3.665 \(\int \frac{\cos (c+d x) (A+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=314 \[ \frac{2 \left (a^2 b^2 (A+9 C)-5 a^4 C+3 A b^4\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}+\frac{2 a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac{2 a \left (-8 a^2 C+A b^2+9 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (a^2 b^2 (A+15 C)-8 a^4 C+3 b^4 (A-C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}} \]

[Out]

(-2*(3*b^4*(A - C) - 8*a^4*C + a^2*b^2*(A + 15*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a +
b)])/(3*b^3*(a^2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*a*(A*b^2 - 8*a^2*C + 9*b^2*C)*Sqrt[(a + b
*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*b^3*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]])
+ (2*a*(A*b^2 + a^2*C)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) + (2*(3*A*b^4 - 5*a^4*C
+ a^2*b^2*(A + 9*C))*Sin[c + d*x])/(3*b^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]])

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Rubi [A]  time = 0.509611, antiderivative size = 314, normalized size of antiderivative = 1., 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 = {3032, 3021, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (a^2 b^2 (A+9 C)-5 a^4 C+3 A b^4\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}+\frac{2 a \left (a^2 C+A b^2\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac{2 a \left (-8 a^2 C+A b^2+9 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (a^2 b^2 (A+15 C)-8 a^4 C+3 b^4 (A-C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^(5/2),x]

[Out]

(-2*(3*b^4*(A - C) - 8*a^4*C + a^2*b^2*(A + 15*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a +
b)])/(3*b^3*(a^2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*a*(A*b^2 - 8*a^2*C + 9*b^2*C)*Sqrt[(a + b
*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*b^3*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]])
+ (2*a*(A*b^2 + a^2*C)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) + (2*(3*A*b^4 - 5*a^4*C
+ a^2*b^2*(A + 9*C))*Sin[c + d*x])/(3*b^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]])

Rule 3032

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1
))/(b^2*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b
*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d)) - ((b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f
*x] + b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(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))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{5/2}} \, dx &=\frac{2 a \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac{2 \int \frac{\frac{3}{2} b \left (A b^2+a^2 C\right )-\frac{1}{2} a \left (A b^2-2 a^2 C+3 b^2 C\right ) \cos (c+d x)-\frac{3}{2} b \left (a^2-b^2\right ) C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=\frac{2 a \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{2 \left (3 A b^4-5 a^4 C+a^2 b^2 (A+9 C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}+\frac{4 \int \frac{\frac{1}{2} a b^2 \left (a^2 C-b^2 (2 A+3 C)\right )-\frac{1}{4} b \left (3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 C)\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{2 a \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{2 \left (3 A b^4-5 a^4 C+a^2 b^2 (A+9 C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}+\frac{\left (a \left (A b^2-8 a^2 C+9 b^2 C\right )\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )}-\frac{\left (3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 C)\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{2 a \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{2 \left (3 A b^4-5 a^4 C+a^2 b^2 (A+9 C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}-\frac{\left (\left (3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 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}{3 b^3 \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\left (a \left (A b^2-8 a^2 C+9 b^2 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}{3 b^3 \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}\\ &=-\frac{2 \left (3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 C)\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 )^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 a \left (A b^2-8 a^2 C+9 b^2 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 )}{3 b^3 \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{2 a \left (A b^2+a^2 C\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{2 \left (3 A b^4-5 a^4 C+a^2 b^2 (A+9 C)\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.13299, size = 227, normalized size = 0.72 \[ \frac{2 \left (\frac{b \sin (c+d x) \left (\left (a^2 b^3 (A+9 C)-5 a^4 b C+3 A b^5\right ) \cos (c+d x)+2 a^3 b^2 (A+4 C)-4 a^5 C+2 a A b^4\right )}{\left (a^2-b^2\right )^2}+\frac{\left (\frac{a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (\left (-a^2 b^2 (A+15 C)+8 a^4 C+3 b^4 (C-A)\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-a (a-b) \left (8 a^2 C-A b^2-9 b^2 C\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )}{(a-b)^2}\right )}{3 b^3 d (a+b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^(5/2),x]

[Out]

(2*((((a + b*Cos[c + d*x])/(a + b))^(3/2)*((8*a^4*C + 3*b^4*(-A + C) - a^2*b^2*(A + 15*C))*EllipticE[(c + d*x)
/2, (2*b)/(a + b)] - a*(a - b)*(-(A*b^2) + 8*a^2*C - 9*b^2*C)*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))/(a - b)^
2 + (b*(2*a*A*b^4 - 4*a^5*C + 2*a^3*b^2*(A + 4*C) + (3*A*b^5 - 5*a^4*b*C + a^2*b^3*(A + 9*C))*Cos[c + d*x])*Si
n[c + d*x])/(a^2 - b^2)^2))/(3*b^3*d*(a + b*Cos[c + d*x])^(3/2))

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Maple [B]  time = 1.764, size = 926, normalized size = 3. \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)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(5/2),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*C/b^3/(-2*b*sin(1/2*d*x+1/2*c)^4+(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)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*El
lipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a+EllipticE(
cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b)+2/b^3*(A*b^2+3*C*a^2)/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*b*sin(1/2*d*x+1/2*c)^4+(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)-2*a*(A*b^2+C*a^2)/b^3*(1/6/b/(a-b)/(a+b)*cos(1/2*d*x
+1/2*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2+1/2*(a-b)/b)^2+8/3*
b*sin(1/2*d*x+1/2*c)^2/(a-b)^2/(a+b)^2*cos(1/2*d*x+1/2*c)*a/(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*
c)^2)^(1/2)+(3*a-b)/(3*a^3+3*a^2*b-3*a*b^2-3*b^3)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)
/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/
(a-b))^(1/2))-4/3*a/(a+b)^2/(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-
2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-E
llipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)/(b*cos(d*x + c) + a)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

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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)*(A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**(5/2),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} + A\right )} \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)/(b*cos(d*x + c) + a)^(5/2), x)