\(\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx\) [894]
Optimal result
Integrand size = 42, antiderivative size = 337 \[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\frac {\left (5 a^2 b B+b^3 B-a^3 C-5 a b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^2 b B+3 b^3 B+a^3 C-7 a b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 b^2 \left (a^2-b^2\right )^2 d}-\frac {\left (3 a^4 b B+10 a^2 b^3 B-b^5 B+a^5 C-10 a^3 b^2 C-3 a b^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a (a-b)^2 b^2 (a+b)^3 d}-\frac {(b B-a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (5 a^2 b B+b^3 B-a^3 C-5 a b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}
\]
[Out]
1/4*(5*B*a^2*b+B*b^3-C*a^3-5*C*a*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))/a/b/(a^2-b^2)^2/d+1/4*(3*B*a^2*b+3*B*b^3+C*a^3-7*C*a*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*(3*B*a^4*b+10*B*a^2*b^3-B*b^5+C*a^5-10*C
*a^3*b^2-3*C*a*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/(a-b)^2/b^2/(a+b)^3/d-1/2*(B*b-C*a)*sin(d*x+c)*cos(d*x+c)^(1/2)/(a^2-b^2)/d/(a+b*cos(d*x+c))^2-1/4*(5
*B*a^2*b+B*b^3-C*a^3-5*C*a*b^2)*sin(d*x+c)*cos(d*x+c)^(1/2)/a/(a^2-b^2)^2/d/(a+b*cos(d*x+c))
Rubi [A] (verified)
Time = 1.12 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.00, number of
steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3108, 3078, 3134, 3138, 2719,
3081, 2720, 2884} \[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=-\frac {(b B-a C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {\left (a^3 C+3 a^2 b B-7 a b^2 C+3 b^3 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 b^2 d \left (a^2-b^2\right )^2}+\frac {\left (a^3 (-C)+5 a^2 b B-5 a b^2 C+b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b d \left (a^2-b^2\right )^2}-\frac {\left (a^3 (-C)+5 a^2 b B-5 a b^2 C+b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 a d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac {\left (a^5 C+3 a^4 b B-10 a^3 b^2 C+10 a^2 b^3 B-3 a b^4 C-b^5 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a b^2 d (a-b)^2 (a+b)^3}
\]
[In]
Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^3),x]
[Out]
((5*a^2*b*B + b^3*B - a^3*C - 5*a*b^2*C)*EllipticE[(c + d*x)/2, 2])/(4*a*b*(a^2 - b^2)^2*d) + ((3*a^2*b*B + 3*
b^3*B + a^3*C - 7*a*b^2*C)*EllipticF[(c + d*x)/2, 2])/(4*b^2*(a^2 - b^2)^2*d) - ((3*a^4*b*B + 10*a^2*b^3*B - b
^5*B + a^5*C - 10*a^3*b^2*C - 3*a*b^4*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(4*a*(a - b)^2*b^2*(a + b)
^3*d) - ((b*B - a*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) - ((5*a^2*b*B +
b^3*B - a^3*C - 5*a*b^2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*a*(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 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 3078
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(B*a - A*b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e
+ f*x])^n/(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 - 1)*Simp[c*(a*A - b*B)*(m + 1) + d*n*(A*b - a*B) + (d*(a*A - b*B)*(m + 1) - c*(A*b - a*B)*
(m + 2))*Sin[e + f*x] - d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x]^2, 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] && LtQ[m, -1] && GtQ[n, 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 3108
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 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}& = \int \frac {\sqrt {\cos (c+d x)} (B+C \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx \\ & = -\frac {(b B-a C) \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 {1}{2} (b B-a C)-2 (a B-b C) \cos (c+d x)+\frac {1}{2} (b B-a C) \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 {(b B-a C) \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 (5 a^2 b B+b^3 B-a^3 C-5 a b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac {\int \frac {\frac {1}{4} \left (7 a^2 b B-b^3 B-3 a^3 C-3 a b^2 C\right )-a \left (2 a^2 B+b^2 B-3 a b C\right ) \cos (c+d x)-\frac {1}{4} \left (5 a^2 b B+b^3 B-a^3 C-5 a b^2 C\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 {(b B-a C) \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 (5 a^2 b B+b^3 B-a^3 C-5 a b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\int \frac {-\frac {1}{4} b \left (7 a^2 b B-b^3 B-3 a^3 C-3 a b^2 C\right )+\frac {1}{4} a \left (3 a^2 b B+3 b^3 B+a^3 C-7 a b^2 C\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 (5 a^2 b B+b^3 B-a^3 C-5 a b^2 C\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 a b \left (a^2-b^2\right )^2} \\ & = \frac {\left (5 a^2 b B+b^3 B-a^3 C-5 a b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b \left (a^2-b^2\right )^2 d}-\frac {(b B-a C) \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 (5 a^2 b B+b^3 B-a^3 C-5 a b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\left (3 a^2 b B+3 b^3 B+a^3 C-7 a b^2 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 b^2 \left (a^2-b^2\right )^2}-\frac {\left (3 a^4 b B+10 a^2 b^3 B-b^5 B+a^5 C-10 a^3 b^2 C-3 a b^4 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{8 a b^2 \left (a^2-b^2\right )^2} \\ & = \frac {\left (5 a^2 b B+b^3 B-a^3 C-5 a b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^2 b B+3 b^3 B+a^3 C-7 a b^2 C\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 (3 a^4 b B+10 a^2 b^3 B-b^5 B+a^5 C-10 a^3 b^2 C-3 a b^4 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a (a-b)^2 b^2 (a+b)^3 d}-\frac {(b B-a C) \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 (5 a^2 b B+b^3 B-a^3 C-5 a b^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \\
\end{align*}
Mathematica [A] (verified)
Time = 4.99 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.08
\[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\frac {\frac {4 \sqrt {\cos (c+d x)} \left (a \left (-7 a^2 b B+b^3 B+3 a^3 C+3 a b^2 C\right )+b \left (-5 a^2 b B-b^3 B+a^3 C+5 a b^2 C\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {\frac {2 \left (-9 a^2 b B+3 b^3 B+5 a^3 C+a b^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {16 a \left (2 a^2 B+b^2 B-3 a b C\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{b (a+b)}-\frac {2 \left (-5 a^2 b B-b^3 B+a^3 C+5 a b^2 C\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{16 a d}
\]
[In]
Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^3),x]
[Out]
((4*Sqrt[Cos[c + d*x]]*(a*(-7*a^2*b*B + b^3*B + 3*a^3*C + 3*a*b^2*C) + b*(-5*a^2*b*B - b^3*B + a^3*C + 5*a*b^2
*C)*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(a + b*Cos[c + d*x])^2) + ((2*(-9*a^2*b*B + 3*b^3*B + 5*a^3*C +
a*b^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (16*a*(2*a^2*B + b^2*B - 3*a*b*C)*((a + b)*Elli
pticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(b*(a + b)) - (2*(-5*a^2*b*B - b^3*B + a
^3*C + 5*a*b^2*C)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c
+ d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sqrt[S
in[c + d*x]^2]))/((a - b)^2*(a + b)^2))/(16*a*d)
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(1849\) vs. \(2(401)=802\).
Time = 13.71 (sec) , antiderivative size = 1850, normalized size of antiderivative =
5.49
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method | result | size |
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default |
\(\text {Expression too large to display}\) |
\(1850\) |
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[In]
int((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+cos(d*x+c)*b)^3/cos(d*x+c)^(1/2),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*C/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/b^2*(B*b-2*C*a)*(-1/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)-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*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*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)))-2*a*(B*b-C*a)/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)*EllipticF(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*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))+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)*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+si
n(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)*El
lipticPi(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))))/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. \[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\text {Timed out}
\]
[In]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3/cos(d*x+c)^(1/2),x, algorithm="fricas")
[Out]
Timed out
Sympy [F(-1)]
Timed out. \[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\text {Timed out}
\]
[In]
integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**3/cos(d*x+c)**(1/2),x)
[Out]
Timed out
Maxima [F(-1)]
Timed out. \[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\text {Timed out}
\]
[In]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3/cos(d*x+c)^(1/2),x, algorithm="maxima")
[Out]
Timed out
Giac [F]
\[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sqrt {\cos \left (d x + c\right )}} \,d x }
\]
[In]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^3/cos(d*x+c)^(1/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/((b*cos(d*x + c) + a)^3*sqrt(cos(d*x + c))), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x
\]
[In]
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(a + b*cos(c + d*x))^3),x)
[Out]
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(a + b*cos(c + d*x))^3), x)