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

Optimal. Leaf size=167 \[ \frac{2 \sqrt{c} g^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \sqrt{g} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} f (c-d) \sqrt{c+d}}-\frac{\sqrt{2} g^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} f (c-d)} \]

[Out]

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

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Rubi [A]  time = 0.582828, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {3974, 3808, 208, 3965} \[ \frac{2 \sqrt{c} g^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \sqrt{g} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} f (c-d) \sqrt{c+d}}-\frac{\sqrt{2} g^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} f (c-d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3974

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(3/2)/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))), x_Symbol] :> -Dist[(a*g)/(b*c - a*d), Int[Sqrt[g*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x],
x] + Dist[(c*g)/(b*c - a*d), Int[(Sqrt[g*Csc[e + f*x]]*Sqrt[a + b*Csc[e + f*x]])/(c + d*Csc[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 3808

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

Rule 208

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

Rule 3965

Int[(Sqrt[csc[(e_.) + (f_.)*(x_)]*(g_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)])/(csc[(e_.) + (f_.)*(x_)]*
(d_.) + (c_)), x_Symbol] :> Dist[(-2*b*g)/f, Subst[Int[1/(b*c + a*d - c*g*x^2), x], x, (b*Cot[e + f*x])/(Sqrt[
g*Csc[e + f*x]]*Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[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+d \sec (e+f x))} \, dx &=-\frac{g \int \frac{\sqrt{g \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}} \, dx}{c-d}+\frac{(c g) \int \frac{\sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}{c+d \sec (e+f x)} \, dx}{a (c-d)}\\ &=\frac{\left (2 g^2\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-g x^2} \, dx,x,-\frac{a \tan (e+f x)}{\sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\right )}{(c-d) f}-\frac{\left (2 c g^2\right ) \operatorname{Subst}\left (\int \frac{1}{a c+a d-c g x^2} \, dx,x,-\frac{a \tan (e+f x)}{\sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\right )}{(c-d) f}\\ &=-\frac{\sqrt{2} 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{a} (c-d) f}+\frac{2 \sqrt{c} g^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \sqrt{g} \tan (e+f x)}{\sqrt{c+d} \sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\right )}{\sqrt{a} (c-d) \sqrt{c+d} f}\\ \end{align*}

Mathematica [A]  time = 0.394346, size = 198, normalized size = 1.19 \[ \frac{g \cos \left (\frac{1}{2} (e+f x)\right ) \sqrt{g \sec (e+f x)} \left (\sqrt{2} \sqrt{c} \left (\log \left (\sqrt{2} \sqrt{c+d}+2 \sqrt{c} \sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sqrt{2} \sqrt{c+d}-2 \sqrt{c} \sin \left (\frac{1}{2} (e+f x)\right )\right )\right )+2 \sqrt{c+d} \log \left (\cos \left (\frac{1}{4} (e+f x)\right )-\sin \left (\frac{1}{4} (e+f x)\right )\right )-2 \sqrt{c+d} \log \left (\sin \left (\frac{1}{4} (e+f x)\right )+\cos \left (\frac{1}{4} (e+f x)\right )\right )\right )}{f (c-d) \sqrt{c+d} \sqrt{a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(g*Cos[(e + f*x)/2]*(2*Sqrt[c + d]*Log[Cos[(e + f*x)/4] - Sin[(e + f*x)/4]] - 2*Sqrt[c + d]*Log[Cos[(e + f*x)/
4] + Sin[(e + f*x)/4]] + Sqrt[2]*Sqrt[c]*(-Log[Sqrt[2]*Sqrt[c + d] - 2*Sqrt[c]*Sin[(e + f*x)/2]] + Log[Sqrt[2]
*Sqrt[c + d] + 2*Sqrt[c]*Sin[(e + f*x)/2]]))*Sqrt[g*Sec[e + f*x]])/((c - d)*Sqrt[c + d]*f*Sqrt[a*(1 + Sec[e +
f*x])])

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Maple [B]  time = 0.338, size = 473, normalized size = 2.8 \begin{align*}{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}}{af \left ( c-d \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{4}} \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 ({\it Arcsinh} \left ({\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \sqrt{2}\sqrt{{\frac{c}{c-d}}}\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }+c\ln \left ( 2\,{\frac{1}{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }\sin \left ( fx+e \right ) +c\cos \left ( fx+e \right ) -d\cos \left ( fx+e \right ) -c+d} \left ( -2\,\sqrt{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1}}\sqrt{{\frac{c}{c-d}}}c\sin \left ( fx+e \right ) +2\,\sqrt{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1}}\sqrt{{\frac{c}{c-d}}}d\sin \left ( fx+e \right ) +\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }\cos \left ( fx+e \right ) -c\sin \left ( fx+e \right ) +d\sin \left ( fx+e \right ) -\sqrt{ \left ( c+d \right ) \left ( c-d \right ) } \right ) } \right ) -c\ln \left ( 2\,{\frac{1}{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }\sin \left ( fx+e \right ) -c\cos \left ( fx+e \right ) +d\cos \left ( fx+e \right ) +c-d} \left ( 2\,\sqrt{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1}}\sqrt{{\frac{c}{c-d}}}c\sin \left ( fx+e \right ) -2\,\sqrt{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1}}\sqrt{{\frac{c}{c-d}}}d\sin \left ( fx+e \right ) +\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }\cos \left ( fx+e \right ) +c\sin \left ( fx+e \right ) -d\sin \left ( fx+e \right ) -\sqrt{ \left ( c+d \right ) \left ( c-d \right ) } \right ) } \right ) \right ){\frac{1}{\sqrt{{\frac{c}{c-d}}}}}{\frac{1}{\sqrt{ \left ( c+d \right ) \left ( c-d \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+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x)

[Out]

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

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Maxima [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+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.56623, size = 2654, normalized size = 15.89 \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+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[-1/2*(sqrt(2)*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
)) + sqrt(c*g/(a*c + a*d))*g*log((c^2*g*cos(f*x + e)^3 - (7*c^2 + 6*c*d)*g*cos(f*x + e)^2 + 4*((c^2 + c*d)*cos
(f*x + e)^2 - (2*c^2 + 3*c*d + d^2)*cos(f*x + e))*sqrt(c*g/(a*c + a*d))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)
)*sqrt(g/cos(f*x + e))*sin(f*x + e) + (2*c*d + d^2)*g*cos(f*x + e) + (8*c^2 + 8*c*d + d^2)*g)/(c^2*cos(f*x + e
)^3 + (c^2 + 2*c*d)*cos(f*x + e)^2 + d^2 + (2*c*d + d^2)*cos(f*x + e))))/((c - d)*f), 1/2*(2*sqrt(2)*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))) - sqrt(c*g/(a*c + a*d))*g*log((c^2*g*cos(f*x + e)^3 - (7*c^2 + 6*c*d)*g*cos(f*x + e)^2 + 4*((c^2 +
c*d)*cos(f*x + e)^2 - (2*c^2 + 3*c*d + d^2)*cos(f*x + e))*sqrt(c*g/(a*c + a*d))*sqrt((a*cos(f*x + e) + a)/cos
(f*x + e))*sqrt(g/cos(f*x + e))*sin(f*x + e) + (2*c*d + d^2)*g*cos(f*x + e) + (8*c^2 + 8*c*d + d^2)*g)/(c^2*co
s(f*x + e)^3 + (c^2 + 2*c*d)*cos(f*x + e)^2 + d^2 + (2*c*d + d^2)*cos(f*x + e))))/((c - d)*f), -1/2*(sqrt(2)*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)) - 2*sqrt(-c*g
/(a*c + a*d))*g*arctan(1/2*(c*cos(f*x + e)^2 - (2*c + d)*cos(f*x + e))*sqrt(-c*g/(a*c + a*d))*sqrt((a*cos(f*x
+ e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))/(c*g*sin(f*x + e))))/((c - d)*f), (sqrt(2)*g*sqrt(-g/a)*arctan(sq
rt(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))) +
sqrt(-c*g/(a*c + a*d))*g*arctan(1/2*(c*cos(f*x + e)^2 - (2*c + d)*cos(f*x + e))*sqrt(-c*g/(a*c + a*d))*sqrt((
a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(g/cos(f*x + e))/(c*g*sin(f*x + e))))/((c - d)*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g \sec{\left (e + f x \right )}\right )^{\frac{3}{2}}}{\sqrt{a \left (\sec{\left (e + f x \right )} + 1\right )} \left (c + d \sec{\left (e + f x \right )}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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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 (d \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+d*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)*(d*sec(f*x + e) + c)), x)