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

Optimal. Leaf size=166 \[ \frac{2 g^2 \sqrt{g \sec (e+f x)} \sqrt{\frac{a \cos (e+f x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (e+f x)|\frac{2 a}{a+b}\right )}{d f \sqrt{a+b \sec (e+f x)}}-\frac{2 c g^2 \sqrt{g \sec (e+f x)} \sqrt{\frac{a \cos (e+f x)+b}{a+b}} \Pi \left (\frac{2 c}{c+d};\frac{1}{2} (e+f x)|\frac{2 a}{a+b}\right )}{d f (c+d) \sqrt{a+b \sec (e+f x)}} \]

[Out]

(2*g^2*Sqrt[(b + a*Cos[e + f*x])/(a + b)]*EllipticPi[2, (e + f*x)/2, (2*a)/(a + b)]*Sqrt[g*Sec[e + f*x]])/(d*f
*Sqrt[a + b*Sec[e + f*x]]) - (2*c*g^2*Sqrt[(b + a*Cos[e + f*x])/(a + b)]*EllipticPi[(2*c)/(c + d), (e + f*x)/2
, (2*a)/(a + b)]*Sqrt[g*Sec[e + f*x]])/(d*(c + d)*f*Sqrt[a + b*Sec[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.855632, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {3979, 3859, 2807, 2805, 3975} \[ \frac{2 g^2 \sqrt{g \sec (e+f x)} \sqrt{\frac{a \cos (e+f x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (e+f x)|\frac{2 a}{a+b}\right )}{d f \sqrt{a+b \sec (e+f x)}}-\frac{2 c g^2 \sqrt{g \sec (e+f x)} \sqrt{\frac{a \cos (e+f x)+b}{a+b}} \Pi \left (\frac{2 c}{c+d};\frac{1}{2} (e+f x)|\frac{2 a}{a+b}\right )}{d f (c+d) \sqrt{a+b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*g^2*Sqrt[(b + a*Cos[e + f*x])/(a + b)]*EllipticPi[2, (e + f*x)/2, (2*a)/(a + b)]*Sqrt[g*Sec[e + f*x]])/(d*f
*Sqrt[a + b*Sec[e + f*x]]) - (2*c*g^2*Sqrt[(b + a*Cos[e + f*x])/(a + b)]*EllipticPi[(2*c)/(c + d), (e + f*x)/2
, (2*a)/(a + b)]*Sqrt[g*Sec[e + f*x]])/(d*(c + d)*f*Sqrt[a + b*Sec[e + f*x]])

Rule 3979

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(5/2)/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))), x_Symbol] :> Dist[g/d, Int[(g*Csc[e + f*x])^(3/2)/Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(c*
g)/d, Int[(g*Csc[e + f*x])^(3/2)/(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] && NeQ[a^2 - b^2, 0]

Rule 3859

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f
*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 3975

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(3/2)/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))), x_Symbol] :> Dist[(g*Sqrt[g*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]],
Int[1/(Sqrt[b + a*Sin[e + f*x]]*(d + c*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{(g \sec (e+f x))^{5/2}}{\sqrt{a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx &=\frac{g \int \frac{(g \sec (e+f x))^{3/2}}{\sqrt{a+b \sec (e+f x)}} \, dx}{d}-\frac{(c g) \int \frac{(g \sec (e+f x))^{3/2}}{\sqrt{a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx}{d}\\ &=\frac{\left (g^2 \sqrt{b+a \cos (e+f x)} \sqrt{g \sec (e+f x)}\right ) \int \frac{\sec (e+f x)}{\sqrt{b+a \cos (e+f x)}} \, dx}{d \sqrt{a+b \sec (e+f x)}}-\frac{\left (c g^2 \sqrt{b+a \cos (e+f x)} \sqrt{g \sec (e+f x)}\right ) \int \frac{1}{\sqrt{b+a \cos (e+f x)} (d+c \cos (e+f x))} \, dx}{d \sqrt{a+b \sec (e+f x)}}\\ &=\frac{\left (g^2 \sqrt{\frac{b+a \cos (e+f x)}{a+b}} \sqrt{g \sec (e+f x)}\right ) \int \frac{\sec (e+f x)}{\sqrt{\frac{b}{a+b}+\frac{a \cos (e+f x)}{a+b}}} \, dx}{d \sqrt{a+b \sec (e+f x)}}-\frac{\left (c g^2 \sqrt{\frac{b+a \cos (e+f x)}{a+b}} \sqrt{g \sec (e+f x)}\right ) \int \frac{1}{\sqrt{\frac{b}{a+b}+\frac{a \cos (e+f x)}{a+b}} (d+c \cos (e+f x))} \, dx}{d \sqrt{a+b \sec (e+f x)}}\\ &=\frac{2 g^2 \sqrt{\frac{b+a \cos (e+f x)}{a+b}} \Pi \left (2;\frac{1}{2} (e+f x)|\frac{2 a}{a+b}\right ) \sqrt{g \sec (e+f x)}}{d f \sqrt{a+b \sec (e+f x)}}-\frac{2 c g^2 \sqrt{\frac{b+a \cos (e+f x)}{a+b}} \Pi \left (\frac{2 c}{c+d};\frac{1}{2} (e+f x)|\frac{2 a}{a+b}\right ) \sqrt{g \sec (e+f x)}}{d (c+d) f \sqrt{a+b \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 4.04738, size = 246, normalized size = 1.48 \[ -\frac{2 i g \cot (e+f x) (g \sec (e+f x))^{3/2} \sqrt{-\frac{a (\cos (e+f x)-1)}{a+b}} \sqrt{\frac{a (\cos (e+f x)+1)}{a-b}} \sqrt{a \cos (e+f x)+b} \left ((a d-b c) \Pi \left (1-\frac{a}{b};i \sinh ^{-1}\left (\sqrt{\frac{1}{a-b}} \sqrt{b+a \cos (e+f x)}\right )|\frac{b-a}{a+b}\right )+b c \Pi \left (\frac{(a-b) c}{a d-b c};i \sinh ^{-1}\left (\sqrt{\frac{1}{a-b}} \sqrt{b+a \cos (e+f x)}\right )|\frac{b-a}{a+b}\right )\right )}{b d f \sqrt{\frac{1}{a-b}} (a d-b c) \sqrt{a+b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*g*Sqrt[-((a*(-1 + Cos[e + f*x]))/(a + b))]*Sqrt[(a*(1 + Cos[e + f*x]))/(a - b)]*Sqrt[b + a*Cos[e + f*x
]]*Cot[e + f*x]*((-(b*c) + a*d)*EllipticPi[1 - a/b, I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[e + f*x]]], (-
a + b)/(a + b)] + b*c*EllipticPi[((a - b)*c)/(-(b*c) + a*d), I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[e + f
*x]]], (-a + b)/(a + b)])*(g*Sec[e + f*x])^(3/2))/(Sqrt[(a - b)^(-1)]*b*d*(-(b*c) + a*d)*f*Sqrt[a + b*Sec[e +
f*x]])

________________________________________________________________________________________

Maple [C]  time = 0.354, size = 344, normalized size = 2.1 \begin{align*}{\frac{-2\,i \left ( -1+\cos \left ( fx+e \right ) \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{fd \left ( c+d \right ) \left ( c-d \right ) \left ( a\cos \left ( fx+e \right ) +b \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ({\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},\sqrt{-{\frac{a-b}{a+b}}} \right ) dc+{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},\sqrt{-{\frac{a-b}{a+b}}} \right ){d}^{2}+2\,{c}^{2}{\it EllipticPi} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},-1,i\sqrt{{\frac{a-b}{a+b}}} \right ) -2\,{\it EllipticPi} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},-1,i\sqrt{{\frac{a-b}{a+b}}} \right ){d}^{2}-2\,{c}^{2}{\it EllipticPi} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},-{\frac{c-d}{c+d}},i\sqrt{{\frac{a-b}{a+b}}} \right ) \right ) \sqrt{{\frac{a\cos \left ( fx+e \right ) +b}{ \left ( a+b \right ) \left ( 1+\cos \left ( fx+e \right ) \right ) }}}\sqrt{{\frac{a\cos \left ( fx+e \right ) +b}{\cos \left ( fx+e \right ) }}} \left ({\frac{g}{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{2}}} \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))^(5/2)/(c+d*sec(f*x+e))/(a+b*sec(f*x+e))^(1/2),x)

[Out]

-2*I/f/d/(c+d)/(c-d)*(EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),(-(a-b)/(a+b))^(1/2))*d*c+EllipticF(I*(-1+cos(f*x
+e))/sin(f*x+e),(-(a-b)/(a+b))^(1/2))*d^2+2*c^2*EllipticPi(I*(-1+cos(f*x+e))/sin(f*x+e),-1,I*((a-b)/(a+b))^(1/
2))-2*EllipticPi(I*(-1+cos(f*x+e))/sin(f*x+e),-1,I*((a-b)/(a+b))^(1/2))*d^2-2*c^2*EllipticPi(I*(-1+cos(f*x+e))
/sin(f*x+e),-(c-d)/(c+d),I*((a-b)/(a+b))^(1/2)))*(1/(a+b)*(a*cos(f*x+e)+b)/(1+cos(f*x+e)))^(1/2)*(1/cos(f*x+e)
*(a*cos(f*x+e)+b))^(1/2)*(-1+cos(f*x+e))*(g/cos(f*x+e))^(5/2)*cos(f*x+e)^3/(a*cos(f*x+e)+b)/(1/(1+cos(f*x+e)))
^(3/2)/sin(f*x+e)^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+b*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+b*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+b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g \sec \left (f x + e\right )\right )^{\frac{5}{2}}}{\sqrt{b \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))^(5/2)/(c+d*sec(f*x+e))/(a+b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((g*sec(f*x + e))^(5/2)/(sqrt(b*sec(f*x + e) + a)*(d*sec(f*x + e) + c)), x)