3.173 \(\int \frac {\sec ^2(e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=86 \[ -\frac {4 c \tan (e+f x)}{15 f \left (a^3 \sec (e+f x)+a^3\right )}+\frac {11 c \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2}-\frac {2 c \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3} \]

[Out]

-2/5*c*tan(f*x+e)/f/(a+a*sec(f*x+e))^3+11/15*c*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^2-4/15*c*tan(f*x+e)/f/(a^3+a^3*
sec(f*x+e))

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Rubi [A]  time = 0.16, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {4008, 4000, 3794} \[ -\frac {4 c \tan (e+f x)}{15 f \left (a^3 \sec (e+f x)+a^3\right )}+\frac {11 c \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2}-\frac {2 c \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) + (11*c*Tan[e + f*x])/(15*a*f*(a + a*Sec[e + f*x])^2) - (4*c*
Tan[e + f*x])/(15*f*(a^3 + a^3*Sec[e + f*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]

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 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)]

Rubi steps

\begin {align*} \int \frac {\sec ^2(e+f x) (c-c \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx &=-\frac {2 c \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac {\int \frac {\sec (e+f x) (-6 a c+5 a c \sec (e+f x))}{(a+a \sec (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac {2 c \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {11 c \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}-\frac {(4 c) \int \frac {\sec (e+f x)}{a+a \sec (e+f x)} \, dx}{15 a^2}\\ &=-\frac {2 c \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {11 c \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}-\frac {4 c \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 43, normalized size = 0.50 \[ -\frac {c (\cos (e+f x)+4) \tan ^3\left (\frac {1}{2} (e+f x)\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{30 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/30*(c*(4 + Cos[e + f*x])*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]^3)/(a^3*f)

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fricas [A]  time = 0.40, size = 78, normalized size = 0.91 \[ \frac {{\left (c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) - 4 \, c\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \]

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))^3,x, algorithm="fricas")

[Out]

1/15*(c*cos(f*x + e)^2 + 3*c*cos(f*x + e) - 4*c)*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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giac [A]  time = 0.82, size = 39, normalized size = 0.45 \[ -\frac {3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 5 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}}{30 \, a^{3} f} \]

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))^3,x, algorithm="giac")

[Out]

-1/30*(3*c*tan(1/2*f*x + 1/2*e)^5 + 5*c*tan(1/2*f*x + 1/2*e)^3)/(a^3*f)

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maple [A]  time = 0.76, size = 37, normalized size = 0.43 \[ \frac {c \left (-\frac {\left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{5}-\frac {\left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{3}\right )}{2 f \,a^{3}} \]

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))^3,x)

[Out]

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

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maxima [A]  time = 0.51, size = 115, normalized size = 1.34 \[ -\frac {\frac {c {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {3 \, c {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \]

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))^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*c*(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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mupad [B]  time = 1.70, size = 35, normalized size = 0.41 \[ -\frac {c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\right )}{30\,a^3\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c/cos(e + f*x))/(cos(e + f*x)^2*(a + a/cos(e + f*x))^3),x)

[Out]

-(c*tan(e/2 + (f*x)/2)^3*(3*tan(e/2 + (f*x)/2)^2 + 5))/(30*a^3*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {c \left (\int \left (- \frac {\sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx\right )}{a^{3}} \]

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))**3,x)

[Out]

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

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