\(\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \, dx\) [804]

Optimal result

Integrand size = 23, antiderivative size = 101 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {2 \left (3 a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {4 a b \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a^2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d} \]

[Out]

2/5*(3*a^2+5*b^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))/d+4/3*
a*b*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/5*a^2*cos(d*x+c)
^(3/2)*sin(d*x+c)/d+4/3*a*b*sin(d*x+c)*cos(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4349, 3873, 3854, 3856, 2720, 4130, 2719} \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {2 \left (3 a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {4 a b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {4 a b \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d} \]

[In]

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

[Out]

(2*(3*a^2 + 5*b^2)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a*b*EllipticF[(c + d*x)/2, 2])/(3*d) + (4*a*b*Sqrt[Co
s[c + d*x]]*Sin[c + d*x])/(3*d) + (2*a^2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d)

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 2720

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

Rule 3854

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

Rule 3873

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[2*a*(b/d
), Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4130

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e
+ f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 4349

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \sec (c+d x))^2}{\sec ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {a^2+b^2 \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx+\left (2 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {4 a b \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a^2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{3} \left (2 a b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \, dx-\frac {1}{5} \left (\left (-3 a^2-5 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {4 a b \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a^2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{3} (2 a b) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} \left (-3 a^2-5 b^2\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = \frac {2 \left (3 a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {4 a b \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a^2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.78 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {6 \left (3 a^2+5 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 a b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 a \sqrt {\cos (c+d x)} (10 b+3 a \cos (c+d x)) \sin (c+d x)}{15 d} \]

[In]

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

[Out]

(6*(3*a^2 + 5*b^2)*EllipticE[(c + d*x)/2, 2] + 20*a*b*EllipticF[(c + d*x)/2, 2] + 2*a*Sqrt[Cos[c + d*x]]*(10*b
+ 3*a*Cos[c + d*x])*Sin[c + d*x])/(15*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(356\) vs. \(2(141)=282\).

Time = 28.58 (sec) , antiderivative size = 357, normalized size of antiderivative = 3.53

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-24 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+24 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+40 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a b -6 a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-20 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a b +10 \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, a b -9 \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, a^{2}-15 \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, b^{2}\right )}{15 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(357\)

[In]

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

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6
+24*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+40*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a*b-6*a^2*sin(1/2*d
*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-20*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a*b+10*(2*sin(1/2*d*x+1/2*c)^2-1)^(1
/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a*b-9*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*
EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2-15*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ell
ipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(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)^(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 = 162, normalized size of antiderivative = 1.60 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {-10 i \, \sqrt {2} a b {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 10 i \, \sqrt {2} a b {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (3 \, a^{2} \cos \left (d x + c\right ) + 10 \, a b\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 3 \, \sqrt {2} {\left (-3 i \, a^{2} - 5 i \, b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (3 i \, a^{2} + 5 i \, b^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{15 \, d} \]

[In]

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

[Out]

1/15*(-10*I*sqrt(2)*a*b*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 10*I*sqrt(2)*a*b*weierstra
ssPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(3*a^2*cos(d*x + c) + 10*a*b)*sqrt(cos(d*x + c))*sin(d*x
+ c) - 3*sqrt(2)*(-3*I*a^2 - 5*I*b^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d
*x + c))) - 3*sqrt(2)*(3*I*a^2 + 5*I*b^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*s
in(d*x + c))))/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \, dx=\int { {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.00 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.01 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {2\,b^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,a\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {4\,a\,b\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d}-\frac {2\,a^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

[In]

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

[Out]

(2*b^2*ellipticE(c/2 + (d*x)/2, 2))/d + (4*a*b*ellipticF(c/2 + (d*x)/2, 2))/(3*d) + (4*a*b*cos(c + d*x)^(1/2)*
sin(c + d*x))/(3*d) - (2*a^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d
*(sin(c + d*x)^2)^(1/2))