3.1040 \(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx\)
Optimal. Leaf size=344 \[ \frac {2 \sin (c+d x) \left (24 a^2 C-28 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{105 b^3 d}+\frac {2 \left (-48 a^3 C+56 a^2 b B-2 a b^2 (35 A+22 C)+63 b^3 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^4 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (-48 a^4 C+56 a^3 b B-2 a^2 b^2 (35 A+16 C)+49 a b^3 B-5 b^4 (7 A+5 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^4 d \sqrt {a+b \cos (c+d x)}}+\frac {2 (7 b B-6 a C) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{35 b^2 d}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{7 b d} \]
[Out]
2/105*(35*A*b^2-28*B*a*b+24*C*a^2+25*C*b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^3/d+2/35*(7*B*b-6*C*a)*cos(d*x
+c)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^2/d+2/7*C*cos(d*x+c)^2*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b/d+2/105*(56
*a^2*b*B+63*b^3*B-48*a^3*C-2*a*b^2*(35*A+22*C))*(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+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/b^4/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2/105*(56*
a^3*b*B+49*a*b^3*B-48*a^4*C-5*b^4*(7*A+5*C)-2*a^2*b^2*(35*A+16*C))*(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/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/b^4/d/(a+b*cos(d*x+c
))^(1/2)
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Rubi [A] time = 0.71, antiderivative size = 344, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used =
{3049, 3023, 2752, 2663, 2661, 2655, 2653} \[ \frac {2 \sin (c+d x) \left (24 a^2 C-28 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)}}{105 b^3 d}-\frac {2 \left (-2 a^2 b^2 (35 A+16 C)+56 a^3 b B-48 a^4 C+49 a b^3 B-5 b^4 (7 A+5 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^4 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (56 a^2 b B-48 a^3 C-2 a b^2 (35 A+22 C)+63 b^3 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^4 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 (7 b B-6 a C) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{35 b^2 d}+\frac {2 C \sin (c+d x) \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)}}{7 b d} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Sqrt[a + b*Cos[c + d*x]],x]
[Out]
(2*(56*a^2*b*B + 63*b^3*B - 48*a^3*C - 2*a*b^2*(35*A + 22*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2,
(2*b)/(a + b)])/(105*b^4*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(56*a^3*b*B + 49*a*b^3*B - 48*a^4*C - 5*b^
4*(7*A + 5*C) - 2*a^2*b^2*(35*A + 16*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a +
b)])/(105*b^4*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(35*A*b^2 - 28*a*b*B + 24*a^2*C + 25*b^2*C)*Sqrt[a + b*Cos[c +
d*x]]*Sin[c + d*x])/(105*b^3*d) + (2*(7*b*B - 6*a*C)*Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(35*b
^2*d) + (2*C*Cos[c + d*x]^2*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(7*b*d)
Rule 2653
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 2655
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] && !GtQ[a + b, 0]
Rule 2661
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 2663
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && !GtQ[a + b, 0]
Rule 2752
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Rule 3023
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Rule 3049
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] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*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] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Rubi steps
\begin {align*} \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx &=\frac {2 C \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{7 b d}+\frac {2 \int \frac {\cos (c+d x) \left (2 a C+\frac {1}{2} b (7 A+5 C) \cos (c+d x)+\frac {1}{2} (7 b B-6 a C) \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx}{7 b}\\ &=\frac {2 (7 b B-6 a C) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{35 b^2 d}+\frac {2 C \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{7 b d}+\frac {4 \int \frac {\frac {1}{2} a (7 b B-6 a C)+\frac {1}{4} b (21 b B+2 a C) \cos (c+d x)+\frac {1}{4} \left (35 A b^2-28 a b B+24 a^2 C+25 b^2 C\right ) \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{35 b^2}\\ &=\frac {2 \left (35 A b^2-28 a b B+24 a^2 C+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b^3 d}+\frac {2 (7 b B-6 a C) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{35 b^2 d}+\frac {2 C \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{7 b d}+\frac {8 \int \frac {\frac {1}{8} b \left (35 A b^2+14 a b B-12 a^2 C+25 b^2 C\right )+\frac {1}{8} \left (56 a^2 b B+63 b^3 B-48 a^3 C-2 a b^2 (35 A+22 C)\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{105 b^3}\\ &=\frac {2 \left (35 A b^2-28 a b B+24 a^2 C+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b^3 d}+\frac {2 (7 b B-6 a C) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{35 b^2 d}+\frac {2 C \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{7 b d}-\frac {\left (56 a^3 b B+49 a b^3 B-48 a^4 C-5 b^4 (7 A+5 C)-2 a^2 b^2 (35 A+16 C)\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{105 b^4}+\frac {\left (56 a^2 b B+63 b^3 B-48 a^3 C-2 a b^2 (35 A+22 C)\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{105 b^4}\\ &=\frac {2 \left (35 A b^2-28 a b B+24 a^2 C+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b^3 d}+\frac {2 (7 b B-6 a C) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{35 b^2 d}+\frac {2 C \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{7 b d}+\frac {\left (\left (56 a^2 b B+63 b^3 B-48 a^3 C-2 a b^2 (35 A+22 C)\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{105 b^4 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (\left (56 a^3 b B+49 a b^3 B-48 a^4 C-5 b^4 (7 A+5 C)-2 a^2 b^2 (35 A+16 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{105 b^4 \sqrt {a+b \cos (c+d x)}}\\ &=\frac {2 \left (56 a^2 b B+63 b^3 B-48 a^3 C-2 a b^2 (35 A+22 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^4 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (56 a^3 b B+49 a b^3 B-48 a^4 C-5 b^4 (7 A+5 C)-2 a^2 b^2 (35 A+16 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^4 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (35 A b^2-28 a b B+24 a^2 C+25 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b^3 d}+\frac {2 (7 b B-6 a C) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{35 b^2 d}+\frac {2 C \cos ^2(c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{7 b d}\\ \end {align*}
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Mathematica [A] time = 1.32, size = 252, normalized size = 0.73 \[ \frac {b (a+b \cos (c+d x)) \left (\sin (c+d x) \left (96 a^2 C-112 a b B+140 A b^2+115 b^2 C\right )+3 b (2 (7 b B-6 a C) \sin (2 (c+d x))+5 b C \sin (3 (c+d x)))\right )+4 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (b^2 \left (-12 a^2 C+14 a b B+35 A b^2+25 b^2 C\right ) F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-\left (48 a^3 C-56 a^2 b B+2 a b^2 (35 A+22 C)-63 b^3 B\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )\right )}{210 b^4 d \sqrt {a+b \cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Sqrt[a + b*Cos[c + d*x]],x]
[Out]
(4*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(35*A*b^2 + 14*a*b*B - 12*a^2*C + 25*b^2*C)*EllipticF[(c + d*x)/2,
(2*b)/(a + b)] - (-56*a^2*b*B - 63*b^3*B + 48*a^3*C + 2*a*b^2*(35*A + 22*C))*((a + b)*EllipticE[(c + d*x)/2, (
2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)])) + b*(a + b*Cos[c + d*x])*((140*A*b^2 - 112*a*b*B + 9
6*a^2*C + 115*b^2*C)*Sin[c + d*x] + 3*b*(2*(7*b*B - 6*a*C)*Sin[2*(c + d*x)] + 5*b*C*Sin[3*(c + d*x)])))/(210*b
^4*d*Sqrt[a + b*Cos[c + d*x]])
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fricas [F] time = 1.09, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {C \cos \left (d x + c\right )^{4} + B \cos \left (d x + c\right )^{3} + A \cos \left (d x + c\right )^{2}}{\sqrt {b \cos \left (d x + c\right ) + a}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
integral((C*cos(d*x + c)^4 + B*cos(d*x + c)^3 + A*cos(d*x + c)^2)/sqrt(b*cos(d*x + c) + a), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{\sqrt {b \cos \left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^2/sqrt(b*cos(d*x + c) + a), x)
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maple [B] time = 3.50, size = 1635, normalized size = 4.75 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x)
[Out]
-2/105*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*C*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^8+(-168*B*b^4+24*C*a*b^3-360*C*b^4)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(140*A*b^4-28*B*a*b^3+168*B*
b^4+24*C*a^2*b^2-24*C*a*b^3+280*C*b^4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-70*A*a*b^3-70*A*b^4+56*B*a^2*
b^2+14*B*a*b^3-42*B*b^4-48*C*a^3*b-12*C*a^2*b^2-44*C*a*b^3-80*C*b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-7
0*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2
*c),(-2*b/(a-b))^(1/2))*a^2*b^2+70*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b)
)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^3+70*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*s
in(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2+35*A*b^4*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b
/(a-b))^(1/2))+56*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE
(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b-56*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)
^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2+63*B*(sin(1/2*d*x+1/2*c)^2)^(1/
2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^3-
63*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/
2*c),(-2*b/(a-b))^(1/2))*b^4-56*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(
1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b-49*a*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*si
n(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3-48*C*(sin(1/2*d*x+1
/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(
1/2))*a^4+48*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(
1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b-44*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a
+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2+44*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-
2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^3+48*C*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),
(-2*b/(a-b))^(1/2))*a^4+32*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*
EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2+25*C*b^4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin
(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)))/b^4/(-2*sin(1/2*d*x+1/2
*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{\sqrt {b \cos \left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^2/sqrt(b*cos(d*x + c) + a), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^(1/2),x)
[Out]
int((cos(c + d*x)^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^(1/2), 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(cos(d*x+c)**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**(1/2),x)
[Out]
Timed out
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