3.289 \(\int \frac{\cos ^3(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=114 \[ -\frac{2 a^3 B \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{B x \left (2 a^2+b^2\right )}{2 b^3}-\frac{a B \sin (c+d x)}{b^2 d}+\frac{B \sin (c+d x) \cos (c+d x)}{2 b d} \]

[Out]

((2*a^2 + b^2)*B*x)/(2*b^3) - (2*a^3*B*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^3*Sq
rt[a + b]*d) - (a*B*Sin[c + d*x])/(b^2*d) + (B*Cos[c + d*x]*Sin[c + d*x])/(2*b*d)

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Rubi [A]  time = 0.21655, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {21, 2793, 3023, 2735, 2659, 205} \[ -\frac{2 a^3 B \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{B x \left (2 a^2+b^2\right )}{2 b^3}-\frac{a B \sin (c+d x)}{b^2 d}+\frac{B \sin (c+d x) \cos (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2 + b^2)*B*x)/(2*b^3) - (2*a^3*B*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^3*Sq
rt[a + b]*d) - (a*B*Sin[c + d*x])/(b^2*d) + (B*Cos[c + d*x]*Sin[c + d*x])/(2*b*d)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 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 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x) (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^2} \, dx &=B \int \frac{\cos ^3(c+d x)}{a+b \cos (c+d x)} \, dx\\ &=\frac{B \cos (c+d x) \sin (c+d x)}{2 b d}+\frac{B \int \frac{a+b \cos (c+d x)-2 a \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 b}\\ &=-\frac{a B \sin (c+d x)}{b^2 d}+\frac{B \cos (c+d x) \sin (c+d x)}{2 b d}+\frac{B \int \frac{a b+\left (2 a^2+b^2\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^2}\\ &=\frac{\left (2 a^2+b^2\right ) B x}{2 b^3}-\frac{a B \sin (c+d x)}{b^2 d}+\frac{B \cos (c+d x) \sin (c+d x)}{2 b d}-\frac{\left (a^3 B\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^3}\\ &=\frac{\left (2 a^2+b^2\right ) B x}{2 b^3}-\frac{a B \sin (c+d x)}{b^2 d}+\frac{B \cos (c+d x) \sin (c+d x)}{2 b d}-\frac{\left (2 a^3 B\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^3 d}\\ &=\frac{\left (2 a^2+b^2\right ) B x}{2 b^3}-\frac{2 a^3 B \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} b^3 \sqrt{a+b} d}-\frac{a B \sin (c+d x)}{b^2 d}+\frac{B \cos (c+d x) \sin (c+d x)}{2 b d}\\ \end{align*}

Mathematica [A]  time = 0.228605, size = 98, normalized size = 0.86 \[ \frac{B \left (2 \left (2 a^2+b^2\right ) (c+d x)+\frac{8 a^3 \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}-4 a b \sin (c+d x)+b^2 \sin (2 (c+d x))\right )}{4 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(2*(2*a^2 + b^2)*(c + d*x) + (8*a^3*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2]
- 4*a*b*Sin[c + d*x] + b^2*Sin[2*(c + d*x)]))/(4*b^3*d)

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Maple [B]  time = 0.122, size = 229, normalized size = 2. \begin{align*} -2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}Ba}{{b}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{B}{bd} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) a}{{b}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{B}{bd}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B{a}^{2}}{d{b}^{3}}}+{\frac{B}{bd}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-2\,{\frac{{a}^{3}B}{d{b}^{3}\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*B*a-1/d/b/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*
c)^3*B-2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)*B*a+1/d/b/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+
1/2*c)*B+2/d/b^3*arctan(tan(1/2*d*x+1/2*c))*B*a^2+1/d/b*arctan(tan(1/2*d*x+1/2*c))*B-2/d*a^3/b^3/((a-b)*(a+b))
^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*B

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60562, size = 759, normalized size = 6.66 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}} B a^{3} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) -{\left (2 \, B a^{4} - B a^{2} b^{2} - B b^{4}\right )} d x +{\left (2 \, B a^{3} b - 2 \, B a b^{3} -{\left (B a^{2} b^{2} - B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{2} b^{3} - b^{5}\right )} d}, -\frac{2 \, \sqrt{a^{2} - b^{2}} B a^{3} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) -{\left (2 \, B a^{4} - B a^{2} b^{2} - B b^{4}\right )} d x +{\left (2 \, B a^{3} b - 2 \, B a b^{3} -{\left (B a^{2} b^{2} - B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{2} b^{3} - b^{5}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(-a^2 + b^2)*B*a^3*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 - 2*sqrt(-a^2 + b^2)*(a*c
os(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - (2*B*a^4 - B*a
^2*b^2 - B*b^4)*d*x + (2*B*a^3*b - 2*B*a*b^3 - (B*a^2*b^2 - B*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^3 - b^5
)*d), -1/2*(2*sqrt(a^2 - b^2)*B*a^3*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (2*B*a^4 -
B*a^2*b^2 - B*b^4)*d*x + (2*B*a^3*b - 2*B*a*b^3 - (B*a^2*b^2 - B*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^3 -
b^5)*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)**3*(a*B+b*B*cos(d*x+c))/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.46652, size = 250, normalized size = 2.19 \begin{align*} -\frac{\frac{4 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )} B a^{3}}{\sqrt{a^{2} - b^{2}} b^{3}} - \frac{{\left (2 \, B a^{2} + B b^{2}\right )}{\left (d x + c\right )}}{b^{3}} + \frac{2 \,{\left (2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/
2*c))/sqrt(a^2 - b^2)))*B*a^3/(sqrt(a^2 - b^2)*b^3) - (2*B*a^2 + B*b^2)*(d*x + c)/b^3 + 2*(2*B*a*tan(1/2*d*x +
1/2*c)^3 + B*b*tan(1/2*d*x + 1/2*c)^3 + 2*B*a*tan(1/2*d*x + 1/2*c) - B*b*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x
+ 1/2*c)^2 + 1)^2*b^2))/d