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

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

[Out]

-1/5*(a+a*sec(f*x+e))*tan(f*x+e)/f/(c-c*sec(f*x+e))^3-1/15*(a+a*sec(f*x+e))*tan(f*x+e)/c/f/(c-c*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} \[ -\frac {\tan (e+f x) (a \sec (e+f x)+a)}{15 c f (c-c \sec (e+f x))^2}-\frac {\tan (e+f x) (a \sec (e+f x)+a)}{5 f (c-c \sec (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.41, size = 87, normalized size = 1.14 \[ -\frac {a \csc \left (\frac {e}{2}\right ) \left (15 \sin \left (e+\frac {f x}{2}\right )-5 \sin \left (e+\frac {3 f x}{2}\right )-15 \sin \left (2 e+\frac {3 f x}{2}\right )+4 \sin \left (2 e+\frac {5 f x}{2}\right )+25 \sin \left (\frac {f x}{2}\right )\right ) \csc ^5\left (\frac {1}{2} (e+f x)\right )}{240 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/240*(a*Csc[e/2]*Csc[(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]))/(c^3*f)

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fricas [A]  time = 0.44, size = 78, normalized size = 1.03 \[ \frac {4 \, a \cos \left (f x + e\right )^{3} + 7 \, a \cos \left (f x + e\right )^{2} + 2 \, a \cos \left (f x + e\right ) - a}{15 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(4*a*cos(f*x + e)^3 + 7*a*cos(f*x + e)^2 + 2*a*cos(f*x + e) - a)/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x
+ e) + c^3*f)*sin(f*x + e))

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giac [A]  time = 0.51, size = 39, normalized size = 0.51 \[ -\frac {5 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3 \, a}{30 \, c^{3} f \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.72, size = 37, normalized size = 0.49 \[ \frac {a \left (-\frac {1}{3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{3}}+\frac {1}{5 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{5}}\right )}{2 f \,c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 117, normalized size = 1.54 \[ -\frac {\frac {a {\left (\frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 3\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}} + \frac {3 \, a {\left (\frac {5 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{5}}{c^{3} \sin \left (f x + e\right )^{5}}}{60 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(a*(10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 3)*(cos(f*x + e) +
1)^5/(c^3*sin(f*x + e)^5) + 3*a*(5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 1)*(cos(f*x + e) + 1)^5/(c^3*sin(f*x
+ e)^5))/f

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mupad [B]  time = 1.71, size = 35, normalized size = 0.46 \[ \frac {a\,{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-5\right )}{30\,c^3\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(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))/c**3

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