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

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

[Out]

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

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Rubi [A]  time = 0.05, 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 = {3769, 3771, 2641} \[ \frac {10 \sin (c+d x)}{21 b^3 d \sqrt {b \sec (c+d x)}}+\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 b^4 d}+\frac {2 \sin (c+d x)}{7 b d (b \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x])^(-7/2),x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 66, normalized size = 0.66 \[ \frac {\sqrt {b \sec (c+d x)} \left (26 \sin (2 (c+d x))+3 \sin (4 (c+d x))+40 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{84 b^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[c + d*x])^(-7/2),x]

[Out]

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

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (d x + c\right )}}{b^{4} \sec \left (d x + c\right )^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sec(d*x + c))/(b^4*sec(d*x + c)^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sec \left (d x + c\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sec(d*x + c))^(-7/2), x)

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maple [C]  time = 0.90, size = 153, normalized size = 1.53 \[ -\frac {2 \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (5 i \sin \left (d x +c \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-3 \left (\cos ^{4}\left (d x +c \right )\right )+3 \left (\cos ^{3}\left (d x +c \right )\right )-5 \left (\cos ^{2}\left (d x +c \right )\right )+5 \cos \left (d x +c \right )\right )}{21 d \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sec \left (d x + c\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sec(d*x + c))^(-7/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/cos(c + d*x))^(7/2),x)

[Out]

int(1/(b/cos(c + d*x))^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*sec(c + d*x))**(-7/2), x)

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