\(\int \sqrt {b \cos (c+d x)} (A+C \cos ^2(c+d x)) \, dx\) [39]

Optimal result

Integrand size = 25, antiderivative size = 77 \[ \int \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 (5 A+3 C) \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 C (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d} \]

[Out]

2/5*C*(b*cos(d*x+c))^(3/2)*sin(d*x+c)/b/d+2/5*(5*A+3*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipt
icE(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 = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3093, 2721, 2719} \[ \int \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d} \]

[In]

Int[Sqrt[b*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2),x]

[Out]

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

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]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.91 \[ \int \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\sqrt {b \cos (c+d x)} \left (2 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+C \sqrt {\cos (c+d x)} \sin (2 (c+d x))\right )}{5 d \sqrt {\cos (c+d x)}} \]

[In]

Integrate[Sqrt[b*Cos[c + d*x]]*(A + C*Cos[c + d*x]^2),x]

[Out]

(Sqrt[b*Cos[c + d*x]]*(2*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2] + C*Sqrt[Cos[c + d*x]]*Sin[2*(c + d*x)]))/(5*d*
Sqrt[Cos[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(260\) vs. \(2(93)=186\).

Time = 9.48 (sec) , antiderivative size = 261, normalized size of antiderivative = 3.39

method	result	size
default	\(\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b \left (8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(261\)
parts	\(\frac {2 A \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}-\frac {2 C \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\)	\(355\)

[In]

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

[Out]

2/5*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*(8*cos(1/2*d*x+1/2*c)*C*sin(1/2*d*x+1/2*c)^6-8
*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+5*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)*E
llipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-b*(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)*b)^(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 = 101, normalized size of antiderivative = 1.31 \[ \int \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \, \sqrt {b \cos \left (d x + c\right )} C \cos \left (d x + c\right ) \sin \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A + 3 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (-5 i \, A - 3 i \, C\right )} \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{5 \, d} \]

[In]

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

[Out]

1/5*(2*sqrt(b*cos(d*x + c))*C*cos(d*x + c)*sin(d*x + c) + sqrt(2)*(5*I*A + 3*I*C)*sqrt(b)*weierstrassZeta(-4,
0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + sqrt(2)*(-5*I*A - 3*I*C)*sqrt(b)*weierstrassZe
ta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d

Sympy [F(-1)]

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

[In]

integrate((b*cos(d*x+c))**(1/2)*(A+C*cos(d*x+c)**2),x)

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + A)*sqrt(b*cos(d*x + c)), x)

Giac [F]

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

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + A)*sqrt(b*cos(d*x + c)), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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