3.102 \(\int \frac {\cos (c+d x) (A+B \cos (c+d x))}{\sqrt {a+a \cos (c+d x)}} \, dx\)

Optimal. Leaf size=118 \[ \frac {2 (3 A-2 B) \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 B \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d} \]

[Out]

-(A-B)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)+2/3*(3*A-2*B)*sin(d*x+
c)/d/(a+a*cos(d*x+c))^(1/2)+2/3*B*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/a/d

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Rubi [A]  time = 0.21, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2968, 3023, 2751, 2649, 206} \[ \frac {2 (3 A-2 B) \sin (c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 B \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*(A - B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d)) + (2*(3*A
- 2*B)*Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (2*B*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*a*d)

Rule 206

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

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2751

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

Rule 2968

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

Rule 3023

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

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 78, normalized size = 0.66 \[ \frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (-3 (A-B) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 A \sin \left (\frac {1}{2} (c+d x)\right )-4 B \sin ^3\left (\frac {1}{2} (c+d x)\right )\right )}{3 d \sqrt {a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*(-3*(A - B)*ArcTanh[Sin[(c + d*x)/2]] + 6*A*Sin[(c + d*x)/2] - 4*B*Sin[(c + d*x)/2]^3))/(3
*d*Sqrt[a*(1 + Cos[c + d*x])])

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fricas [A]  time = 0.63, size = 149, normalized size = 1.26 \[ \frac {4 \, {\left (B \cos \left (d x + c\right ) + 3 \, A - B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) - \frac {3 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(4*(B*cos(d*x + c) + 3*A - B)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c) - 3*sqrt(2)*((A - B)*a*cos(d*x + c) +
(A - B)*a)*log(-(cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3
)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c) + a*d)

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giac [A]  time = 1.56, size = 113, normalized size = 0.96 \[ \frac {\frac {3 \, \sqrt {2} {\left (A - B\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {a}} + \frac {2 \, {\left (\sqrt {2} {\left (3 \, A a - 2 \, B a\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, \sqrt {2} A a\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*sqrt(2)*(A - B)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/sqrt(a) +
2*(sqrt(2)*(3*A*a - 2*B*a)*tan(1/2*d*x + 1/2*c)^2 + 3*sqrt(2)*A*a)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c
)^2 + a)^(3/2))/d

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maple [A]  time = 0.67, size = 194, normalized size = 1.64 \[ \frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-4 B \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 A \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-3 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a +3 B \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{3 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*cos(1/2*d*x+1/2*c)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*B*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*sin
(1/2*d*x+1/2*c)^2+6*A*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-3*A*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d
*x+1/2*c)^2)^(1/2)+a))*a+3*B*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))*a)/a^(3/2)/si
n(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [B]  time = 0.38, size = 160, normalized size = 1.36 \[ \frac {2\,A\,\left (2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |1\right )-\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |1\right )\right )\,\sqrt {\frac {a+a\,\cos \left (c+d\,x\right )}{2\,a}}}{d\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}+\frac {2\,B\,\sin \left (c+d\,x\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{3\,a\,d}-\frac {2\,B\,\left (4\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |1\right )-3\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |1\right )\right )\,\sqrt {\frac {a+a\,\cos \left (c+d\,x\right )}{2\,a}}}{3\,a^2\,d\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*A*(2*ellipticE(c/2 + (d*x)/2, 1) - ellipticF(c/2 + (d*x)/2, 1))*((a + a*cos(c + d*x))/(2*a))^(1/2))/(d*(a +
a*cos(c + d*x))^(1/2)) + (2*B*sin(c + d*x)*(a + a*cos(c + d*x))^(1/2))/(3*a*d) - (2*B*(4*a^2*ellipticE(c/2 +
(d*x)/2, 1) - 3*a^2*ellipticF(c/2 + (d*x)/2, 1))*((a + a*cos(c + d*x))/(2*a))^(1/2))/(3*a^2*d*(a + a*cos(c + d
*x))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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