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]

((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]])

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Rubi [A]  time = 0.151733, antiderivative size = 118, normalized size of antiderivative = 1., 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 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]

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{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*}

Mathematica [A]  time = 0.438449, 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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Maple [B]  time = 0.112, size = 256, normalized size = 2.2 \begin{align*}{\frac{1}{4\,d}\sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 3\,\sqrt{2}\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 ) a \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}B-7\,C\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+8\,C\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}-B\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}+C\sqrt{2}\sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{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{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((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(3/2),x)

[Out]

1/4/cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*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))*a*cos(1/2*d*x+1/2*c)^2*B-7*C*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/co
s(1/2*d*x+1/2*c))*2^(1/2)*cos(1/2*d*x+1/2*c)^2*a+8*C*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-B*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/
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., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Fricas [A]  time = 1.9454, size = 493, normalized size = 4.18 \begin{align*} -\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 )}} \end{align*}

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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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((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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Giac [A]  time = 1.88909, size = 177, normalized size = 1.5 \begin{align*} -\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} \end{align*}

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