\(\int \sqrt {b \cos (c+d x)} \sec ^6(c+d x) \, dx\) [77]

Optimal result

Integrand size = 21, antiderivative size = 123 \[ \int \sqrt {b \cos (c+d x)} \sec ^6(c+d x) \, dx=-\frac {14 \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d \sqrt {\cos (c+d x)}}+\frac {2 b^5 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {14 b^3 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {14 b \sin (c+d x)}{15 d \sqrt {b \cos (c+d x)}} \]

[Out]

2/9*b^5*sin(d*x+c)/d/(b*cos(d*x+c))^(9/2)+14/45*b^3*sin(d*x+c)/d/(b*cos(d*x+c))^(5/2)+14/15*b*sin(d*x+c)/d/(b*
cos(d*x+c))^(1/2)-14/15*(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*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 2716, 2721, 2719} \[ \int \sqrt {b \cos (c+d x)} \sec ^6(c+d x) \, dx=\frac {2 b^5 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {14 b^3 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {14 b \sin (c+d x)}{15 d \sqrt {b \cos (c+d x)}}-\frac {14 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{15 d \sqrt {\cos (c+d x)}} \]

[In]

Int[Sqrt[b*Cos[c + d*x]]*Sec[c + d*x]^6,x]

[Out]

(-14*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*d*Sqrt[Cos[c + d*x]]) + (2*b^5*Sin[c + d*x])/(9*d*(b*
Cos[c + d*x])^(9/2)) + (14*b^3*Sin[c + d*x])/(45*d*(b*Cos[c + d*x])^(5/2)) + (14*b*Sin[c + d*x])/(15*d*Sqrt[b*
Cos[c + d*x]])

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 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

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 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = b^6 \int \frac {1}{(b \cos (c+d x))^{11/2}} \, dx \\ & = \frac {2 b^5 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {1}{9} \left (7 b^4\right ) \int \frac {1}{(b \cos (c+d x))^{7/2}} \, dx \\ & = \frac {2 b^5 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {14 b^3 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {1}{15} \left (7 b^2\right ) \int \frac {1}{(b \cos (c+d x))^{3/2}} \, dx \\ & = \frac {2 b^5 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {14 b^3 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {14 b \sin (c+d x)}{15 d \sqrt {b \cos (c+d x)}}-\frac {7}{15} \int \sqrt {b \cos (c+d x)} \, dx \\ & = \frac {2 b^5 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {14 b^3 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {14 b \sin (c+d x)}{15 d \sqrt {b \cos (c+d x)}}-\frac {\left (7 \sqrt {b \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{15 \sqrt {\cos (c+d x)}} \\ & = -\frac {14 \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d \sqrt {\cos (c+d x)}}+\frac {2 b^5 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {14 b^3 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {14 b \sin (c+d x)}{15 d \sqrt {b \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.64 \[ \int \sqrt {b \cos (c+d x)} \sec ^6(c+d x) \, dx=\frac {\sqrt {b \cos (c+d x)} \sec ^5(c+d x) \left (-336 \cos ^{\frac {9}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+150 \sin (c+d x)+91 \sin (3 (c+d x))+21 \sin (5 (c+d x))\right )}{360 d} \]

[In]

Integrate[Sqrt[b*Cos[c + d*x]]*Sec[c + d*x]^6,x]

[Out]

(Sqrt[b*Cos[c + d*x]]*Sec[c + d*x]^5*(-336*Cos[c + d*x]^(9/2)*EllipticE[(c + d*x)/2, 2] + 150*Sin[c + d*x] + 9
1*Sin[3*(c + d*x)] + 21*Sin[5*(c + d*x)]))/(360*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(413\) vs. \(2(131)=262\).

Time = 3.06 (sec) , antiderivative size = 414, normalized size of antiderivative = 3.37

method	result	size
default	\(-\frac {2 \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}}{144 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{5}}-\frac {7 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}}{180 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{3}}-\frac {14 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{15 \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}+\frac {7 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{15 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}}-\frac {7 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(414\)

[In]

int(sec(d*x+c)^6*(cos(d*x+c)*b)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*(-1/144*cos(1/2*d*x+1/2*c)/b*(-b*(2*sin(1/2*d
*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^5-7/180*cos(1/2*d*x+1/2*c)/b*(-b*(2*sin(1/
2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*
x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos
(1/2*d*x+1/2*c)^2+1)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)*EllipticF(cos(1/2*d*x+1/2*
c),2^(1/2))-7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-si
n(1/2*d*x+1/2*c)^2))^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))))/sin(
1/2*d*x+1/2*c)/((2*cos(1/2*d*x+1/2*c)^2-1)*b)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.04 \[ \int \sqrt {b \cos (c+d x)} \sec ^6(c+d x) \, dx=\frac {-21 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 i \, \sqrt {2} \sqrt {b} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (21 \, \cos \left (d x + c\right )^{4} + 7 \, \cos \left (d x + c\right )^{2} + 5\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{45 \, d \cos \left (d x + c\right )^{5}} \]

[In]

integrate(sec(d*x+c)^6*(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/45*(-21*I*sqrt(2)*sqrt(b)*cos(d*x + c)^5*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*
sin(d*x + c))) + 21*I*sqrt(2)*sqrt(b)*cos(d*x + c)^5*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x
+ c) - I*sin(d*x + c))) + 2*(21*cos(d*x + c)^4 + 7*cos(d*x + c)^2 + 5)*sqrt(b*cos(d*x + c))*sin(d*x + c))/(d*
cos(d*x + c)^5)

Sympy [F(-1)]

Timed out. \[ \int \sqrt {b \cos (c+d x)} \sec ^6(c+d x) \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**6*(b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \sqrt {b \cos (c+d x)} \sec ^6(c+d x) \, dx=\int { \sqrt {b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{6} \,d x } \]

[In]

integrate(sec(d*x+c)^6*(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*cos(d*x + c))*sec(d*x + c)^6, x)

Giac [F]

\[ \int \sqrt {b \cos (c+d x)} \sec ^6(c+d x) \, dx=\int { \sqrt {b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{6} \,d x } \]

[In]

integrate(sec(d*x+c)^6*(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*cos(d*x + c))*sec(d*x + c)^6, x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {b \cos (c+d x)} \sec ^6(c+d x) \, dx=\int \frac {\sqrt {b\,\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^6} \,d x \]

[In]

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

[Out]

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