3.243 \(\int \frac{(g \sec (e+f x))^{5/2}}{\sqrt{a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx\)

Optimal. Leaf size=231 \[ -\frac{2 c^{3/2} g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \sqrt{g} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} d f (c-d) \sqrt{c+d}}+\frac{\sqrt{2} g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} f (c-d)}+\frac{2 g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} d f} \]

[Out]

(2*g^(5/2)*ArcTanh[(Sqrt[a]*Sqrt[g]*Tan[e + f*x])/(Sqrt[g*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[a]*d
*f) + (Sqrt[2]*g^(5/2)*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*c^(3/2)*g^(5/2)*ArcTanh[(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]])])/(Sqrt[a]*(c - d)*d*Sqrt[c + d]*f)

________________________________________________________________________________________

Rubi [A]  time = 0.815862, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3978, 3965, 208, 4023, 3808, 3802} \[ -\frac{2 c^{3/2} g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c} \sqrt{g} \tan (e+f x)}{\sqrt{c+d} \sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} d f (c-d) \sqrt{c+d}}+\frac{\sqrt{2} g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} f (c-d)}+\frac{2 g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a} \sqrt{g \sec (e+f x)}}\right )}{\sqrt{a} d f} \]

Antiderivative was successfully verified.

[In]

Int[(g*Sec[e + f*x])^(5/2)/(Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])),x]

[Out]

(2*g^(5/2)*ArcTanh[(Sqrt[a]*Sqrt[g]*Tan[e + f*x])/(Sqrt[g*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[a]*d
*f) + (Sqrt[2]*g^(5/2)*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*c^(3/2)*g^(5/2)*ArcTanh[(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]])])/(Sqrt[a]*(c - d)*d*Sqrt[c + d]*f)

Rule 3978

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(5/2)/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))), x_Symbol] :> -Dist[(c^2*g^2)/(d*(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] + Dist[g^2/(d*(b*c - a*d)), Int[(Sqrt[g*Csc[e + f*x]]*(a*c + (b*c - a*d)*Csc
[e + f*x]))/Sqrt[a + b*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 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]

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 4023

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Dist[(A*b - a*B)/b, Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n, x], x] + Dist[B
/b, Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A
*b - a*B, 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 3802

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

Rubi steps

\begin{align*} \int \frac{(g \sec (e+f x))^{5/2}}{\sqrt{a+a \sec (e+f x)} (c+d \sec (e+f x))} \, dx &=\frac{g^2 \int \frac{\sqrt{g \sec (e+f x)} (a c+(a c-a d) \sec (e+f x))}{\sqrt{a+a \sec (e+f x)}} \, dx}{a (c-d) d}-\frac{\left (c^2 g^2\right ) \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) d}\\ &=\frac{g^2 \int \frac{\sqrt{g \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}} \, dx}{c-d}+\frac{g^2 \int \sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)} \, dx}{a d}+\frac{\left (2 c^2 g^3\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) d f}\\ &=-\frac{2 c^{3/2} g^{5/2} \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) d \sqrt{c+d} f}-\frac{\left (2 g^3\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 g^3\right ) \operatorname{Subst}\left (\int \frac{1}{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 )}{d f}\\ &=\frac{2 g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{g} \tan (e+f x)}{\sqrt{g \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\right )}{\sqrt{a} d f}+\frac{\sqrt{2} g^{5/2} \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 c^{3/2} g^{5/2} \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) d \sqrt{c+d} f}\\ \end{align*}

Mathematica [A]  time = 0.391366, size = 155, normalized size = 0.67 \[ \frac{2 g^2 \cos \left (\frac{1}{2} (e+f x)\right ) \sqrt{g \sec (e+f x)} \left (\sqrt{2} \left ((c-d) \sqrt{c+d} \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (e+f x)\right )\right )-c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sin \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )\right )+d \sqrt{c+d} \tanh ^{-1}\left (\sin \left (\frac{1}{2} (e+f x)\right )\right )\right )}{d f (c-d) \sqrt{c+d} \sqrt{a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(g*Sec[e + f*x])^(5/2)/(Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x])),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.346, size = 726, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*sec(f*x+e))^(5/2)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x)

[Out]

2/f/a/(c/(c-d))^(1/2)/(-c+d+((c+d)*(c-d))^(1/2))/(c-d+((c+d)*(c-d))^(1/2))/((c+d)*(c-d))^(1/2)*(g/cos(f*x+e))^
(5/2)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*cos(f*x+e)^3*(-1+cos(f*x+e))^3*(arcsinh((-1+cos(f*x+e))/sin(f*x+e)
)*2^(1/2)*(c/(c-d))^(1/2)*((c+d)*(c-d))^(1/2)*d+arctanh(1/2*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)+1-sin(f*x+e))
)*(c/(c-d))^(1/2)*((c+d)*(c-d))^(1/2)*c-arctanh(1/2*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)+1-sin(f*x+e)))*(c/(c-
d))^(1/2)*((c+d)*(c-d))^(1/2)*d-arctanh(1/2*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)+1+sin(f*x+e)))*(c/(c-d))^(1/2
)*((c+d)*(c-d))^(1/2)*c+arctanh(1/2*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)+1+sin(f*x+e)))*(c/(c-d))^(1/2)*((c+d)
*(c-d))^(1/2)*d+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))*c^2-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))*c^2)/sin(
f*x+e)^6/(1/(1+cos(f*x+e)))^(5/2)

________________________________________________________________________________________

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))^(5/2)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [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))^(5/2)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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))**(5/2)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*sec(f*x+e))^(5/2)/(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError