\(\int \cos ^{\frac {7}{2}}(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1174]

Optimal result

Integrand size = 30, antiderivative size = 87 \[ \int \cos ^{\frac {7}{2}}(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {6 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 C \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d} \]

[Out]

6/5*B*(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+2/3*C*(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*B*cos(d*x+c)^(3/2)*sin(d*x+c
)/d+2/3*C*sin(d*x+c)*cos(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4149, 2827, 2715, 2720, 2719} \[ \int \cos ^{\frac {7}{2}}(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {6 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}+\frac {2 C \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 C \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d} \]

[In]

Int[Cos[c + d*x]^(7/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(6*B*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*C*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*C*Sqrt[Cos[c + d*x]]*Sin[c
+ d*x])/(3*d) + (2*B*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d)

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[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 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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 4149

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)
]^2), x_Symbol] :> Dist[b^2, Int[(b*Cos[e + f*x])^(m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; F
reeQ[{b, e, f, A, B, C, m}, x] &&  !IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \cos ^{\frac {3}{2}}(c+d x) (C+B \cos (c+d x)) \, dx \\ & = B \int \cos ^{\frac {5}{2}}(c+d x) \, dx+C \int \cos ^{\frac {3}{2}}(c+d x) \, dx \\ & = \frac {2 C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {1}{5} (3 B) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} C \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {6 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 C \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.76 \[ \int \cos ^{\frac {7}{2}}(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \left (9 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 C \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (5 C+3 B \cos (c+d x)) \sin (c+d x)\right )}{15 d} \]

[In]

Integrate[Cos[c + d*x]^(7/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(2*(9*B*EllipticE[(c + d*x)/2, 2] + 5*C*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(5*C + 3*B*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. \(261\) vs. \(2(127)=254\).

Time = 12.06 (sec) , antiderivative size = 262, normalized size of antiderivative = 3.01

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (24 B +20 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-6 B -10 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-9 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(262\)

[In]

int(cos(d*x+c)^(7/2)*(B*sec(d*x+c)+C*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*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(
24*B+20*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-6*B-10*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-9*B*(si
n(1/2*d*x+1/2*c)^2)^(1/2)*(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))+5*C*(sin(1/2*
d*x+1/2*c)^2)^(1/2)*(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)))/(-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 = 0.11 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.57 \[ \int \cos ^{\frac {7}{2}}(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (3 \, B \cos \left (d x + c\right ) + 5 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 i \, \sqrt {2} C {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} C {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 i \, \sqrt {2} B {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 i \, \sqrt {2} B {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{15 \, d} \]

[In]

integrate(cos(d*x+c)^(7/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/15*(2*(3*B*cos(d*x + c) + 5*C)*sqrt(cos(d*x + c))*sin(d*x + c) - 5*I*sqrt(2)*C*weierstrassPInverse(-4, 0, co
s(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*C*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9*I*s
qrt(2)*B*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 9*I*sqrt(2)*B*wei
erstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {7}{2}}(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(7/2)*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos ^{\frac {7}{2}}(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \cos \left (d x + c\right )^{\frac {7}{2}} \,d x } \]

[In]

integrate(cos(d*x+c)^(7/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*cos(d*x + c)^(7/2), x)

Giac [F]

\[ \int \cos ^{\frac {7}{2}}(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \cos \left (d x + c\right )^{\frac {7}{2}} \,d x } \]

[In]

integrate(cos(d*x+c)^(7/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*cos(d*x + c)^(7/2), x)

Mupad [B] (verification not implemented)

Time = 17.96 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \cos ^{\frac {7}{2}}(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,C\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {2\,C\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d}-\frac {2\,B\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

[In]

int(cos(c + d*x)^(7/2)*(B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(2*C*ellipticF(c/2 + (d*x)/2, 2))/(3*d) + (2*C*cos(c + d*x)^(1/2)*sin(c + d*x))/(3*d) - (2*B*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))