3.113 \(\int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx\)

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

[Out]

2/3*(b*sec(d*x+c))^(3/2)*sin(d*x+c)/b^3/d+2/3*(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^2/d

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Rubi [A]  time = 0.04, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 3768, 3771, 2641} \[ \frac {2 \sin (c+d x) (b \sec (c+d x))^{3/2}}{3 b^3 d}+\frac {2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(b*Sec[c + d*x])^(3/2),x]

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(3*b^2*d) + (2*(b*Sec[c + d*x])^(3/2)*Si
n[c + d*x])/(3*b^3*d)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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 {\sec ^4(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx &=\frac {\int (b \sec (c+d x))^{5/2} \, dx}{b^4}\\ &=\frac {2 (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 b^3 d}+\frac {\int \sqrt {b \sec (c+d x)} \, dx}{3 b^2}\\ &=\frac {2 (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 b^3 d}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 b^2}\\ &=\frac {2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{3 b^2 d}+\frac {2 (b \sec (c+d x))^{3/2} \sin (c+d x)}{3 b^3 d}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 56, normalized size = 0.78 \[ \frac {2 \sec ^3(c+d x) \left (\sin (c+d x)+\cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{3 d (b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(b*Sec[c + d*x])^(3/2),x]

[Out]

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

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (d x + c\right )} \sec \left (d x + c\right )^{2}}{b^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(d*x + c))*sec(d*x + c)^2/b^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{4}}{\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^4/(b*sec(d*x + c))^(3/2), x)

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maple [C]  time = 0.86, size = 125, normalized size = 1.74 \[ -\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}-\cos \left (d x +c \right )+1\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}{3 d \,b^{3} \sin \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(b*sec(d*x+c))^(3/2),x)

[Out]

-2/3/d*(-1+cos(d*x+c))*(I*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*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)*(1+cos(d*x+c))^2*(b/cos(d*x+c))^(3/2)/b^3/sin(d*x+c)^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{4}}{\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^4/(b*sec(d*x + c))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(b/cos(c + d*x))^(3/2)),x)

[Out]

int(1/(cos(c + d*x)^4*(b/cos(c + d*x))^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(b*sec(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)**4/(b*sec(c + d*x))**(3/2), x)

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