3.135 \(\int \frac {\sec (e+f x) \sqrt {c-c \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}} \, dx\)

Optimal. Leaf size=50 \[ -\frac {c \tan (e+f x) \log (\sec (e+f x)+1)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 50, 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 = {3952} \[ -\frac {c \tan (e+f x) \log (\sec (e+f x)+1)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3952

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

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) \sqrt {c-c \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}} \, dx &=-\frac {c \log (1+\sec (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 0.43, size = 140, normalized size = 2.80 \[ \frac {i \left (1+e^{i (e+f x)}\right ) \sqrt {\frac {c \left (-1+e^{i (e+f x)}\right )^2}{1+e^{2 i (e+f x)}}} \left (2 \log \left (1+e^{i (e+f x)}\right )-\log \left (1+e^{2 i (e+f x)}\right )\right )}{f \left (-1+e^{i (e+f x)}\right ) \sqrt {\frac {a \left (1+e^{i (e+f x)}\right )^2}{1+e^{2 i (e+f x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c \sec \left (f x + e\right ) + c} \sec \left (f x + e\right )}{\sqrt {a \sec \left (f x + e\right ) + a}}, x\right ) \]

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))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(-c*sec(f*x + e) + c)*sec(f*x + e)/sqrt(a*sec(f*x + e) + a), 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(sec(f*x+e)*(c-c*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2),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)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/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)))*ln(abs(c*tan(1/2*(f*x+exp(1)))^2-c))/sqrt(-a*c)/f/abs(c)/sign(tan(1/2*(f
*x+exp(1)))^2-1)

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maple [B]  time = 1.98, size = 116, normalized size = 2.32 \[ -\frac {\left (\ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+\ln \left (-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )\right ) \cos \left (f x +e \right ) \sqrt {\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}}{f \sin \left (f x +e \right ) a} \]

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))^(1/2),x)

[Out]

-1/f*(ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+ln(-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e)))*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)/a

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maxima [A]  time = 0.45, size = 64, normalized size = 1.28 \[ -\frac {\frac {\sqrt {c} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{\sqrt {-a}} + \frac {\sqrt {c} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{\sqrt {-a}}}{f} \]

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))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}}{\cos \left (e+f\,x\right )\,\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}} \,d x \]

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))^(1/2)),x)

[Out]

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

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

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))**(1/2),x)

[Out]

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

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