3.100 \(\int \frac{\sec (e+f x) (c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=169 \[ \frac{32 c^4 \tan (e+f x)}{5 a^3 f \sqrt{c-c \sec (e+f x)}}+\frac{16 c^3 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{5 f \left (a^3 \sec (e+f x)+a^3\right )}-\frac{4 c^2 \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{5 a f (a \sec (e+f x)+a)^2}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{5 f (a \sec (e+f x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.402054, antiderivative size = 169, 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 = {3954, 3792} \[ \frac{32 c^4 \tan (e+f x)}{5 a^3 f \sqrt{c-c \sec (e+f x)}}+\frac{16 c^3 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{5 f \left (a^3 \sec (e+f x)+a^3\right )}-\frac{4 c^2 \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{5 a f (a \sec (e+f x)+a)^2}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{5 f (a \sec (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(7/2))/(a + a*Sec[e + f*x])^3,x]

[Out]

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

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 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))^3} \, dx &=\frac{2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{(6 c) \int \frac{\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^2} \, dx}{5 a}\\ &=-\frac{4 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{5 a f (a+a \sec (e+f x))^2}+\frac{2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{\left (8 c^2\right ) \int \frac{\sec (e+f x) (c-c \sec (e+f x))^{3/2}}{a+a \sec (e+f x)} \, dx}{5 a^2}\\ &=\frac{16 c^3 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{5 f \left (a^3+a^3 \sec (e+f x)\right )}-\frac{4 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{5 a f (a+a \sec (e+f x))^2}+\frac{2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{\left (16 c^3\right ) \int \sec (e+f x) \sqrt{c-c \sec (e+f x)} \, dx}{5 a^3}\\ &=\frac{32 c^4 \tan (e+f x)}{5 a^3 f \sqrt{c-c \sec (e+f x)}}+\frac{16 c^3 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{5 f \left (a^3+a^3 \sec (e+f x)\right )}-\frac{4 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{5 a f (a+a \sec (e+f x))^2}+\frac{2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}\\ \end{align*}

Mathematica [A]  time = 0.626385, size = 78, normalized size = 0.46 \[ -\frac{c^3 (249 \cos (e+f x)+110 \cos (2 (e+f x))+23 \cos (3 (e+f x))+130) \cot \left (\frac{1}{2} (e+f x)\right ) \sqrt{c-c \sec (e+f x)}}{10 a^3 f (\cos (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(7/2))/(a + a*Sec[e + f*x])^3,x]

[Out]

-(c^3*(130 + 249*Cos[e + f*x] + 110*Cos[2*(e + f*x)] + 23*Cos[3*(e + f*x)])*Cot[(e + f*x)/2]*Sqrt[c - c*Sec[e
+ f*x]])/(10*a^3*f*(1 + Cos[e + f*x])^3)

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Maple [A]  time = 0.241, size = 85, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 46\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}+110\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+90\,\cos \left ( fx+e \right ) +10 \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{5\,f{a}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{5} \left ( -1+\cos \left ( fx+e \right ) \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)*(c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^3,x)

[Out]

-2/5/a^3/f*(23*cos(f*x+e)^3+55*cos(f*x+e)^2+45*cos(f*x+e)+5)*cos(f*x+e)^3*(c*(-1+cos(f*x+e))/cos(f*x+e))^(7/2)
/sin(f*x+e)^5/(-1+cos(f*x+e))

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Maxima [A]  time = 1.49678, size = 289, normalized size = 1.71 \begin{align*} \frac{2 \,{\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{5 \, \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}} + \frac{\sqrt{2} c^{\frac{7}{2}} \sin \left (f x + e\right )^{12}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{12}}\right )}}{5 \, a^{3} 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))^3,x, algorithm="maxima")

[Out]

2/5*(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 + 5*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 + sqrt(2)*c^(7/2)*sin(f
*x + e)^12/(cos(f*x + e) + 1)^12)/(a^3*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 = 0.484222, size = 261, normalized size = 1.54 \begin{align*} -\frac{2 \,{\left (23 \, c^{3} \cos \left (f x + e\right )^{3} + 55 \, c^{3} \cos \left (f x + e\right )^{2} + 45 \, c^{3} \cos \left (f x + e\right ) + 5 \, c^{3}\right )} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{5 \,{\left (a^{3} f \cos \left (f x + e\right )^{2} + 2 \, a^{3} f \cos \left (f x + e\right ) + a^{3} 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)*(c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

-2/5*(23*c^3*cos(f*x + e)^3 + 55*c^3*cos(f*x + e)^2 + 45*c^3*cos(f*x + e) + 5*c^3)*sqrt((c*cos(f*x + e) - c)/c
os(f*x + e))/((a^3*f*cos(f*x + e)^2 + 2*a^3*f*cos(f*x + e) + a^3*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)*(c-c*sec(f*x+e))**(7/2)/(a+a*sec(f*x+e))**3,x)

[Out]

Timed out

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Giac [A]  time = 6.36034, size = 143, normalized size = 0.85 \begin{align*} -\frac{2 \, \sqrt{2}{\left ({\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{5}{2}} + 5 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{3}{2}} c + 15 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c^{2} - \frac{5 \, c^{3}}{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c}}\right )} c}{5 \, a^{3} 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))^3,x, algorithm="giac")

[Out]

-2/5*sqrt(2)*((c*tan(1/2*f*x + 1/2*e)^2 - c)^(5/2) + 5*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*c + 15*sqrt(c*tan(
1/2*f*x + 1/2*e)^2 - c)*c^2 - 5*c^3/sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c))*c/(a^3*f)