\(\int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx\) [207]

Optimal result

Integrand size = 37, antiderivative size = 145 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx=\frac {2 C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{3/2} d}+\frac {(3 A-5 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}} \]

[Out]

2*C*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d+1/4*(3*A-5*C)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^
(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-1/2*(A+C)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*
cos(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {3121, 3061, 2861, 211, 2853, 222} \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx=\frac {(3 A-5 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {2 C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{3/2} d}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}} \]

[In]

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

[Out]

(2*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(3/2)*d) + ((3*A - 5*C)*ArcTan[(Sqrt[a]*Sin[c
+ d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A + C)*Sqrt[Cos[c +
d*x]]*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2))

Rule 211

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

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 2861

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

Rule 3061

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b - a*B)/b, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]]), x], x] + Dist[B/b, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]], 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]

Rule 3121

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(A+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\frac {1}{2} a (3 A-C)+2 a C \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {(A+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(3 A-5 C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{4 a}+\frac {C \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx}{a^2} \\ & = -\frac {(A+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {(3 A-5 C) \text {Subst}\left (\int \frac {1}{2 a^2+a x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{2 d}-\frac {(2 C) \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 )}{a^2 d} \\ & = \frac {2 C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{3/2} d}+\frac {(3 A-5 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.30 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.18 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx=-\frac {\left (C \arcsin \left (\sqrt {1-\cos (c+d x)}\right ) (1+\cos (c+d x))+5 C \arcsin \left (\sqrt {\cos (c+d x)}\right ) (1+\cos (c+d x))+\sqrt {2} \left ((3 A-5 C) \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )+(A+C) \sqrt {\cos (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )\right ) \sin (c+d x)}{2 d \sqrt {1-\cos (c+d x)} (a (1+\cos (c+d x)))^{3/2}} \]

[In]

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

[Out]

-1/2*((C*ArcSin[Sqrt[1 - Cos[c + d*x]]]*(1 + Cos[c + d*x]) + 5*C*ArcSin[Sqrt[Cos[c + d*x]]]*(1 + Cos[c + d*x])
+ Sqrt[2]*((3*A - 5*C)*ArcTan[Sqrt[Cos[c + d*x]]/Sqrt[Sin[(c + d*x)/2]^2]]*Cos[(c + d*x)/2]^2 + (A + C)*Sqrt[
Cos[c + d*x]*Sin[(c + d*x)/2]^2]))*Sin[c + d*x])/(d*Sqrt[1 - Cos[c + d*x]]*(a*(1 + Cos[c + d*x]))^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(280\) vs. \(2(120)=240\).

Time = 11.75 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.94

method	result	size
default	\(-\frac {\left (3 A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-5 C \cos \left (d x +c \right ) \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+3 A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+2 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-5 C \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+2 C \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-8 C \cos \left (d x +c \right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-8 C \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{4 a^{2} d \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right )^{2}}\)	\(281\)
parts	\(-\frac {A \sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \sqrt {\frac {a}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}}\, \left (\sqrt {-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+3 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}}{4 d \sqrt {-\frac {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1}}\, a^{2}}-\frac {C \left (\sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-4 \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \cos \left (d x +c \right )-5 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-4 \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-5 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {2}}{4 d \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{2}}\)	\(385\)

[In]

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

[Out]

-1/4/a^2/d*(3*A*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)-5*C*cos(d*x+c)*2^(1/2)*arcsin(cot(d*x+c)-csc(
d*x+c))+3*A*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-5*C*2^(1/2)
*arcsin(cot(d*x+c)-csc(d*x+c))+2*C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-8*C*cos(d*x+c)*arctan(tan(d*x+
c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-8*C*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)))*cos(d*x+c)^(1/
2)*(a*(1+cos(d*x+c)))^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))^2

Fricas [A] (verification not implemented)

none

Time = 2.97 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.43 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx=-\frac {\sqrt {2} {\left ({\left (3 \, A - 5 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A - 5 \, C\right )} \cos \left (d x + c\right ) + 3 \, A - 5 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (A + C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 8 \, {\left (C \cos \left (d x + c\right )^{2} + 2 \, C \cos \left (d x + c\right ) + C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

-1/4*(sqrt(2)*((3*A - 5*C)*cos(d*x + c)^2 + 2*(3*A - 5*C)*cos(d*x + c) + 3*A - 5*C)*sqrt(a)*arctan(sqrt(2)*sqr
t(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*sqrt(a*cos(d*x + c) + a)*(A + C)*sqrt(cos
(d*x + c))*sin(d*x + c) + 8*(C*cos(d*x + c)^2 + 2*C*cos(d*x + c) + C)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*
sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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