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

Optimal. Leaf size=72 \[ \frac {4 c^2 \tan (e+f x)}{a f \sqrt {c-c \sec (e+f x)}}+\frac {2 c \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{f (a \sec (e+f x)+a)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {3954, 3792} \[ \frac {4 c^2 \tan (e+f x)}{a f \sqrt {c-c \sec (e+f x)}}+\frac {2 c \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{f (a \sec (e+f x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*c^2*Tan[e + f*x])/(a*f*Sqrt[c - c*Sec[e + f*x]]) + (2*c*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(f*(a + a*Se
c[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 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))^{3/2}}{a+a \sec (e+f x)} \, dx &=\frac {2 c \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{f (a+a \sec (e+f x))}-\frac {(2 c) \int \sec (e+f x) \sqrt {c-c \sec (e+f x)} \, dx}{a}\\ &=\frac {4 c^2 \tan (e+f x)}{a f \sqrt {c-c \sec (e+f x)}}+\frac {2 c \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{f (a+a \sec (e+f x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.24, size = 54, normalized size = 0.75 \[ -\frac {2 c (3 \cos (e+f x)+1) \cot \left (\frac {1}{2} (e+f x)\right ) \sqrt {c-c \sec (e+f x)}}{a f (\cos (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 50, normalized size = 0.69 \[ -\frac {2 \, {\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{a f \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 2.73, size = 62, normalized size = 0.86 \[ -\frac {2 \, \sqrt {2} {\left (\frac {\sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c} c}{a} - \frac {c^{2}}{\sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c} a}\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e)),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 1.85, size = 63, normalized size = 0.88 \[ -\frac {2 \left (3 \cos \left (f x +e \right )+1\right ) \cos \left (f x +e \right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}{a f \sin \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e)),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.86, size = 110, normalized size = 1.53 \[ \frac {2 \, {\left (2 \, \sqrt {2} c^{\frac {3}{2}} - \frac {3 \, \sqrt {2} c^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {\sqrt {2} c^{\frac {3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{a f {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}^{\frac {3}{2}} {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 2.37, size = 77, normalized size = 1.07 \[ -\frac {c\,\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}\,\left (2\,\sin \left (e+f\,x\right )+6\,\sin \left (2\,e+2\,f\,x\right )+2\,\sin \left (3\,e+3\,f\,x\right )+3\,\sin \left (4\,e+4\,f\,x\right )\right )}{a\,f\,{\sin \left (2\,e+2\,f\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c/cos(e + f*x))^(3/2)/(cos(e + f*x)*(a + a/cos(e + f*x))),x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))**(3/2)/(a+a*sec(f*x+e)),x)

[Out]

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

________________________________________________________________________________________