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

Optimal. Leaf size=101 \[ \frac{b^2 \sin (e+f x)}{2 a^2 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )}-\frac{b (4 a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{2 a^{5/2} f (a+b)^{3/2}}+\frac{\sin (e+f x)}{a^2 f} \]

[Out]

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

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Rubi [A]  time = 0.132112, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {4147, 390, 385, 208} \[ \frac{b^2 \sin (e+f x)}{2 a^2 f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )}-\frac{b (4 a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{2 a^{5/2} f (a+b)^{3/2}}+\frac{\sin (e+f x)}{a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(4*a + 3*b)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(2*a^(5/2)*(a + b)^(3/2)*f) + Sin[e + f*x]/(a^2*f
) + (b^2*Sin[e + f*x])/(2*a^2*(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 (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 )^2}{\left (a+b-a x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{a^2}-\frac{b (2 a+b)-2 a b x^2}{a^2 \left (a+b-a x^2\right )^2}\right ) \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\sin (e+f x)}{a^2 f}-\frac{\operatorname{Subst}\left (\int \frac{b (2 a+b)-2 a b x^2}{\left (a+b-a x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{a^2 f}\\ &=\frac{\sin (e+f x)}{a^2 f}+\frac{b^2 \sin (e+f x)}{2 a^2 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )}-\frac{(b (4 a+3 b)) \operatorname{Subst}\left (\int \frac{1}{a+b-a x^2} \, dx,x,\sin (e+f x)\right )}{2 a^2 (a+b) f}\\ &=-\frac{b (4 a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{2 a^{5/2} (a+b)^{3/2} f}+\frac{\sin (e+f x)}{a^2 f}+\frac{b^2 \sin (e+f x)}{2 a^2 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )}\\ \end{align*}

Mathematica [C]  time = 3.47386, size = 945, normalized size = 9.36 \[ \frac{(\cos (2 (e+f x)) a+a+2 b) \sec ^3(e+f x) \left (8 \sqrt{a} \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan (e+f x) b^2-2 i (4 a+3 b) \tan ^{-1}\left (\frac{2 \sin (e) \left (\sin (2 e) a+i a-i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt{a}-i \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt{a}+\sqrt{a+b} \cos (f x) \sqrt{(\cos (e)-i \sin (e))^2} \sqrt{a}-\sqrt{a+b} \cos (2 e+f x) \sqrt{(\cos (e)-i \sin (e))^2} \sqrt{a}+i b+i (a+b) \cos (2 e)+b \sin (2 e)\right )}{i (a+3 b) \cos (e)+i (a+b) \cos (3 e)+i a \cos (e+2 f x)+i a \cos (3 e+2 f x)+3 a \sin (e)+b \sin (e)+a \sin (3 e)+b \sin (3 e)+a \sin (e+2 f x)-a \sin (3 e+2 f x)}\right ) (\cos (2 (e+f x)) a+a+2 b) \sec (e+f x) (\cos (e)-i \sin (e)) b-(4 a+3 b) (\cos (2 (e+f x)) a+a+2 b) \log \left (-\cos (2 (e+f x)) a-2 i \sin (2 e) a+a+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt{a}+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt{a}+2 (a+b) \cos (2 e)-2 i b \sin (2 e)\right ) \sec (e+f x) (\cos (e)-i \sin (e)) b+(4 a+3 b) (\cos (2 (e+f x)) a+a+2 b) \log \left (\cos (2 (e+f x)) a+2 i \sin (2 e) a-a+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt{a}+2 \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt{a}-2 (a+b) \cos (2 e)+2 i b \sin (2 e)\right ) \sec (e+f x) (\cos (e)-i \sin (e)) b+2 (4 a+3 b) \tan ^{-1}\left (\frac{(a+b) \sin (e)}{(a+b) \cos (e)-\sqrt{a} \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} (\cos (2 e)+i \sin (2 e)) \sin (e+f x)}\right ) (\cos (2 (e+f x)) a+a+2 b) \sec (e+f x) (i \cos (e)+\sin (e)) b+8 \sqrt{a} (a+b)^{3/2} \cos (f x) (\cos (2 (e+f x)) a+a+2 b) \sec (e+f x) \sqrt{(\cos (e)-i \sin (e))^2} \sin (e)+8 \sqrt{a} (a+b)^{3/2} \cos (e) (\cos (2 (e+f x)) a+a+2 b) \sec (e+f x) \sqrt{(\cos (e)-i \sin (e))^2} \sin (f x)\right )}{32 a^{5/2} (a+b)^{3/2} f \left (b \sec ^2(e+f x)+a\right )^2 \sqrt{(\cos (e)-i \sin (e))^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^3*((-2*I)*b*(4*a + 3*b)*ArcTan[(2*Sin[e]*(I*a + I*b + I*(a + b)*C
os[2*e] + Sqrt[a]*Sqrt[a + b]*Cos[f*x]*Sqrt[(Cos[e] - I*Sin[e])^2] - Sqrt[a]*Sqrt[a + b]*Cos[2*e + f*x]*Sqrt[(
Cos[e] - I*Sin[e])^2] + a*Sin[2*e] + b*Sin[2*e] - I*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[f*x] -
I*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[2*e + f*x]))/(I*(a + 3*b)*Cos[e] + I*(a + b)*Cos[3*e] +
I*a*Cos[e + 2*f*x] + I*a*Cos[3*e + 2*f*x] + 3*a*Sin[e] + b*Sin[e] + a*Sin[3*e] + b*Sin[3*e] + a*Sin[e + 2*f*x
] - a*Sin[3*e + 2*f*x])]*(a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]*(Cos[e] - I*Sin[e]) - b*(4*a + 3*b)*(a +
2*b + a*Cos[2*(e + f*x)])*Log[a + 2*(a + b)*Cos[2*e] - a*Cos[2*(e + f*x)] - (2*I)*a*Sin[2*e] - (2*I)*b*Sin[2*e
] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[f*x] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e]
)^2]*Sin[2*e + f*x]]*Sec[e + f*x]*(Cos[e] - I*Sin[e]) + b*(4*a + 3*b)*(a + 2*b + a*Cos[2*(e + f*x)])*Log[-a -
2*(a + b)*Cos[2*e] + a*Cos[2*(e + f*x)] + (2*I)*a*Sin[2*e] + (2*I)*b*Sin[2*e] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Co
s[e] - I*Sin[e])^2]*Sin[f*x] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[2*e + f*x]]*Sec[e + f*x]*
(Cos[e] - I*Sin[e]) + 8*Sqrt[a]*(a + b)^(3/2)*Cos[f*x]*(a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]*Sqrt[(Cos[e
] - I*Sin[e])^2]*Sin[e] + 2*b*(4*a + 3*b)*ArcTan[((a + b)*Sin[e])/((a + b)*Cos[e] - Sqrt[a]*Sqrt[a + b]*Sqrt[(
Cos[e] - I*Sin[e])^2]*(Cos[2*e] + I*Sin[2*e])*Sin[e + f*x])]*(a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]*(I*Co
s[e] + Sin[e]) + 8*Sqrt[a]*(a + b)^(3/2)*Cos[e]*(a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]*Sqrt[(Cos[e] - I*S
in[e])^2]*Sin[f*x] + 8*Sqrt[a]*b^2*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[e + f*x]))/(32*a^(5/2)*(a + b)^
(3/2)*f*(a + b*Sec[e + f*x]^2)^2*Sqrt[(Cos[e] - I*Sin[e])^2])

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Maple [A]  time = 0.095, size = 92, normalized size = 0.9 \begin{align*}{\frac{1}{f} \left ({\frac{\sin \left ( fx+e \right ) }{{a}^{2}}}+{\frac{b}{{a}^{2}} \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{4\,a+3\,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)/(a+b*sec(f*x+e)^2)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.589017, size = 867, normalized size = 8.58 \begin{align*} \left [\frac{{\left (4 \, a b^{2} + 3 \, b^{3} +{\left (4 \, a^{2} b + 3 \, a b^{2}\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 (2 \, a^{3} b + 5 \, a^{2} b^{2} + 3 \, a b^{3} + 2 \,{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{4 \,{\left ({\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{5} b + 2 \, a^{4} b^{2} + a^{3} b^{3}\right )} f\right )}}, \frac{{\left (4 \, a b^{2} + 3 \, b^{3} +{\left (4 \, a^{2} b + 3 \, a b^{2}\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 (2 \, a^{3} b + 5 \, a^{2} b^{2} + 3 \, a b^{3} + 2 \,{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{2 \,{\left ({\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{5} b + 2 \, a^{4} b^{2} + a^{3} b^{3}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.19373, size = 158, normalized size = 1.56 \begin{align*} -\frac{\frac{b^{2} \sin \left (f x + e\right )}{{\left (a^{3} + a^{2} b\right )}{\left (a \sin \left (f x + e\right )^{2} - a - b\right )}} - \frac{{\left (4 \, a b + 3 \, b^{2}\right )} \arctan \left (\frac{a \sin \left (f x + e\right )}{\sqrt{-a^{2} - a b}}\right )}{{\left (a^{3} + a^{2} b\right )} \sqrt{-a^{2} - a b}} - \frac{2 \, \sin \left (f x + e\right )}{a^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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