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

Optimal. Leaf size=70 \[ -\frac{c \tanh ^{-1}(\sin (e+f x))}{a^2 f}+\frac{7 c \tan (e+f x)}{3 a^2 f (\sec (e+f x)+1)}-\frac{2 c \tan (e+f x)}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

-((c*ArcTanh[Sin[e + f*x]])/(a^2*f)) + (7*c*Tan[e + f*x])/(3*a^2*f*(1 + Sec[e + f*x])) - (2*c*Tan[e + f*x])/(3
*f*(a + a*Sec[e + f*x])^2)

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Rubi [A]  time = 0.161477, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4008, 3998, 3770, 3794} \[ -\frac{c \tanh ^{-1}(\sin (e+f x))}{a^2 f}+\frac{7 c \tan (e+f x)}{3 a^2 f (\sec (e+f x)+1)}-\frac{2 c \tan (e+f x)}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*ArcTanh[Sin[e + f*x]])/(a^2*f)) + (7*c*Tan[e + f*x])/(3*a^2*f*(1 + Sec[e + f*x])) - (2*c*Tan[e + f*x])/(3
*f*(a + a*Sec[e + f*x])^2)

Rule 4008

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

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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 ^2(e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx &=-\frac{2 c \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac{\int \frac{\sec (e+f x) (-4 a c+3 a c \sec (e+f x))}{a+a \sec (e+f x)} \, dx}{3 a^2}\\ &=-\frac{2 c \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac{c \int \sec (e+f x) \, dx}{a^2}+\frac{(7 c) \int \frac{\sec (e+f x)}{a+a \sec (e+f x)} \, dx}{3 a}\\ &=-\frac{c \tanh ^{-1}(\sin (e+f x))}{a^2 f}-\frac{2 c \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac{7 c \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.451307, size = 335, normalized size = 4.79 \[ \frac{c \sec \left (\frac{e}{2}\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x) \left (-6 \sin \left (e+\frac{f x}{2}\right )+10 \sin \left (e+\frac{3 f x}{2}\right )+3 \cos \left (e+\frac{3 f x}{2}\right ) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+3 \cos \left (2 e+\frac{3 f x}{2}\right ) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+9 \cos \left (\frac{f x}{2}\right ) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )+9 \cos \left (e+\frac{f x}{2}\right ) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )-3 \cos \left (e+\frac{3 f x}{2}\right ) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-3 \cos \left (2 e+\frac{3 f x}{2}\right ) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+24 \sin \left (\frac{f x}{2}\right )\right )}{6 a^2 f (\sec (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Cos[(e + f*x)/2]*Sec[e/2]*Sec[e + f*x]^2*(3*Cos[e + (3*f*x)/2]*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 3
*Cos[2*e + (3*f*x)/2]*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 9*Cos[(f*x)/2]*(Log[Cos[(e + f*x)/2] - Sin[(e
+ f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) + 9*Cos[e + (f*x)/2]*(Log[Cos[(e + f*x)/2] - Sin[(e +
f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) - 3*Cos[e + (3*f*x)/2]*Log[Cos[(e + f*x)/2] + Sin[(e + f*
x)/2]] - 3*Cos[2*e + (3*f*x)/2]*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + 24*Sin[(f*x)/2] - 6*Sin[e + (f*x)/2
] + 10*Sin[e + (3*f*x)/2]))/(6*a^2*f*(1 + Sec[e + f*x])^2)

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Maple [A]  time = 0.062, size = 81, normalized size = 1.2 \begin{align*}{\frac{c}{3\,f{a}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3}}+2\,{\frac{c\tan \left ( 1/2\,fx+e/2 \right ) }{f{a}^{2}}}+{\frac{c}{f{a}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) }-{\frac{c}{f{a}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.977587, size = 194, normalized size = 2.77 \begin{align*} \frac{c{\left (\frac{\frac{9 \, \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}}}{a^{2}} - \frac{6 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac{6 \, \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} + \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)^2*(c-c*sec(f*x+e))/(a+a*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

1/6*(c*((9*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 6*log(sin(f*x + e)/(co
s(f*x + e) + 1) + 1)/a^2 + 6*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/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 = 0.477495, size = 323, normalized size = 4.61 \begin{align*} -\frac{3 \,{\left (c \cos \left (f x + e\right )^{2} + 2 \, c \cos \left (f x + e\right ) + c\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (c \cos \left (f x + e\right )^{2} + 2 \, c \cos \left (f x + e\right ) + c\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \,{\left (5 \, c \cos \left (f x + e\right ) + 7 \, c\right )} \sin \left (f x + e\right )}{6 \,{\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)^2*(c-c*sec(f*x+e))/(a+a*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

-1/6*(3*(c*cos(f*x + e)^2 + 2*c*cos(f*x + e) + c)*log(sin(f*x + e) + 1) - 3*(c*cos(f*x + e)^2 + 2*c*cos(f*x +
e) + c)*log(-sin(f*x + e) + 1) - 2*(5*c*cos(f*x + e) + 7*c)*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{c \left (\int - \frac{\sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{\sec ^{3}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19779, size = 115, normalized size = 1.64 \begin{align*} -\frac{\frac{3 \, c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} - \frac{3 \, c \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} - \frac{a^{4} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 6 \, a^{4} c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{6}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*c*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^2 - 3*c*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^2 - (a^4*c*tan(1/2
*f*x + 1/2*e)^3 + 6*a^4*c*tan(1/2*f*x + 1/2*e))/a^6)/f