3.2.22 \(\int \frac {\sec ^6(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx\) [122]

Optimal. Leaf size=100 \[ -\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {6 \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b^5 d} \]

[Out]

2/5*(b*sec(d*x+c))^(5/2)*sin(d*x+c)/b^5/d-6/5*(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)+6/5*sin(d*x+c)*(b*sec(d*x+c))^(1/2)/b^3/d

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Rubi [A]
time = 0.04, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, 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) (b \sec (c+d x))^{5/2}}{5 b^5 d}+\frac {6 \sin (c+d x) \sqrt {b \sec (c+d x)}}{5 b^3 d}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 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]^6/(b*Sec[c + d*x])^(5/2),x]

[Out]

(-6*EllipticE[(c + d*x)/2, 2])/(5*b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (6*Sqrt[b*Sec[c + d*x]]*Sin
[c + d*x])/(5*b^3*d) + (2*(b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*b^5*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 ^6(c+d x)}{(b \sec (c+d x))^{5/2}} \, dx &=\frac {\int (b \sec (c+d x))^{7/2} \, dx}{b^6}\\ &=\frac {2 (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b^5 d}+\frac {3 \int (b \sec (c+d x))^{3/2} \, dx}{5 b^4}\\ &=\frac {6 \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b^5 d}-\frac {3 \int \frac {1}{\sqrt {b \sec (c+d x)}} \, dx}{5 b^2}\\ &=\frac {6 \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b^5 d}-\frac {3 \int \sqrt {\cos (c+d x)} \, dx}{5 b^2 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\\ &=-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^2 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {6 \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac {2 (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b^5 d}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 64, normalized size = 0.64 \begin {gather*} \frac {-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\sqrt {\cos (c+d x)}}+2 \left (3+\sec ^2(c+d x)\right ) \tan (c+d x)}{5 b^2 d \sqrt {b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6/(b*Sec[c + d*x])^(5/2),x]

[Out]

((-6*EllipticE[(c + d*x)/2, 2])/Sqrt[Cos[c + d*x]] + 2*(3 + Sec[c + d*x]^2)*Tan[c + d*x])/(5*b^2*d*Sqrt[b*Sec[
c + d*x]])

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Maple [C] Result contains complex when optimal does not.
time = 51.20, size = 351, normalized size = 3.51

method	result	size
default	\(-\frac {2 \left (\cos \left (d x +c \right )+1\right )^{2} \left (\cos \left (d x +c \right )-1\right )^{2} \left (3 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right )-3 i \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \EllipticE \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+3 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-3 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \EllipticE \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+3 \left (\cos ^{3}\left (d x +c \right )\right )-2 \left (\cos ^{2}\left (d x +c \right )\right )-1\right ) \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{5 d \,b^{5} \sin \left (d x +c \right )^{5}}\)	\(351\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6/(b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/5/d*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)^2*(3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Ellip
ticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)^3*sin(d*x+c)-3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+
c)+1))^(1/2)*cos(d*x+c)^3*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x+c)+3*I*(1/(cos(d*x+c)+1))^(1/2)*(co
s(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(cos(d*x+c)-1)/sin(d*x+c),I)*cos(d*x+c)^2*sin(d*x+c)-3*I*(1/(cos(d*
x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2*EllipticE(I*(cos(d*x+c)-1)/sin(d*x+c),I)*sin(d*x
+c)+3*cos(d*x+c)^3-2*cos(d*x+c)^2-1)*(b/cos(d*x+c))^(5/2)/b^5/sin(d*x+c)^5

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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)^6/(b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^6/(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 = 0.60, size = 123, normalized size = 1.23 \begin {gather*} \frac {-3 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{2} {\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 ) + 3 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{2} {\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 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{5 \, b^{3} d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/5*(-3*I*sqrt(2)*sqrt(b)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*si
n(d*x + c))) + 3*I*sqrt(2)*sqrt(b)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x +
c) - I*sin(d*x + c))) + 2*(3*cos(d*x + c)^2 + 1)*sqrt(b/cos(d*x + c))*sin(d*x + c))/(b^3*d*cos(d*x + c)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{6}{\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)**6/(b*sec(d*x+c))**(5/2),x)

[Out]

Integral(sec(c + d*x)**6/(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)^6/(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^6/(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 )}^6\,{\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)^6*(b/cos(c + d*x))^(5/2)),x)

[Out]

int(1/(cos(c + d*x)^6*(b/cos(c + d*x))^(5/2)), x)

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