3.457 \(\int \frac {1}{\cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}} \, dx\)

Optimal. Leaf size=492 \[ \frac {2 \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} (a \cos (d+e x)+b+c \sin (d+e x))^2 F\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{3 e \left (a^2-b^2+c^2\right ) \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 b (a \cos (d+e x)+b+c \sin (d+e x))^3 E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{3 e \left (a^2-b^2+c^2\right )^2 \cos ^{\frac {5}{2}}(d+e x) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 (b c \cos (d+e x)-a b \sin (d+e x)) (a \cos (d+e x)+b+c \sin (d+e x))^2}{3 e \left (a^2-b^2+c^2\right )^2 \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}-\frac {2 (c \cos (d+e x)-a \sin (d+e x)) (a \cos (d+e x)+b+c \sin (d+e x))}{3 e \left (a^2-b^2+c^2\right ) \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}} \]

[Out]

-2/3*(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/cos(e*x+d)^(5/2)/(a+b*sec(e*x+d
)+c*tan(e*x+d))^(5/2)+8/3*(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/co
s(e*x+d)^(5/2)/(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))*(b+a*cos(e*x+d)+c*sin(e*x+d))^3/(a^2-b^2+c^2)^2/e/cos(e*x+d)^(5/2)/((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))*(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/cos(e*x+d)^(5/2)/(a+b*sec(e*x+d)+c*tan(e*x+d))^(5/2)

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Rubi [A]  time = 0.49, 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 = {3163, 3129, 3156, 3149, 3119, 2653, 3127, 2661} \[ \frac {2 \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} (a \cos (d+e x)+b+c \sin (d+e x))^2 F\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{3 e \left (a^2-b^2+c^2\right ) \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 b (a \cos (d+e x)+b+c \sin (d+e x))^3 E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{3 e \left (a^2-b^2+c^2\right )^2 \cos ^{\frac {5}{2}}(d+e x) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 (b c \cos (d+e x)-a b \sin (d+e x)) (a \cos (d+e x)+b+c \sin (d+e x))^2}{3 e \left (a^2-b^2+c^2\right )^2 \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}-\frac {2 (c \cos (d+e x)-a \sin (d+e x)) (a \cos (d+e x)+b+c \sin (d+e x))}{3 e \left (a^2-b^2+c^2\right ) \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[d + e*x]^(5/2)*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(5/2)),x]

[Out]

(-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*Cos[d + e*
x]^(5/2)*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(5/2)) + (8*(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*Cos[d + e*x]^(5/2)*(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])]*(b + a*Cos[d +
e*x] + c*Sin[d + e*x])^3)/(3*(a^2 - b^2 + c^2)^2*e*Cos[d + e*x]^(5/2)*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])]*(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*Cos[d + e*x]^(5/2)*(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 3163

Int[cos[(d_.) + (e_.)*(x_)]^(n_)*((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(n_),
x_Symbol] :> Dist[(Cos[d + e*x]^n*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^n)/(b + a*Cos[d + e*x] + c*Sin[d + e*
x])^n, Int[(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n, x], x] /; FreeQ[{a, b, c, d, e}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {1}{\cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}} \, dx &=\frac {(b+a \cos (d+e x)+c \sin (d+e x))^{5/2} \int \frac {1}{(b+a \cos (d+e x)+c \sin (d+e x))^{5/2}} \, dx}{\cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}\\ &=-\frac {2 (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 \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {\left (2 (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 ) \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}\\ &=-\frac {2 (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 \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 (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 \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {\left (4 (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 \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}\\ &=-\frac {2 (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 \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 (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 \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {\left (4 b (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 \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {(b+a \cos (d+e x)+c \sin (d+e x))^{5/2} \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 ) \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}\\ &=-\frac {2 (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 \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 (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 \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {\left (4 b (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 \cos ^{\frac {5}{2}}(d+e x) \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 ((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 ) \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}\\ &=-\frac {2 (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 \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}+\frac {8 (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 \cos ^{\frac {5}{2}}(d+e x) (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 ) (b+a \cos (d+e x)+c \sin (d+e x))^3}{3 \left (a^2-b^2+c^2\right )^2 e \cos ^{\frac {5}{2}}(d+e x) \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 ) (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 \cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}}\\ \end {align*}

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Mathematica [F]  time = 27.74, size = 0, normalized size = 0.00 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(d+e x) (a+b \sec (d+e x)+c \tan (d+e x))^{5/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[1/(Cos[d + e*x]^(5/2)*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(5/2)),x]

[Out]

Integrate[1/(Cos[d + e*x]^(5/2)*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(5/2)), x]

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fricas [F]  time = 1.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a} \sqrt {\cos \left (e x + d\right )}}{b^{3} \cos \left (e x + d\right )^{3} \sec \left (e x + d\right )^{3} + c^{3} \cos \left (e x + d\right )^{3} \tan \left (e x + d\right )^{3} + 3 \, a b^{2} \cos \left (e x + d\right )^{3} \sec \left (e x + d\right )^{2} + 3 \, a^{2} b \cos \left (e x + d\right )^{3} \sec \left (e x + d\right ) + a^{3} \cos \left (e x + d\right )^{3} + 3 \, {\left (b c^{2} \cos \left (e x + d\right )^{3} \sec \left (e x + d\right ) + a c^{2} \cos \left (e x + d\right )^{3}\right )} \tan \left (e x + d\right )^{2} + 3 \, {\left (b^{2} c \cos \left (e x + d\right )^{3} \sec \left (e x + d\right )^{2} + 2 \, a b c \cos \left (e x + d\right )^{3} \sec \left (e x + d\right ) + a^{2} c \cos \left (e x + d\right )^{3}\right )} \tan \left (e x + d\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(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)*sqrt(cos(e*x + d))/(b^3*cos(e*x + d)^3*sec(e*x + d)^3 + c^3
*cos(e*x + d)^3*tan(e*x + d)^3 + 3*a*b^2*cos(e*x + d)^3*sec(e*x + d)^2 + 3*a^2*b*cos(e*x + d)^3*sec(e*x + d) +
a^3*cos(e*x + d)^3 + 3*(b*c^2*cos(e*x + d)^3*sec(e*x + d) + a*c^2*cos(e*x + d)^3)*tan(e*x + d)^2 + 3*(b^2*c*c
os(e*x + d)^3*sec(e*x + d)^2 + 2*a*b*c*cos(e*x + d)^3*sec(e*x + d) + a^2*c*cos(e*x + d)^3)*tan(e*x + d)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a\right )}^{\frac {5}{2}} \cos \left (e x + d\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*sec(e*x + d) + c*tan(e*x + d) + a)^(5/2)*cos(e*x + d)^(5/2)), x)

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maple [C]  time = 1.74, size = 64683, normalized size = 131.47 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(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 \[ \int \frac {1}{{\left (b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a\right )}^{\frac {5}{2}} \cos \left (e x + d\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*sec(e*x + d) + c*tan(e*x + d) + a)^(5/2)*cos(e*x + d)^(5/2)), x)

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

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(1/cos(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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