3.621 \(\int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=392 \[ \frac {2 \sqrt {a+b} \left (a^2-7 a b+9 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 d}+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2 b^2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{d}+\frac {14 b (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 d} \]
[Out]
2/3*a^2*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(3/2)+14/3*(a-b)*b*cot(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)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+se
c(d*x+c))/(a-b))^(1/2)/d+2/3*(a^2-7*a*b+9*b^2)*cot(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)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/d-2*
b^2*cot(d*x+c)*EllipticPi((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(a+b)/b,((-a-b)/(a-b))^(1/2))*(a
+b)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/d
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Rubi [A] time = 0.74, antiderivative size = 392, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used =
{2792, 3053, 2809, 2998, 2816, 2994} \[ \frac {2 \sqrt {a+b} \left (a^2-7 a b+9 b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 d}+\frac {2 a^2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2 b^2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{d}+\frac {14 b (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Cos[c + d*x])^(5/2)/Cos[c + d*x]^(5/2),x]
[Out]
(14*(a - b)*b*Sqrt[a + b]*Cot[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)])/(3*d) + (
2*Sqrt[a + b]*(a^2 - 7*a*b + 9*b^2)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[C
os[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)])
/(3*d) - (2*b^2*Sqrt[a + b]*Cot[c + d*x]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sq
rt[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)])/d + (2*a^2*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))
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 2809
Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(2*b*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)]*EllipticP
i[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/(Sqrt[b*Sin[e + f*x]]*Rt[(c + d)/b, 2])], -((c + d)/(c - d))])/(d
*f), x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/b]
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 3053
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_.) + (b_.)*sin[(e_.) + (f_.
)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[C/b^2, Int[Sqrt[a + b*Sin[e + f
*x]]/Sqrt[c + d*Sin[e + f*x]], x], x] + Dist[1/b^2, Int[(A*b^2 - a^2*C + b*(b*B - 2*a*C)*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, C}, x] && NeQ[b*c - a
*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rubi steps
\begin {align*} \int \frac {(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx &=\frac {2 a^2 \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {\frac {7 a^2 b}{2}+\frac {1}{2} a \left (a^2+9 b^2\right ) \cos (c+d x)+\frac {3}{2} b^3 \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {2 a^2 \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {\frac {7 a^2 b}{2}+\frac {1}{2} a \left (a^2+9 b^2\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx+b^3 \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx\\ &=-\frac {2 b^2 \sqrt {a+b} \cot (c+d x) \Pi \left (\frac {a+b}{b};\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}}}{d}+\frac {2 a^2 \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \left (7 a^2 b\right ) \int \frac {1+\cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx+\frac {1}{3} \left (a \left (a^2-7 a b+9 b^2\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {14 (a-b) b \sqrt {a+b} \cot (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}}}{3 d}+\frac {2 \sqrt {a+b} \left (a^2-7 a b+9 b^2\right ) \cot (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}}}{3 d}-\frac {2 b^2 \sqrt {a+b} \cot (c+d x) \Pi \left (\frac {a+b}{b};\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}}}{d}+\frac {2 a^2 \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\\ \end {align*}
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Mathematica [A] time = 7.04, size = 328, normalized size = 0.84 \[ \frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )} \left (\left (a^3+7 a^2 b+9 a b^2-3 b^3\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \cos (c+d x))}{a+b}} F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )+6 b^3 \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \cos (c+d x))}{a+b}} \Pi \left (-1;\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )-7 a b \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \cos (c+d x))-7 a b (a+b) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \cos (c+d x))}{a+b}} E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )+\frac {2 a \sin (c+d x) (a+b \cos (c+d x)) (a+7 b \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)}}{3 d \sqrt {a+b \cos (c+d x)}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + b*Cos[c + d*x])^(5/2)/Cos[c + d*x]^(5/2),x]
[Out]
((2*a*(a + b*Cos[c + d*x])*(a + 7*b*Cos[c + d*x])*Sin[c + d*x])/Cos[c + d*x]^(3/2) + 2*Sqrt[Cos[(c + d*x)/2]^2
]*(-7*a*b*(a + b)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[((a + b*Cos[c + d*x])*Sec[(c + d*
x)/2]^2)/(a + b)] + (a^3 + 7*a^2*b + 9*a*b^2 - 3*b^3)*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sq
rt[((a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + 6*b^3*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b
)/(a + b)]*Sqrt[((a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - 7*a*b*(a + b*Cos[c + d*x])*Sqrt[Cos[c + d
*x]*Sec[(c + d*x)/2]^2]*Tan[(c + d*x)/2]))/(3*d*Sqrt[a + b*Cos[c + d*x]])
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fricas [F] time = 61.29, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{\cos \left (d x + c\right )^{\frac {5}{2}}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(5/2),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)/cos(d*x + c)^(5/2), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(5/2),x, algorithm="giac")
[Out]
integrate((b*cos(d*x + c) + a)^(5/2)/cos(d*x + c)^(5/2), x)
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maple [B] time = 0.28, size = 1485, normalized size = 3.79 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(5/2),x)
[Out]
-2/3/d*(cos(d*x+c)^2*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+7*cos(d*x+c)^2*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+9*cos(d*x+c)^2*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*b^2-3*(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))*co
s(d*x+c)^2*sin(d*x+c)*b^3+6*(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)*El
lipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*b^3-7*cos(d*x+c)^2*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-7*cos(d*x+c)^2*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^2+cos(d*x+c
)*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+7*cos(d*x+c)*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+9*cos
(d*x+c)*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*b^2-3*cos(d*x+c)*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))*b^3
+6*cos(d*x+c)*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
ticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*b^3-7*cos(d*x+c)*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-7*cos(d*x+c)*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^2+cos(d*x+c)^3*a^2*b+7*cos(d*x+c)^3*a*b^2+
cos(d*x+c)^2*a^3+7*cos(d*x+c)^2*a^2*b-7*cos(d*x+c)^2*a*b^2-8*cos(d*x+c)*a^2*b-a^3)/(a+b*cos(d*x+c))^(1/2)/sin(
d*x+c)/cos(d*x+c)^(3/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(5/2),x, algorithm="maxima")
[Out]
integrate((b*cos(d*x + c) + a)^(5/2)/cos(d*x + c)^(5/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*cos(c + d*x))^(5/2)/cos(c + d*x)^(5/2),x)
[Out]
int((a + b*cos(c + d*x))^(5/2)/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)/cos(d*x+c)**(5/2),x)
[Out]
Timed out
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