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

Optimal. Leaf size=42 \[ -\frac {\tan (e+f x) \sqrt {a \sec (e+f x)+a}}{2 f (c-c \sec (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {3950} \[ -\frac {\tan (e+f x) \sqrt {a \sec (e+f x)+a}}{2 f (c-c \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3950

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

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 62, normalized size = 1.48 \[ \frac {\tan \left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sqrt {a (\sec (e+f x)+1)}}{c f (\sec (e+f x)-1) \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 79, normalized size = 1.88 \[ \frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{{\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)a^2*(1/2-1/2/tan(1
/2*(f*x+exp(1)))^2)*sign(cos(f*x+exp(1)))/sqrt(-a*c)/c/f/abs(a)/sign(tan(1/2*(f*x+exp(1)))^2-1)/sign(tan(1/2*(
f*x+exp(1))))

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maple [A]  time = 2.05, size = 60, normalized size = 1.43 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )}{2 f \cos \left (f x +e \right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.33, size = 514, normalized size = 12.24 \[ -\frac {2 \, {\left ({\left (\sin \left (3 \, f x + 3 \, e\right ) + \sin \left (f x + e\right )\right )} \cos \left (4 \, f x + 4 \, e\right ) - {\left (\cos \left (3 \, f x + 3 \, e\right ) + \cos \left (f x + e\right )\right )} \sin \left (4 \, f x + 4 \, e\right ) + {\left (2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )} \sin \left (3 \, f x + 3 \, e\right ) - 2 \, \cos \left (3 \, f x + 3 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) - 2 \, \cos \left (f x + e\right ) \sin \left (2 \, f x + 2 \, e\right ) + 2 \, \cos \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + \sin \left (f x + e\right )\right )} \sqrt {a} \sqrt {c}}{{\left (c^{2} \cos \left (4 \, f x + 4 \, e\right )^{2} + 4 \, c^{2} \cos \left (3 \, f x + 3 \, e\right )^{2} + 4 \, c^{2} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, c^{2} \cos \left (f x + e\right )^{2} + c^{2} \sin \left (4 \, f x + 4 \, e\right )^{2} + 4 \, c^{2} \sin \left (3 \, f x + 3 \, e\right )^{2} + 4 \, c^{2} \sin \left (2 \, f x + 2 \, e\right )^{2} - 8 \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 4 \, c^{2} \sin \left (f x + e\right )^{2} - 4 \, c^{2} \cos \left (f x + e\right ) + c^{2} - 2 \, {\left (2 \, c^{2} \cos \left (3 \, f x + 3 \, e\right ) - 2 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + 2 \, c^{2} \cos \left (f x + e\right ) - c^{2}\right )} \cos \left (4 \, f x + 4 \, e\right ) - 4 \, {\left (2 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) - 2 \, c^{2} \cos \left (f x + e\right ) + c^{2}\right )} \cos \left (3 \, f x + 3 \, e\right ) - 4 \, {\left (2 \, c^{2} \cos \left (f x + e\right ) - c^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) - 4 \, {\left (c^{2} \sin \left (3 \, f x + 3 \, e\right ) - c^{2} \sin \left (2 \, f x + 2 \, e\right ) + c^{2} \sin \left (f x + e\right )\right )} \sin \left (4 \, f x + 4 \, e\right ) - 8 \, {\left (c^{2} \sin \left (2 \, f x + 2 \, e\right ) - c^{2} \sin \left (f x + e\right )\right )} \sin \left (3 \, f x + 3 \, e\right )\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.00, size = 118, normalized size = 2.81 \[ -\frac {2\,\sqrt {\frac {a\,\left (\cos \left (e+f\,x\right )+1\right )}{\cos \left (e+f\,x\right )}}\,\sqrt {\frac {c\,\left (\cos \left (e+f\,x\right )-1\right )}{\cos \left (e+f\,x\right )}}\,\left (\sin \left (e+f\,x\right )-2\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (3\,e+3\,f\,x\right )\right )}{c^2\,f\,\left (4\,\cos \left (e+f\,x\right )+4\,\cos \left (2\,e+2\,f\,x\right )-4\,\cos \left (3\,e+3\,f\,x\right )+\cos \left (4\,e+4\,f\,x\right )-5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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