\(\int \cos ^{\frac {3}{2}}(c+d x) (A+C \sec ^2(c+d x)) \, dx\) [1078]

Optimal result

Integrand size = 23, antiderivative size = 48 \[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 A \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4151, 3093, 2720} \[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d} \]

[In]

Int[Cos[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(2*(A + 3*C)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*A*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)

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 3093

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos
[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e +
f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rule 4151

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

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.76 (sec) , antiderivative size = 124, normalized size of antiderivative = 2.58 \[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \sqrt {\cos (c+d x)} \left (C+A \cos ^2(c+d x)\right ) \sin (c) \left ((A+3 C) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec (d x-\arctan (\cot (c)))-A \csc (c) \sin (c+d x)\right )}{3 d (A+2 C+A \cos (2 (c+d x)))} \]

[In]

Integrate[Cos[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(-4*Sqrt[Cos[c + d*x]]*(C + A*Cos[c + d*x]^2)*Sin[c]*((A + 3*C)*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^
2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[d*x - ArcTan[Cot[c]]] - A*Csc[c]*Sin[
c + d*x]))/(3*d*(A + 2*C + A*Cos[2*(c + d*x)]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(227\) vs. \(2(68)=136\).

Time = 1.28 (sec) , antiderivative size = 228, normalized size of antiderivative = 4.75

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(228\)

[In]

int(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*A*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*A*
sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))+3*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(co
s(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 = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.69 \[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 3 i \, C\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, A + 3 i \, C\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}{3 \, d} \]

[In]

integrate(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**(3/2)*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)*cos(d*x + c)^(3/2), x)

Giac [F]

\[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*cos(d*x + c)^(3/2), x)

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,A\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {2\,C\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d} \]

[In]

int(cos(c + d*x)^(3/2)*(A + C/cos(c + d*x)^2),x)

[Out]

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