\(\int \frac {(a B+b B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx\) [584]

Optimal result

Integrand size = 36, antiderivative size = 64 \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \]

[Out]

2/3*B*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/3*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {21, 3853, 3856, 2720} \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d} \]

[In]

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

[Out]

(2*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*B*Sec[c + d*x]^(3/2)*Sin[c +
d*x])/(3*d)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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 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*} \text {integral}& = B \int \sec ^{\frac {5}{2}}(c+d x) \, dx \\ & = \frac {2 B \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{3} B \int \sqrt {\sec (c+d x)} \, dx \\ & = \frac {2 B \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {1}{3} \left (B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.73 \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 B \sec ^{\frac {3}{2}}(c+d x) \left (\cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sin (c+d x)\right )}{3 d} \]

[In]

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

[Out]

(2*B*Sec[c + d*x]^(3/2)*(Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + Sin[c + d*x]))/(3*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(213\) vs. \(2(80)=160\).

Time = 3.66 (sec) , antiderivative size = 214, normalized size of antiderivative = 3.34

method	result	size
default	\(-\frac {2 \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) B \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) d}\)	\(214\)

[In]

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

[Out]

-2/3*(-2*(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))*s
in(1/2*d*x+1/2*c)^2-2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*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)))*B*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(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)/(2*cos(1/2*d*x+1/2*c)^2-1)^(3/2)/sin(1/2*d*x+1/2*c)/d

Fricas [C] (verification not implemented)

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

Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.42 \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-i \, \sqrt {2} B \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} B \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + \frac {2 \, B \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{3 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a B+b B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (B\,a+B\,b\,\cos \left (c+d\,x\right )\right )}{a+b\,\cos \left (c+d\,x\right )} \,d x \]

[In]

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

[Out]

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