3.22 \(\int \frac{1}{(c \cos (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=68 \[ \frac{2 \sin (a+b x)}{b c \sqrt{c \cos (a+b x)}}-\frac{2 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{c \cos (a+b x)}}{b c^2 \sqrt{\cos (a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.0360195, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2636, 2640, 2639} \[ \frac{2 \sin (a+b x)}{b c \sqrt{c \cos (a+b x)}}-\frac{2 E\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{c \cos (a+b x)}}{b c^2 \sqrt{\cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Cos[a + b*x])^(-3/2),x]

[Out]

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

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0303218, size = 50, normalized size = 0.74 \[ \frac{2 \left (\sin (a+b x)-\sqrt{\cos (a+b x)} E\left (\left .\frac{1}{2} (a+b x)\right |2\right )\right )}{b c \sqrt{c \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Cos[a + b*x])^(-3/2),x]

[Out]

(2*(-(Sqrt[Cos[a + b*x]]*EllipticE[(a + b*x)/2, 2]) + Sin[a + b*x]))/(b*c*Sqrt[c*Cos[a + b*x]])

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Maple [A]  time = 2.045, size = 168, normalized size = 2.5 \begin{align*} -2\,{\frac{\sqrt{-2\,c \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}+c \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}} \left ( \sqrt{2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,bx+a/2 \right ) ,\sqrt{2} \right ) -2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}\cos \left ( 1/2\,bx+a/2 \right ) \right ) }{c\sqrt{-c \left ( 2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2} \right ) }\sin \left ( 1/2\,bx+a/2 \right ) \sqrt{c \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) }b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*cos(b*x+a))^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (c \cos \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*cos(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate((c*cos(b*x + a))^(-3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c \cos \left (b x + a\right )}}{c^{2} \cos \left (b x + a\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*cos(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*cos(b*x + a))/(c^2*cos(b*x + 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(1/(c*cos(b*x+a))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (c \cos \left (b x + a\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*cos(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate((c*cos(b*x + a))^(-3/2), x)