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

Optimal. Leaf size=164 \[ \frac{2 (15 A-10 B+14 C) \sin (c+d x)}{15 d \sqrt{a \cos (c+d x)+a}}-\frac{\sqrt{2} (A-B+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (5 B-C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{15 a d}+\frac{2 C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt{a \cos (c+d x)+a}} \]

[Out]

-((Sqrt[2]*(A - B + C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d)) + (2*(
15*A - 10*B + 14*C)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*C*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*Sqr
t[a + a*Cos[c + d*x]]) + (2*(5*B - C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(15*a*d)

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Rubi [A]  time = 0.380234, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {3045, 2968, 3023, 2751, 2649, 206} \[ \frac{2 (15 A-10 B+14 C) \sin (c+d x)}{15 d \sqrt{a \cos (c+d x)+a}}-\frac{\sqrt{2} (A-B+C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (5 B-C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{15 a d}+\frac{2 C \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*(A - B + C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d)) + (2*(
15*A - 10*B + 14*C)*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*C*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*Sqr
t[a + a*Cos[c + d*x]]) + (2*(5*B - C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(15*a*d)

Rule 3045

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

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]

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 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 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])

Rubi steps

\begin{align*} \int \frac{\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx &=\frac{2 C \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}+\frac{2 \int \frac{\cos (c+d x) \left (\frac{1}{2} a (5 A+4 C)+\frac{1}{2} a (5 B-C) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{5 a}\\ &=\frac{2 C \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}+\frac{2 \int \frac{\frac{1}{2} a (5 A+4 C) \cos (c+d x)+\frac{1}{2} a (5 B-C) \cos ^2(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{5 a}\\ &=\frac{2 C \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (5 B-C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{15 a d}+\frac{4 \int \frac{\frac{1}{4} a^2 (5 B-C)+\frac{1}{4} a^2 (15 A-10 B+14 C) \cos (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{15 a^2}\\ &=\frac{2 (15 A-10 B+14 C) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 C \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (5 B-C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{15 a d}+(-A+B-C) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{2 (15 A-10 B+14 C) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 C \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (5 B-C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{15 a d}+\frac{(2 (A-B+C)) \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+C) \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 (15 A-10 B+14 C) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 C \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}+\frac{2 (5 B-C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{15 a d}\\ \end{align*}

Mathematica [A]  time = 0.367138, size = 98, normalized size = 0.6 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right ) (30 A+2 (5 B-C) \cos (c+d x)-10 B+3 C \cos (2 (c+d x))+29 C)-15 (A-B+C) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{15 d \sqrt{a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*(-15*(A - B + C)*ArcTanh[Sin[(c + d*x)/2]] + (30*A - 10*B + 29*C + 2*(5*B - C)*Cos[c + d*x
] + 3*C*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(15*d*Sqrt[a*(1 + Cos[c + d*x])])

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Maple [B]  time = 0.133, size = 306, normalized size = 1.9 \begin{align*}{\frac{1}{15\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 24\,C\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-20\,\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( B+C \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-15\,\sqrt{2}\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) aA+15\,\sqrt{2}\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) aB-15\,\sqrt{2}\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) aC+30\,A\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a}+30\,C\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \right ){a}^{-{\frac{3}{2}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{a \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(24*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*si
n(1/2*d*x+1/2*c)^4-20*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*(B+C)*sin(1/2*d*x+1/2*c)^2-15*2^(1/2)*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+15*2^(1/2)*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*B-15*2^(1/2)*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*C+30*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+30*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)
*a^(1/2))/a^(3/2)/sin(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., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [A]  time = 2.00587, size = 466, normalized size = 2.84 \begin{align*} \frac{4 \,{\left (3 \, C \cos \left (d x + c\right )^{2} +{\left (5 \, B - C\right )} \cos \left (d x + c\right ) + 15 \, A - 5 \, B + 13 \, C\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) + \frac{15 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) +{\left (A - B + C\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}}}{30 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(4*(3*C*cos(d*x + c)^2 + (5*B - C)*cos(d*x + c) + 15*A - 5*B + 13*C)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c
) + 15*sqrt(2)*((A - B + C)*a*cos(d*x + c) + (A - B + C)*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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 2.03628, size = 255, normalized size = 1.55 \begin{align*} \frac{\frac{15 \,{\left (\sqrt{2} A - \sqrt{2} B + \sqrt{2} C\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 (15 \, \sqrt{2} A a^{2} + 15 \, \sqrt{2} C a^{2} +{\left (30 \, \sqrt{2} A a^{2} - 10 \, \sqrt{2} B a^{2} + 20 \, \sqrt{2} C a^{2} +{\left (15 \, \sqrt{2} A a^{2} - 10 \, \sqrt{2} B a^{2} + 17 \, \sqrt{2} C a^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\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{5}{2}}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*(sqrt(2)*A - sqrt(2)*B + sqrt(2)*C)*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*(15*sqrt(2)*A*a^2 + 15*sqrt(2)*C*a^2 + (30*sqrt(2)*A*a^2 - 10*sqrt(2)*B*a^2 + 20*sqrt(
2)*C*a^2 + (15*sqrt(2)*A*a^2 - 10*sqrt(2)*B*a^2 + 17*sqrt(2)*C*a^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*
c)^2)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(5/2))/d