3.2.42 \(\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}} \, dx\) [142]

Optimal. Leaf size=95 \[ \frac {\tan (e+f x)}{2 f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}-\frac {\tanh ^{-1}(\cos (e+f x)) \tan (e+f x)}{2 a f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

1/2*tan(f*x+e)/f/(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(1/2)-1/2*arctanh(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 = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {4045, 4044, 3855} \begin {gather*} \frac {\tan (e+f x)}{2 f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{2 a f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4044

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

Rule 4045

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] +
Dist[(m + n + 1)/(a*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n, x], x] /
; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ((ILtQ[m, 0] && ILtQ[n - 1/2, 0
]) || (ILtQ[m - 1/2, 0] && ILtQ[n - 1/2, 0] && LtQ[m, n]))

Rubi steps

\begin {align*} \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}} \, dx &=\frac {\tan (e+f x)}{2 f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {\int \frac {\sec (e+f x)}{\sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \, dx}{2 a}\\ &=\frac {\tan (e+f x)}{2 f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x) \int \csc (e+f x) \, dx}{2 a \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=\frac {\tan (e+f x)}{2 f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}-\frac {\tanh ^{-1}(\cos (e+f x)) \tan (e+f x)}{2 a 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 = 1.39, size = 157, normalized size = 1.65 \begin {gather*} -\frac {\left (1+2 \tanh ^{-1}\left (e^{i (e+f x)}\right ) (1+\cos (e+f x))\right ) \sec ^{\frac {3}{2}}(e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+i \sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin \left (\frac {1}{2} (e+f x)\right )}{\sqrt {2} a \left (1+e^{i (e+f x)}\right ) \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 2.82, size = 123, normalized size = 1.29

method	result	size
default	\(\frac {\sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (-1+\cos \left (f x +e \right )\right )^{2} \left (2 \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+2 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+\cos \left (f x +e \right )-1\right )}{4 f \sqrt {\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )^{3} a^{2}}\)	\(123\)
risch	\(\frac {i \left ({\mathrm e}^{2 i \left (f x +e \right )}-{\mathrm e}^{i \left (f x +e \right )}\right )}{a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) f}-\frac {i \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{2 a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) f}+\frac {i \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{2 a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) f}\)	\(350\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(-1+cos(f*x+e))^2*(2*cos(f*x+e)*ln(-(-1+cos(f*x+e))/sin(f*x+e))+2*ln
(-(-1+cos(f*x+e))/sin(f*x+e))+cos(f*x+e)-1)/(c*(-1+cos(f*x+e))/cos(f*x+e))^(1/2)/sin(f*x+e)^3/a^2

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 433 vs. \(2 (90) = 180\).
time = 0.57, size = 433, normalized size = 4.56 \begin {gather*} -\frac {{\left ({\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 ) - {\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 \, \cos \left (f x + e\right ) \sin \left (2 \, f x + 2 \, e\right ) + 2 \, \cos \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 2 \, \sin \left (f x + e\right )\right )} \sqrt {a} \sqrt {c}}{2 \, {\left (a^{2} c \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a^{2} c \cos \left (f x + e\right )^{2} + a^{2} c \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a^{2} c \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 4 \, a^{2} c \sin \left (f x + e\right )^{2} + 4 \, a^{2} c \cos \left (f x + e\right ) + a^{2} c + 2 \, {\left (2 \, a^{2} c \cos \left (f x + e\right ) + a^{2} c\right )} \cos \left (2 \, f x + 2 \, e\right )\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (90) = 180\).
time = 3.19, size = 412, normalized size = 4.34 \begin {gather*} \left [-\frac {\sqrt {-a c} {\left (\cos \left (f x + e\right ) + 1\right )} \log \left (-\frac {4 \, {\left (2 \, \sqrt {-a c} \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 )^{2} + {\left (a c \cos \left (f x + e\right )^{2} + a c\right )} \sin \left (f x + e\right )\right )}}{{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, \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 )}{4 \, {\left (a^{2} c f \cos \left (f x + e\right ) + a^{2} c f\right )} \sin \left (f x + e\right )}, \frac {\sqrt {a c} {\left (\cos \left (f x + e\right ) + 1\right )} \arctan \left (\frac {\sqrt {a c} \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 )}}}{a c \sin \left (f x + e\right )}\right ) \sin \left (f x + e\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 )}} \cos \left (f x + e\right )}{2 \, {\left (a^{2} c f \cos \left (f x + e\right ) + a^{2} c f\right )} \sin \left (f x + e\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(sqrt(-a*c)*(cos(f*x + e) + 1)*log(-4*(2*sqrt(-a*c)*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)^2 + (a*c*cos(f*x + e)^2 + a*c)*sin(f*x + e))/((cos(f*x + e)^2 - 1)*si
n(f*x + e)))*sin(f*x + e) - 2*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*c*f*cos(f*x + e) + a^2*c*f)*sin(f*x + e)), 1/2*(sqrt(a*c)*(cos(f*x + e) + 1)*arctan(sqrt(a
*c)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))/(a*c*sin(f*x + e)))*sin(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))*cos(f*x + e))/((a^2*c
*f*cos(f*x + e) + a^2*c*f)*sin(f*x + e))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec {\left (e + f x \right )}}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \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(sec(f*x+e)/(a+a*sec(f*x+e))**(3/2)/(c-c*sec(f*x+e))**(1/2),x)

[Out]

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

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Giac [A]
time = 1.89, size = 95, normalized size = 1.00 \begin {gather*} -\frac {c^{2} {\left (\frac {\log \left ({\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}{c} - \frac {\log \left ({\left | c \right |}\right )}{c} - \frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}{c^{2}}\right )}}{4 \, \sqrt {-a c} a f {\left | c \right |} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*c^2*(log(abs(c)*tan(1/2*f*x + 1/2*e)^2)/c - log(abs(c))/c - (c*tan(1/2*f*x + 1/2*e)^2 - c)/c^2)/(sqrt(-a*
c)*a*f*abs(c)*sgn(tan(1/2*f*x + 1/2*e)^3 + tan(1/2*f*x + 1/2*e)))

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

[Out]

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

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