frac{\sin \left ( fx+e \right ) b}{ \left ( 2\,a+2\,b \right ) \left ( -a-b+a \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{8\,a+7\,b}{2\,a+2\,b}{\it Artanh} \left ({\sin \left ( fx+e \right ) a{\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.689115, size = 1304, normalized size = 8.31 \begin{align*} \left [\frac{15 \,{\left (8 \, a b^{4} + 7 \, b^{5} +{\left (8 \, a^{2} b^{3} + 7 \, a b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{a^{2} + a b} \log \left (-\frac{a \cos \left (f x + e\right )^{2} + 2 \, \sqrt{a^{2} + a b} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 2 \,{\left (6 \,{\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} \cos \left (f x + e\right )^{6} + 16 \, a^{5} b - 8 \, a^{4} b^{2} + 26 \, a^{3} b^{3} + 155 \, a^{2} b^{4} + 105 \, a b^{5} + 2 \,{\left (4 \, a^{6} + a^{5} b - 10 \, a^{4} b^{2} - 7 \, a^{3} b^{3}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (8 \, a^{6} + 11 \, a^{4} b^{2} + 54 \, a^{3} b^{3} + 35 \, a^{2} b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{60 \,{\left ({\left (a^{8} + 2 \, a^{7} b + a^{6} b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{7} b + 2 \, a^{6} b^{2} + a^{5} b^{3}\right )} f\right )}}, \frac{15 \,{\left (8 \, a b^{4} + 7 \, b^{5} +{\left (8 \, a^{2} b^{3} + 7 \, a b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{-a^{2} - a b} \arctan \left (\frac{\sqrt{-a^{2} - a b} \sin \left (f x + e\right )}{a + b}\right ) +{\left (6 \,{\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} \cos \left (f x + e\right )^{6} + 16 \, a^{5} b - 8 \, a^{4} b^{2} + 26 \, a^{3} b^{3} + 155 \, a^{2} b^{4} + 105 \, a b^{5} + 2 \,{\left (4 \, a^{6} + a^{5} b - 10 \, a^{4} b^{2} - 7 \, a^{3} b^{3}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (8 \, a^{6} + 11 \, a^{4} b^{2} + 54 \, a^{3} b^{3} + 35 \, a^{2} b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{30 \,{\left ({\left (a^{8} + 2 \, a^{7} b + a^{6} b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{7} b + 2 \, a^{6} b^{2} + a^{5} b^{3}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

[1/60*(15*(8*a*b^4 + 7*b^5 + (8*a^2*b^3 + 7*a*b^4)*cos(f*x + e)^2)*sqrt(a^2 + a*b)*log(-(a*cos(f*x + e)^2 + 2*
sqrt(a^2 + a*b)*sin(f*x + e) - 2*a - b)/(a*cos(f*x + e)^2 + b)) + 2*(6*(a^6 + 2*a^5*b + a^4*b^2)*cos(f*x + e)^
6 + 16*a^5*b - 8*a^4*b^2 + 26*a^3*b^3 + 155*a^2*b^4 + 105*a*b^5 + 2*(4*a^6 + a^5*b - 10*a^4*b^2 - 7*a^3*b^3)*c
os(f*x + e)^4 + 2*(8*a^6 + 11*a^4*b^2 + 54*a^3*b^3 + 35*a^2*b^4)*cos(f*x + e)^2)*sin(f*x + e))/((a^8 + 2*a^7*b
+ a^6*b^2)*f*cos(f*x + e)^2 + (a^7*b + 2*a^6*b^2 + a^5*b^3)*f), 1/30*(15*(8*a*b^4 + 7*b^5 + (8*a^2*b^3 + 7*a*
b^4)*cos(f*x + e)^2)*sqrt(-a^2 - a*b)*arctan(sqrt(-a^2 - a*b)*sin(f*x + e)/(a + b)) + (6*(a^6 + 2*a^5*b + a^4*
b^2)*cos(f*x + e)^6 + 16*a^5*b - 8*a^4*b^2 + 26*a^3*b^3 + 155*a^2*b^4 + 105*a*b^5 + 2*(4*a^6 + a^5*b - 10*a^4*
b^2 - 7*a^3*b^3)*cos(f*x + e)^4 + 2*(8*a^6 + 11*a^4*b^2 + 54*a^3*b^3 + 35*a^2*b^4)*cos(f*x + e)^2)*sin(f*x + e
))/((a^8 + 2*a^7*b + a^6*b^2)*f*cos(f*x + e)^2 + (a^7*b + 2*a^6*b^2 + a^5*b^3)*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)**5/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.18588, size = 266, normalized size = 1.69 \begin{align*} -\frac{\frac{15 \, b^{4} \sin \left (f x + e\right )}{{\left (a^{5} + a^{4} b\right )}{\left (a \sin \left (f x + e\right )^{2} - a - b\right )}} - \frac{15 \,{\left (8 \, a b^{3} + 7 \, b^{4}\right )} \arctan \left (\frac{a \sin \left (f x + e\right )}{\sqrt{-a^{2} - a b}}\right )}{{\left (a^{5} + a^{4} b\right )} \sqrt{-a^{2} - a b}} - \frac{2 \,{\left (3 \, a^{8} \sin \left (f x + e\right )^{5} - 10 \, a^{8} \sin \left (f x + e\right )^{3} + 10 \, a^{7} b \sin \left (f x + e\right )^{3} + 15 \, a^{8} \sin \left (f x + e\right ) - 30 \, a^{7} b \sin \left (f x + e\right ) + 45 \, a^{6} b^{2} \sin \left (f x + e\right )\right )}}{a^{10}}}{30 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

-1/30*(15*b^4*sin(f*x + e)/((a^5 + a^4*b)*(a*sin(f*x + e)^2 - a - b)) - 15*(8*a*b^3 + 7*b^4)*arctan(a*sin(f*x
+ e)/sqrt(-a^2 - a*b))/((a^5 + a^4*b)*sqrt(-a^2 - a*b)) - 2*(3*a^8*sin(f*x + e)^5 - 10*a^8*sin(f*x + e)^3 + 10
*a^7*b*sin(f*x + e)^3 + 15*a^8*sin(f*x + e) - 30*a^7*b*sin(f*x + e) + 45*a^6*b^2*sin(f*x + e))/a^10)/f