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

Optimal. Leaf size=99 \[ \frac {2 A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} (A-B) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d} \]

[Out]

-(A-B)*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))*2^(1/2)/d/a^(1/2)+2*A*si
n(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {2984, 12, 2782, 205} \[ \frac {2 A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} (A-B) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

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

Rule 2782

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 2984

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

Rubi steps

\begin {align*} \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx &=\frac {2 A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 \int -\frac {a (A-B)}{2 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{a}\\ &=\frac {2 A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+(-A+B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx\\ &=\frac {2 A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {(2 a (A-B)) \operatorname {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 )}{d}\\ &=-\frac {\sqrt {2} (A-B) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 1.64, size = 203, normalized size = 2.05 \[ \frac {2 \sin \left (\frac {1}{2} (c+d x)\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (10 B \cos (c+d x)-(A-B) \left (\frac {1}{2} \sin (c+d x) \tan (c+d x) \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )-\frac {5}{4} (4 \cos (c+d x)+\cos (2 (c+d x))+1) \csc ^4\left (\frac {1}{2} (c+d x)\right ) \left (-\cos (c+d x)+\cos (c+d x) \sqrt {2-2 \sec (c+d x)} \tanh ^{-1}\left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right ) (-\sec (c+d x))}\right )+1\right )\right )\right )}{5 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a (\cos (c+d x)+1)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*Sin[(c + d*x)/2]*(10*B*Cos[c + d*x] - (A - B)*((-5*(1 + 4*Cos[c + d*x] + Cos[2*(c + d*x)])
*Csc[(c + d*x)/2]^4*(1 - Cos[c + d*x] + ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]]*Cos[c + d*x]*Sqrt[2
- 2*Sec[c + d*x]]))/4 + (Hypergeometric2F1[2, 5/2, 7/2, -(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]*Sin[c + d*x]*Tan[c
+ d*x])/2)))/(5*d*Cos[c + d*x]^(3/2)*Sqrt[a*(1 + Cos[c + d*x])])

________________________________________________________________________________________

fricas [A]  time = 1.18, size = 143, normalized size = 1.44 \[ \frac {2 \, \sqrt {a \cos \left (d x + c\right ) + a} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \frac {\sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right )^{2} + {\left (A - B\right )} a \cos \left (d x + c\right )\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a}}\right )}{\sqrt {a}}}{a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \cos \left (d x + c\right ) + A}{\sqrt {a \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

maple [B]  time = 0.26, size = 230, normalized size = 2.32 \[ \frac {\sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )-B \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+A \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-B \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+2 A \sin \left (d x +c \right )\right )}{d a \left (1+\cos \left (d x +c \right )\right ) \sqrt {\cos \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \cos {\left (c + d x \right )}}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________