3.87 \(\int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{a+a \sec (e+f x)} \, dx\)

Optimal. Leaf size=108 \[ \frac {32 c^3 \tan (e+f x)}{3 a f \sqrt {c-c \sec (e+f x)}}+\frac {8 c^2 \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{3 a f}+\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{f (a \sec (e+f x)+a)} \]

[Out]

2*c*(c-c*sec(f*x+e))^(3/2)*tan(f*x+e)/f/(a+a*sec(f*x+e))+32/3*c^3*tan(f*x+e)/a/f/(c-c*sec(f*x+e))^(1/2)+8/3*c^
2*(c-c*sec(f*x+e))^(1/2)*tan(f*x+e)/a/f

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Rubi [A]  time = 0.19, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {3954, 3793, 3792} \[ \frac {32 c^3 \tan (e+f x)}{3 a f \sqrt {c-c \sec (e+f x)}}+\frac {8 c^2 \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{3 a f}+\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{f (a \sec (e+f x)+a)} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(5/2))/(a + a*Sec[e + f*x]),x]

[Out]

(32*c^3*Tan[e + f*x])/(3*a*f*Sqrt[c - c*Sec[e + f*x]]) + (8*c^2*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(3*a*f)
+ (2*c*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(f*(a + a*Sec[e + f*x]))

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]

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 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)]

Rubi steps

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

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Mathematica [A]  time = 0.44, size = 74, normalized size = 0.69 \[ -\frac {c^2 (20 \cos (e+f x)+23 \cos (2 (e+f x))+21) \cot \left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sqrt {c-c \sec (e+f x)}}{3 a f (\cos (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(5/2))/(a + a*Sec[e + f*x]),x]

[Out]

-1/3*(c^2*(21 + 20*Cos[e + f*x] + 23*Cos[2*(e + f*x)])*Cot[(e + f*x)/2]*Sec[e + f*x]*Sqrt[c - c*Sec[e + f*x]])
/(a*f*(1 + Cos[e + f*x]))

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fricas [A]  time = 0.43, size = 77, normalized size = 0.71 \[ -\frac {2 \, {\left (23 \, c^{2} \cos \left (f x + e\right )^{2} + 10 \, c^{2} \cos \left (f x + e\right ) - c^{2}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{3 \, a f \cos \left (f x + e\right ) \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e)),x, algorithm="fricas")

[Out]

-2/3*(23*c^2*cos(f*x + e)^2 + 10*c^2*cos(f*x + e) - c^2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(a*f*cos(f*x
+ e)*sin(f*x + e))

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giac [A]  time = 2.51, size = 87, normalized size = 0.81 \[ -\frac {4 \, \sqrt {2} c^{2} {\left (\frac {3 \, \sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}}{a} - \frac {6 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c + c^{2}}{{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {3}{2}} a}\right )}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e)),x, algorithm="giac")

[Out]

-4/3*sqrt(2)*c^2*(3*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/a - (6*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c + c^2)/((c*tan(
1/2*f*x + 1/2*e)^2 - c)^(3/2)*a))/f

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maple [A]  time = 1.89, size = 73, normalized size = 0.68 \[ -\frac {2 \left (23 \left (\cos ^{2}\left (f x +e \right )\right )+10 \cos \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}{3 a f \sin \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e)),x)

[Out]

-2/3/a/f*(23*cos(f*x+e)^2+10*cos(f*x+e)-1)*cos(f*x+e)*(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)/sin(f*x+e)/(-1+cos(
f*x+e))^2

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maxima [A]  time = 0.64, size = 137, normalized size = 1.27 \[ -\frac {4 \, {\left (8 \, \sqrt {2} c^{\frac {5}{2}} - \frac {20 \, \sqrt {2} c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {15 \, \sqrt {2} c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {3 \, \sqrt {2} c^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )}}{3 \, a f {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e)),x, algorithm="maxima")

[Out]

-4/3*(8*sqrt(2)*c^(5/2) - 20*sqrt(2)*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 15*sqrt(2)*c^(5/2)*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 - 3*sqrt(2)*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)/(a*f*(sin(f*x + e)/(cos(f
*x + e) + 1) + 1)^(5/2)*(sin(f*x + e)/(cos(f*x + e) + 1) - 1)^(5/2))

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mupad [B]  time = 4.23, size = 125, normalized size = 1.16 \[ \frac {2\,c^2\,\sqrt {\frac {c\,\left (\cos \left (e+f\,x\right )-1\right )}{\cos \left (e+f\,x\right )}}\,\left (2\,\sin \left (e+f\,x\right )-44\,\sin \left (2\,e+2\,f\,x\right )+25\,\sin \left (3\,e+3\,f\,x\right )-26\,\sin \left (4\,e+4\,f\,x\right )+23\,\sin \left (5\,e+5\,f\,x\right )\right )}{3\,a\,f\,\left (\cos \left (3\,e+3\,f\,x\right )-2\,\cos \left (e+f\,x\right )-2\,\cos \left (4\,e+4\,f\,x\right )+\cos \left (5\,e+5\,f\,x\right )+2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c/cos(e + f*x))^(5/2)/(cos(e + f*x)*(a + a/cos(e + f*x))),x)

[Out]

(2*c^2*((c*(cos(e + f*x) - 1))/cos(e + f*x))^(1/2)*(2*sin(e + f*x) - 44*sin(2*e + 2*f*x) + 25*sin(3*e + 3*f*x)
- 26*sin(4*e + 4*f*x) + 23*sin(5*e + 5*f*x)))/(3*a*f*(cos(3*e + 3*f*x) - 2*cos(e + f*x) - 2*cos(4*e + 4*f*x)
+ cos(5*e + 5*f*x) + 2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )}}{\sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {2 c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{3}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))**(5/2)/(a+a*sec(f*x+e)),x)

[Out]

(Integral(c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)/(sec(e + f*x) + 1), x) + Integral(-2*c**2*sqrt(-c*sec(e
+ f*x) + c)*sec(e + f*x)**2/(sec(e + f*x) + 1), x) + Integral(c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)**3/(
sec(e + f*x) + 1), x))/a

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