3.31 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^5} \, dx\)

Optimal. Leaf size=80 \[ -\frac {\tan (e+f x) (a \sec (e+f x)+a)^3}{63 c f (c-c \sec (e+f x))^4}-\frac {\tan (e+f x) (a \sec (e+f x)+a)^3}{9 f (c-c \sec (e+f x))^5} \]

[Out]

-1/9*(a+a*sec(f*x+e))^3*tan(f*x+e)/f/(c-c*sec(f*x+e))^5-1/63*(a+a*sec(f*x+e))^3*tan(f*x+e)/c/f/(c-c*sec(f*x+e)
)^4

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Rubi [A]  time = 0.15, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {3951, 3950} \[ -\frac {\tan (e+f x) (a \sec (e+f x)+a)^3}{63 c f (c-c \sec (e+f x))^4}-\frac {\tan (e+f x) (a \sec (e+f x)+a)^3}{9 f (c-c \sec (e+f x))^5} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x])^5,x]

[Out]

-((a + a*Sec[e + f*x])^3*Tan[e + f*x])/(9*f*(c - c*Sec[e + f*x])^5) - ((a + a*Sec[e + f*x])^3*Tan[e + f*x])/(6
3*c*f*(c - c*Sec[e + f*x])^4)

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))^3}{(c-c \sec (e+f x))^5} \, dx &=-\frac {(a+a \sec (e+f x))^3 \tan (e+f x)}{9 f (c-c \sec (e+f x))^5}+\frac {\int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^4} \, dx}{9 c}\\ &=-\frac {(a+a \sec (e+f x))^3 \tan (e+f x)}{9 f (c-c \sec (e+f x))^5}-\frac {(a+a \sec (e+f x))^3 \tan (e+f x)}{63 c f (c-c \sec (e+f x))^4}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 141, normalized size = 1.76 \[ -\frac {a^3 \csc \left (\frac {e}{2}\right ) \left (315 \sin \left (e+\frac {f x}{2}\right )-189 \sin \left (e+\frac {3 f x}{2}\right )-483 \sin \left (2 e+\frac {3 f x}{2}\right )+225 \sin \left (2 e+\frac {5 f x}{2}\right )+63 \sin \left (3 e+\frac {5 f x}{2}\right )-9 \sin \left (3 e+\frac {7 f x}{2}\right )-63 \sin \left (4 e+\frac {7 f x}{2}\right )+8 \sin \left (4 e+\frac {9 f x}{2}\right )+693 \sin \left (\frac {f x}{2}\right )\right ) \csc ^9\left (\frac {1}{2} (e+f x)\right )}{16128 c^5 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x])^5,x]

[Out]

-1/16128*(a^3*Csc[e/2]*Csc[(e + f*x)/2]^9*(693*Sin[(f*x)/2] + 315*Sin[e + (f*x)/2] - 189*Sin[e + (3*f*x)/2] -
483*Sin[2*e + (3*f*x)/2] + 225*Sin[2*e + (5*f*x)/2] + 63*Sin[3*e + (5*f*x)/2] - 9*Sin[3*e + (7*f*x)/2] - 63*Si
n[4*e + (7*f*x)/2] + 8*Sin[4*e + (9*f*x)/2]))/(c^5*f)

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fricas [A]  time = 0.45, size = 140, normalized size = 1.75 \[ \frac {8 \, a^{3} \cos \left (f x + e\right )^{5} + 31 \, a^{3} \cos \left (f x + e\right )^{4} + 44 \, a^{3} \cos \left (f x + e\right )^{3} + 26 \, a^{3} \cos \left (f x + e\right )^{2} + 4 \, a^{3} \cos \left (f x + e\right ) - a^{3}}{63 \, {\left (c^{5} f \cos \left (f x + e\right )^{4} - 4 \, c^{5} f \cos \left (f x + e\right )^{3} + 6 \, c^{5} f \cos \left (f x + e\right )^{2} - 4 \, c^{5} f \cos \left (f x + e\right ) + c^{5} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^5,x, algorithm="fricas")

[Out]

1/63*(8*a^3*cos(f*x + e)^5 + 31*a^3*cos(f*x + e)^4 + 44*a^3*cos(f*x + e)^3 + 26*a^3*cos(f*x + e)^2 + 4*a^3*cos
(f*x + e) - a^3)/((c^5*f*cos(f*x + e)^4 - 4*c^5*f*cos(f*x + e)^3 + 6*c^5*f*cos(f*x + e)^2 - 4*c^5*f*cos(f*x +
e) + c^5*f)*sin(f*x + e))

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giac [A]  time = 0.89, size = 43, normalized size = 0.54 \[ -\frac {9 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 7 \, a^{3}}{126 \, c^{5} f \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^5,x, algorithm="giac")

[Out]

-1/126*(9*a^3*tan(1/2*f*x + 1/2*e)^2 - 7*a^3)/(c^5*f*tan(1/2*f*x + 1/2*e)^9)

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maple [A]  time = 0.95, size = 39, normalized size = 0.49 \[ \frac {a^{3} \left (-\frac {1}{7 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{7}}+\frac {1}{9 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{9}}\right )}{2 f \,c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^5,x)

[Out]

1/2/f*a^3/c^5*(-1/7/tan(1/2*e+1/2*f*x)^7+1/9/tan(1/2*e+1/2*f*x)^9)

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maxima [B]  time = 0.37, size = 357, normalized size = 4.46 \[ -\frac {\frac {a^{3} {\left (\frac {180 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {378 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {420 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {315 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 35\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{c^{5} \sin \left (f x + e\right )^{9}} + \frac {15 \, a^{3} {\left (\frac {18 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {42 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {63 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 7\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{c^{5} \sin \left (f x + e\right )^{9}} - \frac {5 \, a^{3} {\left (\frac {18 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {42 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {63 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + 7\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{c^{5} \sin \left (f x + e\right )^{9}} + \frac {21 \, a^{3} {\left (\frac {18 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {45 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 5\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{c^{5} \sin \left (f x + e\right )^{9}}}{5040 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^5,x, algorithm="maxima")

[Out]

-1/5040*(a^3*(180*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 378*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 420*sin(f*x
+ e)^6/(cos(f*x + e) + 1)^6 - 315*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 35)*(cos(f*x + e) + 1)^9/(c^5*sin(f*x
+ e)^9) + 15*a^3*(18*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 42*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 63*sin(f*x
+ e)^8/(cos(f*x + e) + 1)^8 - 7)*(cos(f*x + e) + 1)^9/(c^5*sin(f*x + e)^9) - 5*a^3*(18*sin(f*x + e)^2/(cos(f*
x + e) + 1)^2 - 42*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 63*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 7)*(cos(f*x
+ e) + 1)^9/(c^5*sin(f*x + e)^9) + 21*a^3*(18*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 45*sin(f*x + e)^8/(cos(f*x
+ e) + 1)^8 - 5)*(cos(f*x + e) + 1)^9/(c^5*sin(f*x + e)^9))/f

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mupad [B]  time = 1.62, size = 37, normalized size = 0.46 \[ \frac {a^3\,{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (7\,{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-9\right )}{126\,c^5\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^3/(cos(e + f*x)*(c - c/cos(e + f*x))^5),x)

[Out]

(a^3*cot(e/2 + (f*x)/2)^7*(7*cot(e/2 + (f*x)/2)^2 - 9))/(126*c^5*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {a^{3} \left (\int \frac {\sec {\left (e + f x \right )}}{\sec ^{5}{\left (e + f x \right )} - 5 \sec ^{4}{\left (e + f x \right )} + 10 \sec ^{3}{\left (e + f x \right )} - 10 \sec ^{2}{\left (e + f x \right )} + 5 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {3 \sec ^{2}{\left (e + f x \right )}}{\sec ^{5}{\left (e + f x \right )} - 5 \sec ^{4}{\left (e + f x \right )} + 10 \sec ^{3}{\left (e + f x \right )} - 10 \sec ^{2}{\left (e + f x \right )} + 5 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {3 \sec ^{3}{\left (e + f x \right )}}{\sec ^{5}{\left (e + f x \right )} - 5 \sec ^{4}{\left (e + f x \right )} + 10 \sec ^{3}{\left (e + f x \right )} - 10 \sec ^{2}{\left (e + f x \right )} + 5 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {\sec ^{4}{\left (e + f x \right )}}{\sec ^{5}{\left (e + f x \right )} - 5 \sec ^{4}{\left (e + f x \right )} + 10 \sec ^{3}{\left (e + f x \right )} - 10 \sec ^{2}{\left (e + f x \right )} + 5 \sec {\left (e + f x \right )} - 1}\, dx\right )}{c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e))**5,x)

[Out]

-a**3*(Integral(sec(e + f*x)/(sec(e + f*x)**5 - 5*sec(e + f*x)**4 + 10*sec(e + f*x)**3 - 10*sec(e + f*x)**2 +
5*sec(e + f*x) - 1), x) + Integral(3*sec(e + f*x)**2/(sec(e + f*x)**5 - 5*sec(e + f*x)**4 + 10*sec(e + f*x)**3
- 10*sec(e + f*x)**2 + 5*sec(e + f*x) - 1), x) + Integral(3*sec(e + f*x)**3/(sec(e + f*x)**5 - 5*sec(e + f*x)
**4 + 10*sec(e + f*x)**3 - 10*sec(e + f*x)**2 + 5*sec(e + f*x) - 1), x) + Integral(sec(e + f*x)**4/(sec(e + f*
x)**5 - 5*sec(e + f*x)**4 + 10*sec(e + f*x)**3 - 10*sec(e + f*x)**2 + 5*sec(e + f*x) - 1), x))/c**5

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