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

Optimal. Leaf size=158 \[ -\frac {2 \tan (e+f x) (a \sec (e+f x)+a)}{315 c f \left (c^2-c^2 \sec (e+f x)\right )^2}-\frac {2 \tan (e+f x) (a \sec (e+f x)+a)}{105 c^2 f (c-c \sec (e+f x))^3}-\frac {\tan (e+f x) (a \sec (e+f x)+a)}{21 c f (c-c \sec (e+f x))^4}-\frac {\tan (e+f x) (a \sec (e+f x)+a)}{9 f (c-c \sec (e+f x))^5} \]

[Out]

-1/9*(a+a*sec(f*x+e))*tan(f*x+e)/f/(c-c*sec(f*x+e))^5-1/21*(a+a*sec(f*x+e))*tan(f*x+e)/c/f/(c-c*sec(f*x+e))^4-
2/105*(a+a*sec(f*x+e))*tan(f*x+e)/c^2/f/(c-c*sec(f*x+e))^3-2/315*(a+a*sec(f*x+e))*tan(f*x+e)/c/f/(c^2-c^2*sec(
f*x+e))^2

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Rubi [A]  time = 0.21, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3951, 3950} \[ -\frac {2 \tan (e+f x) (a \sec (e+f x)+a)}{315 c f \left (c^2-c^2 \sec (e+f x)\right )^2}-\frac {2 \tan (e+f x) (a \sec (e+f x)+a)}{105 c^2 f (c-c \sec (e+f x))^3}-\frac {\tan (e+f x) (a \sec (e+f x)+a)}{21 c f (c-c \sec (e+f x))^4}-\frac {\tan (e+f x) (a \sec (e+f x)+a)}{9 f (c-c \sec (e+f x))^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + a*Sec[e + f*x])*Tan[e + f*x])/(9*f*(c - c*Sec[e + f*x])^5) - ((a + a*Sec[e + f*x])*Tan[e + f*x])/(21*c*
f*(c - c*Sec[e + f*x])^4) - (2*(a + a*Sec[e + f*x])*Tan[e + f*x])/(105*c^2*f*(c - c*Sec[e + f*x])^3) - (2*(a +
a*Sec[e + f*x])*Tan[e + f*x])/(315*c*f*(c^2 - c^2*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))^5} \, dx &=-\frac {(a+a \sec (e+f x)) \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))}{(c-c \sec (e+f x))^4} \, dx}{3 c}\\ &=-\frac {(a+a \sec (e+f x)) \tan (e+f x)}{9 f (c-c \sec (e+f x))^5}-\frac {(a+a \sec (e+f x)) \tan (e+f x)}{21 c f (c-c \sec (e+f x))^4}+\frac {2 \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^3} \, dx}{21 c^2}\\ &=-\frac {(a+a \sec (e+f x)) \tan (e+f x)}{9 f (c-c \sec (e+f x))^5}-\frac {(a+a \sec (e+f x)) \tan (e+f x)}{21 c f (c-c \sec (e+f x))^4}-\frac {2 (a+a \sec (e+f x)) \tan (e+f x)}{105 c^2 f (c-c \sec (e+f x))^3}+\frac {2 \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^2} \, dx}{105 c^3}\\ &=-\frac {(a+a \sec (e+f x)) \tan (e+f x)}{9 f (c-c \sec (e+f x))^5}-\frac {(a+a \sec (e+f x)) \tan (e+f x)}{21 c f (c-c \sec (e+f x))^4}-\frac {2 (a+a \sec (e+f x)) \tan (e+f x)}{105 c^2 f (c-c \sec (e+f x))^3}-\frac {2 (a+a \sec (e+f x)) \tan (e+f x)}{315 c^3 f (c-c \sec (e+f x))^2}\\ \end {align*}

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Mathematica [A]  time = 0.36, size = 139, normalized size = 0.88 \[ -\frac {a \csc \left (\frac {e}{2}\right ) \left (3465 \sin \left (e+\frac {f x}{2}\right )-2247 \sin \left (e+\frac {3 f x}{2}\right )-2625 \sin \left (2 e+\frac {3 f x}{2}\right )+1143 \sin \left (2 e+\frac {5 f x}{2}\right )+945 \sin \left (3 e+\frac {5 f x}{2}\right )-207 \sin \left (3 e+\frac {7 f x}{2}\right )-315 \sin \left (4 e+\frac {7 f x}{2}\right )+58 \sin \left (4 e+\frac {9 f x}{2}\right )+3843 \sin \left (\frac {f x}{2}\right )\right ) \csc ^9\left (\frac {1}{2} (e+f x)\right )}{80640 c^5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/80640*(a*Csc[e/2]*Csc[(e + f*x)/2]^9*(3843*Sin[(f*x)/2] + 3465*Sin[e + (f*x)/2] - 2247*Sin[e + (3*f*x)/2] -
2625*Sin[2*e + (3*f*x)/2] + 1143*Sin[2*e + (5*f*x)/2] + 945*Sin[3*e + (5*f*x)/2] - 207*Sin[3*e + (7*f*x)/2] -
315*Sin[4*e + (7*f*x)/2] + 58*Sin[4*e + (9*f*x)/2]))/(c^5*f)

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fricas [A]  time = 0.42, size = 128, normalized size = 0.81 \[ \frac {58 \, a \cos \left (f x + e\right )^{5} + 83 \, a \cos \left (f x + e\right )^{4} + 4 \, a \cos \left (f x + e\right )^{3} - 11 \, a \cos \left (f x + e\right )^{2} + 8 \, a \cos \left (f x + e\right ) - 2 \, a}{315 \, {\left (c^{5} f \cos \left (f x + e\right )^{4} - 4 \, c^{5} f \cos \left (f x + e\right )^{3} + 6 \, c^{5} f \cos \left (f x + e\right )^{2} - 4 \, c^{5} f \cos \left (f x + e\right ) + c^{5} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(58*a*cos(f*x + e)^5 + 83*a*cos(f*x + e)^4 + 4*a*cos(f*x + e)^3 - 11*a*cos(f*x + e)^2 + 8*a*cos(f*x + e)
- 2*a)/((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 = 0.47, size = 69, normalized size = 0.44 \[ -\frac {105 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 189 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 135 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 35 \, a}{2520 \, c^{5} f \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2520*(105*a*tan(1/2*f*x + 1/2*e)^6 - 189*a*tan(1/2*f*x + 1/2*e)^4 + 135*a*tan(1/2*f*x + 1/2*e)^2 - 35*a)/(c
^5*f*tan(1/2*f*x + 1/2*e)^9)

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maple [A]  time = 0.77, size = 63, normalized size = 0.40 \[ \frac {a \left (-\frac {3}{7 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{7}}+\frac {1}{9 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{9}}-\frac {1}{3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{3}}+\frac {3}{5 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )^{5}}\right )}{8 f \,c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/f*a/c^5*(-3/7/tan(1/2*e+1/2*f*x)^7+1/9/tan(1/2*e+1/2*f*x)^9-1/3/tan(1/2*e+1/2*f*x)^3+3/5/tan(1/2*e+1/2*f*x
)^5)

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maxima [A]  time = 0.70, size = 197, normalized size = 1.25 \[ -\frac {\frac {a {\left (\frac {180 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {378 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {420 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {315 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 35\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{c^{5} \sin \left (f x + e\right )^{9}} + \frac {5 \, a {\left (\frac {18 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {42 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {63 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - 7\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{9}}{c^{5} \sin \left (f x + e\right )^{9}}}{5040 \, f} \]

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))^5,x, algorithm="maxima")

[Out]

-1/5040*(a*(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) + 5*a*(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))/f

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mupad [B]  time = 1.84, size = 106, normalized size = 0.67 \[ \frac {a\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (35\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-135\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+189\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-105\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\right )}{2520\,c^5\,f\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9} \]

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))^5),x)

[Out]

(a*cos(e/2 + (f*x)/2)^3*(35*cos(e/2 + (f*x)/2)^6 - 105*sin(e/2 + (f*x)/2)^6 + 189*cos(e/2 + (f*x)/2)^2*sin(e/2
+ (f*x)/2)^4 - 135*cos(e/2 + (f*x)/2)^4*sin(e/2 + (f*x)/2)^2))/(2520*c^5*f*sin(e/2 + (f*x)/2)^9)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(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*s
ec(e + f*x) - 1), x) + Integral(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))/c**5

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