3.2.81 \(\int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))} \, dx\) [181]

Optimal. Leaf size=116 \[ -\frac {g^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {g} \tan (e+f x)}{\sqrt {2} \sqrt {g \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}\right )}{\sqrt {2} \sqrt {a} c f}+\frac {g \cot (e+f x) \sqrt {g \sec (e+f x)} \sqrt {a+a \sec (e+f x)}}{a c f} \]

[Out]

-1/2*g^(3/2)*arctanh(1/2*a^(1/2)*g^(1/2)*tan(f*x+e)*2^(1/2)/(g*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2))/c/f*2
^(1/2)/a^(1/2)+g*cot(f*x+e)*(g*sec(f*x+e))^(1/2)*(a+a*sec(f*x+e))^(1/2)/a/c/f

________________________________________________________________________________________

Rubi [A]
time = 0.19, antiderivative size = 150, normalized size of antiderivative = 1.29, number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4049, 96, 95, 211} \begin {gather*} \frac {g^{3/2} \tan (e+f x) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {g \sec (e+f x)}}{\sqrt {g} \sqrt {c-c \sec (e+f x)}}\right )}{\sqrt {2} \sqrt {c} f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {g \tan (e+f x) \sqrt {g \sec (e+f x)}}{f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((g*Sqrt[g*Sec[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x]))) + (g^(3/2)*ArcTan[(
Sqrt[2]*Sqrt[c]*Sqrt[g*Sec[e + f*x]])/(Sqrt[g]*Sqrt[c - c*Sec[e + f*x]])]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c]*f*Sqr
t[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Dist[n*((d*e - c*f)/((m + 1)*(b*e - a*f
))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 4049

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(236\) vs. \(2(116)=232\).
time = 2.83, size = 236, normalized size = 2.03 \begin {gather*} -\frac {a \cos \left (\frac {1}{2} (e+f x)\right ) (g \sec (e+f x))^{5/2} \sin ^3\left (\frac {1}{2} (e+f x)\right ) \left (-4-4 \sec (e+f x)+\frac {\left (\log \left (1-2 \sec (e+f x)-3 \sec ^2(e+f x)-2 \sqrt {2} \sqrt {\sec (e+f x)} \sqrt {1+\sec (e+f x)} \sqrt {\tan ^2(e+f x)}\right )-\log \left (1-2 \sec (e+f x)-3 \sec ^2(e+f x)+2 \sqrt {2} \sqrt {\sec (e+f x)} \sqrt {1+\sec (e+f x)} \sqrt {\tan ^2(e+f x)}\right )\right ) \sqrt {\tan ^2(e+f x)}}{\sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )}}\right )}{c f g (-1+\sec (e+f x))^2 (a (1+\sec (e+f x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Cos[(e + f*x)/2]*(g*Sec[e + f*x])^(5/2)*Sin[(e + f*x)/2]^3*(-4 - 4*Sec[e + f*x] + ((Log[1 - 2*Sec[e + f*x
] - 3*Sec[e + f*x]^2 - 2*Sqrt[2]*Sqrt[Sec[e + f*x]]*Sqrt[1 + Sec[e + f*x]]*Sqrt[Tan[e + f*x]^2]] - Log[1 - 2*S
ec[e + f*x] - 3*Sec[e + f*x]^2 + 2*Sqrt[2]*Sqrt[Sec[e + f*x]]*Sqrt[1 + Sec[e + f*x]]*Sqrt[Tan[e + f*x]^2]])*Sq
rt[Tan[e + f*x]^2])/Sqrt[Sec[(e + f*x)/2]^2]))/(c*f*g*(-1 + Sec[e + f*x])^2*(a*(1 + Sec[e + f*x]))^(3/2)))

________________________________________________________________________________________

Maple [A]
time = 3.98, size = 152, normalized size = 1.31

method	result	size
default	\(\frac {\left (\arcsinh \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \sqrt {2}\, \cos \left (f x +e \right )-\arcsinh \left (\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \sqrt {2}-2 \sin \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\right ) \left (\frac {g}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \left (-1+\cos \left (f x +e \right )\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}}{2 c f \left (\frac {1}{\cos \left (f x +e \right )+1}\right )^{\frac {3}{2}} \sin \left (f x +e \right )^{4} a}\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 576 vs. \(2 (103) = 206\).
time = 0.60, size = 576, normalized size = 4.97 \begin {gather*} \frac {{\left (4 \, g \cos \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 4 \, g \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - {\left (g \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + g \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} - 2 \, g \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + g\right )} \log \left (\cos \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) + {\left (g \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + g \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} - 2 \, g \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + g\right )} \log \left (\cos \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) + 4 \, g \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {g}}{2 \, {\left (\sqrt {2} c \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sqrt {2} c \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} - 2 \, \sqrt {2} c \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + \sqrt {2} c\right )} \sqrt {a} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 1.86, size = 356, normalized size = 3.07 \begin {gather*} \left [\frac {\sqrt {2} a g \sqrt {\frac {g}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {g}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + g \cos \left (f x + e\right )^{2} - 2 \, g \cos \left (f x + e\right ) - 3 \, g}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 4 \, g \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{4 \, a c f \sin \left (f x + e\right )}, \frac {\sqrt {2} a g \sqrt {-\frac {g}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {-\frac {g}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{g \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 2 \, g \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {g}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{2 \, a c f \sin \left (f x + e\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(sqrt(2)*a*g*sqrt(g/a)*log(-(2*sqrt(2)*sqrt(g/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x +
e))*cos(f*x + e)*sin(f*x + e) + g*cos(f*x + e)^2 - 2*g*cos(f*x + e) - 3*g)/(cos(f*x + e)^2 + 2*cos(f*x + e) +
1))*sin(f*x + e) + 4*g*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))*cos(f*x + e))/(a*c*f*sin(
f*x + e)), 1/2*(sqrt(2)*a*g*sqrt(-g/a)*arctan(sqrt(2)*sqrt(-g/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(
g/cos(f*x + e))*cos(f*x + e)/(g*sin(f*x + e)))*sin(f*x + e) + 2*g*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt
(g/cos(f*x + e))*cos(f*x + e))/(a*c*f*sin(f*x + e))]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

integrate(-(g*sec(f*x + e))^(3/2)/(sqrt(a*sec(f*x + e) + a)*(c*sec(f*x + e) - c)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________