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

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

[Out]

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

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Rubi [A]  time = 0.302711, antiderivative size = 150, normalized size of antiderivative = 1.29, number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3964, 94, 93, 205} \[ \frac{g^{3/2} \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{g \sec (e+f x)}}{\sqrt{g} \sqrt{c-c \sec (e+f x)}}\right )}{\sqrt{2} \sqrt{c} f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{g \tan (e+f x) \sqrt{g \sec (e+f x)}}{f \sqrt{a \sec (e+f x)+a} (c-c \sec (e+f x))} \]

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 3964

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]

Rule 94

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[p, 1] &&  !SumSimplerQ[m, 1])

Rule 93

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 205

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

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)) \operatorname{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 ) \operatorname{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 ) \operatorname{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]  time = 2.94445, size = 236, normalized size = 2.03 \[ -\frac{a \sin ^3\left (\frac{1}{2} (e+f x)\right ) \cos \left (\frac{1}{2} (e+f x)\right ) (g \sec (e+f x))^{5/2} \left (-4 \sec (e+f x)+\frac{\sqrt{\tan ^2(e+f x)} \left (\log \left (-3 \sec ^2(e+f x)-2 \sec (e+f x)-2 \sqrt{2} \sqrt{\tan ^2(e+f x)} \sqrt{\sec (e+f x)+1} \sqrt{\sec (e+f x)}+1\right )-\log \left (-3 \sec ^2(e+f x)-2 \sec (e+f x)+2 \sqrt{2} \sqrt{\tan ^2(e+f x)} \sqrt{\sec (e+f x)+1} \sqrt{\sec (e+f x)}+1\right )\right )}{\sqrt{\sec ^2\left (\frac{1}{2} (e+f x)\right )}}-4\right )}{c f g (\sec (e+f x)-1)^2 (a (\sec (e+f x)+1))^{3/2}} \]

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)))

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Maple [A]  time = 0.305, size = 152, normalized size = 1.3 \begin{align*} -{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{2\,fca \left ( \sin \left ( fx+e \right ) \right ) ^{4}} \left ( -\cos \left ( fx+e \right ){\it Arcsinh} \left ({\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \sqrt{2}+2\,\sin \left ( fx+e \right ) \sqrt{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1}}+{\it Arcsinh} \left ({\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \sqrt{2} \right ) \left ({\frac{g}{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ( \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) ^{-{\frac{3}{2}}}} \end{align*}

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)

[Out]

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

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Maxima [B]  time = 1.9542, size = 724, normalized size = 6.24 \begin{align*} \text{result too large to display} \end{align*}

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)

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Fricas [A]  time = 0.509675, size = 818, normalized size = 7.05 \begin{align*} \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{align*}

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))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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

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)