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

Optimal. Leaf size=262 \[ -\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-8 a^2 B+6 a A b-b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^3 d \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b^2 d} \]

[Out]

-2*a^2*(A*b-B*a)*sin(d*x+c)/b^2/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)+2/3*B*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^2
/d+2/3*(6*A*a^2*b-3*A*b^3-8*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)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/b^3/(a^2-b^2)/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2/3
*(6*A*a*b-8*B*a^2-B*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/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/b^3/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.49, antiderivative size = 262, 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 = {2988, 3023, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-8 a^2 B+6 a A b-b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (6 a^2 A b-8 a^3 B+5 a b^2 B-3 A b^3\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^3 d \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 B \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(6*a^2*A*b - 3*A*b^3 - 8*a^3*B + 5*a*b^2*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])
/(3*b^3*(a^2 - b^2)*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(6*a*A*b - 8*a^2*B - b^2*B)*Sqrt[(a + b*Cos[c +
d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*b^3*d*Sqrt[a + b*Cos[c + d*x]]) - (2*a^2*(A*b - a*B)
*Sin[c + d*x])/(b^2*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) + (2*B*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*b
^2*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 2988

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/
(f*d^2*(n + 1)*(c^2 - d^2)), x] - Dist[1/(d^2*(n + 1)*(c^2 - d^2)), Int[(c + d*Sin[e + f*x])^(n + 1)*Simp[d*(n
+ 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c - 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n +
1))) + 2*a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 1)*(c^2 - d^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[n, -1]

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]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx &=-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 \int \frac {\frac {1}{2} a b (A b-a B)+\frac {1}{2} \left (2 a^2-b^2\right ) (A b-a B) \cos (c+d x)+\frac {1}{2} b \left (a^2-b^2\right ) B \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 B \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^2 d}+\frac {4 \int \frac {\frac {1}{4} b^2 \left (3 a A b-2 a^2 B-b^2 B\right )+\frac {1}{4} b \left (6 a^2 A b-3 A b^3-8 a^3 B+5 a b^2 B\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )}\\ &=-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 B \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^2 d}-\frac {\left (6 a A b-8 a^2 B-b^2 B\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 b^3}+\frac {\left (6 a^2 A b-3 A b^3-8 a^3 B+5 a b^2 B\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{3 b^3 \left (a^2-b^2\right )}\\ &=-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 B \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^2 d}+\frac {\left (\left (6 a^2 A b-3 A b^3-8 a^3 B+5 a b^2 B\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{3 b^3 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (\left (6 a A b-8 a^2 B-b^2 B\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}{3 b^3 \sqrt {a+b \cos (c+d x)}}\\ &=\frac {2 \left (6 a^2 A b-3 A b^3-8 a^3 B+5 a b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^3 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (6 a A b-8 a^2 B-b^2 B\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 )}{3 b^3 d \sqrt {a+b \cos (c+d x)}}-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 B \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.51, size = 189, normalized size = 0.72 \[ \frac {2 \left (b \sin (c+d x) \left (\frac {a \left (-4 a^2 B+3 a A b+b^2 B\right )}{b^2-a^2}+b B \cos (c+d x)\right )+\frac {\sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left ((a-b) \left (8 a^2 B-6 a A b+b^2 B\right ) F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+\left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )}{a-b}\right )}{3 b^3 d \sqrt {a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((Sqrt[(a + b*Cos[c + d*x])/(a + b)]*((6*a^2*A*b - 3*A*b^3 - 8*a^3*B + 5*a*b^2*B)*EllipticE[(c + d*x)/2, (2
*b)/(a + b)] + (a - b)*(-6*a*A*b + 8*a^2*B + b^2*B)*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))/(a - b) + b*((a*(3
*a*A*b - 4*a^2*B + b^2*B))/(-a^2 + b^2) + b*B*Cos[c + d*x])*Sin[c + d*x]))/(3*b^3*d*Sqrt[a + b*Cos[c + d*x]])

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fricas [F]  time = 1.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \cos \left (d x + c\right )^{3} + A \cos \left (d x + c\right )^{2}\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 4.67, size = 954, normalized size = 3.64 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/3/b^3*(4*B*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*
x+1/2*c)^4+(-2*B*a*b-2*B*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-6*A*a*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))+3*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*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))*b^2+8*a^2*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))+b^2*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))-5*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+5*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)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a^
2*(A*b-B*a)/b^3/sin(1/2*d*x+1/2*c)^2/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)/(a^2-b^2)*(-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)^(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-(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+2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+
1/2*c)^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 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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