3.1.24 \(\int \frac {1}{(c \sec (a+b x))^{7/2}} \, dx\) [24]

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

[Out]

2/7*sin(b*x+a)/b/c/(c*sec(b*x+a))^(5/2)+10/21*sin(b*x+a)/b/c^3/(c*sec(b*x+a))^(1/2)+10/21*(cos(1/2*a+1/2*b*x)^
2)^(1/2)/cos(1/2*a+1/2*b*x)*EllipticF(sin(1/2*a+1/2*b*x),2^(1/2))*cos(b*x+a)^(1/2)*(c*sec(b*x+a))^(1/2)/b/c^4

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Rubi [A]
time = 0.04, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3854, 3856, 2720} \begin {gather*} \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 {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}} \end {gather*}

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 2720

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

Rule 3854

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 3856

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]

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*}

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Mathematica [A]
time = 0.11, size = 66, normalized size = 0.66 \begin {gather*} \frac {\sqrt {c \sec (a+b x)} \left (40 \sqrt {\cos (a+b x)} F\left (\left .\frac {1}{2} (a+b x)\right |2\right )+26 \sin (2 (a+b x))+3 \sin (4 (a+b x))\right )}{84 b c^4} \end {gather*}

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] Result contains complex when optimal does not.
time = 27.34, size = 153, normalized size = 1.53

method	result	size
default	\(-\frac {2 \left (\cos \left (b x +a \right )+1\right )^{2} \left (-1+\cos \left (b x +a \right )\right ) \left (5 i \EllipticF \left (\frac {i \left (-1+\cos \left (b x +a \right )\right )}{\sin \left (b x +a \right )}, i\right ) \sin \left (b x +a \right ) \sqrt {\frac {1}{\cos \left (b x +a \right )+1}}\, \sqrt {\frac {\cos \left (b x +a \right )}{\cos \left (b x +a \right )+1}}-3 \left (\cos ^{4}\left (b x +a \right )\right )+3 \left (\cos ^{3}\left (b x +a \right )\right )-5 \left (\cos ^{2}\left (b x +a \right )\right )+5 \cos \left (b x +a \right )\right )}{21 b \sin \left (b x +a \right )^{3} \left (\frac {c}{\cos \left (b x +a \right )}\right )^{\frac {7}{2}} \cos \left (b x +a \right )^{4}}\)	\(153\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.03, size = 100, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (3 \, \cos \left (b x + a\right )^{3} + 5 \, \cos \left (b x + a\right )\right )} \sqrt {\frac {c}{\cos \left (b x + a\right )}} \sin \left (b x + a\right ) - 5 i \, \sqrt {2} \sqrt {c} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + 5 i \, \sqrt {2} \sqrt {c} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )}{21 \, b c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21*(2*(3*cos(b*x + a)^3 + 5*cos(b*x + a))*sqrt(c/cos(b*x + a))*sin(b*x + a) - 5*I*sqrt(2)*sqrt(c)*weierstras
sPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a)) + 5*I*sqrt(2)*sqrt(c)*weierstrassPInverse(-4, 0, cos(b*x + a)
- I*sin(b*x + a)))/(b*c^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (c \sec {\left (a + b x \right )}\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*sec(a + b*x))**(-7/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (\frac {c}{\cos \left (a+b\,x\right )}\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c/cos(a + b*x))^(7/2),x)

[Out]

int(1/(c/cos(a + b*x))^(7/2), x)

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