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

Optimal. Leaf size=142 \[ \frac{128 c^4 \tan (e+f x)}{5 a f \sqrt{c-c \sec (e+f x)}}+\frac{32 c^3 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{5 a f}+\frac{12 c^2 \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{5 a f}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{f (a \sec (e+f x)+a)} \]

[Out]

(128*c^4*Tan[e + f*x])/(5*a*f*Sqrt[c - c*Sec[e + f*x]]) + (32*c^3*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(5*a*
f) + (12*c^2*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(5*a*f) + (2*c*(c - c*Sec[e + f*x])^(5/2)*Tan[e + f*x])/
(f*(a + a*Sec[e + f*x]))

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Rubi [A]  time = 0.233667, antiderivative size = 142, 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} \[ \frac{128 c^4 \tan (e+f x)}{5 a f \sqrt{c-c \sec (e+f x)}}+\frac{32 c^3 \tan (e+f x) \sqrt{c-c \sec (e+f x)}}{5 a f}+\frac{12 c^2 \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{5 a f}+\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{f (a \sec (e+f x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(128*c^4*Tan[e + f*x])/(5*a*f*Sqrt[c - c*Sec[e + f*x]]) + (32*c^3*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(5*a*
f) + (12*c^2*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(5*a*f) + (2*c*(c - c*Sec[e + f*x])^(5/2)*Tan[e + f*x])/
(f*(a + a*Sec[e + f*x]))

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)} \, dx &=\frac{2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{(6 c) \int \sec (e+f x) (c-c \sec (e+f x))^{5/2} \, dx}{a}\\ &=\frac{12 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{5 a f}+\frac{2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{\left (48 c^2\right ) \int \sec (e+f x) (c-c \sec (e+f x))^{3/2} \, dx}{5 a}\\ &=\frac{32 c^3 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{5 a f}+\frac{12 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{5 a f}+\frac{2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac{\left (64 c^3\right ) \int \sec (e+f x) \sqrt{c-c \sec (e+f x)} \, dx}{5 a}\\ &=\frac{128 c^4 \tan (e+f x)}{5 a f \sqrt{c-c \sec (e+f x)}}+\frac{32 c^3 \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{5 a f}+\frac{12 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{5 a f}+\frac{2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{f (a+a \sec (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.756324, size = 86, normalized size = 0.61 \[ -\frac{c^3 (245 \cos (e+f x)+86 \cos (2 (e+f x))+91 \cos (3 (e+f x))+90) \cot \left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x) \sqrt{c-c \sec (e+f x)}}{10 a f (\cos (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c^3*(90 + 245*Cos[e + f*x] + 86*Cos[2*(e + f*x)] + 91*Cos[3*(e + f*x)])*Cot[(e + f*x)/2]*Sec[e + f*x]^2*Sqrt
[c - c*Sec[e + f*x]])/(10*a*f*(1 + Cos[e + f*x]))

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Maple [A]  time = 0.187, size = 83, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 182\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}+86\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-14\,\cos \left ( fx+e \right ) +2 \right ) \cos \left ( fx+e \right ) }{5\,fa\sin \left ( fx+e \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) ^{3}} \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)),x)

[Out]

-2/5/a/f*(91*cos(f*x+e)^3+43*cos(f*x+e)^2-7*cos(f*x+e)+1)*cos(f*x+e)*(c*(-1+cos(f*x+e))/cos(f*x+e))^(7/2)/sin(
f*x+e)/(-1+cos(f*x+e))^3

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Maxima [A]  time = 1.56788, size = 220, normalized size = 1.55 \begin{align*} \frac{8 \,{\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}}\right )}}{5 \, a 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)),x, algorithm="maxima")

[Out]

8/5*(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)/(a*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 = 0.481039, size = 209, normalized size = 1.47 \begin{align*} -\frac{2 \,{\left (91 \, c^{3} \cos \left (f x + e\right )^{3} + 43 \, c^{3} \cos \left (f x + e\right )^{2} - 7 \, 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 )}}}{5 \, a f \cos \left (f x + e\right )^{2} \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)),x, algorithm="fricas")

[Out]

-2/5*(91*c^3*cos(f*x + e)^3 + 43*c^3*cos(f*x + e)^2 - 7*c^3*cos(f*x + e) + c^3)*sqrt((c*cos(f*x + e) - c)/cos(
f*x + e))/(a*f*cos(f*x + e)^2*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)),x)

[Out]

Timed out

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Giac [A]  time = 2.06639, size = 151, normalized size = 1.06 \begin{align*} -\frac{8 \, \sqrt{2}{\left (5 \, \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c} c - \frac{15 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{2} c^{2} + 5 \,{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )} c^{3} + c^{4}}{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c\right )}^{\frac{5}{2}}}\right )} c^{2}}{5 \, a 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)),x, algorithm="giac")

[Out]

-8/5*sqrt(2)*(5*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c - (15*(c*tan(1/2*f*x + 1/2*e)^2 - c)^2*c^2 + 5*(c*tan(1/2
*f*x + 1/2*e)^2 - c)*c^3 + c^4)/(c*tan(1/2*f*x + 1/2*e)^2 - c)^(5/2))*c^2/(a*f)