3.1096 \(\int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=210 \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b B-3 a^3 C-a b^2 (3 A+C)+b^3 B\right )}{3 b^4 d}+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}+\frac{2 a^2 \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^4 d (a+b)}+\frac{2 (b B-a C) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 b^2 d}+\frac{2 C \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 b d} \]

[Out]

(2*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3*b^2*C)*EllipticE[(c + d*x)/2, 2])/(5*b^3*d) + (2*(3*a^2*b*B + b^3*B - 3*a^
3*C - a*b^2*(3*A + C))*EllipticF[(c + d*x)/2, 2])/(3*b^4*d) + (2*a^2*(A*b^2 - a*(b*B - a*C))*EllipticPi[(2*b)/
(a + b), (c + d*x)/2, 2])/(b^4*(a + b)*d) + (2*(b*B - a*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b^2*d) + (2*C*C
os[c + d*x]^(3/2)*Sin[c + d*x])/(5*b*d)

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Rubi [A]  time = 0.878859, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {3049, 3059, 2639, 3002, 2641, 2805} \[ \frac{2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (3 a^2 b B-3 a^3 C-a b^2 (3 A+C)+b^3 B\right )}{3 b^4 d}+\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}+\frac{2 a^2 \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^4 d (a+b)}+\frac{2 (b B-a C) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 b^2 d}+\frac{2 C \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3*b^2*C)*EllipticE[(c + d*x)/2, 2])/(5*b^3*d) + (2*(3*a^2*b*B + b^3*B - 3*a^
3*C - a*b^2*(3*A + C))*EllipticF[(c + d*x)/2, 2])/(3*b^4*d) + (2*a^2*(A*b^2 - a*(b*B - a*C))*EllipticPi[(2*b)/
(a + b), (c + d*x)/2, 2])/(b^4*(a + b)*d) + (2*(b*B - a*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b^2*d) + (2*C*C
os[c + d*x]^(3/2)*Sin[c + d*x])/(5*b*d)

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 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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (A+B \cos (c+d x)+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} (b B-a C) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{5 b}\\ &=\frac{2 (b B-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 (b B-a C)+\frac{1}{4} b (5 b B+4 a C) \cos (c+d x)+\frac{3}{4} \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 b^2}\\ &=\frac{2 (b B-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 b (b B-a C)-\frac{5}{4} \left (3 a^2 b B+b^3 B-3 a^3 C-a b^2 (3 A+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 b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 b^3}\\ &=\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d}+\frac{2 (b B-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 (3 a^2 b B+b^3 B-3 a^3 C-a b^2 (3 A+C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 b^4}+\frac{\left (a^2 \left (A b^2-a (b B-a C)\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{b^4}\\ &=\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 b^3 d}+\frac{2 \left (3 a^2 b B+b^3 B-3 a^3 C-a b^2 (3 A+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 (b B-a 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 (b B-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*}

Mathematica [A]  time = 2.3118, size = 276, normalized size = 1.31 \[ \frac{\frac{2 b^2 \left (5 a^2 C-5 a b B+15 A b^2+9 b^2 C\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right ) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a \sqrt{\sin ^2(c+d x)}}+4 b^2 \sin (c+d x) \sqrt{\cos (c+d x)} (-5 a C+5 b B+3 b C \cos (c+d x))+2 b^2 (4 a C+5 b B) \left (2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-\frac{2 a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{30 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*b^2*(15*A*b^2 - 5*a*b*B + 5*a^2*C + 9*b^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + 2*b^2*(5
*b*B + 4*a*C)*(2*EllipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)) + 4*b^2*
Sqrt[Cos[c + d*x]]*(5*b*B - 5*a*C + 3*b*C*Cos[c + d*x])*Sin[c + d*x] + (6*(5*A*b^2 - 5*a*b*B + 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*Sqrt[Sin[c + d*x]^2]))/(
30*b^4*d)

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Maple [B]  time = 2.404, size = 803, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4/5*C/b*(-4*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c
)+14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2)-6*sin(1/2*d*x+1/2*c)^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)+4/3/b^2*(B*b-C*a-3*C*b)*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+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))-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/
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)-sin(1/2*d*x+1/2*c)^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^3*(A*b^2-B*a*b-2*B*b^2+C*a^2+2*C*a*b+3*C*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)*(Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*(A*a*b^2+A*b^3-B*a^2*b-B*a*b^2-B*b
^3+C*a^3+C*a^2*b+C*a*b^2+C*b^3)/b^4*(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))-4*a^2*(A*b^2-B*a*b+C*a^2)/b^3/
(-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)))/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., size = 0, normalized size = 0. \begin{align*} \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 )^{\frac{3}{2}}}{b \cos \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )^{\frac{3}{2}}}{b \cos \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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