3.24 \(\int \frac{1}{(c \sec (a+b x))^{7/2}} \, dx\)

Optimal. Leaf size=100 \[ \frac{10 \sqrt{\cos (a+b x)} \text{EllipticF}\left (\frac{1}{2} (a+b x),2\right ) \sqrt{c \sec (a+b x)}}{21 b c^4}+\frac{10 \sin (a+b x)}{21 b c^3 \sqrt{c \sec (a+b x)}}+\frac{2 \sin (a+b x)}{7 b c (c \sec (a+b x))^{5/2}} \]

[Out]

(10*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2]*Sqrt[c*Sec[a + b*x]])/(21*b*c^4) + (2*Sin[a + b*x])/(7*b*c*(c
*Sec[a + b*x])^(5/2)) + (10*Sin[a + b*x])/(21*b*c^3*Sqrt[c*Sec[a + b*x]])

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Rubi [A]  time = 0.059252, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3769, 3771, 2641} \[ \frac{10 \sin (a+b x)}{21 b c^3 \sqrt{c \sec (a+b x)}}+\frac{10 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{c \sec (a+b x)}}{21 b c^4}+\frac{2 \sin (a+b x)}{7 b c (c \sec (a+b x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sec[a + b*x])^(-7/2),x]

[Out]

(10*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2]*Sqrt[c*Sec[a + b*x]])/(21*b*c^4) + (2*Sin[a + b*x])/(7*b*c*(c
*Sec[a + b*x])^(5/2)) + (10*Sin[a + b*x])/(21*b*c^3*Sqrt[c*Sec[a + b*x]])

Rule 3769

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

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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 \sec (a+b x))^{7/2}} \, dx &=\frac{2 \sin (a+b x)}{7 b c (c \sec (a+b x))^{5/2}}+\frac{5 \int \frac{1}{(c \sec (a+b x))^{3/2}} \, dx}{7 c^2}\\ &=\frac{2 \sin (a+b x)}{7 b c (c \sec (a+b x))^{5/2}}+\frac{10 \sin (a+b x)}{21 b c^3 \sqrt{c \sec (a+b x)}}+\frac{5 \int \sqrt{c \sec (a+b x)} \, dx}{21 c^4}\\ &=\frac{2 \sin (a+b x)}{7 b c (c \sec (a+b x))^{5/2}}+\frac{10 \sin (a+b x)}{21 b c^3 \sqrt{c \sec (a+b x)}}+\frac{\left (5 \sqrt{\cos (a+b x)} \sqrt{c \sec (a+b x)}\right ) \int \frac{1}{\sqrt{\cos (a+b x)}} \, dx}{21 c^4}\\ &=\frac{10 \sqrt{\cos (a+b x)} F\left (\left .\frac{1}{2} (a+b x)\right |2\right ) \sqrt{c \sec (a+b x)}}{21 b c^4}+\frac{2 \sin (a+b x)}{7 b c (c \sec (a+b x))^{5/2}}+\frac{10 \sin (a+b x)}{21 b c^3 \sqrt{c \sec (a+b x)}}\\ \end{align*}

Mathematica [A]  time = 0.0960934, size = 66, normalized size = 0.66 \[ \frac{\sqrt{c \sec (a+b x)} \left (40 \sqrt{\cos (a+b x)} \text{EllipticF}\left (\frac{1}{2} (a+b x),2\right )+26 \sin (2 (a+b x))+3 \sin (4 (a+b x))\right )}{84 b c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Sec[a + b*x])^(-7/2),x]

[Out]

(Sqrt[c*Sec[a + b*x]]*(40*Sqrt[Cos[a + b*x]]*EllipticF[(a + b*x)/2, 2] + 26*Sin[2*(a + b*x)] + 3*Sin[4*(a + b*
x)]))/(84*b*c^4)

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Maple [C]  time = 0.19, size = 153, normalized size = 1.5 \begin{align*} -{\frac{2\, \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2} \left ( -1+\cos \left ( bx+a \right ) \right ) }{21\,b \left ( \cos \left ( bx+a \right ) \right ) ^{4} \left ( \sin \left ( bx+a \right ) \right ) ^{3}} \left ( 5\,i{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( bx+a \right ) \right ) }{\sin \left ( bx+a \right ) }},i \right ) \sqrt{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( bx+a \right ) }{\cos \left ( bx+a \right ) +1}}}\sin \left ( bx+a \right ) -3\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}+3\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}-5\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}+5\,\cos \left ( bx+a \right ) \right ) \left ({\frac{c}{\cos \left ( bx+a \right ) }} \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*sec(b*x+a))^(7/2),x)

[Out]

-2/21/b*(cos(b*x+a)+1)^2*(-1+cos(b*x+a))*(5*I*EllipticF(I*(-1+cos(b*x+a))/sin(b*x+a),I)*(1/(cos(b*x+a)+1))^(1/
2)*(cos(b*x+a)/(cos(b*x+a)+1))^(1/2)*sin(b*x+a)-3*cos(b*x+a)^4+3*cos(b*x+a)^3-5*cos(b*x+a)^2+5*cos(b*x+a))/(c/
cos(b*x+a))^(7/2)/cos(b*x+a)^4/sin(b*x+a)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (c \sec \left (b x + a\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*sec(b*x+a))^(7/2),x, algorithm="maxima")

[Out]

integrate((c*sec(b*x + a))^(-7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c \sec \left (b x + a\right )}}{c^{4} \sec \left (b x + a\right )^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*sec(b*x+a))^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*sec(b*x + a))/(c^4*sec(b*x + a)^4), 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*sec(b*x+a))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (c \sec \left (b x + a\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*sec(b*x+a))^(7/2),x, algorithm="giac")

[Out]

integrate((c*sec(b*x + a))^(-7/2), x)