\(\int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [170]

Optimal result

Integrand size = 35, antiderivative size = 76 \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \sqrt {a} B \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {2 a A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \]

[Out]

2*B*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))*a^(1/2)/d+2*a*A*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d
*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3059, 2853, 222} \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 a A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {2 \sqrt {a} B \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d} \]

[In]

Int[(Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(3/2),x]

[Out]

(2*Sqrt[a]*B*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (2*a*A*Sin[c + d*x])/(d*Sqrt[Cos[c +
d*x]]*Sqrt[a + a*Cos[c + d*x]])

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 2853

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 3059

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n +
1)*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+B \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 a A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}-\frac {(2 B) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a}}} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d} \\ & = \frac {2 \sqrt {a} B \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {2 a A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (\sqrt {2} B \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+2 A \sin \left (\frac {1}{2} (c+d x)\right )\right )}{d \sqrt {\cos (c+d x)}} \]

[In]

Integrate[(Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(3/2),x]

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(Sqrt[2]*B*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*Sqrt[Cos[c + d*x]] +
2*A*Sin[(c + d*x)/2]))/(d*Sqrt[Cos[c + d*x]])

Maple [A] (verified)

Time = 7.56 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.53

method	result	size
parts	\(\frac {2 A \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sin \left (d x +c \right )}{d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}}+\frac {2 B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )}{d \sqrt {\cos \left (d x +c \right )}}\)	\(116\)
default	\(\frac {2 \left (B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \cos \left (d x +c \right )+B \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+A \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}}\)	\(147\)

[In]

int((a+cos(d*x+c)*a)^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2*A/d*(a*(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/(1+cos(d*x+c))/cos(d*x+c)^(1/2)+2*B/d/cos(d*x+c)^(1/2)*(cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)*(a*(1+cos(d*x+c)))^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.43 \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \, {\left (\sqrt {a \cos \left (d x + c\right ) + a} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left (B \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )\right )}}{d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )} \]

[In]

integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

2*(sqrt(a*cos(d*x + c) + a)*A*sqrt(cos(d*x + c))*sin(d*x + c) - (B*cos(d*x + c)^2 + B*cos(d*x + c))*sqrt(a)*ar
ctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))))/(d*cos(d*x + c)^2 + d*cos(d*x + c))

Sympy [F]

\[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \left (A + B \cos {\left (c + d x \right )}\right )}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

[In]

integrate((a+a*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)**(3/2),x)

[Out]

Integral(sqrt(a*(cos(c + d*x) + 1))*(A + B*cos(c + d*x))/cos(c + d*x)**(3/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (66) = 132\).

Time = 0.39 (sec) , antiderivative size = 245, normalized size of antiderivative = 3.22 \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {B \sqrt {a} \arctan \left ({\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + \sin \left (d x + c\right ), {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + \cos \left (d x + c\right )\right ) + \frac {2 \, A {\left (\frac {\sqrt {2} \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {3}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {3}{2}}}}{d} \]

[In]

integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

(B*sqrt(a)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(si
n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x
+ 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + cos(d*x + c)) + 2*A*(sqrt(2)*sqrt
(a)*sin(d*x + c)/(cos(d*x + c) + 1) - sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/((sin(d*x + c)/(cos
(d*x + c) + 1) + 1)^(3/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(3/2)))/d

Giac [F]

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

[In]

integrate((a+a*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*sqrt(a*cos(d*x + c) + a)/cos(d*x + c)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \]

[In]

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(1/2))/cos(c + d*x)^(3/2),x)

[Out]

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^(1/2))/cos(c + d*x)^(3/2), x)