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

Optimal. Leaf size=42 \[ \frac {2 F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{3 b}+\frac {2 \sin (a+b x)}{3 b \cos ^{\frac {3}{2}}(a+b x)} \]

[Out]

2/3*(cos(1/2*b*x+1/2*a)^2)^(1/2)/cos(1/2*b*x+1/2*a)*EllipticF(sin(1/2*b*x+1/2*a),2^(1/2))/b+2/3*sin(b*x+a)/b/c
os(b*x+a)^(3/2)

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Rubi [A]  time = 0.02, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2636, 2641} \[ \frac {2 F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{3 b}+\frac {2 \sin (a+b x)}{3 b \cos ^{\frac {3}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^(-5/2),x]

[Out]

(2*EllipticF[(a + b*x)/2, 2])/(3*b) + (2*Sin[a + b*x])/(3*b*Cos[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 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}{\cos ^{\frac {5}{2}}(a+b x)} \, dx &=\frac {2 \sin (a+b x)}{3 b \cos ^{\frac {3}{2}}(a+b x)}+\frac {1}{3} \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx\\ &=\frac {2 F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{3 b}+\frac {2 \sin (a+b x)}{3 b \cos ^{\frac {3}{2}}(a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 36, normalized size = 0.86 \[ \frac {2 \left (F\left (\left .\frac {1}{2} (a+b x)\right |2\right )+\frac {\sin (a+b x)}{\cos ^{\frac {3}{2}}(a+b x)}\right )}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^(-5/2),x]

[Out]

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

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\cos \left (b x + a\right )^{\frac {5}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(cos(b*x + a)^(-5/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\cos \left (b x + a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(b*x + a)^(-5/2), x)

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maple [B]  time = 0.12, size = 213, normalized size = 5.07 \[ -\frac {2 \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )}\, \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )^{\frac {3}{2}} \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/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))*((2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)/(
-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)^(3/2)/sin(1/2*b*x+1/2*a)/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\cos \left (b x + a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(b*x + a)^(-5/2), x)

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mupad [B]  time = 0.27, size = 42, normalized size = 1.00 \[ \frac {2\,\sin \left (a+b\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (a+b\,x\right )}^2\right )}{3\,b\,{\cos \left (a+b\,x\right )}^{3/2}\,\sqrt {{\sin \left (a+b\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(a + b*x)^(5/2),x)

[Out]

(2*sin(a + b*x)*hypergeom([-3/4, 1/2], 1/4, cos(a + b*x)^2))/(3*b*cos(a + b*x)^(3/2)*(sin(a + b*x)^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\cos ^{\frac {5}{2}}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(b*x+a)**(5/2),x)

[Out]

Integral(cos(a + b*x)**(-5/2), x)

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