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

Optimal. Leaf size=76 \[ \frac {\tan (e+f x) (c-c \sec (e+f x))}{15 a f (a \sec (e+f x)+a)^2}+\frac {\tan (e+f x) (c-c \sec (e+f x))}{5 f (a \sec (e+f x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3951, 3950} \[ \frac {\tan (e+f x) (c-c \sec (e+f x))}{15 a f (a \sec (e+f x)+a)^2}+\frac {\tan (e+f x) (c-c \sec (e+f x))}{5 f (a \sec (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3950

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

Rule 3951

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))
^(n_.), x_Symbol] :> Simp[(b*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(c + d*Csc[e + f*x])^n)/(a*f*(2*m + 1)), x] +
Dist[(m + n + 1)/(a*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n, x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[m + n + 1, 0] && NeQ[2
*m + 1, 0] &&  !LtQ[n, 0] &&  !(IGtQ[n + 1/2, 0] && LtQ[n + 1/2, -(m + n)])

Rubi steps

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

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Mathematica [A]  time = 0.35, size = 87, normalized size = 1.14 \[ \frac {c \sec \left (\frac {e}{2}\right ) \left (-15 \sin \left (e+\frac {f x}{2}\right )+5 \sin \left (e+\frac {3 f x}{2}\right )-15 \sin \left (2 e+\frac {3 f x}{2}\right )+4 \sin \left (2 e+\frac {5 f x}{2}\right )+25 \sin \left (\frac {f x}{2}\right )\right ) \sec ^5\left (\frac {1}{2} (e+f x)\right )}{240 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sec[e/2]*Sec[(e + f*x)/2]^5*(25*Sin[(f*x)/2] - 15*Sin[e + (f*x)/2] + 5*Sin[e + (3*f*x)/2] - 15*Sin[2*e + (3
*f*x)/2] + 4*Sin[2*e + (5*f*x)/2]))/(240*a^3*f)

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fricas [A]  time = 0.43, size = 79, normalized size = 1.04 \[ \frac {{\left (4 \, c \cos \left (f x + e\right )^{2} - 3 \, c \cos \left (f x + e\right ) - c\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))/(a+a*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(4*c*cos(f*x + e)^2 - 3*c*cos(f*x + e) - 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 = 0.40, size = 39, normalized size = 0.51 \[ \frac {3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 5 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}}{30 \, a^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(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 = 0.80, size = 37, normalized size = 0.49 \[ \frac {c \left (\frac {\left (\tan ^{5}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{5}-\frac {\left (\tan ^{3}\left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{3}\right )}{2 f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(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 = 0.34, size = 115, normalized size = 1.51 \[ \frac {\frac {c {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {3 \, c {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(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 = 1.58, size = 35, normalized size = 0.46 \[ \frac {c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-5\right )}{30\,a^3\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c/cos(e + f*x))/(cos(e + f*x)*(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 \[ - \frac {c \left (\int \left (- \frac {\sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx\right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))/(a+a*sec(f*x+e))**3,x)

[Out]

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