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

Optimal. Leaf size=155 \[ -\frac{64 c^4 \tan (e+f x)}{3 a^2 f \sqrt{c-c \sec (e+f x)}}-\frac{16 c^3 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{3 a^2 f}-\frac{4 c^2 \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{f \left (a^2 \sec (e+f x)+a^2\right )}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

(-64*c^4*Tan[e + f*x])/(3*a^2*f*Sqrt[c - c*Sec[e + f*x]]) - (16*c^3*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(3*
a^2*f) - (4*c^2*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(f*(a^2 + a^2*Sec[e + f*x])) + (2*c*(c - c*Sec[e + f*
x])^(5/2)*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2)

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Rubi [A]  time = 0.317446, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {3954, 3793, 3792} \[ -\frac{64 c^4 \tan (e+f x)}{3 a^2 f \sqrt{c-c \sec (e+f x)}}-\frac{16 c^3 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{3 a^2 f}-\frac{4 c^2 \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{f \left (a^2 \sec (e+f x)+a^2\right )}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*c^4*Tan[e + f*x])/(3*a^2*f*Sqrt[c - c*Sec[e + f*x]]) - (16*c^3*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(3*
a^2*f) - (4*c^2*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(f*(a^2 + a^2*Sec[e + f*x])) + (2*c*(c - c*Sec[e + f*
x])^(5/2)*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2)

Rule 3954

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

Rule 3793

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

Rule 3792

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

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^2} \, dx &=\frac{2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac{(2 c) \int \frac{\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{a+a \sec (e+f x)} \, dx}{a}\\ &=-\frac{4 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f \left (a^2+a^2 \sec (e+f x)\right )}+\frac{2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac{\left (8 c^2\right ) \int \sec (e+f x) (c-c \sec (e+f x))^{3/2} \, dx}{a^2}\\ &=-\frac{16 c^3 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{3 a^2 f}-\frac{4 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f \left (a^2+a^2 \sec (e+f x)\right )}+\frac{2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac{\left (32 c^3\right ) \int \sec (e+f x) \sqrt{c-c \sec (e+f x)} \, dx}{3 a^2}\\ &=-\frac{64 c^4 \tan (e+f x)}{3 a^2 f \sqrt{c-c \sec (e+f x)}}-\frac{16 c^3 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{3 a^2 f}-\frac{4 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f \left (a^2+a^2 \sec (e+f x)\right )}+\frac{2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 0.94085, size = 84, normalized size = 0.54 \[ \frac{c^3 (195 \cos (e+f x)+138 \cos (2 (e+f x))+45 \cos (3 (e+f x))+134) \cot \left (\frac{1}{2} (e+f x)\right ) \sec (e+f x) \sqrt{c-c \sec (e+f x)}}{6 a^2 f (\cos (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*(134 + 195*Cos[e + f*x] + 138*Cos[2*(e + f*x)] + 45*Cos[3*(e + f*x)])*Cot[(e + f*x)/2]*Sec[e + f*x]*Sqrt[
c - c*Sec[e + f*x]])/(6*a^2*f*(1 + Cos[e + f*x])^2)

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Maple [A]  time = 0.193, size = 85, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 6\,\cos \left ( fx+e \right ) +2 \right ) \left ( 15\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+18\,\cos \left ( fx+e \right ) -1 \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{3\,f{a}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{3} \left ( -1+\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)*(c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^2,x)

[Out]

-2/3/a^2/f*(3*cos(f*x+e)+1)*(15*cos(f*x+e)^2+18*cos(f*x+e)-1)*cos(f*x+e)^2*(c*(-1+cos(f*x+e))/cos(f*x+e))^(7/2
)/sin(f*x+e)^3/(-1+cos(f*x+e))^2

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Maxima [A]  time = 1.72648, size = 254, normalized size = 1.64 \begin{align*} -\frac{4 \,{\left (16 \, \sqrt{2} c^{\frac{7}{2}} - \frac{56 \, \sqrt{2} c^{\frac{7}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{70 \, \sqrt{2} c^{\frac{7}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{35 \, \sqrt{2} c^{\frac{7}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac{4 \, \sqrt{2} c^{\frac{7}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + \frac{\sqrt{2} c^{\frac{7}{2}} \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}}\right )}}{3 \, a^{2} f{\left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}^{\frac{7}{2}}{\left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*(16*sqrt(2)*c^(7/2) - 56*sqrt(2)*c^(7/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 70*sqrt(2)*c^(7/2)*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 - 35*sqrt(2)*c^(7/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 4*sqrt(2)*c^(7/2)*sin(
f*x + e)^8/(cos(f*x + e) + 1)^8 + sqrt(2)*c^(7/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)/(a^2*f*(sin(f*x + e)/
(cos(f*x + e) + 1) + 1)^(7/2)*(sin(f*x + e)/(cos(f*x + e) + 1) - 1)^(7/2))

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Fricas [A]  time = 0.487042, size = 243, normalized size = 1.57 \begin{align*} \frac{2 \,{\left (45 \, c^{3} \cos \left (f x + e\right )^{3} + 69 \, c^{3} \cos \left (f x + e\right )^{2} + 15 \, c^{3} \cos \left (f x + e\right ) - c^{3}\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{3 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f \cos \left (f x + e\right )\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)*(c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

2/3*(45*c^3*cos(f*x + e)^3 + 69*c^3*cos(f*x + e)^2 + 15*c^3*cos(f*x + e) - c^3)*sqrt((c*cos(f*x + e) - c)/cos(
f*x + e))/((a^2*f*cos(f*x + e)^2 + a^2*f*cos(f*x + e))*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)*(c-c*sec(f*x+e))**(7/2)/(a+a*sec(f*x+e))**2,x)

[Out]

Timed out

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Giac [A]  time = 4.21786, size = 144, normalized size = 0.93 \begin{align*} \frac{4 \, \sqrt{2}{\left ({\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} + 9 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c - \frac{9 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} c^{2} + c^{3}}{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}}}\right )} c^{2}}{3 \, a^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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