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

Optimal. Leaf size=126 \[ -\frac{b^3 \sin (e+f x)}{2 a^3 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{b^2 (6 a+5 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{2 a^{7/2} f (a+b)^{3/2}}+\frac{(a-2 b) \sin (e+f x)}{a^3 f}-\frac{\sin ^3(e+f x)}{3 a^2 f} \]

[Out]

(b^2*(6*a + 5*b)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(2*a^(7/2)*(a + b)^(3/2)*f) + ((a - 2*b)*Sin[e +
f*x])/(a^3*f) - Sin[e + f*x]^3/(3*a^2*f) - (b^3*Sin[e + f*x])/(2*a^3*(a + b)*f*(a + b - a*Sin[e + f*x]^2))

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Rubi [A]  time = 0.155194, antiderivative size = 126, 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} \[ -\frac{b^3 \sin (e+f x)}{2 a^3 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{b^2 (6 a+5 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{2 a^{7/2} f (a+b)^{3/2}}+\frac{(a-2 b) \sin (e+f x)}{a^3 f}-\frac{\sin ^3(e+f x)}{3 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*(6*a + 5*b)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(2*a^(7/2)*(a + b)^(3/2)*f) + ((a - 2*b)*Sin[e +
f*x])/(a^3*f) - Sin[e + f*x]^3/(3*a^2*f) - (b^3*Sin[e + f*x])/(2*a^3*(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 ^3(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 )^3}{\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 b}{a^3}-\frac{x^2}{a^2}+\frac{b^2 (3 a+2 b)-3 a b^2 x^2}{a^3 \left (a+b-a x^2\right )^2}\right ) \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{(a-2 b) \sin (e+f x)}{a^3 f}-\frac{\sin ^3(e+f x)}{3 a^2 f}+\frac{\operatorname{Subst}\left (\int \frac{b^2 (3 a+2 b)-3 a b^2 x^2}{\left (a+b-a x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{a^3 f}\\ &=\frac{(a-2 b) \sin (e+f x)}{a^3 f}-\frac{\sin ^3(e+f x)}{3 a^2 f}-\frac{b^3 \sin (e+f x)}{2 a^3 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )}+\frac{\left (b^2 (6 a+5 b)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b-a x^2} \, dx,x,\sin (e+f x)\right )}{2 a^3 (a+b) f}\\ &=\frac{b^2 (6 a+5 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{2 a^{7/2} (a+b)^{3/2} f}+\frac{(a-2 b) \sin (e+f x)}{a^3 f}-\frac{\sin ^3(e+f x)}{3 a^2 f}-\frac{b^3 \sin (e+f x)}{2 a^3 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.15593, size = 139, normalized size = 1.1 \[ \frac{a^{3/2} \sin (3 (e+f x))-\frac{3 b^2 (6 a+5 b) \left (\log \left (\sqrt{a+b}-\sqrt{a} \sin (e+f x)\right )-\log \left (\sqrt{a+b}+\sqrt{a} \sin (e+f x)\right )\right )}{(a+b)^{3/2}}+3 \sqrt{a} \sin (e+f x) \left (-\frac{4 b^3}{(a+b) (a \cos (2 (e+f x))+a+2 b)}+3 a-8 b\right )}{12 a^{7/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*b^2*(6*a + 5*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) + 3*Sqrt[a]*(3*a - 8*b - (4*b^3)/((a + b)*(a + 2*b + a*Cos[2*(e + f*x)])))*Sin[e + f*x] + a^(3/2)*Si
n[3*(e + f*x)])/(12*a^(7/2)*f)

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Maple [A]  time = 0.101, size = 120, normalized size = 1. \begin{align*}{\frac{1}{f} \left ( -{\frac{1}{{a}^{3}} \left ({\frac{a \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{3}}-\sin \left ( fx+e \right ) a+2\,\sin \left ( fx+e \right ) b \right ) }-{\frac{{b}^{2}}{{a}^{3}} \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{6\,a+5\,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)^3/(a+b*sec(f*x+e)^2)^2,x)

[Out]

1/f*(-1/a^3*(1/3*a*sin(f*x+e)^3-sin(f*x+e)*a+2*sin(f*x+e)*b)-b^2/a^3*(-1/2*b/(a+b)*sin(f*x+e)/(-a-b+a*sin(f*x+
e)^2)-1/2*(6*a+5*b)/(a+b)/((a+b)*a)^(1/2)*arctanh(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)^3/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.651429, size = 1077, normalized size = 8.55 \begin{align*} \left [\frac{3 \,{\left (6 \, a b^{3} + 5 \, b^{4} +{\left (6 \, a^{2} b^{2} + 5 \, a b^{3}\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 (4 \, a^{4} b - 4 \, a^{3} b^{2} - 23 \, a^{2} b^{3} - 15 \, a b^{4} + 2 \,{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (2 \, a^{5} - a^{4} b - 8 \, a^{3} b^{2} - 5 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{12 \,{\left ({\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{6} b + 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} f\right )}}, -\frac{3 \,{\left (6 \, a b^{3} + 5 \, b^{4} +{\left (6 \, a^{2} b^{2} + 5 \, a b^{3}\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 (4 \, a^{4} b - 4 \, a^{3} b^{2} - 23 \, a^{2} b^{3} - 15 \, a b^{4} + 2 \,{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (2 \, a^{5} - a^{4} b - 8 \, a^{3} b^{2} - 5 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{6 \,{\left ({\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{6} b + 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*(6*a*b^3 + 5*b^4 + (6*a^2*b^2 + 5*a*b^3)*cos(f*x + e)^2)*sqrt(a^2 + a*b)*log(-(a*cos(f*x + e)^2 - 2*s
qrt(a^2 + a*b)*sin(f*x + e) - 2*a - b)/(a*cos(f*x + e)^2 + b)) + 2*(4*a^4*b - 4*a^3*b^2 - 23*a^2*b^3 - 15*a*b^
4 + 2*(a^5 + 2*a^4*b + a^3*b^2)*cos(f*x + e)^4 + 2*(2*a^5 - a^4*b - 8*a^3*b^2 - 5*a^2*b^3)*cos(f*x + e)^2)*sin
(f*x + e))/((a^7 + 2*a^6*b + a^5*b^2)*f*cos(f*x + e)^2 + (a^6*b + 2*a^5*b^2 + a^4*b^3)*f), -1/6*(3*(6*a*b^3 +
5*b^4 + (6*a^2*b^2 + 5*a*b^3)*cos(f*x + e)^2)*sqrt(-a^2 - a*b)*arctan(sqrt(-a^2 - a*b)*sin(f*x + e)/(a + b)) -
(4*a^4*b - 4*a^3*b^2 - 23*a^2*b^3 - 15*a*b^4 + 2*(a^5 + 2*a^4*b + a^3*b^2)*cos(f*x + e)^4 + 2*(2*a^5 - a^4*b
- 8*a^3*b^2 - 5*a^2*b^3)*cos(f*x + e)^2)*sin(f*x + e))/((a^7 + 2*a^6*b + a^5*b^2)*f*cos(f*x + e)^2 + (a^6*b +
2*a^5*b^2 + a^4*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)**3/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.15722, size = 205, normalized size = 1.63 \begin{align*} \frac{\frac{3 \, b^{3} \sin \left (f x + e\right )}{{\left (a^{4} + a^{3} b\right )}{\left (a \sin \left (f x + e\right )^{2} - a - b\right )}} - \frac{3 \,{\left (6 \, a b^{2} + 5 \, b^{3}\right )} \arctan \left (\frac{a \sin \left (f x + e\right )}{\sqrt{-a^{2} - a b}}\right )}{{\left (a^{4} + a^{3} b\right )} \sqrt{-a^{2} - a b}} - \frac{2 \,{\left (a^{4} \sin \left (f x + e\right )^{3} - 3 \, a^{4} \sin \left (f x + e\right ) + 6 \, a^{3} b \sin \left (f x + e\right )\right )}}{a^{6}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*b^3*sin(f*x + e)/((a^4 + a^3*b)*(a*sin(f*x + e)^2 - a - b)) - 3*(6*a*b^2 + 5*b^3)*arctan(a*sin(f*x + e)
/sqrt(-a^2 - a*b))/((a^4 + a^3*b)*sqrt(-a^2 - a*b)) - 2*(a^4*sin(f*x + e)^3 - 3*a^4*sin(f*x + e) + 6*a^3*b*sin
(f*x + e))/a^6)/f