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

Optimal. Leaf size=94 \[ \frac {c \tan (e+f x) \log (\cos (e+f x)+1)}{a f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {c \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3907, 3911, 31} \[ \frac {c \tan (e+f x) \log (\cos (e+f x)+1)}{a f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {c \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3907

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[
(-2*a*Cot[e + f*x]*(c + d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[1/c, Int[Sqrt[a +
b*Csc[e + f*x]]*(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] && LtQ[n, -2^(-1)]

Rule 3911

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

Rubi steps

\begin {align*} \int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{3/2}} \, dx &=-\frac {c \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {\int \frac {\sqrt {c-c \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}} \, dx}{a}\\ &=-\frac {c \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {(c \tan (e+f x)) \operatorname {Subst}\left (\int \frac {1}{a+a x} \, dx,x,\cos (e+f x)\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {c \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {c \log (1+\cos (e+f x)) \tan (e+f x)}{a f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 0.63, size = 106, normalized size = 1.13 \[ \frac {i \cot \left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sqrt {c-c \sec (e+f x)} \left (2 i \log \left (1+e^{i (e+f x)}\right )+\left (f x+2 i \log \left (1+e^{i (e+f x)}\right )\right ) \cos (e+f x)+f x+i\right )}{f (a (\sec (e+f x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \sec \left (f x + e\right ) + a} \sqrt {-c \sec \left (f x + e\right ) + c}}{a^{2} \sec \left (f x + e\right )^{2} + 2 \, a^{2} \sec \left (f x + e\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(a^2*sec(f*x + e)^2 + 2*a^2*sec(f*x + e) + a^2), x
)

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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((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: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)2*sqrt(2)*(-1/4*sq
rt(-a*c)*(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))))/sqrt(2)/a^2/abs(c)+
1/2*c*sqrt(-a*c)*sign(tan(1/2*(f*x+exp(1)))^3+tan(1/2*(f*x+exp(1))))*ln(abs(2*(c*tan(1/2*(f*x+exp(1)))^2-c)+4*
c))/sqrt(2)/a^2/abs(c))*sign(cos(f*x+exp(1)))/f

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maple [A]  time = 2.30, size = 119, normalized size = 1.27 \[ -\frac {\left (2 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+\cos ^{2}\left (f x +e \right )-2 \cos \left (f x +e \right )-2 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+1\right ) \cos \left (f x +e \right ) \sqrt {\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}}{2 f \sin \left (f x +e \right )^{3} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.96, size = 395, normalized size = 4.20 \[ -\frac {{\left ({\left (f x + e\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, {\left (f x + e\right )} \cos \left (f x + e\right )^{2} + {\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, {\left (f x + e\right )} \sin \left (f x + e\right )^{2} + f x - 2 \, {\left (2 \, {\left (2 \, \cos \left (f x + e\right ) + 1\right )} \cos \left (2 \, f x + 2 \, e\right ) + \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, \cos \left (f x + e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 4 \, \sin \left (f x + e\right )^{2} + 4 \, \cos \left (f x + e\right ) + 1\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) + 1\right ) + 2 \, {\left (f x + 2 \, {\left (f x + e\right )} \cos \left (f x + e\right ) + e - \sin \left (f x + e\right )\right )} \cos \left (2 \, f x + 2 \, e\right ) + 4 \, {\left (f x + e\right )} \cos \left (f x + e\right ) + 2 \, {\left (2 \, {\left (f x + e\right )} \sin \left (f x + e\right ) + \cos \left (f x + e\right )\right )} \sin \left (2 \, f x + 2 \, e\right ) + e - 2 \, \sin \left (f x + e\right )\right )} \sqrt {a} \sqrt {c}}{{\left (a^{2} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2} \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a^{2} \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 4 \, a^{2} \sin \left (f x + e\right )^{2} + 4 \, a^{2} \cos \left (f x + e\right ) + a^{2} + 2 \, {\left (2 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((f*x + e)*cos(2*f*x + 2*e)^2 + 4*(f*x + e)*cos(f*x + e)^2 + (f*x + e)*sin(2*f*x + 2*e)^2 + 4*(f*x + e)*sin(f
*x + e)^2 + f*x - 2*(2*(2*cos(f*x + e) + 1)*cos(2*f*x + 2*e) + cos(2*f*x + 2*e)^2 + 4*cos(f*x + e)^2 + sin(2*f
*x + 2*e)^2 + 4*sin(2*f*x + 2*e)*sin(f*x + e) + 4*sin(f*x + e)^2 + 4*cos(f*x + e) + 1)*arctan2(sin(f*x + e), c
os(f*x + e) + 1) + 2*(f*x + 2*(f*x + e)*cos(f*x + e) + e - sin(f*x + e))*cos(2*f*x + 2*e) + 4*(f*x + e)*cos(f*
x + e) + 2*(2*(f*x + e)*sin(f*x + e) + cos(f*x + e))*sin(2*f*x + 2*e) + e - 2*sin(f*x + e))*sqrt(a)*sqrt(c)/((
a^2*cos(2*f*x + 2*e)^2 + 4*a^2*cos(f*x + e)^2 + a^2*sin(2*f*x + 2*e)^2 + 4*a^2*sin(2*f*x + 2*e)*sin(f*x + e) +
4*a^2*sin(f*x + e)^2 + 4*a^2*cos(f*x + e) + a^2 + 2*(2*a^2*cos(f*x + e) + a^2)*cos(2*f*x + 2*e))*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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