3.1.90 \(\int \frac {\sqrt {a+a \sec (e+f x)}}{\sqrt {c-c \sec (e+f x)}} \, dx\) [90]

Optimal. Leaf size=51 \[ \frac {a \log (1-\cos (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

a*ln(1-cos(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 = 0.06, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3996, 31} \begin {gather*} \frac {a \tan (e+f x) \log (1-\cos (e+f x))}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Log[1 - Cos[e + f*x]]*Tan[e + f*x])/(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 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Dist
[(-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 {a+a \sec (e+f x)}}{\sqrt {c-c \sec (e+f x)}} \, dx &=\frac {(a c \tan (e+f x)) \text {Subst}\left (\int \frac {1}{-c+c x} \, dx,x,\cos (e+f x)\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=\frac {a \log (1-\cos (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] Result contains complex when optimal does not.
time = 0.93, size = 86, normalized size = 1.69 \begin {gather*} -\frac {\left (-1+e^{i (e+f x)}\right ) \left (f x+2 i \log \left (1-e^{i (e+f x)}\right )\right ) \sqrt {a (1+\sec (e+f x))}}{\left (1+e^{i (e+f x)}\right ) f \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(99\) vs. \(2(47)=94\).
time = 0.27, size = 100, normalized size = 1.96

method	result	size
default	\(-\frac {\sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (2 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )\right ) \sqrt {\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right )}{f \sin \left (f x +e \right ) c}\)	\(100\)
risch	\(\frac {\sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) x}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}}-\frac {2 \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \left (f x +e \right )}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}-\frac {2 i \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(284\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 69, normalized size = 1.35 \begin {gather*} \frac {\frac {2 \, \sqrt {-a} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{\sqrt {c}} - \frac {\sqrt {-a} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt {c}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.45, size = 72, normalized size = 1.41 \begin {gather*} -\frac {\frac {\sqrt {-a c} a \log \left ({\left | a \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}{c {\left | a \right |}} - \frac {\sqrt {-a c} a \log \left ({\left | -a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a \right |}\right )}{c {\left | a \right |}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sqrt(-a*c)*a*log(abs(a)*tan(1/2*f*x + 1/2*e)^2)/(c*abs(a)) - sqrt(-a*c)*a*log(abs(-a*tan(1/2*f*x + 1/2*e)^2
- a))/(c*abs(a)))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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