3.706 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+C \cos ^2(c+d x))}{a+b \cos (c+d x)} \, dx\)
Optimal. Leaf size=181 \[ -\frac {2 a \left (C \left (3 a^2+b^2\right )+3 A b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^4 d}+\frac {2 a^2 \left (a^2 C+A b^2\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^4 d (a+b)}+\frac {2 \left (5 a^2 C+b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^3 d}-\frac {2 a C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b^2 d}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d} \]
[Out]
2/5*(5*a^2*C+b^2*(5*A+3*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^3/d-2/3*a*(3*A*b^2+(3*a^2+b^2)*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^4/d+2*a^2*(A*b^2+C*a^2)*(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))/b^4/(a+b)/d+2/5*C*cos(d*x+c)^(3/2)*sin(d*x+c)/b/d-2/3*a*C*sin(d*x+c)*cos(d*x+c)^(
1/2)/b^2/d
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Rubi [A] time = 0.79, antiderivative size = 181, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used =
{3050, 3049, 3059, 2639, 3002, 2641, 2805} \[ -\frac {2 a \left (C \left (3 a^2+b^2\right )+3 A b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^4 d}+\frac {2 \left (5 a^2 C+b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^3 d}+\frac {2 a^2 \left (a^2 C+A b^2\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^4 d (a+b)}-\frac {2 a C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b^2 d}+\frac {2 C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 b d} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^(3/2)*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x]),x]
[Out]
(2*(5*a^2*C + b^2*(5*A + 3*C))*EllipticE[(c + d*x)/2, 2])/(5*b^3*d) - (2*a*(3*A*b^2 + (3*a^2 + b^2)*C)*Ellipti
cF[(c + d*x)/2, 2])/(3*b^4*d) + (2*a^2*(A*b^2 + a^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b^4*(a + b)
*d) - (2*a*C*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b^2*d) + (2*C*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*b*d)
Rule 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rule 2641
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]
Rule 2805
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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 3002
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 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])))
Rule 3050
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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])))
Rule 3059
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 {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx &=\frac {2 C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac {2 \int \frac {\sqrt {\cos (c+d x)} \left (\frac {3 a C}{2}+\frac {1}{2} b (5 A+3 C) \cos (c+d x)-\frac {5}{2} a C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{5 b}\\ &=-\frac {2 a C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^2 d}+\frac {2 C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac {4 \int \frac {-\frac {5 a^2 C}{4}+a b C \cos (c+d x)+\frac {3}{4} \left (5 a^2 C+b^2 (5 A+3 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 b^2}\\ &=-\frac {2 a C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^2 d}+\frac {2 C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b d}-\frac {4 \int \frac {\frac {5}{4} a^2 b C+\frac {5}{4} a \left (3 A b^2+\left (3 a^2+b^2\right ) C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 b^3}+\frac {\left (5 a^2 C+b^2 (5 A+3 C)\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 b^3}\\ &=\frac {2 \left (5 a^2 C+b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^3 d}-\frac {2 a C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^2 d}+\frac {2 C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac {\left (a^2 \left (A b^2+a^2 C\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{b^4}-\frac {\left (a \left (3 A b^2+\left (3 a^2+b^2\right ) C\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 b^4}\\ &=\frac {2 \left (5 a^2 C+b^2 (5 A+3 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^3 d}-\frac {2 a \left (3 A b^2+\left (3 a^2+b^2\right ) C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 b^4 d}+\frac {2 a^2 \left (A b^2+a^2 C\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^4 (a+b) d}-\frac {2 a C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b^2 d}+\frac {2 C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 b d}\\ \end {align*}
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Mathematica [A] time = 2.24, size = 244, normalized size = 1.35 \[ \frac {\frac {2 \left (5 a^2 C+15 A b^2+9 b^2 C\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {6 \left (5 a^2 C+5 A b^2+3 b^2 C\right ) \sin (c+d x) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b^2 \sqrt {\sin ^2(c+d x)}}+4 C \sin (c+d x) \sqrt {\cos (c+d x)} (3 b \cos (c+d x)-5 a)+8 a C \left (2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-\frac {2 a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}\right )}{30 b^2 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^(3/2)*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x]),x]
[Out]
((2*(15*A*b^2 + 5*a^2*C + 9*b^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + 8*a*C*(2*EllipticF[(c
+ d*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)) + 4*C*Sqrt[Cos[c + d*x]]*(-5*a + 3*b*C
os[c + d*x])*Sin[c + d*x] + (6*(5*A*b^2 + 5*a^2*C + 3*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[Sin[c + d*x]^2]))/(30*b^2*d)
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fricas [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)^(3/2)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="fricas")
[Out]
Timed out
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{b \cos \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^(3/2)/(b*cos(d*x + c) + a), x)
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maple [B] time = 2.61, size = 948, normalized size = 5.24 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(3/2)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x)
[Out]
2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((24*C*a*b^3-24*C*b^4)*cos(1/2*d*x+1/2*c)*sin(1/2
*d*x+1/2*c)^6+(20*C*a^2*b^2-44*C*a*b^3+24*C*b^4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*C*a^2*b^2+16*C*a
*b^3-6*C*b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+15*A*a^2*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x
+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-15*a*A*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*
x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2
*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^3-15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/
2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^4-15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2
*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a^2*b^2+15*a^4*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)
*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-15*a^3*b*C*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+5*C*a^2*b^2*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-5*C*a*b^3*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+15*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(
2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b-15*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2+9*C*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^3-9*C*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^4-15*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)
*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a^4)/b^4/(a-b)/(-2*sin(1/2
*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^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 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{b \cos \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^(3/2)/(b*cos(d*x + c) + a), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{a+b\,\cos \left (c+d\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)^(3/2)*(A + C*cos(c + d*x)^2))/(a + b*cos(c + d*x)),x)
[Out]
int((cos(c + d*x)^(3/2)*(A + C*cos(c + d*x)^2))/(a + b*cos(c + d*x)), 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)**(3/2)*(A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c)),x)
[Out]
Timed out
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