3.743 \(\int (a+b \cos (c+d x))^{5/2} \sec ^{\frac {11}{2}}(c+d x) \, dx\)
Optimal. Leaf size=494 \[ \frac {2 \left (49 a^2+75 b^2\right ) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{315 d}+\frac {2 b \left (163 a^2+5 b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{315 a d}+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{9 d}-\frac {2 (a-b) \sqrt {a+b} \left (147 a^3-114 a^2 b+165 a b^2+10 b^3\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{315 a^2 d \sqrt {\sec (c+d x)}}+\frac {2 (a-b) \sqrt {a+b} \left (147 a^4+279 a^2 b^2-10 b^4\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{315 a^3 d \sqrt {\sec (c+d x)}}+\frac {38 a b \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{63 d} \]
[Out]
2/315*b*(163*a^2+5*b^2)*sec(d*x+c)^(3/2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/a/d+2/315*(49*a^2+75*b^2)*sec(d*x+c
)^(5/2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d+38/63*a*b*sec(d*x+c)^(7/2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d+2/9
*a^2*sec(d*x+c)^(9/2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d+2/315*(a-b)*(147*a^4+279*a^2*b^2-10*b^4)*csc(d*x+c)*
EllipticE((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/
2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^3/d/sec(d*x+c)^(1/2)-2/315*(a-b)*(147*a^3-1
14*a^2*b+165*a*b^2+10*b^3)*csc(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a
-b))^(1/2))*(a+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/d/s
ec(d*x+c)^(1/2)
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Rubi [A] time = 1.52, antiderivative size = 494, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used =
{4222, 2792, 3055, 2998, 2816, 2994} \[ \frac {2 \left (49 a^2+75 b^2\right ) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{315 d}+\frac {2 b \left (163 a^2+5 b^2\right ) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{315 a d}-\frac {2 (a-b) \sqrt {a+b} \left (-114 a^2 b+147 a^3+165 a b^2+10 b^3\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{315 a^2 d \sqrt {\sec (c+d x)}}+\frac {2 (a-b) \sqrt {a+b} \left (279 a^2 b^2+147 a^4-10 b^4\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{315 a^3 d \sqrt {\sec (c+d x)}}+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {9}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{9 d}+\frac {38 a b \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}{63 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(11/2),x]
[Out]
(2*(a - b)*Sqrt[a + b]*(147*a^4 + 279*a^2*b^2 - 10*b^4)*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticE[ArcSin[Sqrt[
a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)
]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(315*a^3*d*Sqrt[Sec[c + d*x]]) - (2*(a - b)*Sqrt[a + b]*(147*a^3 - 114
*a^2*b + 165*a*b^2 + 10*b^3)*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a
+ b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]
))/(a - b)])/(315*a^2*d*Sqrt[Sec[c + d*x]]) + (2*b*(163*a^2 + 5*b^2)*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^(3/
2)*Sin[c + d*x])/(315*a*d) + (2*(49*a^2 + 75*b^2)*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(3
15*d) + (38*a*b*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(63*d) + (2*a^2*Sqrt[a + b*Cos[c + d
*x]]*Sec[c + d*x]^(9/2)*Sin[c + d*x])/(9*d)
Rule 2792
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], 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[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Rule 2816
Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*
Tan[e + f*x]*Rt[(a + b)/d, 2]*Sqrt[(a*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*Ellipt
icF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/(Sqrt[d*Sin[e + f*x]]*Rt[(a + b)/d, 2])], -((a + b)/(a - b))])/(a*f), x] /
; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]
Rule 2994
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*A*(c - d)*Tan[e + f*x]*Rt[(c + d)/b, 2]*Sqrt[(c*(1 + Csc[e + f*x]))/(c
- d)]*Sqrt[(c*(1 - Csc[e + f*x]))/(c + d)]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/(Sqrt[b*Sin[e + f*x]]*Rt[
(c + d)/b, 2])], -((c + d)/(c - d))])/(f*b*c^2), x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] &&
EqQ[A, B] && PosQ[(c + d)/b]
Rule 2998
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*s
in[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A - B)/(a - b), Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e
+ f*x]]), x], x] - Dist[(A*b - a*B)/(a - b), Int[(1 + Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin
[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && NeQ[A, B]
Rule 3055
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))
Rule 4222
Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u,
x]
Rubi steps
\begin {align*} \int (a+b \cos (c+d x))^{5/2} \sec ^{\frac {11}{2}}(c+d x) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\\ &=\frac {2 a^2 \sqrt {a+b \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {1}{9} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {19 a^2 b}{2}+\frac {1}{2} a \left (7 a^2+27 b^2\right ) \cos (c+d x)+\frac {3}{2} b \left (2 a^2+3 b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {38 a b \sqrt {a+b \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 a^2 \sqrt {a+b \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} a^2 \left (49 a^2+75 b^2\right )+\frac {1}{4} a b \left (137 a^2+63 b^2\right ) \cos (c+d x)+19 a^2 b^2 \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{63 a}\\ &=\frac {2 \left (49 a^2+75 b^2\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {38 a b \sqrt {a+b \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 a^2 \sqrt {a+b \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{8} a^2 b \left (163 a^2+5 b^2\right )+\frac {1}{8} a^3 \left (147 a^2+605 b^2\right ) \cos (c+d x)+\frac {1}{4} a^2 b \left (49 a^2+75 b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{315 a^2}\\ &=\frac {2 b \left (163 a^2+5 b^2\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 a d}+\frac {2 \left (49 a^2+75 b^2\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {38 a b \sqrt {a+b \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 a^2 \sqrt {a+b \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac {\left (16 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{16} a^2 \left (147 a^4+279 a^2 b^2-10 b^4\right )+\frac {3}{16} a^3 b \left (261 a^2+155 b^2\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{945 a^3}\\ &=\frac {2 b \left (163 a^2+5 b^2\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 a d}+\frac {2 \left (49 a^2+75 b^2\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {38 a b \sqrt {a+b \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 a^2 \sqrt {a+b \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}-\frac {\left ((a-b) \left (147 a^3-114 a^2 b+165 a b^2+10 b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{315 a}+\frac {\left (\left (147 a^4+279 a^2 b^2-10 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1+\cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{315 a}\\ &=\frac {2 (a-b) \sqrt {a+b} \left (147 a^4+279 a^2 b^2-10 b^4\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{315 a^3 d \sqrt {\sec (c+d x)}}-\frac {2 (a-b) \sqrt {a+b} \left (147 a^3-114 a^2 b+165 a b^2+10 b^3\right ) \sqrt {\cos (c+d x)} \csc (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{315 a^2 d \sqrt {\sec (c+d x)}}+\frac {2 b \left (163 a^2+5 b^2\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{315 a d}+\frac {2 \left (49 a^2+75 b^2\right ) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{315 d}+\frac {38 a b \sqrt {a+b \cos (c+d x)} \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 a^2 \sqrt {a+b \cos (c+d x)} \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{9 d}\\ \end {align*}
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Mathematica [A] time = 16.29, size = 521, normalized size = 1.05 \[ \frac {\sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)} \left (\frac {2 \sec (c+d x) \left (163 a^2 b \sin (c+d x)+5 b^3 \sin (c+d x)\right )}{315 a}+\frac {2}{315} \sec ^2(c+d x) \left (49 a^2 \sin (c+d x)+75 b^2 \sin (c+d x)\right )+\frac {2}{9} a^2 \tan (c+d x) \sec ^3(c+d x)+\frac {2 \left (147 a^4+279 a^2 b^2-10 b^4\right ) \sin (c+d x)}{315 a^2}+\frac {38}{63} a b \tan (c+d x) \sec ^2(c+d x)\right )}{d}+\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (-\left (\left (147 a^4+279 a^2 b^2-10 b^4\right ) \cos (c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \cos (c+d x))\right )+2 a \left (147 a^4+261 a^3 b+279 a^2 b^2+155 a b^3-10 b^4\right ) \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (\cos (c+d x)+1)}} F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )-2 \left (147 a^5+147 a^4 b+279 a^3 b^2+279 a^2 b^3-10 a b^4-10 b^5\right ) \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{315 a^2 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {a+b \cos (c+d x)}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + b*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(11/2),x]
[Out]
(2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(-2*(147*a^5 + 147*a^4*b + 279*a^3*b^2 + 279*a^2*b^3 - 10*a*b^4 - 10*
b^5)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[A
rcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + 2*a*(147*a^4 + 261*a^3*b + 279*a^2*b^2 + 155*a*b^3 - 10*b^4)*Sqrt
[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan
[(c + d*x)/2]], (-a + b)/(a + b)] - (147*a^4 + 279*a^2*b^2 - 10*b^4)*Cos[c + d*x]*(a + b*Cos[c + d*x])*Sec[(c
+ d*x)/2]^2*Tan[(c + d*x)/2]))/(315*a^2*d*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) + (Sqrt[a + b*Cos
[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(147*a^4 + 279*a^2*b^2 - 10*b^4)*Sin[c + d*x])/(315*a^2) + (2*Sec[c + d*x]^2
*(49*a^2*Sin[c + d*x] + 75*b^2*Sin[c + d*x]))/315 + (2*Sec[c + d*x]*(163*a^2*b*Sin[c + d*x] + 5*b^3*Sin[c + d*
x]))/(315*a) + (38*a*b*Sec[c + d*x]^2*Tan[c + d*x])/63 + (2*a^2*Sec[c + d*x]^3*Tan[c + d*x])/9))/d
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fricas [F] time = 1.26, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {11}{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(5/2)*sec(d*x+c)^(11/2),x, algorithm="fricas")
[Out]
integral((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*sqrt(b*cos(d*x + c) + a)*sec(d*x + c)^(11/2), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{\frac {11}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(5/2)*sec(d*x+c)^(11/2),x, algorithm="giac")
[Out]
integrate((b*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^(11/2), x)
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maple [B] time = 0.37, size = 2512, normalized size = 5.09 \[ \text {Expression too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*cos(d*x+c))^(5/2)*sec(d*x+c)^(11/2),x)
[Out]
-2/315/d*(-80*cos(d*x+c)^3*a^2*b^3+147*cos(d*x+c)^5*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x
+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^5+65*cos(d*x+c)^
5*a^4*b+279*cos(d*x+c)^5*a^3*b^2-199*cos(d*x+c)^5*a^2*b^3-10*cos(d*x+c)^5*a*b^4-272*cos(d*x+c)^4*a^3*b^2+5*cos
(d*x+c)^4*a*b^4-82*cos(d*x+c)^3*a^4*b-170*cos(d*x+c)^2*a^3*b^2-130*cos(d*x+c)*a^4*b+147*cos(d*x+c)^6*a^4*b+163
*cos(d*x+c)^6*a^3*b^2+279*cos(d*x+c)^6*a^2*b^3+5*cos(d*x+c)^6*a*b^4-35*a^5-10*cos(d*x+c)^6*b^5+147*cos(d*x+c)^
5*a^5+10*cos(d*x+c)^5*b^5-98*cos(d*x+c)^4*a^5-14*cos(d*x+c)^2*a^5-10*cos(d*x+c)^5*sin(d*x+c)*(cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a
+b))^(1/2))*a*b^4-147*cos(d*x+c)^5*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c
))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^4*b-279*cos(d*x+c)^5*sin(d*x+c)*(
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d
*x+c),(-(a-b)/(a+b))^(1/2))*a^3*b^2-279*cos(d*x+c)^5*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*
x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b^3+10*cos(d*
x+c)^5*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-
1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^4+261*cos(d*x+c)^4*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(
1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*
a^4*b+279*cos(d*x+c)^4*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1
/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3*b^2+155*cos(d*x+c)^4*sin(d*x+c)*(cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a
-b)/(a+b))^(1/2))*a^2*b^3-10*cos(d*x+c)^4*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+co
s(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^4-147*cos(d*x+c)^4*sin(d
*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c)
)/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^4*b-279*cos(d*x+c)^4*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*c
os(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3*b^2-279*
cos(d*x+c)^4*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipt
icE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b^3+10*cos(d*x+c)^4*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^
(1/2))*a*b^4+261*cos(d*x+c)^5*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a
+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^4*b+279*cos(d*x+c)^5*sin(d*x+c)*(cos(d
*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c)
,(-(a-b)/(a+b))^(1/2))*a^3*b^2+155*cos(d*x+c)^5*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))
/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b^3-147*cos(d*x+c)
^5*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+co
s(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^5+10*cos(d*x+c)^5*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((
a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*b^5+147
*cos(d*x+c)^4*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellip
ticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^5-147*cos(d*x+c)^4*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1
/2))*a^5+10*cos(d*x+c)^4*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^
(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*b^5)*cos(d*x+c)/(a+b*cos(d*x+c))^(1/2)*(1/cos
(d*x+c))^(11/2)/sin(d*x+c)/a^2
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{\frac {11}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(5/2)*sec(d*x+c)^(11/2),x, algorithm="maxima")
[Out]
integrate((b*cos(d*x + c) + a)^(5/2)*sec(d*x + c)^(11/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{11/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((1/cos(c + d*x))^(11/2)*(a + b*cos(c + d*x))^(5/2),x)
[Out]
int((1/cos(c + d*x))^(11/2)*(a + b*cos(c + d*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((a+b*cos(d*x+c))**(5/2)*sec(d*x+c)**(11/2),x)
[Out]
Timed out
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