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

Optimal. Leaf size=43 \[ \frac {c \tan (e+f x)}{2 f (a \sec (e+f x)+a)^{5/2} \sqrt {c-c \sec (e+f x)}} \]

[Out]

1/2*c*tan(f*x+e)/f/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 43, 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 = {3953} \[ \frac {c \tan (e+f x)}{2 f (a \sec (e+f x)+a)^{5/2} \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3953

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

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 71, normalized size = 1.65 \[ \frac {(2 \cos (e+f x)+1) \csc \left (\frac {1}{2} (e+f x)\right ) \sec ^3\left (\frac {1}{2} (e+f x)\right ) \sqrt {c-c \sec (e+f x)}}{8 a^2 f \sqrt {a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + 2*Cos[e + f*x])*Csc[(e + f*x)/2]*Sec[(e + f*x)/2]^3*Sqrt[c - c*Sec[e + f*x]])/(8*a^2*f*Sqrt[a*(1 + Sec[e
+ f*x])])

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fricas [B]  time = 0.45, size = 104, normalized size = 2.42 \[ \frac {{\left (2 \, \cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{2 \, {\left (a^{3} f \cos \left (f x + e\right )^{2} + 2 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )} \sin \left (f x + e\right )} \]

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))^(5/2),x, algorithm="fricas")

[Out]

1/2*(2*cos(f*x + e)^2 + cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*
x + e))/((a^3*f*cos(f*x + e)^2 + 2*a^3*f*cos(f*x + e) + a^3*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))^(5/2),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)1/8*(c*tan(1/2*(f*
x+exp(1)))^2-c)^2*sign(tan(1/2*(f*x+exp(1)))^3+tan(1/2*(f*x+exp(1))))*sign(cos(f*x+exp(1)))/a^2/sqrt(-a*c)/f/a
bs(c)/sign(tan(1/2*(f*x+exp(1)))^2-1)

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maple [B]  time = 2.17, size = 83, normalized size = 1.93 \[ -\frac {\sqrt {\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \left (3 \cos \left (f x +e \right )+1\right ) \left (-1+\cos \left (f x +e \right )\right )^{3}}{8 f \sin \left (f x +e \right )^{5} a^{3}} \]

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))^(5/2),x)

[Out]

-1/8/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)*(3*cos(f*x+e)+1)*(-
1+cos(f*x+e))^3/sin(f*x+e)^5/a^3

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maxima [A]  time = 0.46, size = 58, normalized size = 1.35 \[ -\frac {\sqrt {c} {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}^{2} {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}^{2}}{8 \, \sqrt {-a} a^{2} f} \]

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))^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.25, size = 120, normalized size = 2.79 \[ -\frac {2\,\left (3\,\sin \left (e+f\,x\right )+3\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (3\,e+3\,f\,x\right )\right )\,\sqrt {\frac {c\,\left (\cos \left (e+f\,x\right )-1\right )}{\cos \left (e+f\,x\right )}}}{a^2\,f\,\sqrt {\frac {a\,\left (\cos \left (e+f\,x\right )+1\right )}{\cos \left (e+f\,x\right )}}\,\left (4\,\cos \left (2\,e+2\,f\,x\right )-4\,\cos \left (e+f\,x\right )+4\,\cos \left (3\,e+3\,f\,x\right )+\cos \left (4\,e+4\,f\,x\right )-5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c/cos(e + f*x))^(1/2)/(cos(e + f*x)*(a + a/cos(e + f*x))^(5/2)),x)

[Out]

-(2*(3*sin(e + f*x) + 3*sin(2*e + 2*f*x) + sin(3*e + 3*f*x))*((c*(cos(e + f*x) - 1))/cos(e + f*x))^(1/2))/(a^2
*f*((a*(cos(e + f*x) + 1))/cos(e + f*x))^(1/2)*(4*cos(2*e + 2*f*x) - 4*cos(e + 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 \[ \int \frac {\sqrt {- c \left (\sec {\left (e + f x \right )} - 1\right )} \sec {\left (e + f x \right )}}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

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))**(5/2),x)

[Out]

Integral(sqrt(-c*(sec(e + f*x) - 1))*sec(e + f*x)/(a*(sec(e + f*x) + 1))**(5/2), x)

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