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

Optimal. Leaf size=247 \[ -\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a^2-3 a b \sin (c+d x)-5 b^2\right )}{35 b^3 d}+\frac{8 \left (-9 a^2 b^2+4 a^4+5 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{35 b^4 d \sqrt{a+b \sin (c+d x)}}-\frac{32 a \left (a^2-2 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{35 b^4 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{7 b d} \]

[Out]

(2*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]])/(7*b*d) - (32*a*(a^2 - 2*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/
(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(35*b^4*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (8*(4*a^4 - 9*a^2*b^2 + 5*b
^4)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(35*b^4*d*Sqrt[a + b*Sin[
c + d*x]]) - (4*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(4*a^2 - 5*b^2 - 3*a*b*Sin[c + d*x]))/(35*b^3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.36545, antiderivative size = 247, 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, 2865, 2752, 2663, 2661, 2655, 2653} \[ -\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a^2-3 a b \sin (c+d x)-5 b^2\right )}{35 b^3 d}+\frac{8 \left (-9 a^2 b^2+4 a^4+5 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{35 b^4 d \sqrt{a+b \sin (c+d x)}}-\frac{32 a \left (a^2-2 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{35 b^4 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{7 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]])/(7*b*d) - (32*a*(a^2 - 2*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/
(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(35*b^4*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (8*(4*a^4 - 9*a^2*b^2 + 5*b
^4)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(35*b^4*d*Sqrt[a + b*Sin[
c + d*x]]) - (4*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(4*a^2 - 5*b^2 - 3*a*b*Sin[c + d*x]))/(35*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 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 \frac{\cos ^4(c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx &=\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{7 b d}+\frac{6 \int \frac{\cos ^2(c+d x) (b+a \sin (c+d x))}{\sqrt{a+b \sin (c+d x)}} \, dx}{7 b}\\ &=\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{7 b d}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a^2-5 b^2-3 a b \sin (c+d x)\right )}{35 b^3 d}+\frac{8 \int \frac{-\frac{1}{2} b \left (a^2-5 b^2\right )-2 a \left (a^2-2 b^2\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{35 b^3}\\ &=\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{7 b d}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a^2-5 b^2-3 a b \sin (c+d x)\right )}{35 b^3 d}-\frac{\left (16 a \left (a^2-2 b^2\right )\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{35 b^4}+\frac{\left (4 \left (4 a^4-9 a^2 b^2+5 b^4\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{35 b^4}\\ &=\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{7 b d}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a^2-5 b^2-3 a b \sin (c+d x)\right )}{35 b^3 d}-\frac{\left (16 a \left (a^2-2 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}{35 b^4 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (4 \left (4 a^4-9 a^2 b^2+5 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}{35 b^4 \sqrt{a+b \sin (c+d x)}}\\ &=\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{7 b d}-\frac{32 a \left (a^2-2 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)}}{35 b^4 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{8 \left (4 a^4-9 a^2 b^2+5 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}}}{35 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^2-5 b^2-3 a b \sin (c+d x)\right )}{35 b^3 d}\\ \end{align*}

Mathematica [A]  time = 1.03655, size = 219, normalized size = 0.89 \[ \frac{b \cos (c+d x) \left (\left (45 b^3-8 a^2 b\right ) \sin (c+d x)-32 a^3-2 a b^2 \cos (2 (c+d x))+62 a b^2+5 b^3 \sin (3 (c+d x))\right )-16 \left (-9 a^2 b^2+4 a^4+5 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+64 a \left (a^2 b+a^3-2 a b^2-2 b^3\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{70 b^4 d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*a*(a^3 + a^2*b - 2*a*b^2 - 2*b^3)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x]
)/(a + b)] - 16*(4*a^4 - 9*a^2*b^2 + 5*b^4)*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]*(-32*a^3 + 62*a*b^2 - 2*a*b^2*Cos[2*(c + d*x)] + (-8*a^2*b + 45*b^3)*Sin[c +
d*x] + 5*b^3*Sin[3*(c + d*x)]))/(70*b^4*d*Sqrt[a + b*Sin[c + d*x]])

________________________________________________________________________________________

Maple [B]  time = 0.547, size = 942, 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(cos(d*x+c)^4/(a+b*sin(d*x+c))^(1/2),x)

[Out]

-2/35*(-5*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^4*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^3*b^2-36*((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^3+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*b^4+20*((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^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)*EllipticE(((a+b*sin(d*x+c))/(a-b)
)^(1/2),((a-b)/(a+b))^(1/2))*a^5+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)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^2-32*((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*b^4+a*b^4*sin(d*x+c)^4-2*a^2*b^3*sin(d*x+c)^3+20*b^5*sin(d*x+c)^3-8*a^3*b^2*s
in(d*x+c)^2+14*a*b^4*sin(d*x+c)^2+2*a^2*b^3*sin(d*x+c)-15*b^5*sin(d*x+c)+8*a^3*b^2-15*a*b^4)/b^5/cos(d*x+c)/(a
+b*sin(d*x+c))^(1/2)/d

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{4}}{\sqrt{b \sin \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)^4/(a+b*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\cos \left (d x + c\right )^{4}}{\sqrt{b \sin \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)^4/(a+b*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{4}{\left (c + d x \right )}}{\sqrt{a + b \sin{\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(cos(c + d*x)**4/sqrt(a + b*sin(c + d*x)), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{4}}{\sqrt{b \sin \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)^4/(a+b*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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