3.23 \(\int \frac{1}{(c \cos (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=72 \[ \frac{2 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{3 b c^2 \sqrt{c \cos (a+b x)}}+\frac{2 \sin (a+b x)}{3 b c (c \cos (a+b x))^{3/2}} \]

[Out]

(2*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])/(3*b*c^2*Sqrt[c*Cos[a + b*x]]) + (2*Sin[a + b*x])/(3*b*c*(c*C
os[a + b*x])^(3/2))

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Rubi [A]  time = 0.0353468, antiderivative size = 72, 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, 2642, 2641} \[ \frac{2 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{3 b c^2 \sqrt{c \cos (a+b x)}}+\frac{2 \sin (a+b x)}{3 b c (c \cos (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Cos[a + b*x])^(-5/2),x]

[Out]

(2*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2])/(3*b*c^2*Sqrt[c*Cos[a + b*x]]) + (2*Sin[a + b*x])/(3*b*c*(c*C
os[a + b*x])^(3/2))

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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(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))^{5/2}} \, dx &=\frac{2 \sin (a+b x)}{3 b c (c \cos (a+b x))^{3/2}}+\frac{\int \frac{1}{\sqrt{c \cos (a+b x)}} \, dx}{3 c^2}\\ &=\frac{2 \sin (a+b x)}{3 b c (c \cos (a+b x))^{3/2}}+\frac{\sqrt{\cos (a+b x)} \int \frac{1}{\sqrt{\cos (a+b x)}} \, dx}{3 c^2 \sqrt{c \cos (a+b x)}}\\ &=\frac{2 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{3 b c^2 \sqrt{c \cos (a+b x)}}+\frac{2 \sin (a+b x)}{3 b c (c \cos (a+b x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0591189, size = 51, normalized size = 0.71 \[ \frac{2 \left (\tan (a+b x)+\sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right )\right )}{3 b c^2 \sqrt{c \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Cos[a + b*x])^(-5/2),x]

[Out]

(2*(Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2] + Tan[a + b*x]))/(3*b*c^2*Sqrt[c*Cos[a + b*x]])

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Maple [B]  time = 1.763, size = 241, normalized size = 3.4 \begin{align*} -{\frac{2}{3\,{c}^{2}b} \left ( -2\,\sqrt{ \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,bx+a/2 \right ) ,\sqrt{2} \right ) \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}+\sqrt{ \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ({\frac{bx}{2}}+{\frac{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 ) \sqrt{c \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}}{\frac{1}{\sqrt{-c \left ( 2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{4}- \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2} \right ) }}} \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) ^{-1} \left ( \sin \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{c \left ( 2\, \left ( \cos \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*cos(b*x+a))^(5/2),x)

[Out]

-2/3*(-2*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(cos(1/2*b*x+1/2*a),2^(1/2))*s
in(1/2*b*x+1/2*a)^2+(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(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*(c*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(
1/2)/(-c*(2*sin(1/2*b*x+1/2*a)^4-sin(1/2*b*x+1/2*a)^2))^(1/2)/(2*cos(1/2*b*x+1/2*a)^2-1)/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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*cos(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate((c*cos(b*x + a))^(-5/2), 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^{3} \cos \left (b x + a\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*cos(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*cos(b*x + a))/(c^3*cos(b*x + a)^3), 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))**(5/2),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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*cos(b*x+a))^(5/2),x, algorithm="giac")

[Out]

integrate((c*cos(b*x + a))^(-5/2), x)