3.112 \(\int \frac{\cos ^3(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=119 \[ \frac{b \cos ^2(c+d x)}{2 d \left (a^2+b^2\right )}+\frac{a \sin (c+d x) \cos (c+d x)}{2 d \left (a^2+b^2\right )}+\frac{b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{a b^2 x}{\left (a^2+b^2\right )^2}+\frac{a x}{2 \left (a^2+b^2\right )} \]

[Out]

(a*b^2*x)/(a^2 + b^2)^2 + (a*x)/(2*(a^2 + b^2)) + (b*Cos[c + d*x]^2)/(2*(a^2 + b^2)*d) + (b^3*Log[a*Cos[c + d*
x] + b*Sin[c + d*x]])/((a^2 + b^2)^2*d) + (a*Cos[c + d*x]*Sin[c + d*x])/(2*(a^2 + b^2)*d)

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Rubi [A]  time = 0.129146, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3100, 2635, 8, 3098, 3133} \[ \frac{b \cos ^2(c+d x)}{2 d \left (a^2+b^2\right )}+\frac{a \sin (c+d x) \cos (c+d x)}{2 d \left (a^2+b^2\right )}+\frac{b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^2}+\frac{a b^2 x}{\left (a^2+b^2\right )^2}+\frac{a x}{2 \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b^2*x)/(a^2 + b^2)^2 + (a*x)/(2*(a^2 + b^2)) + (b*Cos[c + d*x]^2)/(2*(a^2 + b^2)*d) + (b^3*Log[a*Cos[c + d*
x] + b*Sin[c + d*x]])/((a^2 + b^2)^2*d) + (a*Cos[c + d*x]*Sin[c + d*x])/(2*(a^2 + b^2)*d)

Rule 3100

Int[cos[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
Simp[(b*Cos[c + d*x]^(m - 1))/(d*(a^2 + b^2)*(m - 1)), x] + (Dist[a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1), x]
, x] + Dist[b^2/(a^2 + b^2), Int[Cos[c + d*x]^(m - 2)/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a,
b, c, d}, x] && NeQ[a^2 + b^2, 0] && GtQ[m, 1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3098

Int[cos[(c_.) + (d_.)*(x_)]/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp
[(a*x)/(a^2 + b^2), x] + Dist[b/(a^2 + b^2), Int[(b*Cos[c + d*x] - a*Sin[c + d*x])/(a*Cos[c + d*x] + b*Sin[c +
d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]

Rule 3133

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[((b*B + c*C)*x)/(b^2 + c^2), x] + Simp[((c*B - b*C)*L
og[a + b*Cos[d + e*x] + c*Sin[d + e*x]])/(e*(b^2 + c^2)), x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[b^2
+ c^2, 0] && EqQ[A*(b^2 + c^2) - a*(b*B + c*C), 0]

Rubi steps

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

Mathematica [C]  time = 0.220539, size = 143, normalized size = 1.2 \[ \frac{b \left (a^2+b^2\right ) \cos (2 (c+d x))+a^3 \sin (2 (c+d x))+2 a^3 c+2 a^3 d x+a b^2 \sin (2 (c+d x))+2 b^3 \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )+6 a b^2 c+6 a b^2 d x-4 i b^3 \tan ^{-1}(\tan (c+d x))+4 i b^3 c+4 i b^3 d x}{4 d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*c + 6*a*b^2*c + (4*I)*b^3*c + 2*a^3*d*x + 6*a*b^2*d*x + (4*I)*b^3*d*x - (4*I)*b^3*ArcTan[Tan[c + d*x]]
+ b*(a^2 + b^2)*Cos[2*(c + d*x)] + 2*b^3*Log[(a*Cos[c + d*x] + b*Sin[c + d*x])^2] + a^3*Sin[2*(c + d*x)] + a*b
^2*Sin[2*(c + d*x)])/(4*(a^2 + b^2)^2*d)

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Maple [B]  time = 0.122, size = 236, normalized size = 2. \begin{align*}{\frac{{b}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\tan \left ( dx+c \right ){a}^{3}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }}+{\frac{\tan \left ( dx+c \right ) a{b}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }}+{\frac{{a}^{2}b}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }}+{\frac{{b}^{3}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }}-{\frac{{b}^{3}\ln \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{3\,\arctan \left ( \tan \left ( dx+c \right ) \right ) a{b}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{3}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.7359, size = 383, normalized size = 3.22 \begin{align*} \frac{\frac{b^{3} \log \left (-a - \frac{2 \, b \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{b^{3} \log \left (\frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{3} + 3 \, a b^{2}\right )} \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{\frac{a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{2 \, b \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2} + b^{2} + \frac{2 \,{\left (a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{{\left (a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^3*log(-a - 2*b*sin(d*x + c)/(cos(d*x + c) + 1) + a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/(a^4 + 2*a^2*b^2 +
b^4) - b^3*log(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) + (a^3 + 3*a*b^2)*arctan(sin(d
*x + c)/(cos(d*x + c) + 1))/(a^4 + 2*a^2*b^2 + b^4) + (a*sin(d*x + c)/(cos(d*x + c) + 1) - 2*b*sin(d*x + c)^2/
(cos(d*x + c) + 1)^2 - a*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + b^2 + 2*(a^2 + b^2)*sin(d*x + c)^2/(cos(d
*x + c) + 1)^2 + (a^2 + b^2)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4))/d

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Fricas [A]  time = 0.52543, size = 278, normalized size = 2.34 \begin{align*} \frac{b^{3} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) +{\left (a^{3} + 3 \, a b^{2}\right )} d x +{\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{2} +{\left (a^{3} + a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^3*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) + (a^3 + 3*a*b^2)*d*x + (a^2*
b + b^3)*cos(d*x + c)^2 + (a^3 + a*b^2)*cos(d*x + c)*sin(d*x + c))/((a^4 + 2*a^2*b^2 + b^4)*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*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.14924, size = 246, normalized size = 2.07 \begin{align*} \frac{\frac{2 \, b^{4} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac{b^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{3} + 3 \, a b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{b^{3} \tan \left (d x + c\right )^{2} + a^{3} \tan \left (d x + c\right ) + a b^{2} \tan \left (d x + c\right ) + a^{2} b + 2 \, b^{3}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\tan \left (d x + c\right )^{2} + 1\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^4*log(abs(b*tan(d*x + c) + a))/(a^4*b + 2*a^2*b^3 + b^5) - b^3*log(tan(d*x + c)^2 + 1)/(a^4 + 2*a^2*b
^2 + b^4) + (a^3 + 3*a*b^2)*(d*x + c)/(a^4 + 2*a^2*b^2 + b^4) + (b^3*tan(d*x + c)^2 + a^3*tan(d*x + c) + a*b^2
*tan(d*x + c) + a^2*b + 2*b^3)/((a^4 + 2*a^2*b^2 + b^4)*(tan(d*x + c)^2 + 1)))/d