\(\int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx\) [598]

Optimal result

Integrand size = 23, antiderivative size = 244 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=-\frac {\left (a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b \left (a^2-b^2\right )^2 d}+\frac {a \left (a^2-7 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 b^2 \left (a^2-b^2\right )^2 d}-\frac {\left (a^4-10 a^2 b^2-3 b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 (a-b)^2 b^2 (a+b)^3 d}+\frac {a \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\left (a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \]

[Out]

-1/4*(a^2+5*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b/(a^2-
b^2)^2/d+1/4*a*(a^2-7*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2
))/b^2/(a^2-b^2)^2/d-1/4*(a^4-10*a^2*b^2-3*b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin
(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/(a-b)^2/b^2/(a+b)^3/d+1/2*a*sin(d*x+c)*cos(d*x+c)^(1/2)/(a^2-b^2)/d/(a+b*co
s(d*x+c))^2+1/4*(a^2+5*b^2)*sin(d*x+c)*cos(d*x+c)^(1/2)/(a^2-b^2)^2/d/(a+b*cos(d*x+c))

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2878, 3134, 3138, 2719, 3081, 2720, 2884} \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {a \left (a^2-7 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 b^2 d \left (a^2-b^2\right )^2}-\frac {\left (a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b d \left (a^2-b^2\right )^2}+\frac {\left (a^2+5 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac {\left (a^4-10 a^2 b^2-3 b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 b^2 d (a-b)^2 (a+b)^3} \]

[In]

Int[Cos[c + d*x]^(3/2)/(a + b*Cos[c + d*x])^3,x]

[Out]

-1/4*((a^2 + 5*b^2)*EllipticE[(c + d*x)/2, 2])/(b*(a^2 - b^2)^2*d) + (a*(a^2 - 7*b^2)*EllipticF[(c + d*x)/2, 2
])/(4*b^2*(a^2 - b^2)^2*d) - ((a^4 - 10*a^2*b^2 - 3*b^4)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(4*(a - b)
^2*b^2*(a + b)^3*d) + (a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + ((a^2 + 5
*b^2)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2878

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-(b*c - a*d))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n - 1)/(f*(m + 1)*(a^2 - b^
2))), x] + Dist[1/((m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)*Simp[c*
(a*c - b*d)*(m + 1) + d*(b*c - a*d)*(n - 1) + (d*(a*c - b*d)*(m + 1) - c*(b*c - a*d)*(m + 2))*Sin[e + f*x] - d
*(b*c - a*d)*(m + n + 1)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && LtQ[1, n, 2] && IntegersQ[2*m, 2*n]

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], 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[c + d, 0]

Rule 3081

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3134

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] + D
ist[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] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3138

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(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*} \text {integral}& = \frac {a \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\int \frac {-\frac {a}{2}+2 b \cos (c+d x)-\frac {1}{2} a \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )} \\ & = \frac {a \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\left (a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac {\int \frac {-\frac {3}{4} a \left (a^2+b^2\right )+3 a^2 b \cos (c+d x)+\frac {1}{4} a \left (a^2+5 b^2\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a \left (a^2-b^2\right )^2} \\ & = \frac {a \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\left (a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\int \frac {\frac {3}{4} a b \left (a^2+b^2\right )+\frac {1}{4} a^2 \left (a^2-7 b^2\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a b \left (a^2-b^2\right )^2}-\frac {\left (a^2+5 b^2\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 b \left (a^2-b^2\right )^2} \\ & = -\frac {\left (a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b \left (a^2-b^2\right )^2 d}+\frac {a \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\left (a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\left (a \left (a^2-7 b^2\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 b^2 \left (a^2-b^2\right )^2}-\frac {\left (a^4-10 a^2 b^2-3 b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{8 b^2 \left (a^2-b^2\right )^2} \\ & = -\frac {\left (a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b \left (a^2-b^2\right )^2 d}+\frac {a \left (a^2-7 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 b^2 \left (a^2-b^2\right )^2 d}-\frac {\left (a^4-10 a^2 b^2-3 b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 (a-b)^2 b^2 (a+b)^3 d}+\frac {a \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac {\left (a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.35 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\frac {\frac {4 \sqrt {\cos (c+d x)} \left (3 a \left (a^2+b^2\right )+b \left (a^2+5 b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}-\frac {-\frac {2 \left (5 a^2+b^2\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+24 a \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )+\frac {2 \left (a^2+5 b^2\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{16 d} \]

[In]

Integrate[Cos[c + d*x]^(3/2)/(a + b*Cos[c + d*x])^3,x]

[Out]

((4*Sqrt[Cos[c + d*x]]*(3*a*(a^2 + b^2) + b*(a^2 + 5*b^2)*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(a + b*Co
s[c + d*x])^2) - ((-2*(5*a^2 + b^2)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + 24*a*(2*EllipticF[(c
+ d*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)) + (2*(a^2 + 5*b^2)*(-2*a*b*EllipticE[A
rcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*Ellipt
icPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sqrt[Sin[c + d*x]^2]))/((a - b)^2*(a + b)^2
))/(16*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1835\) vs. \(2(308)=616\).

Time = 9.60 (sec) , antiderivative size = 1836, normalized size of antiderivative = 7.52

method	result	size
default	\(\text {Expression too large to display}\)	\(1836\)

[In]

int(cos(d*x+c)^(3/2)/(a+cos(d*x+c)*b)^3,x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4/b/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(
-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1
/2*c),-2*b/(a-b),2^(1/2))+2*a^2/b^2*(-1/2/a*b^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*
d*x+1/2*c)^2)^(1/2)/(2*b*cos(1/2*d*x+1/2*c)^2+a-b)^2-3/4*b^2*(3*a^2-b^2)/a^2/(a^2-b^2)^2*cos(1/2*d*x+1/2*c)*(-
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*b*cos(1/2*d*x+1/2*c)^2+a-b)-7/8/(a+b)/(a^2-b^2)*(sin(1/2
*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/4/(a+b)/(a^2-b^2)/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^
2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b+3/8/(a
+b)/(a^2-b^2)/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(
1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-9/8*b/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/
2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*
x+1/2*c),2^(1/2))+3/8*b^3/a^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*s
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9/8*b/(a^2-b^2)^2*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*
EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3/8*b^3/a^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2
*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-15/4
*a^2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d
*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+3/2/(a^2-b^2)^2/(-2*
a*b+2*b^2)*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2
*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-3/4/a^2/(a^2-b^2)^2/(-2*a*b+2*b^2)*b^5*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^
(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))-4*a/b^2*(-1/a*b^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*si
n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*b*cos(1/2*d*x+1/2*c)^2+a-b)-1/2/a/(a+b)*(sin(1/2*d*x+1/2*c)^
2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(
1/2*d*x+1/2*c),2^(1/2))-1/2/(a^2-b^2)*b/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*s
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2/(a^2-b^2)*b/a*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*
EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+2*b^2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*
d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/
(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+2*b^2)*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)
/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))))/sin(
1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(3/2)/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate(cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^(3/2)/(b*cos(d*x + c) + a)^3, x)

Giac [F]

\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate(cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(3/2)/(b*cos(d*x + c) + a)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^3} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{3/2}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]

[In]

int(cos(c + d*x)^(3/2)/(a + b*cos(c + d*x))^3,x)

[Out]

int(cos(c + d*x)^(3/2)/(a + b*cos(c + d*x))^3, x)