3.524 \(\int \frac{\cos ^2(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=160 \[ \frac{4 a \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{b^2 d \sqrt{a+b \sin (c+d x)}}-\frac{4 \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{b^2 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 \cos (c+d x)}{b d \sqrt{a+b \sin (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.187307, antiderivative size = 160, normalized size of antiderivative = 1., 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 = {2693, 2752, 2663, 2661, 2655, 2653} \[ \frac{4 a \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{b^2 d \sqrt{a+b \sin (c+d x)}}-\frac{4 \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{b^2 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 \cos (c+d x)}{b d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

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

Rule 2693

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 + 1)), x] + Dist[(g^2*(p - 1))/(b*(m + 1)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a
^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && 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]

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

Mathematica [A]  time = 2.73182, size = 125, normalized size = 0.78 \[ \frac{4 (a+b) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-2 \left (2 a \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+b \cos (c+d x)\right )}{b^2 d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(a + b*Sin[c + d*x])^(3/2),x]

[Out]

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

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Maple [B]  time = 0.549, size = 434, normalized size = 2.7 \begin{align*} 2\,{\frac{1}{{b}^{3}\cos \left ( dx+c \right ) \sqrt{a+b\sin \left ( dx+c \right ) }d} \left ( 2\,\sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}}\sqrt{-{\frac{ \left ( \sin \left ( dx+c \right ) -1 \right ) b}{a+b}}}\sqrt{-{\frac{ \left ( 1+\sin \left ( dx+c \right ) \right ) b}{a-b}}}{\it EllipticE} \left ( \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ){a}^{2}-2\,\sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}}\sqrt{-{\frac{ \left ( \sin \left ( dx+c \right ) -1 \right ) b}{a+b}}}\sqrt{-{\frac{ \left ( 1+\sin \left ( dx+c \right ) \right ) b}{a-b}}}{\it EllipticE} \left ( \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ){b}^{2}-2\,a\sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}}\sqrt{-{\frac{ \left ( \sin \left ( dx+c \right ) -1 \right ) b}{a+b}}}\sqrt{-{\frac{ \left ( 1+\sin \left ( dx+c \right ) \right ) b}{a-b}}}{\it EllipticF} \left ( \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ) b+2\,\sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}}\sqrt{-{\frac{ \left ( \sin \left ( dx+c \right ) -1 \right ) b}{a+b}}}\sqrt{-{\frac{ \left ( 1+\sin \left ( dx+c \right ) \right ) b}{a-b}}}{\it EllipticF} \left ( \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ){b}^{2}+ \left ( \sin \left ( dx+c \right ) \right ) ^{2}{b}^{2}-{b}^{2} \right ) } \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))^(3/2),x)

[Out]

2*(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^2-2*((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))*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)*EllipticF(((
a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*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^2+sin(d
*x+c)^2*b^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 \frac{\cos \left (d x + c\right )^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{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))^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^2/(b*sin(d*x + c) + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{b \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{2}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{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))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{2}{\left (c + d x \right )}}{\left (a + b \sin{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, 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))**(3/2),x)

[Out]

Integral(cos(c + d*x)**2/(a + b*sin(c + d*x))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{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))^(3/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^2/(b*sin(d*x + c) + a)^(3/2), x)