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

Optimal. Leaf size=312 \[ \frac {g^2 \sin (e+f x) \sqrt {g \sec (e+f x)} (a \cos (e+f x)+b)}{f (a-b) (c \cos (e+f x)+c) \sqrt {a+b \sec (e+f x)}}-\frac {g^2 \sqrt {g \sec (e+f x)} \sqrt {\frac {a \cos (e+f x)+b}{a+b}} F\left (\frac {1}{2} (e+f x)|\frac {2 a}{a+b}\right )}{c f \sqrt {a+b \sec (e+f x)}}-\frac {g^2 \sqrt {g \sec (e+f x)} (a \cos (e+f x)+b) E\left (\frac {1}{2} (e+f x)|\frac {2 a}{a+b}\right )}{c f (a-b) \sqrt {\frac {a \cos (e+f x)+b}{a+b}} \sqrt {a+b \sec (e+f x)}}+\frac {2 g^2 \sqrt {g \sec (e+f x)} \sqrt {\frac {a \cos (e+f x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (e+f x)|\frac {2 a}{a+b}\right )}{c f \sqrt {a+b \sec (e+f x)}} \]

[Out]

g^2*(b+a*cos(f*x+e))*sin(f*x+e)*(g*sec(f*x+e))^(1/2)/(a-b)/f/(c+c*cos(f*x+e))/(a+b*sec(f*x+e))^(1/2)-g^2*(b+a*
cos(f*x+e))*(cos(1/2*e+1/2*f*x)^2)^(1/2)/cos(1/2*e+1/2*f*x)*EllipticE(sin(1/2*e+1/2*f*x),2^(1/2)*(a/(a+b))^(1/
2))*(g*sec(f*x+e))^(1/2)/(a-b)/c/f/((b+a*cos(f*x+e))/(a+b))^(1/2)/(a+b*sec(f*x+e))^(1/2)-g^2*(cos(1/2*e+1/2*f*
x)^2)^(1/2)/cos(1/2*e+1/2*f*x)*EllipticF(sin(1/2*e+1/2*f*x),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)/c/f/(a+b*sec(f*x+e))^(1/2)+2*g^2*(cos(1/2*e+1/2*f*x)^2)^(1/2)/cos(1/2*e+1/2*f*x)*El
lipticPi(sin(1/2*e+1/2*f*x),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)/c/f
/(a+b*sec(f*x+e))^(1/2)

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Rubi [A]  time = 0.89, antiderivative size = 312, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.282, Rules used = {3979, 3859, 2807, 2805, 3975, 2768, 2752, 2663, 2661, 2655, 2653} \[ \frac {g^2 \sin (e+f x) \sqrt {g \sec (e+f x)} (a \cos (e+f x)+b)}{f (a-b) (c \cos (e+f x)+c) \sqrt {a+b \sec (e+f x)}}-\frac {g^2 \sqrt {g \sec (e+f x)} \sqrt {\frac {a \cos (e+f x)+b}{a+b}} F\left (\frac {1}{2} (e+f x)|\frac {2 a}{a+b}\right )}{c f \sqrt {a+b \sec (e+f x)}}-\frac {g^2 \sqrt {g \sec (e+f x)} (a \cos (e+f x)+b) E\left (\frac {1}{2} (e+f x)|\frac {2 a}{a+b}\right )}{c f (a-b) \sqrt {\frac {a \cos (e+f x)+b}{a+b}} \sqrt {a+b \sec (e+f x)}}+\frac {2 g^2 \sqrt {g \sec (e+f x)} \sqrt {\frac {a \cos (e+f x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (e+f x)|\frac {2 a}{a+b}\right )}{c f \sqrt {a+b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2653

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

Rule 2655

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

Rule 2661

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

Rule 2663

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

Rule 2752

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

Rule 2768

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b
^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(b*c - a*d)*(a + b*Sin[e + f*x])), x] + Dist[d/(a*(b*c - a*
d)), Int[(c + d*Sin[e + f*x])^n*(a*n - b*(n + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 0] && (IntegerQ[2*n] || EqQ[c, 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 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 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 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]

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]

Rubi steps

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

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Mathematica [F]  time = 15.55, size = 0, normalized size = 0.00 \[ \int \frac {(g \sec (e+f x))^{5/2}}{\sqrt {a+b \sec (e+f x)} (c+c \sec (e+f x))} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \sec \left (f x + e\right )\right )^{\frac {5}{2}}}{\sqrt {b \sec \left (f x + e\right ) + a} {\left (c \sec \left (f x + e\right ) + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [C]  time = 2.50, size = 355, normalized size = 1.14 \[ \frac {i \left (4 a \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, \sqrt {-\frac {a -b}{a +b}}\right )-2 b \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, \sqrt {-\frac {a -b}{a +b}}\right )-a \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, \sqrt {-\frac {a -b}{a +b}}\right )-b \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, \sqrt {-\frac {a -b}{a +b}}\right )-4 \EllipticPi \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, -1, i \sqrt {\frac {a -b}{a +b}}\right ) a +4 \EllipticPi \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, -1, i \sqrt {\frac {a -b}{a +b}}\right ) b \right ) \sqrt {\frac {b +a \cos \left (f x +e \right )}{\left (1+\cos \left (f x +e \right )\right ) \left (a +b \right )}}\, \sqrt {\frac {b +a \cos \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \left (-1+\cos \left (f x +e \right )\right ) \left (\frac {g}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \left (\cos ^{3}\left (f x +e \right )\right )}{c f \left (b +a \cos \left (f x +e \right )\right ) \left (\frac {1}{1+\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sin \left (f x +e \right )^{2} \left (a -b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I/c/f*(4*a*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),(-(a-b)/(a+b))^(1/2))-2*b*EllipticF(I*(-1+cos(f*x+e))/sin(f*
x+e),(-(a-b)/(a+b))^(1/2))-a*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),(-(a-b)/(a+b))^(1/2))-b*EllipticE(I*(-1+co
s(f*x+e))/sin(f*x+e),(-(a-b)/(a+b))^(1/2))-4*EllipticPi(I*(-1+cos(f*x+e))/sin(f*x+e),-1,I*((a-b)/(a+b))^(1/2))
*a+4*EllipticPi(I*(-1+cos(f*x+e))/sin(f*x+e),-1,I*((a-b)/(a+b))^(1/2))*b)*((b+a*cos(f*x+e))/(1+cos(f*x+e))/(a+
b))^(1/2)*((b+a*cos(f*x+e))/cos(f*x+e))^(1/2)*(-1+cos(f*x+e))*(g/cos(f*x+e))^(5/2)*cos(f*x+e)^3/(b+a*cos(f*x+e
))/(1/(1+cos(f*x+e)))^(3/2)/sin(f*x+e)^2/(a-b)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (\frac {g}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{\sqrt {a+\frac {b}{\cos \left (e+f\,x\right )}}\,\left (c+\frac {c}{\cos \left (e+f\,x\right )}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*sec(f*x+e))**(5/2)/(c+c*sec(f*x+e))/(a+b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

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