3.1.91 \(\int \sec (c+d x) (b \sec (c+d x))^{5/2} \, dx\) [91]

Optimal. Leaf size=97 \[ -\frac {6 b^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {6 b^2 \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {16, 3853, 3856, 2719} \begin {gather*} -\frac {6 b^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {6 b^2 \sin (c+d x) \sqrt {b \sec (c+d x)}}{5 d}+\frac {2 \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]
time = 0.18, size = 61, normalized size = 0.63 \begin {gather*} \frac {(b \sec (c+d x))^{5/2} \left (-12 \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+7 \sin (c+d x)+3 \sin (3 (c+d x))\right )}{10 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Sec[c + d*x])^(5/2)*(-12*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 7*Sin[c + d*x] + 3*Sin[3*(c + d*x)
]))/(10*d)

________________________________________________________________________________________

Maple [C] Result contains complex when optimal does not.
time = 44.94, size = 348, normalized size = 3.59

method	result	size
default	\(-\frac {2 \left (\cos \left (d x +c \right )+1\right )^{2} \left (\cos \left (d x +c \right )-1\right )^{2} \left (3 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right )-3 i \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \EllipticE \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+3 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-3 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \EllipticE \left (\frac {i \left (\cos \left (d x +c \right )-1\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )+3 \left (\cos ^{3}\left (d x +c \right )\right )-2 \left (\cos ^{2}\left (d x +c \right )\right )-1\right ) \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{5 d \sin \left (d x +c \right )^{5}}\)	\(348\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.79, size = 125, normalized size = 1.29 \begin {gather*} \frac {-3 i \, \sqrt {2} b^{\frac {5}{2}} \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} b^{\frac {5}{2}} \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 \, b^{2} \cos \left (d x + c\right )^{2} + b^{2}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{5 \, 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)*(b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________