3.479 \(\int \cos ^4(c+d x) \sqrt{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=298 \[ -\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a \left (a^2-3 b^2\right )-3 b \left (a^2+7 b^2\right ) \sin (c+d x)\right )}{315 b^3 d}+\frac{32 a \left (-4 a^2 b^2+a^4+3 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^4 d \sqrt{a+b \sin (c+d x)}}-\frac{8 \left (-15 a^2 b^2+4 a^4-21 b^4\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^4 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{2 \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 b d}-\frac{4 a \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{21 b d} \]

[Out]

(-4*a*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]])/(21*b*d) + (2*Cos[c + d*x]^3*(a + b*Sin[c + d*x])^(3/2))/(9*b*d
) - (8*(4*a^4 - 15*a^2*b^2 - 21*b^4)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(3
15*b^4*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (32*a*(a^4 - 4*a^2*b^2 + 3*b^4)*EllipticF[(c - Pi/2 + d*x)/2, (
2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(315*b^4*d*Sqrt[a + b*Sin[c + d*x]]) - (4*Cos[c + d*x]*Sqrt[
a + b*Sin[c + d*x]]*(4*a*(a^2 - 3*b^2) - 3*b*(a^2 + 7*b^2)*Sin[c + d*x]))/(315*b^3*d)

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Rubi [A]  time = 0.565301, antiderivative size = 298, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {2695, 2862, 2865, 2752, 2663, 2661, 2655, 2653} \[ -\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a \left (a^2-3 b^2\right )-3 b \left (a^2+7 b^2\right ) \sin (c+d x)\right )}{315 b^3 d}+\frac{32 a \left (-4 a^2 b^2+a^4+3 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^4 d \sqrt{a+b \sin (c+d x)}}-\frac{8 \left (-15 a^2 b^2+4 a^4-21 b^4\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{315 b^4 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{2 \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 b d}-\frac{4 a \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{21 b d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(-4*a*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]])/(21*b*d) + (2*Cos[c + d*x]^3*(a + b*Sin[c + d*x])^(3/2))/(9*b*d
) - (8*(4*a^4 - 15*a^2*b^2 - 21*b^4)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(3
15*b^4*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (32*a*(a^4 - 4*a^2*b^2 + 3*b^4)*EllipticF[(c - Pi/2 + d*x)/2, (
2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(315*b^4*d*Sqrt[a + b*Sin[c + d*x]]) - (4*Cos[c + d*x]*Sqrt[
a + b*Sin[c + d*x]]*(4*a*(a^2 - 3*b^2) - 3*b*(a^2 + 7*b^2)*Sin[c + d*x]))/(315*b^3*d)

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 2862

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[1/(m + p + 1), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(m + p + 1) + b*d*m + (a*d*
m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && Gt
Q[m, 0] &&  !LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x])

Rule 2865

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 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 ^4(c+d x) \sqrt{a+b \sin (c+d x)} \, dx &=\frac{2 \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 b d}+\frac{2 \int \cos ^2(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)} \, dx}{3 b}\\ &=-\frac{4 a \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{21 b d}+\frac{2 \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 b d}+\frac{4 \int \frac{\cos ^2(c+d x) \left (4 a b+\frac{1}{2} \left (a^2+7 b^2\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{21 b}\\ &=-\frac{4 a \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{21 b d}+\frac{2 \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 b d}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a \left (a^2-3 b^2\right )-3 b \left (a^2+7 b^2\right ) \sin (c+d x)\right )}{315 b^3 d}+\frac{16 \int \frac{-\frac{1}{4} a b \left (a^2-33 b^2\right )-\frac{1}{4} \left (4 a^4-15 a^2 b^2-21 b^4\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{315 b^3}\\ &=-\frac{4 a \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{21 b d}+\frac{2 \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 b d}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a \left (a^2-3 b^2\right )-3 b \left (a^2+7 b^2\right ) \sin (c+d x)\right )}{315 b^3 d}+\frac{\left (16 a \left (a^4-4 a^2 b^2+3 b^4\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{315 b^4}+\frac{\left (4 \left (-4 a^4+15 a^2 b^2+21 b^4\right )\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{315 b^4}\\ &=-\frac{4 a \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{21 b d}+\frac{2 \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 b d}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a \left (a^2-3 b^2\right )-3 b \left (a^2+7 b^2\right ) \sin (c+d x)\right )}{315 b^3 d}+\frac{\left (4 \left (-4 a^4+15 a^2 b^2+21 b^4\right ) \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{315 b^4 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (16 a \left (a^4-4 a^2 b^2+3 b^4\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}{315 b^4 \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{4 a \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{21 b d}+\frac{2 \cos ^3(c+d x) (a+b \sin (c+d x))^{3/2}}{9 b d}-\frac{8 \left (4 a^4-15 a^2 b^2-21 b^4\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)}}{315 b^4 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{32 a \left (a^4-4 a^2 b^2+3 b^4\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}}}{315 b^4 d \sqrt{a+b \sin (c+d x)}}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a \left (a^2-3 b^2\right )-3 b \left (a^2+7 b^2\right ) \sin (c+d x)\right )}{315 b^3 d}\\ \end{align*}

Mathematica [A]  time = 0.838995, size = 233, normalized size = 0.78 \[ \frac{2 b \cos (c+d x) (a+b \sin (c+d x)) \left (b \left (24 a^2+203 b^2\right ) \sin (c+d x)-32 a^3+10 a b^2 \cos (2 (c+d x))+106 a b^2+35 b^3 \sin (3 (c+d x))\right )+32 \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \left (a b^2 \left (a^2-33 b^2\right ) F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+\left (-15 a^2 b^2+4 a^4-21 b^4\right ) \left ((a+b) E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )\right )\right )}{1260 b^4 d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(32*(a*b^2*(a^2 - 33*b^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)] + (4*a^4 - 15*a^2*b^2 - 21*b^4)*((a
+ b)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)] - a*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]))*Sqr
t[(a + b*Sin[c + d*x])/(a + b)] + 2*b*Cos[c + d*x]*(a + b*Sin[c + d*x])*(-32*a^3 + 106*a*b^2 + 10*a*b^2*Cos[2*
(c + d*x)] + b*(24*a^2 + 203*b^2)*Sin[c + d*x] + 35*b^3*Sin[3*(c + d*x)]))/(1260*b^4*d*Sqrt[a + b*Sin[c + d*x]
])

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Maple [B]  time = 0.542, size = 1189, normalized size = 4. \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)^4*(a+b*sin(d*x+c))^(1/2),x)

[Out]

-2/315*(-35*b^6*sin(d*x+c)^6-40*a*b^5*sin(d*x+c)^5+16*((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))*a^5*b-12*
((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))*a^4*b^2-64*((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))*a^3*b
^3-72*((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))*a^2*b^4+48*((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))
*a*b^5+84*((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)*Ellip
ticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^6-16*((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^6+76*((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*b^2+24*((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^4-84*((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)*E
llipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^6+a^2*b^4*sin(d*x+c)^4+112*b^6*sin(d*x+c)^4-2*a
^3*b^3*sin(d*x+c)^3+146*a*b^5*sin(d*x+c)^3-8*a^4*b^2*sin(d*x+c)^2+28*a^2*b^4*sin(d*x+c)^2-77*b^6*sin(d*x+c)^2+
2*a^3*b^3*sin(d*x+c)-106*a*b^5*sin(d*x+c)+8*a^4*b^2-29*a^2*b^4)/b^5/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 )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sin(d*x + c) + a)*cos(d*x + c)^4, 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 )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sin(d*x + c) + a)*cos(d*x + c)^4, 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 ^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+b*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a + b*sin(c + d*x))*cos(c + d*x)**4, 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 )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sin(d*x + c) + a)*cos(d*x + c)^4, x)