3.4.79 \(\int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx\) [379]
Optimal. Leaf size=337 \[ \frac {\left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^2 \left (a^2-b^2\right )^2 d}-\frac {\left (3 a^4 A b+10 a^2 A b^3-A b^5+a^5 B-10 a^3 b^2 B-3 a b^4 B\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a (a-b)^2 b^2 (a+b)^3 d}-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))} \]
[Out]
1/4*(5*A*a^2*b+A*b^3-B*a^3-5*B*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*A*a^2*b+3*A*b^3+B*a^3-7*B*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*A*a^4*b+10*A*a^2*b^3-A*b^5+B*a^5-10*B
*a^3*b^2-3*B*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*(A*b-B*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
*A*a^2*b+A*b^3-B*a^3-5*B*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))
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Rubi [A]
time = 0.61, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3078, 3134,
3138, 2719, 3081, 2720, 2884} \begin {gather*} -\frac {(A b-a B) \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 B+3 a^2 A b-7 a b^2 B+3 A b^3\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^2 d \left (a^2-b^2\right )^2}+\frac {\left (a^3 (-B)+5 a^2 A b-5 a b^2 B+A b^3\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 (-B)+5 a^2 A b-5 a b^2 B+A b^3\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 B+3 a^4 A b-10 a^3 b^2 B+10 a^2 A b^3-3 a b^4 B-A b^5\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b^2 d (a-b)^2 (a+b)^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(Sqrt[Cos[c + d*x]]*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^3,x]
[Out]
((5*a^2*A*b + A*b^3 - a^3*B - 5*a*b^2*B)*EllipticE[(c + d*x)/2, 2])/(4*a*b*(a^2 - b^2)^2*d) + ((3*a^2*A*b + 3*
A*b^3 + a^3*B - 7*a*b^2*B)*EllipticF[(c + d*x)/2, 2])/(4*b^2*(a^2 - b^2)^2*d) - ((3*a^4*A*b + 10*a^2*A*b^3 - A
*b^5 + a^5*B - 10*a^3*b^2*B - 3*a*b^4*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(4*a*(a - b)^2*b^2*(a + b)
^3*d) - ((A*b - a*B)*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*A*b +
A*b^3 - a^3*B - 5*a*b^2*B)*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 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*} \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{(a+b \cos (c+d x))^3} \, dx &=-\frac {(A b-a B) \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} (A b-a B)-2 (a A-b B) \cos (c+d x)+\frac {1}{2} (A b-a B) \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 b-a B) \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 A b+A b^3-a^3 B-5 a b^2 B\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 A b-A b^3-3 a^3 B-3 a b^2 B\right )-a \left (2 a^2 A+A b^2-3 a b B\right ) \cos (c+d x)-\frac {1}{4} \left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\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 b-a B) \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 A b+A b^3-a^3 B-5 a b^2 B\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 A b-A b^3-3 a^3 B-3 a b^2 B\right )+\frac {1}{4} a \left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\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 A b+A b^3-a^3 B-5 a b^2 B\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 a b \left (a^2-b^2\right )^2}\\ &=\frac {\left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\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 {(A b-a B) \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 A b+A b^3-a^3 B-5 a b^2 B\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 A b+3 A b^3+a^3 B-7 a b^2 B\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 A b+10 a^2 A b^3-A b^5+a^5 B-10 a^3 b^2 B-3 a b^4 B\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 A b+A b^3-a^3 B-5 a b^2 B\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 A b+3 A b^3+a^3 B-7 a b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 b^2 \left (a^2-b^2\right )^2 d}-\frac {\left (3 a^4 A b+10 a^2 A b^3-A b^5+a^5 B-10 a^3 b^2 B-3 a b^4 B\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a (a-b)^2 b^2 (a+b)^3 d}-\frac {(A b-a B) \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 A b+A b^3-a^3 B-5 a b^2 B\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*}
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Mathematica [A]
time = 14.52, size = 365, normalized size = 1.08 \begin {gather*} \frac {\frac {4 \sqrt {\cos (c+d x)} \left (a \left (-7 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right )+b \left (-5 a^2 A b-A b^3+a^3 B+5 a b^2 B\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 A b+3 A b^3+5 a^3 B+a b^2 B\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {16 a \left (2 a^2 A+A b^2-3 a b B\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{b (a+b)}-\frac {2 \left (-5 a^2 A b-A b^3+a^3 B+5 a b^2 B\right ) \left (-2 a b E\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+\left (-2 a^2+b^2\right ) \Pi \left (-\frac {b}{a};\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\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} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(Sqrt[Cos[c + d*x]]*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^3,x]
[Out]
((4*Sqrt[Cos[c + d*x]]*(a*(-7*a^2*A*b + A*b^3 + 3*a^3*B + 3*a*b^2*B) + b*(-5*a^2*A*b - A*b^3 + a^3*B + 5*a*b^2
*B)*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(a + b*Cos[c + d*x])^2) + ((2*(-9*a^2*A*b + 3*A*b^3 + 5*a^3*B +
a*b^2*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (16*a*(2*a^2*A + A*b^2 - 3*a*b*B)*((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*A*b - A*b^3 + a
^3*B + 5*a*b^2*B)*(-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)
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1849\) vs.
\(2(401)=802\).
time = 1.22, size = 1850, normalized size = 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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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^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/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*b-2*B*a)/b^2*(-b^2/a/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*b+a-b)-1/2/(a+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*b/a/(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+s
in(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*b/a/(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)*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*(A*b-B*a)/b^2*(-1/2*b^
2/a/(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*cos(1/2*d*x+1/2*c)^2*
b+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*cos(1/2*d*x+1/2*c)^2*b+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*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*(s
in(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+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)*Elli
pticPi(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
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^3,x, algorithm="maxima")
[Out]
integrate((B*cos(d*x + c) + A)*sqrt(cos(d*x + c))/(b*cos(d*x + c) + a)^3, x)
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^3,x, algorithm="fricas")
[Out]
Timed out
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**(1/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))**3,x)
[Out]
Exception raised: SystemError >> excessive stack use: stack is 3066 deep
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(1/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^3,x, algorithm="giac")
[Out]
integrate((B*cos(d*x + c) + A)*sqrt(cos(d*x + c))/(b*cos(d*x + c) + a)^3, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)^(1/2)*(A + B*cos(c + d*x)))/(a + b*cos(c + d*x))^3,x)
[Out]
int((cos(c + d*x)^(1/2)*(A + B*cos(c + d*x)))/(a + b*cos(c + d*x))^3, x)
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