3.71 \(\int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{7/2} \, dx\)

Optimal. Leaf size=171 \[ -\frac{256 c^4 \tan (e+f x) (a \sec (e+f x)+a)^2}{1155 f \sqrt{c-c \sec (e+f x)}}-\frac{64 c^3 \tan (e+f x) (a \sec (e+f x)+a)^2 \sqrt{c-c \sec (e+f x)}}{231 f}-\frac{8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}}{33 f}-\frac{2 c \tan (e+f x) (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{5/2}}{11 f} \]

[Out]

(-256*c^4*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(1155*f*Sqrt[c - c*Sec[e + f*x]]) - (64*c^3*(a + a*Sec[e + f*x]
)^2*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(231*f) - (8*c^2*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(3/2)*
Tan[e + f*x])/(33*f) - (2*c*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(11*f)

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Rubi [A]  time = 0.447941, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {3955, 3953} \[ -\frac{256 c^4 \tan (e+f x) (a \sec (e+f x)+a)^2}{1155 f \sqrt{c-c \sec (e+f x)}}-\frac{64 c^3 \tan (e+f x) (a \sec (e+f x)+a)^2 \sqrt{c-c \sec (e+f x)}}{231 f}-\frac{8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}}{33 f}-\frac{2 c \tan (e+f x) (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{5/2}}{11 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(7/2),x]

[Out]

(-256*c^4*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(1155*f*Sqrt[c - c*Sec[e + f*x]]) - (64*c^3*(a + a*Sec[e + f*x]
)^2*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(231*f) - (8*c^2*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(3/2)*
Tan[e + f*x])/(33*f) - (2*c*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(11*f)

Rule 3955

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

Rule 3953

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

Rubi steps

\begin{align*} \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{7/2} \, dx &=-\frac{2 c (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{11 f}+\frac{1}{11} (12 c) \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2} \, dx\\ &=-\frac{8 c^2 (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{33 f}-\frac{2 c (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{11 f}+\frac{1}{33} \left (32 c^2\right ) \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2} \, dx\\ &=-\frac{64 c^3 (a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{231 f}-\frac{8 c^2 (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{33 f}-\frac{2 c (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{11 f}+\frac{1}{231} \left (128 c^3\right ) \int \sec (e+f x) (a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)} \, dx\\ &=-\frac{256 c^4 (a+a \sec (e+f x))^2 \tan (e+f x)}{1155 f \sqrt{c-c \sec (e+f x)}}-\frac{64 c^3 (a+a \sec (e+f x))^2 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{231 f}-\frac{8 c^2 (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{33 f}-\frac{2 c (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{11 f}\\ \end{align*}

Mathematica [A]  time = 1.59485, size = 88, normalized size = 0.51 \[ \frac{2 a^2 c^3 \cos ^4\left (\frac{1}{2} (e+f x)\right ) (3419 \cos (e+f x)-1510 \cos (2 (e+f x))+533 \cos (3 (e+f x))-1930) \cot \left (\frac{1}{2} (e+f x)\right ) \sec ^5(e+f x) \sqrt{c-c \sec (e+f x)}}{1155 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(7/2),x]

[Out]

(2*a^2*c^3*Cos[(e + f*x)/2]^4*(-1930 + 3419*Cos[e + f*x] - 1510*Cos[2*(e + f*x)] + 533*Cos[3*(e + f*x)])*Cot[(
e + f*x)/2]*Sec[e + f*x]^5*Sqrt[c - c*Sec[e + f*x]])/(1155*f)

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Maple [A]  time = 0.2, size = 85, normalized size = 0.5 \begin{align*} -{\frac{2\,{a}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{5} \left ( 533\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}-755\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+455\,\cos \left ( fx+e \right ) -105 \right ) }{1155\,f \left ( -1+\cos \left ( fx+e \right ) \right ) ^{6} \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^(7/2),x)

[Out]

-2/1155*a^2/f*(c*(-1+cos(f*x+e))/cos(f*x+e))^(7/2)*sin(f*x+e)^5*(533*cos(f*x+e)^3-755*cos(f*x+e)^2+455*cos(f*x
+e)-105)/(-1+cos(f*x+e))^6/cos(f*x+e)^2

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Maxima [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))^2*(c-c*sec(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.489223, size = 359, normalized size = 2.1 \begin{align*} \frac{2 \,{\left (533 \, a^{2} c^{3} \cos \left (f x + e\right )^{6} + 844 \, a^{2} c^{3} \cos \left (f x + e\right )^{5} - 211 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} - 472 \, a^{2} c^{3} \cos \left (f x + e\right )^{3} + 295 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} + 140 \, a^{2} c^{3} \cos \left (f x + e\right ) - 105 \, a^{2} c^{3}\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{1155 \, f \cos \left (f x + e\right )^{5} \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))^2*(c-c*sec(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

2/1155*(533*a^2*c^3*cos(f*x + e)^6 + 844*a^2*c^3*cos(f*x + e)^5 - 211*a^2*c^3*cos(f*x + e)^4 - 472*a^2*c^3*cos
(f*x + e)^3 + 295*a^2*c^3*cos(f*x + e)^2 + 140*a^2*c^3*cos(f*x + e) - 105*a^2*c^3)*sqrt((c*cos(f*x + e) - c)/c
os(f*x + e))/(f*cos(f*x + e)^5*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))**2*(c-c*sec(f*x+e))**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 4.10614, size = 153, normalized size = 0.89 \begin{align*} -\frac{64 \, \sqrt{2}{\left (231 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{3} c^{4} + 495 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{2} c^{5} + 385 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} c^{6} + 105 \, c^{7}\right )} a^{2} c^{2}}{1155 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{11}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^(7/2),x, algorithm="giac")

[Out]

-64/1155*sqrt(2)*(231*(c*tan(1/2*f*x + 1/2*e)^2 - c)^3*c^4 + 495*(c*tan(1/2*f*x + 1/2*e)^2 - c)^2*c^5 + 385*(c
*tan(1/2*f*x + 1/2*e)^2 - c)*c^6 + 105*c^7)*a^2*c^2/((c*tan(1/2*f*x + 1/2*e)^2 - c)^(11/2)*f)