3.452 \(\int \frac {\sec ^{\frac {5}{2}}(d+e x)}{(a+b \sec (d+e x)+c \tan (d+e x))^{5/2}} \, dx\)
Optimal. Leaf size=492 \[ \frac {2 \sec ^{\frac {5}{2}}(d+e x) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} (a \cos (d+e x)+b+c \sin (d+e x))^2 F\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{3 e \left (a^2-b^2+c^2\right ) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 b \sec ^{\frac {5}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^3 E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{3 e \left (a^2-b^2+c^2\right )^2 \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 \sec ^{\frac {5}{2}}(d+e x) (b c \cos (d+e x)-a b \sin (d+e x)) (a \cos (d+e x)+b+c \sin (d+e x))^2}{3 e \left (a^2-b^2+c^2\right )^2 (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}-\frac {2 \sec ^{\frac {5}{2}}(d+e x) (c \cos (d+e x)-a \sin (d+e x)) (a \cos (d+e x)+b+c \sin (d+e x))}{3 e \left (a^2-b^2+c^2\right ) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}} \]
[Out]
-2/3*sec(e*x+d)^(5/2)*(c*cos(e*x+d)-a*sin(e*x+d))*(b+a*cos(e*x+d)+c*sin(e*x+d))/(a^2-b^2+c^2)/e/(a+b*sec(e*x+d
)+c*tan(e*x+d))^(5/2)+8/3*sec(e*x+d)^(5/2)*(b*c*cos(e*x+d)-a*b*sin(e*x+d))*(b+a*cos(e*x+d)+c*sin(e*x+d))^2/(a^
2-b^2+c^2)^2/e/(a+b*sec(e*x+d)+c*tan(e*x+d))^(5/2)+8/3*b*(cos(1/2*d+1/2*e*x-1/2*arctan(a,c))^2)^(1/2)/cos(1/2*
d+1/2*e*x-1/2*arctan(a,c))*EllipticE(sin(1/2*d+1/2*e*x-1/2*arctan(a,c)),2^(1/2)*((a^2+c^2)^(1/2)/(b+(a^2+c^2)^
(1/2)))^(1/2))*sec(e*x+d)^(5/2)*(b+a*cos(e*x+d)+c*sin(e*x+d))^3/(a^2-b^2+c^2)^2/e/((b+a*cos(e*x+d)+c*sin(e*x+d
))/(b+(a^2+c^2)^(1/2)))^(1/2)/(a+b*sec(e*x+d)+c*tan(e*x+d))^(5/2)+2/3*(cos(1/2*d+1/2*e*x-1/2*arctan(a,c))^2)^(
1/2)/cos(1/2*d+1/2*e*x-1/2*arctan(a,c))*EllipticF(sin(1/2*d+1/2*e*x-1/2*arctan(a,c)),2^(1/2)*((a^2+c^2)^(1/2)/
(b+(a^2+c^2)^(1/2)))^(1/2))*sec(e*x+d)^(5/2)*(b+a*cos(e*x+d)+c*sin(e*x+d))^2*((b+a*cos(e*x+d)+c*sin(e*x+d))/(b
+(a^2+c^2)^(1/2)))^(1/2)/(a^2-b^2+c^2)/e/(a+b*sec(e*x+d)+c*tan(e*x+d))^(5/2)
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Rubi [A] time = 0.52, antiderivative size = 492, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used =
{3167, 3129, 3156, 3149, 3119, 2653, 3127, 2661} \[ \frac {2 \sec ^{\frac {5}{2}}(d+e x) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} (a \cos (d+e x)+b+c \sin (d+e x))^2 F\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{3 e \left (a^2-b^2+c^2\right ) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 b \sec ^{\frac {5}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^3 E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{3 e \left (a^2-b^2+c^2\right )^2 \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 \sec ^{\frac {5}{2}}(d+e x) (b c \cos (d+e x)-a b \sin (d+e x)) (a \cos (d+e x)+b+c \sin (d+e x))^2}{3 e \left (a^2-b^2+c^2\right )^2 (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}-\frac {2 \sec ^{\frac {5}{2}}(d+e x) (c \cos (d+e x)-a \sin (d+e x)) (a \cos (d+e x)+b+c \sin (d+e x))}{3 e \left (a^2-b^2+c^2\right ) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[Sec[d + e*x]^(5/2)/(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(5/2),x]
[Out]
(-2*Sec[d + e*x]^(5/2)*(c*Cos[d + e*x] - a*Sin[d + e*x])*(b + a*Cos[d + e*x] + c*Sin[d + e*x]))/(3*(a^2 - b^2
+ c^2)*e*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(5/2)) + (8*Sec[d + e*x]^(5/2)*(b*c*Cos[d + e*x] - a*b*Sin[d +
e*x])*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^2)/(3*(a^2 - b^2 + c^2)^2*e*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^
(5/2)) + (8*b*EllipticE[(d + e*x - ArcTan[a, c])/2, (2*Sqrt[a^2 + c^2])/(b + Sqrt[a^2 + c^2])]*Sec[d + e*x]^(5
/2)*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^3)/(3*(a^2 - b^2 + c^2)^2*e*Sqrt[(b + a*Cos[d + e*x] + c*Sin[d + e*x
])/(b + Sqrt[a^2 + c^2])]*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(5/2)) + (2*EllipticF[(d + e*x - ArcTan[a, c])
/2, (2*Sqrt[a^2 + c^2])/(b + Sqrt[a^2 + c^2])]*Sec[d + e*x]^(5/2)*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^2*Sqrt
[(b + a*Cos[d + e*x] + c*Sin[d + e*x])/(b + Sqrt[a^2 + c^2])])/(3*(a^2 - b^2 + c^2)*e*(a + b*Sec[d + e*x] + c*
Tan[d + e*x])^(5/2))
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 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 3119
Int[Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*C
os[d + e*x] + c*Sin[d + e*x]]/Sqrt[(a + b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])], Int[Sqrt[a/(a
+ Sqrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]])/(a + Sqrt[b^2 + c^2])], x], x] /; FreeQ[{a
, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[b^2 + c^2, 0] && !GtQ[a + Sqrt[b^2 + c^2], 0]
Rule 3127
Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a +
b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])]/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]], Int[1/Sqrt[
a/(a + Sqrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]])/(a + Sqrt[b^2 + c^2])], x], x] /; Free
Q[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[b^2 + c^2, 0] && !GtQ[a + Sqrt[b^2 + c^2], 0]
Rule 3129
Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_), x_Symbol] :> Simp[((-(c*Cos[d
+ e*x]) + b*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1))/(e*(n + 1)*(a^2 - b^2 - c^2)), x] +
Dist[1/((n + 1)*(a^2 - b^2 - c^2)), Int[(a*(n + 1) - b*(n + 2)*Cos[d + e*x] - c*(n + 2)*Sin[d + e*x])*(a + b*C
os[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && LtQ[n
, -1] && NeQ[n, -3/2]
Rule 3149
Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.)
+ (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)]], x_Symbol] :> Dist[B/b, Int[Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]
, x], x] + Dist[(A*b - a*B)/b, Int[1/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]], x], x] /; FreeQ[{a, b, c, d, e
, A, B, C}, x] && EqQ[B*c - b*C, 0] && NeQ[A*b - a*B, 0]
Rule 3156
Int[((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_)*((A_.) + cos[(d_.) + (e_.)*(x
_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> -Simp[((c*B - b*C - (a*C - c*A)*Cos[d + e*x] + (a*B -
b*A)*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1))/(e*(n + 1)*(a^2 - b^2 - c^2)), x] + Dist[1/
((n + 1)*(a^2 - b^2 - c^2)), Int[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1)*Simp[(n + 1)*(a*A - b*B - c*C)
+ (n + 2)*(a*B - b*A)*Cos[d + e*x] + (n + 2)*(a*C - c*A)*Sin[d + e*x], x], x], x] /; FreeQ[{a, b, c, d, e, A,
B, C}, x] && LtQ[n, -1] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[n, -2]
Rule 3167
Int[sec[(d_.) + (e_.)*(x_)]^(n_.)*((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(m_)
, x_Symbol] :> Dist[(Sec[d + e*x]^n*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n)/(a + b*Sec[d + e*x] + c*Tan[d + e
*x])^n, Int[1/(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[m + n, 0] &&
!IntegerQ[n]
Rubi steps
\begin {align*} \int \frac {\sec ^{\frac {5}{2}}(d+e x)}{(a+b \sec (d+e x)+c \tan (d+e x))^{5/2}} \, dx &=\frac {\left (\sec ^{\frac {5}{2}}(d+e x) (b+a \cos (d+e x)+c \sin (d+e x))^{5/2}\right ) \int \frac {1}{(b+a \cos (d+e x)+c \sin (d+e x))^{5/2}} \, dx}{(a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}\\ &=-\frac {2 \sec ^{\frac {5}{2}}(d+e x) (c \cos (d+e x)-a \sin (d+e x)) (b+a \cos (d+e x)+c \sin (d+e x))}{3 \left (a^2-b^2+c^2\right ) e (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {\left (2 \sec ^{\frac {5}{2}}(d+e x) (b+a \cos (d+e x)+c \sin (d+e x))^{5/2}\right ) \int \frac {-\frac {3 b}{2}+\frac {1}{2} a \cos (d+e x)+\frac {1}{2} c \sin (d+e x)}{(b+a \cos (d+e x)+c \sin (d+e x))^{3/2}} \, dx}{3 \left (a^2-b^2+c^2\right ) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}\\ &=-\frac {2 \sec ^{\frac {5}{2}}(d+e x) (c \cos (d+e x)-a \sin (d+e x)) (b+a \cos (d+e x)+c \sin (d+e x))}{3 \left (a^2-b^2+c^2\right ) e (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 \sec ^{\frac {5}{2}}(d+e x) (b c \cos (d+e x)-a b \sin (d+e x)) (b+a \cos (d+e x)+c \sin (d+e x))^2}{3 \left (a^2-b^2+c^2\right )^2 e (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {\left (4 \sec ^{\frac {5}{2}}(d+e x) (b+a \cos (d+e x)+c \sin (d+e x))^{5/2}\right ) \int \frac {\frac {1}{4} \left (a^2+3 b^2+c^2\right )+a b \cos (d+e x)+b c \sin (d+e x)}{\sqrt {b+a \cos (d+e x)+c \sin (d+e x)}} \, dx}{3 \left (a^2-b^2+c^2\right )^2 (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}\\ &=-\frac {2 \sec ^{\frac {5}{2}}(d+e x) (c \cos (d+e x)-a \sin (d+e x)) (b+a \cos (d+e x)+c \sin (d+e x))}{3 \left (a^2-b^2+c^2\right ) e (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 \sec ^{\frac {5}{2}}(d+e x) (b c \cos (d+e x)-a b \sin (d+e x)) (b+a \cos (d+e x)+c \sin (d+e x))^2}{3 \left (a^2-b^2+c^2\right )^2 e (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {\left (4 b \sec ^{\frac {5}{2}}(d+e x) (b+a \cos (d+e x)+c \sin (d+e x))^{5/2}\right ) \int \sqrt {b+a \cos (d+e x)+c \sin (d+e x)} \, dx}{3 \left (a^2-b^2+c^2\right )^2 (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {\left (\sec ^{\frac {5}{2}}(d+e x) (b+a \cos (d+e x)+c \sin (d+e x))^{5/2}\right ) \int \frac {1}{\sqrt {b+a \cos (d+e x)+c \sin (d+e x)}} \, dx}{3 \left (a^2-b^2+c^2\right ) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}\\ &=-\frac {2 \sec ^{\frac {5}{2}}(d+e x) (c \cos (d+e x)-a \sin (d+e x)) (b+a \cos (d+e x)+c \sin (d+e x))}{3 \left (a^2-b^2+c^2\right ) e (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 \sec ^{\frac {5}{2}}(d+e x) (b c \cos (d+e x)-a b \sin (d+e x)) (b+a \cos (d+e x)+c \sin (d+e x))^2}{3 \left (a^2-b^2+c^2\right )^2 e (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {\left (4 b \sec ^{\frac {5}{2}}(d+e x) (b+a \cos (d+e x)+c \sin (d+e x))^3\right ) \int \sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \cos \left (d+e x-\tan ^{-1}(a,c)\right )}{b+\sqrt {a^2+c^2}}} \, dx}{3 \left (a^2-b^2+c^2\right )^2 \sqrt {\frac {b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt {a^2+c^2}}} (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {\left (\sec ^{\frac {5}{2}}(d+e x) (b+a \cos (d+e x)+c \sin (d+e x))^2 \sqrt {\frac {b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt {a^2+c^2}}}\right ) \int \frac {1}{\sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \cos \left (d+e x-\tan ^{-1}(a,c)\right )}{b+\sqrt {a^2+c^2}}}} \, dx}{3 \left (a^2-b^2+c^2\right ) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}\\ &=-\frac {2 \sec ^{\frac {5}{2}}(d+e x) (c \cos (d+e x)-a \sin (d+e x)) (b+a \cos (d+e x)+c \sin (d+e x))}{3 \left (a^2-b^2+c^2\right ) e (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 \sec ^{\frac {5}{2}}(d+e x) (b c \cos (d+e x)-a b \sin (d+e x)) (b+a \cos (d+e x)+c \sin (d+e x))^2}{3 \left (a^2-b^2+c^2\right )^2 e (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 b E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right ) \sec ^{\frac {5}{2}}(d+e x) (b+a \cos (d+e x)+c \sin (d+e x))^3}{3 \left (a^2-b^2+c^2\right )^2 e \sqrt {\frac {b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt {a^2+c^2}}} (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {2 F\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right ) \sec ^{\frac {5}{2}}(d+e x) (b+a \cos (d+e x)+c \sin (d+e x))^2 \sqrt {\frac {b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt {a^2+c^2}}}}{3 \left (a^2-b^2+c^2\right ) e (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}\\ \end {align*}
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Mathematica [C] time = 6.55, size = 2708, normalized size = 5.50 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sec[d + e*x]^(5/2)/(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(5/2),x]
[Out]
(Sec[d + e*x]^(5/2)*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^3*((8*b*(a^2 + c^2))/(3*a*c*(-a^2 + b^2 - c^2)^2) +
(2*(b*c + a^2*Sin[d + e*x] + c^2*Sin[d + e*x]))/(3*a*(a^2 - b^2 + c^2)*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^2
) - (2*(a^2*c + 3*b^2*c + c^3 + 4*a^2*b*Sin[d + e*x] + 4*b*c^2*Sin[d + e*x]))/(3*a*(a^2 - b^2 + c^2)^2*(b + a*
Cos[d + e*x] + c*Sin[d + e*x]))))/(e*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(5/2)) + (2*a^2*AppellF1[1/2, 1/2,
1/2, 3/2, -((b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2]*(1 - b/(Sqrt[1 + a^2/c^2]*
c))*c)), -((b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2]*(-1 - b/(Sqrt[1 + a^2/c^2]*
c))*c))]*Sec[d + e*x]^(5/2)*Sec[d + e*x + ArcTan[a/c]]*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^(5/2)*Sqrt[(c*Sqr
t[(a^2 + c^2)/c^2] - c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]])/(b + c*Sqrt[(a^2 + c^2)/c^2])]*Sqrt[b
+ c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]]]*Sqrt[(c*Sqrt[(a^2 + c^2)/c^2] + c*Sqrt[(a^2 + c^2)/c^2]
*Sin[d + e*x + ArcTan[a/c]])/(-b + c*Sqrt[(a^2 + c^2)/c^2])])/(3*Sqrt[1 + a^2/c^2]*c*(a^2 - b^2 + c^2)^2*e*(a
+ b*Sec[d + e*x] + c*Tan[d + e*x])^(5/2)) + (2*b^2*AppellF1[1/2, 1/2, 1/2, 3/2, -((b + Sqrt[1 + a^2/c^2]*c*Sin
[d + e*x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2]*(1 - b/(Sqrt[1 + a^2/c^2]*c))*c)), -((b + Sqrt[1 + a^2/c^2]*c*Sin[
d + e*x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2]*(-1 - b/(Sqrt[1 + a^2/c^2]*c))*c))]*Sec[d + e*x]^(5/2)*Sec[d + e*x
+ ArcTan[a/c]]*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^(5/2)*Sqrt[(c*Sqrt[(a^2 + c^2)/c^2] - c*Sqrt[(a^2 + c^2)/
c^2]*Sin[d + e*x + ArcTan[a/c]])/(b + c*Sqrt[(a^2 + c^2)/c^2])]*Sqrt[b + c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x +
ArcTan[a/c]]]*Sqrt[(c*Sqrt[(a^2 + c^2)/c^2] + c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]])/(-b + c*Sqr
t[(a^2 + c^2)/c^2])])/(Sqrt[1 + a^2/c^2]*c*(a^2 - b^2 + c^2)^2*e*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(5/2))
+ (2*c*AppellF1[1/2, 1/2, 1/2, 3/2, -((b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2]*
(1 - b/(Sqrt[1 + a^2/c^2]*c))*c)), -((b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2]*(
-1 - b/(Sqrt[1 + a^2/c^2]*c))*c))]*Sec[d + e*x]^(5/2)*Sec[d + e*x + ArcTan[a/c]]*(b + a*Cos[d + e*x] + c*Sin[d
+ e*x])^(5/2)*Sqrt[(c*Sqrt[(a^2 + c^2)/c^2] - c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]])/(b + c*Sqrt
[(a^2 + c^2)/c^2])]*Sqrt[b + c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]]]*Sqrt[(c*Sqrt[(a^2 + c^2)/c^2]
+ c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]])/(-b + c*Sqrt[(a^2 + c^2)/c^2])])/(3*Sqrt[1 + a^2/c^2]*(
a^2 - b^2 + c^2)^2*e*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(5/2)) + (4*a^2*b*Sec[d + e*x]^(5/2)*(b + a*Cos[d +
e*x] + c*Sin[d + e*x])^(5/2)*(-((c*AppellF1[-1/2, -1/2, -1/2, 1/2, -((b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - A
rcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]*(1 - b/(a*Sqrt[1 + c^2/a^2])))), -((b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - Ar
cTan[c/a]])/(a*Sqrt[1 + c^2/a^2]*(-1 - b/(a*Sqrt[1 + c^2/a^2]))))]*Sin[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2
/a^2]*Sqrt[(a*Sqrt[(a^2 + c^2)/a^2] - a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]])/(b + a*Sqrt[(a^2 + c
^2)/a^2])]*Sqrt[b + a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]]]*Sqrt[(a*Sqrt[(a^2 + c^2)/a^2] + a*Sqrt
[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]])/(-b + a*Sqrt[(a^2 + c^2)/a^2])])) - ((2*a*(b + a*Sqrt[1 + c^2/a^
2]*Cos[d + e*x - ArcTan[c/a]]))/(a^2 + c^2) - (c*Sin[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]))/Sqrt[b + a
*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]]]))/(3*c*(a^2 - b^2 + c^2)^2*e*(a + b*Sec[d + e*x] + c*Tan[d + e*
x])^(5/2)) + (4*b*c*Sec[d + e*x]^(5/2)*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^(5/2)*(-((c*AppellF1[-1/2, -1/2,
-1/2, 1/2, -((b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]*(1 - b/(a*Sqrt[1 + c^2/
a^2])))), -((b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]*(-1 - b/(a*Sqrt[1 + c^2/
a^2]))))]*Sin[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]*Sqrt[(a*Sqrt[(a^2 + c^2)/a^2] - a*Sqrt[(a^2 + c^2)/
a^2]*Cos[d + e*x - ArcTan[c/a]])/(b + a*Sqrt[(a^2 + c^2)/a^2])]*Sqrt[b + a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x -
ArcTan[c/a]]]*Sqrt[(a*Sqrt[(a^2 + c^2)/a^2] + a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]])/(-b + a*Sqr
t[(a^2 + c^2)/a^2])])) - ((2*a*(b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]]))/(a^2 + c^2) - (c*Sin[d +
e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]))/Sqrt[b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]]]))/(3*(a^2
- b^2 + c^2)^2*e*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(5/2))
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fricas [F] time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a} \sec \left (e x + d\right )^{\frac {5}{2}}}{b^{3} \sec \left (e x + d\right )^{3} + c^{3} \tan \left (e x + d\right )^{3} + 3 \, a b^{2} \sec \left (e x + d\right )^{2} + 3 \, a^{2} b \sec \left (e x + d\right ) + a^{3} + 3 \, {\left (b c^{2} \sec \left (e x + d\right ) + a c^{2}\right )} \tan \left (e x + d\right )^{2} + 3 \, {\left (b^{2} c \sec \left (e x + d\right )^{2} + 2 \, a b c \sec \left (e x + d\right ) + a^{2} c\right )} \tan \left (e x + d\right )}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(e*x+d)^(5/2)/(a+b*sec(e*x+d)+c*tan(e*x+d))^(5/2),x, algorithm="fricas")
[Out]
integral(sqrt(b*sec(e*x + d) + c*tan(e*x + d) + a)*sec(e*x + d)^(5/2)/(b^3*sec(e*x + d)^3 + c^3*tan(e*x + d)^3
+ 3*a*b^2*sec(e*x + d)^2 + 3*a^2*b*sec(e*x + d) + a^3 + 3*(b*c^2*sec(e*x + d) + a*c^2)*tan(e*x + d)^2 + 3*(b^
2*c*sec(e*x + d)^2 + 2*a*b*c*sec(e*x + d) + a^2*c)*tan(e*x + d)), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (e x + d\right )^{\frac {5}{2}}}{{\left (b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(e*x+d)^(5/2)/(a+b*sec(e*x+d)+c*tan(e*x+d))^(5/2),x, algorithm="giac")
[Out]
integrate(sec(e*x + d)^(5/2)/(b*sec(e*x + d) + c*tan(e*x + d) + a)^(5/2), x)
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maple [C] time = 2.55, size = 64693, normalized size = 131.49 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(e*x+d)^(5/2)/(a+b*sec(e*x+d)+c*tan(e*x+d))^(5/2),x)
[Out]
result too large to display
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (e x + d\right )^{\frac {5}{2}}}{{\left (b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(e*x+d)^(5/2)/(a+b*sec(e*x+d)+c*tan(e*x+d))^(5/2),x, algorithm="maxima")
[Out]
integrate(sec(e*x + d)^(5/2)/(b*sec(e*x + d) + c*tan(e*x + d) + a)^(5/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (\frac {1}{\cos \left (d+e\,x\right )}\right )}^{5/2}}{{\left (a+c\,\mathrm {tan}\left (d+e\,x\right )+\frac {b}{\cos \left (d+e\,x\right )}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((1/cos(d + e*x))^(5/2)/(a + c*tan(d + e*x) + b/cos(d + e*x))^(5/2),x)
[Out]
int((1/cos(d + e*x))^(5/2)/(a + c*tan(d + e*x) + b/cos(d + e*x))^(5/2), 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(sec(e*x+d)**(5/2)/(a+b*sec(e*x+d)+c*tan(e*x+d))**(5/2),x)
[Out]
Timed out
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