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

Optimal. Leaf size=175 \[ -\frac {4 \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}}+\frac {4 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 b d} \]

[Out]

2/3*cos(d*x+c)*(a+b*sin(d*x+c))^(1/2)/b/d-4/3*a*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*
EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*(a+b*sin(d*x+c))^(1/2)/b^2/d/((a+b*sin(d*x+c))/(a
+b))^(1/2)+4/3*(a^2-b^2)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4
*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2)/b^2/d/(a+b*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2695, 2752, 2663, 2661, 2655, 2653} \[ -\frac {4 \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}}+\frac {4 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.82, size = 145, normalized size = 0.83 \[ \frac {4 \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+2 b \cos (c+d x) (a+b \sin (c+d x))-4 a (a+b) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{3 b^2 d \sqrt {a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\cos \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right ) + a}}, x\right ) \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right ) + a}}\,{d x} \]

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

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maple [B]  time = 0.74, size = 462, normalized size = 2.64 \[ \frac {\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b}{3}-\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{3}}{3}-\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticE \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3}}{3}+\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticE \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{2}}{3}-\frac {2 b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {2 a \,b^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{3}+\frac {2 b^{3} \sin \left (d x +c \right )}{3}+\frac {2 a \,b^{2}}{3}}{b^{3} \cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d} \]

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]

2/3*(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)*Elliptic
F(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b-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
^3-2*((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)*EllipticE(
((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3+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)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a*b^2
-b^3*sin(d*x+c)^3-a*b^2*sin(d*x+c)^2+b^3*sin(d*x+c)+a*b^2)/b^3/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right ) + a}}\,{d x} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^2}{\sqrt {a+b\,\sin \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{2}{\left (c + d x \right )}}{\sqrt {a + b \sin {\left (c + d x \right )}}}\, dx \]

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

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