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

Optimal. Leaf size=163 \[ -\frac{256 c^4 \tan (e+f x) (a \sec (e+f x)+a)}{315 f \sqrt{c-c \sec (e+f x)}}-\frac{64 c^3 \tan (e+f x) (a \sec (e+f x)+a) \sqrt{c-c \sec (e+f x)}}{105 f}-\frac{8 c^2 \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{21 f}-\frac{2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{5/2}}{9 f} \]

[Out]

(-256*c^4*(a + a*Sec[e + f*x])*Tan[e + f*x])/(315*f*Sqrt[c - c*Sec[e + f*x]]) - (64*c^3*(a + a*Sec[e + f*x])*S
qrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(105*f) - (8*c^2*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2)*Tan[e +
f*x])/(21*f) - (2*c*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(9*f)

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Rubi [A]  time = 0.277657, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {3955, 3953} \[ -\frac{256 c^4 \tan (e+f x) (a \sec (e+f x)+a)}{315 f \sqrt{c-c \sec (e+f x)}}-\frac{64 c^3 \tan (e+f x) (a \sec (e+f x)+a) \sqrt{c-c \sec (e+f x)}}{105 f}-\frac{8 c^2 \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{3/2}}{21 f}-\frac{2 c \tan (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x))^{5/2}}{9 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-256*c^4*(a + a*Sec[e + f*x])*Tan[e + f*x])/(315*f*Sqrt[c - c*Sec[e + f*x]]) - (64*c^3*(a + a*Sec[e + f*x])*S
qrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(105*f) - (8*c^2*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(3/2)*Tan[e +
f*x])/(21*f) - (2*c*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(9*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)) (c-c \sec (e+f x))^{7/2} \, dx &=-\frac{2 c (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{9 f}+\frac{1}{3} (4 c) \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2} \, dx\\ &=-\frac{8 c^2 (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{21 f}-\frac{2 c (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{9 f}+\frac{1}{21} \left (32 c^2\right ) \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \, dx\\ &=-\frac{64 c^3 (a+a \sec (e+f x)) \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{105 f}-\frac{8 c^2 (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{21 f}-\frac{2 c (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{9 f}+\frac{1}{105} \left (128 c^3\right ) \int \sec (e+f x) (a+a \sec (e+f x)) \sqrt{c-c \sec (e+f x)} \, dx\\ &=-\frac{256 c^4 (a+a \sec (e+f x)) \tan (e+f x)}{315 f \sqrt{c-c \sec (e+f x)}}-\frac{64 c^3 (a+a \sec (e+f x)) \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{105 f}-\frac{8 c^2 (a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{21 f}-\frac{2 c (a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{9 f}\\ \end{align*}

Mathematica [A]  time = 0.890051, size = 86, normalized size = 0.53 \[ \frac{a c^3 \cos ^2\left (\frac{1}{2} (e+f x)\right ) (1617 \cos (e+f x)-642 \cos (2 (e+f x))+319 \cos (3 (e+f x))-782) \cot \left (\frac{1}{2} (e+f x)\right ) \sec ^4(e+f x) \sqrt{c-c \sec (e+f x)}}{315 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c^3*Cos[(e + f*x)/2]^2*(-782 + 1617*Cos[e + f*x] - 642*Cos[2*(e + f*x)] + 319*Cos[3*(e + f*x)])*Cot[(e + f*
x)/2]*Sec[e + f*x]^4*Sqrt[c - c*Sec[e + f*x]])/(315*f)

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Maple [A]  time = 0.186, size = 83, normalized size = 0.5 \begin{align*}{\frac{2\,a \left ( \sin \left ( fx+e \right ) \right ) ^{3} \left ( 319\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}-321\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+165\,\cos \left ( fx+e \right ) -35 \right ) }{315\,f \left ( -1+\cos \left ( fx+e \right ) \right ) ^{5}\cos \left ( fx+e \right ) } \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))*(c-c*sec(f*x+e))^(7/2),x)

[Out]

2/315*a/f*(c*(-1+cos(f*x+e))/cos(f*x+e))^(7/2)*sin(f*x+e)^3*(319*cos(f*x+e)^3-321*cos(f*x+e)^2+165*cos(f*x+e)-
35)/(-1+cos(f*x+e))^5/cos(f*x+e)

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

[Out]

Timed out

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Fricas [A]  time = 0.48252, size = 298, normalized size = 1.83 \begin{align*} \frac{2 \,{\left (319 \, a c^{3} \cos \left (f x + e\right )^{5} + 317 \, a c^{3} \cos \left (f x + e\right )^{4} - 158 \, a c^{3} \cos \left (f x + e\right )^{3} - 26 \, a c^{3} \cos \left (f x + e\right )^{2} + 95 \, a c^{3} \cos \left (f x + e\right ) - 35 \, a c^{3}\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{315 \, f \cos \left (f x + e\right )^{4} \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))*(c-c*sec(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

2/315*(319*a*c^3*cos(f*x + e)^5 + 317*a*c^3*cos(f*x + e)^4 - 158*a*c^3*cos(f*x + e)^3 - 26*a*c^3*cos(f*x + e)^
2 + 95*a*c^3*cos(f*x + e) - 35*a*c^3)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(f*cos(f*x + e)^4*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))*(c-c*sec(f*x+e))**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 2.53527, size = 150, normalized size = 0.92 \begin{align*} \frac{32 \, \sqrt{2}{\left (105 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{3} c^{3} + 189 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{2} c^{4} + 135 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} c^{5} + 35 \, c^{6}\right )} a c^{2}}{315 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{9}{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))*(c-c*sec(f*x+e))^(7/2),x, algorithm="giac")

[Out]

32/315*sqrt(2)*(105*(c*tan(1/2*f*x + 1/2*e)^2 - c)^3*c^3 + 189*(c*tan(1/2*f*x + 1/2*e)^2 - c)^2*c^4 + 135*(c*t
an(1/2*f*x + 1/2*e)^2 - c)*c^5 + 35*c^6)*a*c^2/((c*tan(1/2*f*x + 1/2*e)^2 - c)^(9/2)*f)