3.124 \(\int \frac{\sec ^4(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=68 \[ \frac{2 \sin (c+d x) \sqrt{b \sec (c+d x)}}{b^3 d}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0383202, antiderivative size = 68, 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, 2639} \[ \frac{2 \sin (c+d x) \sqrt{b \sec (c+d x)}}{b^3 d}-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(b*Sec[c + d*x])^(5/2),x]

[Out]

(-2*EllipticE[(c + d*x)/2, 2])/(b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*Sqrt[b*Sec[c + d*x]]*Sin[c
+ d*x])/(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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(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))^{5/2}} \, dx &=\frac{\int (b \sec (c+d x))^{3/2} \, dx}{b^4}\\ &=\frac{2 \sqrt{b \sec (c+d x)} \sin (c+d x)}{b^3 d}-\frac{\int \frac{1}{\sqrt{b \sec (c+d x)}} \, dx}{b^2}\\ &=\frac{2 \sqrt{b \sec (c+d x)} \sin (c+d x)}{b^3 d}-\frac{\int \sqrt{\cos (c+d x)} \, dx}{b^2 \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}\\ &=-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 \sqrt{b \sec (c+d x)} \sin (c+d x)}{b^3 d}\\ \end{align*}

Mathematica [A]  time = 0.0525358, size = 51, normalized size = 0.75 \[ \frac{2 \tan (c+d x)-\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{\sqrt{\cos (c+d x)}}}{b^2 d \sqrt{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(b*Sec[c + d*x])^(5/2),x]

[Out]

((-2*EllipticE[(c + d*x)/2, 2])/Sqrt[Cos[c + d*x]] + 2*Tan[c + d*x])/(b^2*d*Sqrt[b*Sec[c + d*x]])

________________________________________________________________________________________

Maple [C]  time = 0.234, size = 322, normalized size = 4.7 \begin{align*} -2\,{\frac{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{5} \left ( \cos \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 ) -i\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) +i{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -i\sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) +\cos \left ( dx+c \right ) -1 \right ) \left ({\frac{b}{\cos \left ( dx+c \right ) }} \right ) ^{-5/2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(b*sec(d*x+c))^(5/2),x)

[Out]

-2/d*(cos(d*x+c)+1)^2*(-1+cos(d*x+c))^2*(I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Elliptic
F(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)*sin(d*x+c)-I*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)
*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+I*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c
),I)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-I*EllipticE(I*(-1+cos(d*x+c))/sin(d
*x+c),I)*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+cos(d*x+c)-1)/sin(d*x+c)^5/cos(
d*x+c)^3/(b/cos(d*x+c))^(5/2)

________________________________________________________________________________________

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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^4/(b*sec(d*x + c))^(5/2), x)

________________________________________________________________________________________

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 )}{b^{3}}, 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))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(d*x + c))*sec(d*x + c)/b^3, x)

________________________________________________________________________________________

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{5}{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))**(5/2),x)

[Out]

Integral(sec(c + d*x)**4/(b*sec(c + d*x))**(5/2), x)

________________________________________________________________________________________

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{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^4/(b*sec(d*x + c))^(5/2), x)