3.117 \(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=169 \[ \frac {(19 A-75 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-9 B) \sin (c+d x)}{4 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}-\frac {(5 A-13 B) \sin (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

1/4*(A-B)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)-1/16*(5*A-13*B)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/
2)+1/32*(19*A-75*B)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)-1/4*(A-9*
B)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.42, antiderivative size = 169, normalized size of antiderivative = 1.00, 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, 3019, 2751, 2649, 206} \[ -\frac {(A-9 B) \sin (c+d x)}{4 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(19 A-75 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}-\frac {(5 A-13 B) \sin (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

((19*A - 75*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + ((
A - B)*Cos[c + d*x]^2*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - ((5*A - 13*B)*Sin[c + d*x])/(16*a*d*(a
+ a*Cos[c + d*x])^(3/2)) - ((A - 9*B)*Sin[c + d*x])/(4*a^2*d*Sqrt[a + a*Cos[c + d*x]])

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 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 3019

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[((A*b - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx &=\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {\cos (c+d x) \left (2 a (A-B)-\frac {1}{2} a (A-9 B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {2 a (A-B) \cos (c+d x)-\frac {1}{2} a (A-9 B) \cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(5 A-13 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {\int \frac {-\frac {3}{4} a^2 (5 A-13 B)+a^2 (A-9 B) \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(5 A-13 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(A-9 B) \sin (c+d x)}{4 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {(19 A-75 B) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(5 A-13 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(A-9 B) \sin (c+d x)}{4 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(19 A-75 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 )}{16 a^2 d}\\ &=\frac {(19 A-75 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(5 A-13 B) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}-\frac {(A-9 B) \sin (c+d x)}{4 a^2 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.78, size = 100, normalized size = 0.59 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) ((85 B-13 A) \cos (c+d x)-9 A+16 B \cos (2 (c+d x))+65 B)+2 (19 A-75 B) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{16 a d (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(2*(19*A - 75*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 + (-9*A + 65*B + (-13*A + 85*B)*Cos[c + d*x] + 1
6*B*Cos[2*(c + d*x)])*Tan[(c + d*x)/2])/(16*a*d*(a*(1 + Cos[c + d*x]))^(3/2))

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fricas [A]  time = 0.74, size = 237, normalized size = 1.40 \[ -\frac {\sqrt {2} {\left ({\left (19 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A - 75 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (19 \, A - 75 \, B\right )} \cos \left (d x + c\right ) + 19 \, A - 75 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (32 \, B \cos \left (d x + c\right )^{2} - {\left (13 \, A - 85 \, B\right )} \cos \left (d x + c\right ) - 9 \, A + 49 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

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))^(5/2),x, algorithm="fricas")

[Out]

-1/64*(sqrt(2)*((19*A - 75*B)*cos(d*x + c)^3 + 3*(19*A - 75*B)*cos(d*x + c)^2 + 3*(19*A - 75*B)*cos(d*x + c) +
19*A - 75*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*c
os(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*(32*B*cos(d*x + c)^2 - (13*A - 85*B)*cos(d*x + c
) - 9*A + 49*B)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*
d*cos(d*x + c) + a^3*d)

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giac [A]  time = 2.61, size = 181, normalized size = 1.07 \[ \frac {\frac {{\left ({\left (\frac {2 \, {\left (\sqrt {2} A a^{6} - \sqrt {2} B a^{6}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8}} - \frac {9 \, \sqrt {2} A a^{6} - 17 \, \sqrt {2} B a^{6}}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {11 \, \sqrt {2} A a^{6} - 83 \, \sqrt {2} B a^{6}}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {{\left (19 \, \sqrt {2} A - 75 \, \sqrt {2} B\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {5}{2}}}}{32 \, d} \]

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))^(5/2),x, algorithm="giac")

[Out]

1/32*(((2*(sqrt(2)*A*a^6 - sqrt(2)*B*a^6)*tan(1/2*d*x + 1/2*c)^2/a^8 - (9*sqrt(2)*A*a^6 - 17*sqrt(2)*B*a^6)/a^
8)*tan(1/2*d*x + 1/2*c)^2 - (11*sqrt(2)*A*a^6 - 83*sqrt(2)*B*a^6)/a^8)*tan(1/2*d*x + 1/2*c)/sqrt(a*tan(1/2*d*x
+ 1/2*c)^2 + a) - (19*sqrt(2)*A - 75*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^(5/2))/d

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maple [B]  time = 0.86, size = 327, normalized size = 1.93 \[ \frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (19 A \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -75 B \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +64 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-13 A \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+21 B \sqrt {a}\, \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 A \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-2 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x)

[Out]

1/32*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(19*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)^4*a-75*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)^4*a+64*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)^4-13*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+21*B*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+2*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*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)^3/a^(7/2)/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 = 0.00, 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)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^2*(A + B*cos(c + d*x)))/(a + a*cos(c + d*x))^(5/2),x)

[Out]

int((cos(c + d*x)^2*(A + B*cos(c + d*x)))/(a + a*cos(c + d*x))^(5/2), x)

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sympy [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)**2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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