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

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

[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)

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Rubi [A]  time = 0.0377128, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, 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 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]

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{\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*}

Mathematica [A]  time = 0.0968243, size = 56, normalized size = 0.78 \[ \frac{2 \sec ^3(c+d x) \left (\cos ^{\frac{3}{2}}(c+d x) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+\sin (c+d x)\right )}{3 d (b \sec (c+d x))^{3/2}} \]

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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Maple [C]  time = 0.17, size = 125, normalized size = 1.7 \begin{align*} -{\frac{ \left ( -2+2\,\cos \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}{3\,d{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}} \left ( i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) \left ({\frac{b}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}}} \end{align*}

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{4}}{\left (b \sec \left (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sec \left (d x + c\right )} \sec \left (d x + c\right )^{2}}{b^{2}}, x\right ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{4}{\left (c + d x \right )}}{\left (b \sec{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{4}}{\left (b \sec \left (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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)