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

Optimal. Leaf size=170 \[ \frac{2 a \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{2 a \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{b d \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{b d \sqrt{a+b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.182183, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2754, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 a \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{2 a \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{b d \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{b d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*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] && LtQ[m, -1] && 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 (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx &=\frac{2 a \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{2 \int \frac{\frac{b}{2}+\frac{1}{2} a \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{a^2-b^2}\\ &=\frac{2 a \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}+\frac{\int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{b}-\frac{a \int \sqrt{a+b \cos (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{2 a \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{\left (a \sqrt{a+b \cos (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}} \, dx}{b \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\sqrt{\frac{a+b \cos (c+d x)}{a+b}} \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{b \sqrt{a+b \cos (c+d x)}}\\ &=-\frac{2 a \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{b \left (a^2-b^2\right ) d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{b d \sqrt{a+b \cos (c+d x)}}+\frac{2 a \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.507921, size = 137, normalized size = 0.81 \[ \frac{2 \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+2 a b \sin (c+d x)-2 a (a+b) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{b d (a-b) (a+b) \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 3.72, size = 373, normalized size = 2.2 \begin{align*} -2\,{\frac{1}{ \left ( a+b \right ) \left ( a-b \right ) b\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b}d} \left ( \sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\,{\frac{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b}{a-b}}+{\frac{a+b}{a-b}}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{-2\,{\frac{b}{a-b}}} \right ){a}^{2}-{b}^{2}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\,{\frac{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b}{a-b}}+{\frac{a+b}{a-b}}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{-2\,{\frac{b}{a-b}}} \right ) -\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\,{\frac{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b}{a-b}}+{\frac{a+b}{a-b}}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{-2\,{\frac{b}{a-b}}} \right ){a}^{2}+\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\,{\frac{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b}{a-b}}+{\frac{a+b}{a-b}}}b{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{-2\,{\frac{b}{a-b}}} \right ) a-2\,ab\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(cos(d*x + c)/(b*cos(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 \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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