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

Optimal. Leaf size=218 \[ -\frac{2 \left (2 a^2 C-3 b^2 (5 A+3 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{4 a C \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}-\frac{4 a C \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{15 b d} \]

[Out]

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

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Rubi [A]  time = 0.325578, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {3024, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 \left (2 a^2 C-3 b^2 (5 A+3 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{4 a C \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}-\frac{4 a C \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{15 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3024

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (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) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] &&  !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 \sqrt{a+b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac{2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}+\frac{2 \int \sqrt{a+b \cos (c+d x)} \left (\frac{1}{2} b (5 A+3 C)-a C \cos (c+d x)\right ) \, dx}{5 b}\\ &=-\frac{4 a C \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac{2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}+\frac{4 \int \frac{\frac{1}{4} a b (15 A+7 C)-\frac{1}{4} \left (2 a^2 C-3 b^2 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{15 b}\\ &=-\frac{4 a C \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac{2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}+\frac{\left (2 a \left (a^2-b^2\right ) C\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{15 b^2}+\frac{1}{15} \left (15 A+\left (9-\frac{2 a^2}{b^2}\right ) C\right ) \int \sqrt{a+b \cos (c+d x)} \, dx\\ &=-\frac{4 a C \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac{2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}+\frac{\left (\left (15 A+\left (9-\frac{2 a^2}{b^2}\right ) 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}{15 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\left (2 a \left (a^2-b^2\right ) C \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 (15 A+\left (9-\frac{2 a^2}{b^2}\right ) C\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 ) C \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 C \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac{2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}\\ \end{align*}

Mathematica [A]  time = 0.92953, size = 181, normalized size = 0.83 \[ \frac{-2 (a+b) \left (2 a^2 C-15 A b^2-9 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+b C \sin (c+d x) \left (2 a^2+8 a b \cos (c+d x)+3 b^2 \cos (2 (c+d x))+3 b^2\right )+4 a C \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b)*(-15*A*b^2 + 2*a^2*C - 9*b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(a
+ b)] + 4*a*(a^2 - b^2)*C*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + b*C*(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 = 0.355, size = 821, 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((A+C*cos(d*x+c)^2)*(a+b*cos(d*x+c))^(1/2),x)

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(24*C*cos(1/2*d*x+1/2*c)^7*b^3+16*C*cos(1/2*
d*x+1/2*c)^5*a*b^2-48*C*cos(1/2*d*x+1/2*c)^5*b^3+15*A*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2-15*A*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+2*C*cos(1/2*d*x
+1/2*c)^3*a^2*b-24*C*cos(1/2*d*x+1/2*c)^3*a*b^2+30*C*cos(1/2*d*x+1/2*c)^3*b^3+2*C*(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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3-2*a*C*(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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))
^(1/2))*b^2-2*C*(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)*EllipticE(cos(1/2*d*
x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+2*C*(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)
*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b+9*C*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2-9*C*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3-2*C*cos(1/
2*d*x+1/2*c)*a^2*b+8*C*cos(1/2*d*x+1/2*c)*a*b^2-6*C*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{\left (C \cos \left (d x + c\right )^{2} + A\right )} \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((A+C*cos(d*x+c)^2)*(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*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 ({\left (C \cos \left (d x + c\right )^{2} + A\right )} \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((A+C*cos(d*x+c)^2)*(a+b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + A)*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((A+C*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((A+C*cos(d*x+c)^2)*(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Timed out