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

Optimal. Leaf size=171 \[ -\frac{(3 A-7 B) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{6 a^2 d}-\frac{(7 A-11 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(A-B) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}+\frac{(9 A-13 B) \sin (c+d x)}{3 a d \sqrt{a \cos (c+d x)+a}} \]

[Out]

-((7*A - 11*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + ((A
- B)*Cos[c + d*x]^2*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((9*A - 13*B)*Sin[c + d*x])/(3*a*d*Sqrt[
a + a*Cos[c + d*x]]) - ((3*A - 7*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(6*a^2*d)

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Rubi [A]  time = 0.420456, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2977, 2968, 3023, 2751, 2649, 206} \[ -\frac{(3 A-7 B) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{6 a^2 d}-\frac{(7 A-11 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(A-B) \sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}+\frac{(9 A-13 B) \sin (c+d x)}{3 a d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((7*A - 11*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + ((A
- B)*Cos[c + d*x]^2*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((9*A - 13*B)*Sin[c + d*x])/(3*a*d*Sqrt[
a + a*Cos[c + d*x]]) - ((3*A - 7*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(6*a^2*d)

Rule 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], 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] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 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 ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\cos (c+d x) \left (2 a (A-B)-\frac{1}{2} a (3 A-7 B) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{2 a (A-B) \cos (c+d x)-\frac{1}{2} a (3 A-7 B) \cos ^2(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(3 A-7 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}+\frac{\int \frac{-\frac{1}{4} a^2 (3 A-7 B)+\frac{1}{2} a^2 (9 A-13 B) \cos (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{3 a^3}\\ &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(9 A-13 B) \sin (c+d x)}{3 a d \sqrt{a+a \cos (c+d x)}}-\frac{(3 A-7 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}-\frac{(7 A-11 B) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{4 a}\\ &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(9 A-13 B) \sin (c+d x)}{3 a d \sqrt{a+a \cos (c+d x)}}-\frac{(3 A-7 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}+\frac{(7 A-11 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 )}{2 a d}\\ &=-\frac{(7 A-11 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(9 A-13 B) \sin (c+d x)}{3 a d \sqrt{a+a \cos (c+d x)}}-\frac{(3 A-7 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.704158, size = 97, normalized size = 0.57 \[ \frac{\tan \left (\frac{1}{2} (c+d x)\right ) (12 (A-B) \cos (c+d x)+15 A+2 B \cos (2 (c+d x))-17 B)-3 (7 A-11 B) \cos \left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{6 a d \sqrt{a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(7*A - 11*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2] + (15*A - 17*B + 12*(A - B)*Cos[c + d*x] + 2*B*Cos
[2*(c + d*x)])*Tan[(c + d*x)/2])/(6*a*d*Sqrt[a*(1 + Cos[c + d*x])])

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Maple [B]  time = 2.164, size = 327, normalized size = 1.9 \begin{align*} -{\frac{1}{12\,d}\sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( -16\,B\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+21\,A\ln \left ( 2\,{\frac{2\,\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+2\,a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) \sqrt{2} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-33\,B\ln \left ( 2\,{\frac{2\,\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+2\,a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) \sqrt{2} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-24\,A\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+40\,B\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-3\,A\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+3\,B\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{a}^{-{\frac{5}{2}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{ \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-16*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^
4+21*A*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^2*a-
33*B*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^2*a-24
*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)^2+40*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(
1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^2-3*A*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+3*B*2^(1/2)*(a*sin(1/2*d*
x+1/2*c)^2)^(1/2)*a^(1/2))/cos(1/2*d*x+1/2*c)/a^(5/2)/sin(1/2*d*x+1/2*c)/(cos(1/2*d*x+1/2*c)^2*a)^(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)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.68249, size = 544, normalized size = 3.18 \begin{align*} -\frac{3 \, \sqrt{2}{\left ({\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right ) + 7 \, A - 11 \, B\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \,{\left (4 \, B \cos \left (d x + c\right )^{2} + 12 \,{\left (A - B\right )} \cos \left (d x + c\right ) + 15 \, A - 19 \, B\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{24 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*sqrt(2)*((7*A - 11*B)*cos(d*x + c)^2 + 2*(7*A - 11*B)*cos(d*x + c) + 7*A - 11*B)*sqrt(a)*log(-(a*cos(
d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2
+ 2*cos(d*x + c) + 1)) - 4*(4*B*cos(d*x + c)^2 + 12*(A - B)*cos(d*x + c) + 15*A - 19*B)*sqrt(a*cos(d*x + c) +
a)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*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)**2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.91589, size = 227, normalized size = 1.33 \begin{align*} \frac{\frac{3 \,{\left (7 \, \sqrt{2} A - 11 \, \sqrt{2} 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 )}{a^{\frac{3}{2}}} + \frac{{\left ({\left (\frac{3 \,{\left (\sqrt{2} A a - \sqrt{2} B a\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a} + \frac{2 \,{\left (15 \, \sqrt{2} A a - 23 \, \sqrt{2} B a\right )}}{a}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \frac{27 \,{\left (\sqrt{2} A a - \sqrt{2} B a\right )}}{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}}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(7*sqrt(2)*A - 11*sqrt(2)*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
)))/a^(3/2) + ((3*(sqrt(2)*A*a - sqrt(2)*B*a)*tan(1/2*d*x + 1/2*c)^2/a + 2*(15*sqrt(2)*A*a - 23*sqrt(2)*B*a)/a
)*tan(1/2*d*x + 1/2*c)^2 + 27*(sqrt(2)*A*a - sqrt(2)*B*a)/a)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 +
a)^(3/2))/d