3.221 \(\int \frac{\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=65 \[ \frac{(c+2 d) \tan (e+f x)}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac{(c-d) \tan (e+f x)}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

((c - d)*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2) + ((c + 2*d)*Tan[e + f*x])/(3*f*(a^2 + a^2*Sec[e + f*x]))

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Rubi [A]  time = 0.0865492, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {4000, 3794} \[ \frac{(c+2 d) \tan (e+f x)}{3 f \left (a^2 \sec (e+f x)+a^2\right )}+\frac{(c-d) \tan (e+f x)}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - d)*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2) + ((c + 2*d)*Tan[e + f*x])/(3*f*(a^2 + a^2*Sec[e + f*x]))

Rule 4000

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(a*B*m + A*b*
(m + 1))/(a*b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, A, B, e, f}, x
] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] && LtQ[m, -2^(-1)]

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx &=\frac{(c-d) \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac{(c+2 d) \int \frac{\sec (e+f x)}{a+a \sec (e+f x)} \, dx}{3 a}\\ &=\frac{(c-d) \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac{(c+2 d) \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.210091, size = 76, normalized size = 1.17 \[ \frac{\sec \left (\frac{e}{2}\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \left ((2 c+d) \sin \left (e+\frac{3 f x}{2}\right )+3 (c+d) \sin \left (\frac{f x}{2}\right )-3 c \sin \left (e+\frac{f x}{2}\right )\right )}{3 a^2 f (\cos (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(e + f*x)/2]*Sec[e/2]*(3*(c + d)*Sin[(f*x)/2] - 3*c*Sin[e + (f*x)/2] + (2*c + d)*Sin[e + (3*f*x)/2]))/(3*
a^2*f*(1 + Cos[e + f*x])^2)

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Maple [A]  time = 0.051, size = 60, normalized size = 0.9 \begin{align*}{\frac{1}{2\,f{a}^{2}} \left ( -{\frac{c}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+{\frac{d}{3} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+c\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) d \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^2,x)

[Out]

1/2/f/a^2*(-1/3*tan(1/2*f*x+1/2*e)^3*c+1/3*tan(1/2*f*x+1/2*e)^3*d+c*tan(1/2*f*x+1/2*e)+tan(1/2*f*x+1/2*e)*d)

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Maxima [A]  time = 0.97264, size = 126, normalized size = 1.94 \begin{align*} \frac{\frac{d{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}} + \frac{c{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

1/6*(d*(3*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 + c*(3*sin(f*x + e)/(cos(
f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2)/f

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Fricas [A]  time = 0.435586, size = 144, normalized size = 2.22 \begin{align*} \frac{{\left ({\left (2 \, c + d\right )} \cos \left (f x + e\right ) + c + 2 \, d\right )} \sin \left (f x + e\right )}{3 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((2*c + d)*cos(f*x + e) + c + 2*d)*sin(f*x + e)/(a^2*f*cos(f*x + e)^2 + 2*a^2*f*cos(f*x + e) + a^2*f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c \sec{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{d \sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))/(a+a*sec(f*x+e))**2,x)

[Out]

(Integral(c*sec(e + f*x)/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(d*sec(e + f*x)**2/(sec(e + f*x)
**2 + 2*sec(e + f*x) + 1), x))/a**2

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Giac [A]  time = 1.14865, size = 86, normalized size = 1.32 \begin{align*} -\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 3 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 3 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{6 \, a^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(c*tan(1/2*f*x + 1/2*e)^3 - d*tan(1/2*f*x + 1/2*e)^3 - 3*c*tan(1/2*f*x + 1/2*e) - 3*d*tan(1/2*f*x + 1/2*e
))/(a^2*f)