3.487 \(\int \cos ^2(c+d x) \sqrt{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=207 \[ \frac{4 a \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (2 a^2-9 b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}-\frac{4 a \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{15 b d} \]

[Out]

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

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Rubi [A]  time = 0.28356, antiderivative size = 207, 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 = {2791, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{4 a \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (2 a^2-9 b^2\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}-\frac{4 a \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{15 b d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*Sqrt[a + b*Cos[c + d*x]],x]

[Out]

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

Rule 2791

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

Rule 2753

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

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 \cos ^2(c+d x) \sqrt{a+b \cos (c+d x)} \, dx &=\frac{2 (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}+\frac{2 \int \left (\frac{3 b}{2}-a \cos (c+d x)\right ) \sqrt{a+b \cos (c+d x)} \, dx}{5 b}\\ &=-\frac{4 a \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac{2 (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}+\frac{4 \int \frac{\frac{7 a b}{4}-\frac{1}{4} \left (2 a^2-9 b^2\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{15 b}\\ &=-\frac{4 a \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac{2 (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}+\frac{1}{15} \left (9-\frac{2 a^2}{b^2}\right ) \int \sqrt{a+b \cos (c+d x)} \, dx+\frac{\left (2 a \left (a^2-b^2\right )\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{15 b^2}\\ &=-\frac{4 a \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac{2 (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}+\frac{\left (\left (9-\frac{2 a^2}{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 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\left (2 a \left (a^2-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^2 \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 \left (9-\frac{2 a^2}{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 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{4 a \left (a^2-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^2 d \sqrt{a+b \cos (c+d x)}}-\frac{4 a \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac{2 (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}\\ \end{align*}

Mathematica [A]  time = 0.791228, size = 180, normalized size = 0.87 \[ \frac{b \sin (c+d x) \left (2 a^2+8 a b \cos (c+d x)+3 b^2 \cos (2 (c+d x))+3 b^2\right )+4 a \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-2 \left (2 a^2 b+2 a^3-9 a b^2-9 b^3\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*Sqrt[a + b*Cos[c + d*x]],x]

[Out]

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

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Maple [B]  time = 2.934, size = 665, normalized size = 3.2 \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)^2*(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+16*cos(1/2*d*x+
1/2*c)^5*a*b^2-48*cos(1/2*d*x+1/2*c)^5*b^3+2*cos(1/2*d*x+1/2*c)^3*a^2*b-24*cos(1/2*d*x+1/2*c)^3*a*b^2+30*cos(1
/2*d*x+1/2*c)^3*b^3+2*(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^3-2*(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^2-2*(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+2*(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^3-2*cos(1/2*d*x+1/2*c)*a^2*b+8*cos(1/2*d*x+1/2*c)*a*b^2-6*cos(1/2*d*x+1/2*c)*b^3)/
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*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 \sqrt{b \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*cos(d*x + c) + a)*cos(d*x + c)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [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)^2*(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Timed out