3.32 \(\int \frac{\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^6} \, dx\)

Optimal. Leaf size=121 \[ -\frac{2 \tan (e+f x) (a \sec (e+f x)+a)^3}{693 c^2 f (c-c \sec (e+f x))^4}-\frac{2 \tan (e+f x) (a \sec (e+f x)+a)^3}{99 c f (c-c \sec (e+f x))^5}-\frac{\tan (e+f x) (a \sec (e+f x)+a)^3}{11 f (c-c \sec (e+f x))^6} \]

[Out]

-((a + a*Sec[e + f*x])^3*Tan[e + f*x])/(11*f*(c - c*Sec[e + f*x])^6) - (2*(a + a*Sec[e + f*x])^3*Tan[e + f*x])
/(99*c*f*(c - c*Sec[e + f*x])^5) - (2*(a + a*Sec[e + f*x])^3*Tan[e + f*x])/(693*c^2*f*(c - c*Sec[e + f*x])^4)

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Rubi [A]  time = 0.230325, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {3951, 3950} \[ -\frac{2 \tan (e+f x) (a \sec (e+f x)+a)^3}{693 c^2 f (c-c \sec (e+f x))^4}-\frac{2 \tan (e+f x) (a \sec (e+f x)+a)^3}{99 c f (c-c \sec (e+f x))^5}-\frac{\tan (e+f x) (a \sec (e+f x)+a)^3}{11 f (c-c \sec (e+f x))^6} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x])^6,x]

[Out]

-((a + a*Sec[e + f*x])^3*Tan[e + f*x])/(11*f*(c - c*Sec[e + f*x])^6) - (2*(a + a*Sec[e + f*x])^3*Tan[e + f*x])
/(99*c*f*(c - c*Sec[e + f*x])^5) - (2*(a + a*Sec[e + f*x])^3*Tan[e + f*x])/(693*c^2*f*(c - c*Sec[e + f*x])^4)

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)])

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]

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^6} \, dx &=-\frac{(a+a \sec (e+f x))^3 \tan (e+f x)}{11 f (c-c \sec (e+f x))^6}+\frac{2 \int \frac{\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^5} \, dx}{11 c}\\ &=-\frac{(a+a \sec (e+f x))^3 \tan (e+f x)}{11 f (c-c \sec (e+f x))^6}-\frac{2 (a+a \sec (e+f x))^3 \tan (e+f x)}{99 c f (c-c \sec (e+f x))^5}+\frac{2 \int \frac{\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^4} \, dx}{99 c^2}\\ &=-\frac{(a+a \sec (e+f x))^3 \tan (e+f x)}{11 f (c-c \sec (e+f x))^6}-\frac{2 (a+a \sec (e+f x))^3 \tan (e+f x)}{99 c f (c-c \sec (e+f x))^5}-\frac{2 (a+a \sec (e+f x))^3 \tan (e+f x)}{693 c^2 f (c-c \sec (e+f x))^4}\\ \end{align*}

Mathematica [A]  time = 0.677084, size = 167, normalized size = 1.38 \[ -\frac{a^3 \csc \left (\frac{e}{2}\right ) \left (21252 \sin \left (e+\frac{f x}{2}\right )-15444 \sin \left (e+\frac{3 f x}{2}\right )-10626 \sin \left (2 e+\frac{3 f x}{2}\right )+4950 \sin \left (2 e+\frac{5 f x}{2}\right )+8085 \sin \left (3 e+\frac{5 f x}{2}\right )-2959 \sin \left (3 e+\frac{7 f x}{2}\right )-1386 \sin \left (4 e+\frac{7 f x}{2}\right )+176 \sin \left (4 e+\frac{9 f x}{2}\right )+693 \sin \left (5 e+\frac{9 f x}{2}\right )-79 \sin \left (5 e+\frac{11 f x}{2}\right )+15246 \sin \left (\frac{f x}{2}\right )\right ) \csc ^{11}\left (\frac{1}{2} (e+f x)\right )}{709632 c^6 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x])^6,x]

[Out]

-(a^3*Csc[e/2]*Csc[(e + f*x)/2]^11*(15246*Sin[(f*x)/2] + 21252*Sin[e + (f*x)/2] - 15444*Sin[e + (3*f*x)/2] - 1
0626*Sin[2*e + (3*f*x)/2] + 4950*Sin[2*e + (5*f*x)/2] + 8085*Sin[3*e + (5*f*x)/2] - 2959*Sin[3*e + (7*f*x)/2]
- 1386*Sin[4*e + (7*f*x)/2] + 176*Sin[4*e + (9*f*x)/2] + 693*Sin[5*e + (9*f*x)/2] - 79*Sin[5*e + (11*f*x)/2]))
/(709632*c^6*f)

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Maple [A]  time = 0.125, size = 52, normalized size = 0.4 \begin{align*}{\frac{{a}^{3}}{4\,f{c}^{6}} \left ( -{\frac{1}{11} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-11}}-{\frac{1}{7} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-7}}+{\frac{2}{9} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{-9}} \right ) } \end{align*}

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))^6,x)

[Out]

1/4/f*a^3/c^6*(-1/11/tan(1/2*f*x+1/2*e)^11-1/7/tan(1/2*f*x+1/2*e)^7+2/9/tan(1/2*f*x+1/2*e)^9)

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Maxima [B]  time = 1.09299, size = 699, normalized size = 5.78 \begin{align*} \text{result too large to display} \end{align*}

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))^6,x, algorithm="maxima")

[Out]

1/110880*(3*a^3*(385*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 990*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 1386*sin(
f*x + e)^6/(cos(f*x + e) + 1)^6 - 1155*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 3465*sin(f*x + e)^10/(cos(f*x + e
) + 1)^10 - 315)*(cos(f*x + e) + 1)^11/(c^6*sin(f*x + e)^11) + 9*a^3*(385*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
- 330*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 462*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 1155*sin(f*x + e)^8/(cos
(f*x + e) + 1)^8 - 1155*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 - 105)*(cos(f*x + e) + 1)^11/(c^6*sin(f*x + e)^1
1) + 5*a^3*(385*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 990*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 1386*sin(f*x +
e)^6/(cos(f*x + e) + 1)^6 - 1155*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 693*sin(f*x + e)^10/(cos(f*x + e) + 1)
^10 - 63)*(cos(f*x + e) + 1)^11/(c^6*sin(f*x + e)^11) - a^3*(385*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 990*sin
(f*x + e)^4/(cos(f*x + e) + 1)^4 - 1386*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 1155*sin(f*x + e)^8/(cos(f*x + e
) + 1)^8 + 3465*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 315)*(cos(f*x + e) + 1)^11/(c^6*sin(f*x + e)^11))/f

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Fricas [A]  time = 0.466873, size = 413, normalized size = 3.41 \begin{align*} \frac{79 \, a^{3} \cos \left (f x + e\right )^{6} + 298 \, a^{3} \cos \left (f x + e\right )^{5} + 404 \, a^{3} \cos \left (f x + e\right )^{4} + 216 \, a^{3} \cos \left (f x + e\right )^{3} + 19 \, a^{3} \cos \left (f x + e\right )^{2} - 10 \, a^{3} \cos \left (f x + e\right ) + 2 \, a^{3}}{693 \,{\left (c^{6} f \cos \left (f x + e\right )^{5} - 5 \, c^{6} f \cos \left (f x + e\right )^{4} + 10 \, c^{6} f \cos \left (f x + e\right )^{3} - 10 \, c^{6} f \cos \left (f x + e\right )^{2} + 5 \, c^{6} f \cos \left (f x + e\right ) - c^{6} f\right )} \sin \left (f x + e\right )} \end{align*}

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))^6,x, algorithm="fricas")

[Out]

1/693*(79*a^3*cos(f*x + e)^6 + 298*a^3*cos(f*x + e)^5 + 404*a^3*cos(f*x + e)^4 + 216*a^3*cos(f*x + e)^3 + 19*a
^3*cos(f*x + e)^2 - 10*a^3*cos(f*x + e) + 2*a^3)/((c^6*f*cos(f*x + e)^5 - 5*c^6*f*cos(f*x + e)^4 + 10*c^6*f*co
s(f*x + e)^3 - 10*c^6*f*cos(f*x + e)^2 + 5*c^6*f*cos(f*x + e) - c^6*f)*sin(f*x + e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**6,x)

[Out]

Timed out

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Giac [A]  time = 1.29698, size = 81, normalized size = 0.67 \begin{align*} -\frac{99 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 154 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 63 \, a^{3}}{2772 \, c^{6} f \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{11}} \end{align*}

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))^6,x, algorithm="giac")

[Out]

-1/2772*(99*a^3*tan(1/2*f*x + 1/2*e)^4 - 154*a^3*tan(1/2*f*x + 1/2*e)^2 + 63*a^3)/(c^6*f*tan(1/2*f*x + 1/2*e)^
11)