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

Optimal. Leaf size=215 \[ -\frac{4 a \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{a+b \sin (c+d x)}}+\frac{4 \left (a^2+3 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac{4 a \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{15 b d} \]

[Out]

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

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Rubi [A]  time = 0.260553, 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 = {2695, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac{4 a \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{a+b \sin (c+d x)}}+\frac{4 \left (a^2+3 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{2 \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 b d}-\frac{4 a \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{15 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2695

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(b*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

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

Mathematica [A]  time = 0.798058, size = 185, normalized size = 0.86 \[ \frac{b \cos (c+d x) \left (2 a^2+8 a b \sin (c+d x)-3 b^2 \cos (2 (c+d x))+3 b^2\right )+4 a \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-4 \left (a^2 b+a^3+3 a b^2+3 b^3\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.389, size = 792, normalized size = 3.7 \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*sin(d*x+c))^(1/2),x)

[Out]

2/15*(2*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Ellipti
cF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b+6*a^2*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c
)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/
2))*b^2-2*a*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Ell
ipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^3-6*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-
1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2)
)*b^4-2*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Ellipti
cE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4-4*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b
/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^
2*b^2+6*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Ellipti
cE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^4-3*b^4*sin(d*x+c)^4-4*a*b^3*sin(d*x+c)^3-a^2*b^2*sin
(d*x+c)^2+3*b^4*sin(d*x+c)^2+4*a*b^3*sin(d*x+c)+a^2*b^2)/b^3/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sin \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*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sin(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 \sin \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*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+b*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a + b*sin(c + d*x))*cos(c + d*x)**2, x)

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

[Out]

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