3.230 \(\int \frac{\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=102 \[ \frac{(2 c+3 d) \tan (e+f x)}{15 f \left (a^3 \sec (e+f x)+a^3\right )}+\frac{(2 c+3 d) \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2}+\frac{(c-d) \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3} \]

[Out]

((c - d)*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) + ((2*c + 3*d)*Tan[e + f*x])/(15*a*f*(a + a*Sec[e + f*x])^
2) + ((2*c + 3*d)*Tan[e + f*x])/(15*f*(a^3 + a^3*Sec[e + f*x]))

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Rubi [A]  time = 0.115281, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {4000, 3796, 3794} \[ \frac{(2 c+3 d) \tan (e+f x)}{15 f \left (a^3 \sec (e+f x)+a^3\right )}+\frac{(2 c+3 d) \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2}+\frac{(c-d) \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c + d*Sec[e + f*x]))/(a + a*Sec[e + f*x])^3,x]

[Out]

((c - d)*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) + ((2*c + 3*d)*Tan[e + f*x])/(15*a*f*(a + a*Sec[e + f*x])^
2) + ((2*c + 3*d)*Tan[e + f*x])/(15*f*(a^3 + a^3*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 3796

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

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))^3} \, dx &=\frac{(c-d) \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{(2 c+3 d) \int \frac{\sec (e+f x)}{(a+a \sec (e+f x))^2} \, dx}{5 a}\\ &=\frac{(c-d) \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{(2 c+3 d) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{(2 c+3 d) \int \frac{\sec (e+f x)}{a+a \sec (e+f x)} \, dx}{15 a^2}\\ &=\frac{(c-d) \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{(2 c+3 d) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{(2 c+3 d) \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.337327, size = 135, normalized size = 1.32 \[ \frac{\sec \left (\frac{e}{2}\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \left (-15 (2 c+d) \sin \left (e+\frac{f x}{2}\right )+5 (8 c+3 d) \sin \left (\frac{f x}{2}\right )+20 c \sin \left (e+\frac{3 f x}{2}\right )-15 c \sin \left (2 e+\frac{3 f x}{2}\right )+7 c \sin \left (2 e+\frac{5 f x}{2}\right )+15 d \sin \left (e+\frac{3 f x}{2}\right )+3 d \sin \left (2 e+\frac{5 f x}{2}\right )\right )}{30 a^3 f (\cos (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c + d*Sec[e + f*x]))/(a + a*Sec[e + f*x])^3,x]

[Out]

(Cos[(e + f*x)/2]*Sec[e/2]*(5*(8*c + 3*d)*Sin[(f*x)/2] - 15*(2*c + d)*Sin[e + (f*x)/2] + 20*c*Sin[e + (3*f*x)/
2] + 15*d*Sin[e + (3*f*x)/2] - 15*c*Sin[2*e + (3*f*x)/2] + 7*c*Sin[2*e + (5*f*x)/2] + 3*d*Sin[2*e + (5*f*x)/2]
))/(30*a^3*f*(1 + Cos[e + f*x])^3)

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Maple [A]  time = 0.053, size = 64, normalized size = 0.6 \begin{align*}{\frac{1}{4\,f{a}^{3}} \left ({\frac{c-d}{5} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{5}}-{\frac{2\,c}{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))^3,x)

[Out]

1/4/f/a^3*(1/5*(c-d)*tan(1/2*f*x+1/2*e)^5-2/3*tan(1/2*f*x+1/2*e)^3*c+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 = 1.01488, size = 155, normalized size = 1.52 \begin{align*} \frac{\frac{c{\left (\frac{15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac{3 \, d{\left (\frac{5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, 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))^3,x, algorithm="maxima")

[Out]

1/60*(c*(15*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f
*x + e) + 1)^5)/a^3 + 3*d*(5*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3)/f

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Fricas [A]  time = 0.451149, size = 227, normalized size = 2.23 \begin{align*} \frac{{\left ({\left (7 \, c + 3 \, d\right )} \cos \left (f x + e\right )^{2} + 3 \,{\left (2 \, c + 3 \, d\right )} \cos \left (f x + e\right ) + 2 \, c + 3 \, d\right )} \sin \left (f x + e\right )}{15 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} 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))^3,x, algorithm="fricas")

[Out]

1/15*((7*c + 3*d)*cos(f*x + e)^2 + 3*(2*c + 3*d)*cos(f*x + e) + 2*c + 3*d)*sin(f*x + e)/(a^3*f*cos(f*x + e)^3
+ 3*a^3*f*cos(f*x + e)^2 + 3*a^3*f*cos(f*x + e) + a^3*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 ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{d \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx}{a^{3}} \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))**3,x)

[Out]

(Integral(c*sec(e + f*x)/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(d*sec(e + f
*x)**2/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x))/a**3

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Giac [A]  time = 1.2609, size = 108, normalized size = 1.06 \begin{align*} \frac{3 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 3 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 10 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 15 \, c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 15 \, d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{60 \, a^{3} 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))^3,x, algorithm="giac")

[Out]

1/60*(3*c*tan(1/2*f*x + 1/2*e)^5 - 3*d*tan(1/2*f*x + 1/2*e)^5 - 10*c*tan(1/2*f*x + 1/2*e)^3 + 15*c*tan(1/2*f*x
+ 1/2*e) + 15*d*tan(1/2*f*x + 1/2*e))/(a^3*f)