3.2.24 \(\int \frac {\sec ^4(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\) [124]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 68, 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, 3853, 3856, 2719} \begin {gather*} \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)}} \end {gather*}

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 2719

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

Rule 3853

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 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 {\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*}

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Mathematica [A]
time = 0.06, size = 51, normalized size = 0.75 \begin {gather*} \frac {-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\sqrt {\cos (c+d x)}}+2 \tan (c+d x)}{b^2 d \sqrt {b \sec (c+d x)}} \end {gather*}

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]])

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Maple [C] Result contains complex when optimal does not.
time = 53.60, size = 322, normalized size = 4.74

method	result	size
default	\(-\frac {2 \left (\cos \left (d x +c \right )+1\right )^{2} \left (\cos \left (d x +c \right )-1\right )^{2} \left (i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right )-i \cos \left (d x +c \right ) \EllipticE \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right )+i \EllipticF \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right )-i \EllipticE \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\cos \left (d x +c \right )-1\right )}{d \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )^{5} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}\)	\(322\)

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,method=_RETURNVERBOSE)

[Out]

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

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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(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)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.73, size = 86, normalized size = 1.26 \begin {gather*} \frac {-i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + i \, \sqrt {2} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{b^{3} d} \end {gather*}

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]

(-I*sqrt(2)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + I*sqrt
(2)*sqrt(b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*sqrt(b/cos(d
*x + c))*sin(d*x + c))/(b^3*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

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)

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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(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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(b/cos(c + d*x))^(5/2)),x)

[Out]

int(1/(cos(c + d*x)^4*(b/cos(c + d*x))^(5/2)), x)

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