3.523 \(\int \frac{\cos ^3(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=215 \[ -\frac{2 a \left (8 a^2+7 b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^3 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (8 a^2+9 b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{8 a \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{15 b^2 d}+\frac{2 \sin (c+d x) \cos (c+d x) \sqrt{a+b \cos (c+d x)}}{5 b d} \]

[Out]

(2*(8*a^2 + 9*b^2)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(15*b^3*d*Sqrt[(a + b*Cos[c
+ d*x])/(a + b)]) - (2*a*(8*a^2 + 7*b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a +
b)])/(15*b^3*d*Sqrt[a + b*Cos[c + d*x]]) - (8*a*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*b^2*d) + (2*Cos[c
+ d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*b*d)

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Rubi [A]  time = 0.287047, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {2793, 3023, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 a \left (8 a^2+7 b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^3 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (8 a^2+9 b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{8 a \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{15 b^2 d}+\frac{2 \sin (c+d x) \cos (c+d x) \sqrt{a+b \cos (c+d x)}}{5 b d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/Sqrt[a + b*Cos[c + d*x]],x]

[Out]

(2*(8*a^2 + 9*b^2)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(15*b^3*d*Sqrt[(a + b*Cos[c
+ d*x])/(a + b)]) - (2*a*(8*a^2 + 7*b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a +
b)])/(15*b^3*d*Sqrt[a + b*Cos[c + d*x]]) - (8*a*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*b^2*d) + (2*Cos[c
+ d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*b*d)

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 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 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 ^3(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx &=\frac{2 \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 b d}+\frac{2 \int \frac{a+\frac{3}{2} b \cos (c+d x)-2 a \cos ^2(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{5 b}\\ &=-\frac{8 a \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac{2 \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 b d}+\frac{4 \int \frac{\frac{a b}{2}+\frac{1}{4} \left (8 a^2+9 b^2\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{15 b^2}\\ &=-\frac{8 a \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac{2 \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 b d}-\frac{\left (a \left (8 a^2+7 b^2\right )\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{15 b^3}+\frac{\left (8 a^2+9 b^2\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{15 b^3}\\ &=-\frac{8 a \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac{2 \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 b d}+\frac{\left (\left (8 a^2+9 b^2\right ) \sqrt{a+b \cos (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}} \, dx}{15 b^3 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (a \left (8 a^2+7 b^2\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}{15 b^3 \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 \left (8 a^2+9 b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 a \left (8 a^2+7 b^2\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 )}{15 b^3 d \sqrt{a+b \cos (c+d x)}}-\frac{8 a \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac{2 \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{5 b d}\\ \end{align*}

Mathematica [A]  time = 0.932382, size = 182, normalized size = 0.85 \[ \frac{b \sin (c+d x) \left (-8 a^2-2 a b \cos (c+d x)+3 b^2 \cos (2 (c+d x))+3 b^2\right )-2 a \left (8 a^2+7 b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+2 \left (8 a^2 b+8 a^3+9 a b^2+9 b^3\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^3 d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/Sqrt[a + b*Cos[c + d*x]],x]

[Out]

(2*(8*a^3 + 8*a^2*b + 9*a*b^2 + 9*b^3)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(a + b)
] - 2*a*(8*a^2 + 7*b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + b*(-8*a^2 +
3*b^2 - 2*a*b*Cos[c + d*x] + 3*b^2*Cos[2*(c + d*x)])*Sin[c + d*x])/(15*b^3*d*Sqrt[a + b*Cos[c + d*x]])

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Maple [B]  time = 3.274, size = 665, normalized size = 3.1 \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/(a+b*cos(d*x+c))^(1/2),x)

[Out]

-2/15*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(24*cos(1/2*d*x+1/2*c)^7*b^3-4*cos(1/2*d*x+1
/2*c)^5*a*b^2-48*cos(1/2*d*x+1/2*c)^5*b^3-8*cos(1/2*d*x+1/2*c)^3*a^2*b+6*cos(1/2*d*x+1/2*c)^3*a*b^2+30*cos(1/2
*d*x+1/2*c)^3*b^3-8*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(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^3-7*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(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*b^2+8*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(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^3-8*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((
2*b*cos(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+9*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*((2*b*cos(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^2-9*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(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^3+8*cos(1/2*d*x+1/2*c)*a^2*b-2*cos(1/2*d*x+1/2*c)*a*b^2-6*cos(1/2*d*x+1/2*c)*b^3)/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)/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{\cos \left (d x + c\right )^{3}}{\sqrt{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/(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^3/sqrt(b*cos(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(cos(d*x + c)^3/sqrt(b*cos(d*x + c) + a), 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)**3/(a+b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{3}}{\sqrt{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/(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^3/sqrt(b*cos(d*x + c) + a), x)