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

Optimal. Leaf size=118 \[ \frac {(3 B-7 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(B-C) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}+\frac {2 C \sin (c+d x)}{a d \sqrt {a \cos (c+d x)+a}} \]

[Out]

-1/2*(B-C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)+1/4*(3*B-7*C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d
*x+c))^(1/2))/a^(3/2)/d*2^(1/2)+2*C*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3019, 2751, 2649, 206} \[ \frac {(3 B-7 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(B-C) \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}+\frac {2 C \sin (c+d x)}{a d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*B - 7*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((B -
C)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + (2*C*Sin[c + d*x])/(a*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 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 {B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx &=-\frac {(B-C) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {\int \frac {-\frac {3}{2} a (B-C)-2 a C \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(B-C) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {2 C \sin (c+d x)}{a d \sqrt {a+a \cos (c+d x)}}+\frac {(3 B-7 C) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a}\\ &=-\frac {(B-C) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {2 C \sin (c+d x)}{a d \sqrt {a+a \cos (c+d x)}}-\frac {(3 B-7 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 )}{2 a d}\\ &=\frac {(3 B-7 C) \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 {(B-C) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {2 C \sin (c+d x)}{a d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.44, size = 104, normalized size = 0.88 \[ \frac {\sin \left (\frac {1}{2} (c+d x)\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) (B-4 C \cos (c+d x)-5 C)-(3 B-7 C) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d \left (\sin ^2\left (\frac {1}{2} (c+d x)\right )-1\right ) (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((3*B - 7*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5) + Cos[(c + d*x)/2]^3*(B - 5*C - 4*C*Cos[c + d*x])
*Sin[(c + d*x)/2])/(d*(a*(1 + Cos[c + d*x]))^(3/2)*(-1 + Sin[(c + d*x)/2]^2))

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fricas [A]  time = 0.45, size = 189, normalized size = 1.60 \[ -\frac {\sqrt {2} {\left ({\left (3 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, B - 7 \, C\right )} \cos \left (d x + c\right ) + 3 \, B - 7 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (4 \, C \cos \left (d x + c\right ) - B + 5 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(sqrt(2)*((3*B - 7*C)*cos(d*x + c)^2 + 2*(3*B - 7*C)*cos(d*x + c) + 3*B - 7*C)*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*c
os(d*x + c) + 1)) - 4*(4*C*cos(d*x + c) - B + 5*C)*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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giac [A]  time = 1.18, size = 131, normalized size = 1.11 \[ -\frac {\frac {{\left (\frac {\sqrt {2} {\left (B a^{2} - C a^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{3}} + \frac {\sqrt {2} {\left (B a^{2} - 9 \, C a^{2}\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} + \frac {\sqrt {2} {\left (3 \, B - 7 \, C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {3}{2}}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((sqrt(2)*(B*a^2 - C*a^2)*tan(1/2*d*x + 1/2*c)^2/a^3 + sqrt(2)*(B*a^2 - 9*C*a^2)/a^3)*tan(1/2*d*x + 1/2*c
)/sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) + sqrt(2)*(3*B - 7*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)))/a^(3/2))/d

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maple [B]  time = 1.28, size = 256, normalized size = 2.17 \[ \frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (3 B \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \sqrt {2}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -7 C \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +8 C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}+C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*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-7*C*2^(1/2)*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))*cos(1/2*d*x+1/2*c)^2*a+8*C*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-B*2^(1/2
)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+C*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)/(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((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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