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

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

[Out]

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

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Rubi [A]  time = 0.13, 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} \[ \frac {\tan (e+f x) \sqrt {c-c \sec (e+f x)}}{2 f (a \sec (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.19, size = 42, normalized size = 1.00 \[ \frac {\csc (e+f x) \sqrt {c-c \sec (e+f x)}}{a f \sqrt {a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.45, size = 78, normalized size = 1.86 \[ \frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{{\left (a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)-1/2*c^2*(c*tan(1/
2*(f*x+exp(1)))^2-c)*sign(tan(1/2*(f*x+exp(1)))^3+tan(1/2*(f*x+exp(1))))*sign(cos(f*x+exp(1)))/sqrt(-a*c)/a/c/
f/abs(c)/sign(tan(1/2*(f*x+exp(1)))^2-1)

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maple [A]  time = 2.07, size = 73, normalized size = 1.74 \[ \frac {\sqrt {\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right )^{2}}{2 f \sin \left (f x +e \right )^{3} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.47, size = 54, normalized size = 1.29 \[ \frac {\sqrt {c} {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )} {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}}{2 \, \sqrt {-a} a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.55, size = 50, normalized size = 1.19 \[ \frac {\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}}{a\,f\,\sin \left (e+f\,x\right )\,\sqrt {\frac {a\,\left (\cos \left (e+f\,x\right )+1\right )}{\cos \left (e+f\,x\right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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