3.622 \(\int \frac{(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=338 \[ -\frac{2 (a-b) \sqrt{a+b} \left (9 a^2-8 a b+15 b^2\right ) \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{15 a d}+\frac{2 (a-b) \sqrt{a+b} \left (9 a^2+23 b^2\right ) \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{15 a d}+\frac{2 a^2 \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{22 a b \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{15 d \cos ^{\frac{3}{2}}(c+d x)} \]
[Out]
(2*(a - b)*Sqrt[a + b]*(9*a^2 + 23*b^2)*Cot[c + d*x]*EllipticE[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)])/(15*a*d) - (2*(a - b)*Sqrt[a + b]*(9*a^2 - 8*a*b + 15*b^2)*Cot[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)])/(15*a*d) + (2*a^2*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2))
+ (22*a*b*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2))
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Rubi [A] time = 0.764616, antiderivative size = 338, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{2792, 3055, 2998, 2816, 2994} \[ -\frac{2 (a-b) \sqrt{a+b} \left (9 a^2-8 a b+15 b^2\right ) \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{15 a d}+\frac{2 (a-b) \sqrt{a+b} \left (9 a^2+23 b^2\right ) \cot (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{15 a d}+\frac{2 a^2 \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{22 a b \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{15 d \cos ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Cos[c + d*x])^(5/2)/Cos[c + d*x]^(7/2),x]
[Out]
(2*(a - b)*Sqrt[a + b]*(9*a^2 + 23*b^2)*Cot[c + d*x]*EllipticE[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)])/(15*a*d) - (2*(a - b)*Sqrt[a + b]*(9*a^2 - 8*a*b + 15*b^2)*Cot[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)])/(15*a*d) + (2*a^2*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2))
+ (22*a*b*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*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 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 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 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]
Rubi steps
\begin{align*} \int \frac{(a+b \cos (c+d x))^{5/2}}{\cos ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{2 a^2 \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{\frac{11 a^2 b}{2}+\frac{3}{2} a \left (a^2+5 b^2\right ) \cos (c+d x)+\frac{1}{2} b \left (2 a^2+5 b^2\right ) \cos ^2(c+d x)}{\cos ^{\frac{5}{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)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{22 a b \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 \int \frac{\frac{1}{4} a^2 \left (9 a^2+23 b^2\right )+\frac{1}{4} a b \left (17 a^2+15 b^2\right ) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx}{15 a}\\ &=\frac{2 a^2 \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{22 a b \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{1}{15} \left ((a-b) \left (9 a^2-8 a b+15 b^2\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+b \cos (c+d x)}} \, dx+\frac{1}{15} \left (a \left (9 a^2+23 b^2\right )\right ) \int \frac{1+\cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx\\ &=\frac{2 (a-b) \sqrt{a+b} \left (9 a^2+23 b^2\right ) \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}}}{15 a d}-\frac{2 (a-b) \sqrt{a+b} \left (9 a^2-8 a b+15 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}}}{15 a d}+\frac{2 a^2 \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{22 a b \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}\\ \end{align*}
Mathematica [A] time = 11.4606, size = 386, normalized size = 1.14 \[ \frac{\tan (c+d x) (a+b \cos (c+d x)) \left (\left (9 a^2+23 b^2\right ) \cos (2 (c+d x))+15 a^2+22 a b \cos (c+d x)+23 b^2\right )-4 \cos ^2\left (\frac{1}{2} (c+d x)\right )^{5/2} \left (\frac{\cos (c+d x)}{\cos (c+d x)+1}\right )^{3/2} \sqrt{\cos (c+d x)+1} \left (\left (9 a^2+23 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos (c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right )\right )^{3/2} (a+b \cos (c+d x))-\left (17 a^2 b+9 a^3+23 a b^2+15 b^3\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a+b \cos (c+d x))}{a+b}} F\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )+\left (9 a^2 b+9 a^3+23 a b^2+23 b^3\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a+b \cos (c+d x))}{a+b}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )\right )}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + b*Cos[c + d*x])^(5/2)/Cos[c + d*x]^(7/2),x]
[Out]
(-4*(Cos[(c + d*x)/2]^2)^(5/2)*(Cos[c + d*x]/(1 + Cos[c + d*x]))^(3/2)*Sqrt[1 + Cos[c + d*x]]*((9*a^3 + 9*a^2*
b + 23*a*b^2 + 23*b^3)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((a + b*C
os[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - (9*a^3 + 17*a^2*b + 23*a*b^2 + 15*b^3)*EllipticF[ArcSin[Tan[(c + d
*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + (9*a^2
+ 23*b^2)*(a + b*Cos[c + d*x])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sec[c + d*x]*Tan[(c + d*x)/2]) + (a +
b*Cos[c + d*x])*(15*a^2 + 23*b^2 + 22*a*b*Cos[c + d*x] + (9*a^2 + 23*b^2)*Cos[2*(c + d*x)])*Tan[c + d*x])/(15*
d*Cos[c + d*x]^(3/2)*Sqrt[a + b*Cos[c + d*x]])
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Maple [B] time = 0.486, size = 1750, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(7/2),x)
[Out]
2/15/d*(3*a^3-15*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF((
-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)^3*sin(d*x+c)*b^3+9*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*c
os(d*x+c)^3*sin(d*x+c)*a^3+34*a*b^2*cos(d*x+c)^2-9*cos(d*x+c)^4*a^2*b-11*cos(d*x+c)^4*a*b^2-5*cos(d*x+c)^3*a^2
*b-23*cos(d*x+c)^3*a*b^2+14*cos(d*x+c)*a^2*b+23*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1
+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*b^3-9*(
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))/sin
(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*a^3-23*cos(d*x+c)^4*b^3-9*cos(d*x+c)^3*a^3+23*cos(d*x+c)
^3*b^3+6*cos(d*x+c)^2*a^3+9*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*
EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)^3*sin(d*x+c)*a^2*b+23*(cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(
a+b))^(1/2))*cos(d*x+c)^3*sin(d*x+c)*a*b^2-17*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+c
os(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)^3*sin(d*x+c)*a^2*b-23*
(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))/si
n(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)^3*sin(d*x+c)*a*b^2+9*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b
*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin
(d*x+c)*a^2*b+23*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE((
-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*a*b^2-17*(cos(d*x+c)/(1+cos(d*x+c)))^(
1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2)
)*cos(d*x+c)^2*sin(d*x+c)*a^2*b-23*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))
^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*a*b^2+23*(cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-
(a-b)/(a+b))^(1/2))*cos(d*x+c)^3*sin(d*x+c)*b^3-9*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/
(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)^3*sin(d*x+c)*a^3+9
*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))/s
in(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*a^3-15*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b
*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin
(d*x+c)*b^3)/(a+b*cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)^(5/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(7/2),x, algorithm="maxima")
[Out]
integrate((b*cos(d*x + c) + a)^(5/2)/cos(d*x + c)^(7/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\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{7}{2}}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(7/2),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)^(7/2), x)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))**(5/2)/cos(d*x+c)**(7/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^(5/2)/cos(d*x+c)^(7/2),x, algorithm="giac")
[Out]
integrate((b*cos(d*x + c) + a)^(5/2)/cos(d*x + c)^(7/2), x)