\(\int \frac {(g \sec (e+f x))^{5/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx\) [284]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [C] (verified)
   Fricas [F(-1)]
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 39, antiderivative size = 166 \[ \int \frac {(g \sec (e+f x))^{5/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\frac {2 g^2 \sqrt {\frac {b+a \cos (e+f x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (e+f x),\frac {2 a}{a+b}\right ) \sqrt {g \sec (e+f x)}}{d f \sqrt {a+b \sec (e+f x)}}-\frac {2 c g^2 \sqrt {\frac {b+a \cos (e+f x)}{a+b}} \operatorname {EllipticPi}\left (\frac {2 c}{c+d},\frac {1}{2} (e+f x),\frac {2 a}{a+b}\right ) \sqrt {g \sec (e+f x)}}{d (c+d) f \sqrt {a+b \sec (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {4064, 3944, 2886, 2884, 4060} \[ \int \frac {(g \sec (e+f x))^{5/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\frac {2 g^2 \sqrt {g \sec (e+f x)} \sqrt {\frac {a \cos (e+f x)+b}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (e+f x),\frac {2 a}{a+b}\right )}{d f \sqrt {a+b \sec (e+f x)}}-\frac {2 c g^2 \sqrt {g \sec (e+f x)} \sqrt {\frac {a \cos (e+f x)+b}{a+b}} \operatorname {EllipticPi}\left (\frac {2 c}{c+d},\frac {1}{2} (e+f x),\frac {2 a}{a+b}\right )}{d f (c+d) \sqrt {a+b \sec (e+f x)}} \]

[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 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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 2886

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/
(c + d))*Sin[e + f*x]]), 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 3944

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[d*Sqrt
[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 4060

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]

Rule 4064

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]

Rubi steps \begin{align*} \text {integral}& = \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}} \operatorname {EllipticPi}\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}} \operatorname {EllipticPi}\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] (verified)

Result contains complex when optimal does not.

Time = 25.39 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.48 \[ \int \frac {(g \sec (e+f x))^{5/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=-\frac {2 i g \sqrt {-\frac {a (-1+\cos (e+f x))}{a+b}} \sqrt {\frac {a (1+\cos (e+f x))}{a-b}} \sqrt {b+a \cos (e+f x)} \cot (e+f x) \left ((-b c+a d) \operatorname {EllipticPi}\left (1-\frac {a}{b},i \text {arcsinh}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (e+f x)}\right ),\frac {-a+b}{a+b}\right )+b c \operatorname {EllipticPi}\left (\frac {(a-b) c}{-b c+a d},i \text {arcsinh}\left (\sqrt {\frac {1}{a-b}} \sqrt {b+a \cos (e+f x)}\right ),\frac {-a+b}{a+b}\right )\right ) (g \sec (e+f x))^{3/2}}{\sqrt {\frac {1}{a-b}} b d (-b c+a d) f \sqrt {a+b \sec (e+f x)}} \]

[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] (verified)

Result contains complex when optimal does not.

Time = 9.42 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.89

method result size
default \(\frac {2 i \left (\operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), \sqrt {-\frac {a -b}{a +b}}\right ) d c +\operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), \sqrt {-\frac {a -b}{a +b}}\right ) d^{2}+2 c^{2} \operatorname {EllipticPi}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), -1, i \sqrt {\frac {a -b}{a +b}}\right )-2 \operatorname {EllipticPi}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), -1, i \sqrt {\frac {a -b}{a +b}}\right ) d^{2}-2 c^{2} \operatorname {EllipticPi}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), -\frac {c -d}{c +d}, i \sqrt {\frac {a -b}{a +b}}\right )\right ) \sqrt {\frac {b +a \cos \left (f x +e \right )}{\left (a +b \right ) \left (\cos \left (f x +e \right )+1\right )}}\, \sqrt {a +b \sec \left (f x +e \right )}\, \sqrt {g \sec \left (f x +e \right )}\, g^{2} \cos \left (f x +e \right )}{f d \left (c +d \right ) \left (c -d \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \left (b +a \cos \left (f x +e \right )\right )}\) \(313\)

[In]

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

[Out]

2*I/f/d/(c+d)/(c-d)*(EllipticF(I*(cot(f*x+e)-csc(f*x+e)),(-(a-b)/(a+b))^(1/2))*d*c+EllipticF(I*(cot(f*x+e)-csc
(f*x+e)),(-(a-b)/(a+b))^(1/2))*d^2+2*c^2*EllipticPi(I*(cot(f*x+e)-csc(f*x+e)),-1,I*((a-b)/(a+b))^(1/2))-2*Elli
pticPi(I*(cot(f*x+e)-csc(f*x+e)),-1,I*((a-b)/(a+b))^(1/2))*d^2-2*c^2*EllipticPi(I*(cot(f*x+e)-csc(f*x+e)),-(c-
d)/(c+d),I*((a-b)/(a+b))^(1/2)))*(1/(a+b)*(b+a*cos(f*x+e))/(cos(f*x+e)+1))^(1/2)*(a+b*sec(f*x+e))^(1/2)*(g*sec
(f*x+e))^(1/2)*g^2/(1/(cos(f*x+e)+1))^(1/2)/(b+a*cos(f*x+e))*cos(f*x+e)

Fricas [F(-1)]

Timed out. \[ \int \frac {(g \sec (e+f x))^{5/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\text {Timed out} \]

[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)]

Timed out. \[ \int \frac {(g \sec (e+f x))^{5/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\text {Timed out} \]

[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

Maxima [F]

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

[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]

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

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(g \sec (e+f x))^{5/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\int \frac {{\left (\frac {g}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{\sqrt {a+\frac {b}{\cos \left (e+f\,x\right )}}\,\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )} \,d x \]

[In]

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

[Out]

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