3.111 \(\int \frac{\cos ^4(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=166 \[ -\frac{a \sin ^3(c+d x)}{3 d \left (a^2+b^2\right )}+\frac{a b^2 \sin (c+d x)}{d \left (a^2+b^2\right )^2}+\frac{a \sin (c+d x)}{d \left (a^2+b^2\right )}+\frac{b \cos ^3(c+d x)}{3 d \left (a^2+b^2\right )}+\frac{b^3 \cos (c+d x)}{d \left (a^2+b^2\right )^2}-\frac{b^4 \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}} \]

[Out]

-((b^4*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(5/2)*d)) + (b^3*Cos[c + d*x])
/((a^2 + b^2)^2*d) + (b*Cos[c + d*x]^3)/(3*(a^2 + b^2)*d) + (a*b^2*Sin[c + d*x])/((a^2 + b^2)^2*d) + (a*Sin[c
+ d*x])/((a^2 + b^2)*d) - (a*Sin[c + d*x]^3)/(3*(a^2 + b^2)*d)

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Rubi [A]  time = 0.17495, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3100, 2633, 2637, 3074, 206} \[ -\frac{a \sin ^3(c+d x)}{3 d \left (a^2+b^2\right )}+\frac{a b^2 \sin (c+d x)}{d \left (a^2+b^2\right )^2}+\frac{a \sin (c+d x)}{d \left (a^2+b^2\right )}+\frac{b \cos ^3(c+d x)}{3 d \left (a^2+b^2\right )}+\frac{b^3 \cos (c+d x)}{d \left (a^2+b^2\right )^2}-\frac{b^4 \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

-((b^4*ArcTanh[(b*Cos[c + d*x] - a*Sin[c + d*x])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(5/2)*d)) + (b^3*Cos[c + d*x])
/((a^2 + b^2)^2*d) + (b*Cos[c + d*x]^3)/(3*(a^2 + b^2)*d) + (a*b^2*Sin[c + d*x])/((a^2 + b^2)^2*d) + (a*Sin[c
+ d*x])/((a^2 + b^2)*d) - (a*Sin[c + d*x]^3)/(3*(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 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[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{\cos ^4(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx &=\frac{b \cos ^3(c+d x)}{3 \left (a^2+b^2\right ) d}+\frac{a \int \cos ^3(c+d x) \, dx}{a^2+b^2}+\frac{b^2 \int \frac{\cos ^2(c+d x)}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{a^2+b^2}\\ &=\frac{b^3 \cos (c+d x)}{\left (a^2+b^2\right )^2 d}+\frac{b \cos ^3(c+d x)}{3 \left (a^2+b^2\right ) d}+\frac{\left (a b^2\right ) \int \cos (c+d x) \, dx}{\left (a^2+b^2\right )^2}+\frac{b^4 \int \frac{1}{a \cos (c+d x)+b \sin (c+d x)} \, dx}{\left (a^2+b^2\right )^2}-\frac{a \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{\left (a^2+b^2\right ) d}\\ &=\frac{b^3 \cos (c+d x)}{\left (a^2+b^2\right )^2 d}+\frac{b \cos ^3(c+d x)}{3 \left (a^2+b^2\right ) d}+\frac{a b^2 \sin (c+d x)}{\left (a^2+b^2\right )^2 d}+\frac{a \sin (c+d x)}{\left (a^2+b^2\right ) d}-\frac{a \sin ^3(c+d x)}{3 \left (a^2+b^2\right ) d}-\frac{b^4 \operatorname{Subst}\left (\int \frac{1}{a^2+b^2-x^2} \, dx,x,b \cos (c+d x)-a \sin (c+d x)\right )}{\left (a^2+b^2\right )^2 d}\\ &=-\frac{b^4 \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2} d}+\frac{b^3 \cos (c+d x)}{\left (a^2+b^2\right )^2 d}+\frac{b \cos ^3(c+d x)}{3 \left (a^2+b^2\right ) d}+\frac{a b^2 \sin (c+d x)}{\left (a^2+b^2\right )^2 d}+\frac{a \sin (c+d x)}{\left (a^2+b^2\right ) d}-\frac{a \sin ^3(c+d x)}{3 \left (a^2+b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 1.01504, size = 137, normalized size = 0.83 \[ \frac{\sqrt{a^2+b^2} \left (3 b \left (a^2+5 b^2\right ) \cos (c+d x)+b \left (a^2+b^2\right ) \cos (3 (c+d x))+2 a \sin (c+d x) \left (\left (a^2+b^2\right ) \cos (2 (c+d x))+5 a^2+11 b^2\right )\right )+24 b^4 \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{12 d \left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

(24*b^4*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]] + Sqrt[a^2 + b^2]*(3*b*(a^2 + 5*b^2)*Cos[c + d*x] +
b*(a^2 + b^2)*Cos[3*(c + d*x)] + 2*a*(5*a^2 + 11*b^2 + (a^2 + b^2)*Cos[2*(c + d*x)])*Sin[c + d*x]))/(12*(a^2
+ b^2)^(5/2)*d)

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Maple [A]  time = 0.135, size = 221, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( 2\,{\frac{{b}^{4}}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-2\,{\frac{ \left ( -{a}^{3}-2\,a{b}^{2} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+ \left ( -{a}^{2}b-2\,{b}^{3} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( -2/3\,{a}^{3}-8/3\,a{b}^{2} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-2\,{b}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+ \left ( -{a}^{3}-2\,a{b}^{2} \right ) \tan \left ( 1/2\,dx+c/2 \right ) -1/3\,{a}^{2}b-4/3\,{b}^{3}}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

1/d*(2*b^4/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))-2/(a^
4+2*a^2*b^2+b^4)*((-a^3-2*a*b^2)*tan(1/2*d*x+1/2*c)^5+(-a^2*b-2*b^3)*tan(1/2*d*x+1/2*c)^4+(-2/3*a^3-8/3*a*b^2)
*tan(1/2*d*x+1/2*c)^3-2*b^3*tan(1/2*d*x+1/2*c)^2+(-a^3-2*a*b^2)*tan(1/2*d*x+1/2*c)-1/3*a^2*b-4/3*b^3)/(1+tan(1
/2*d*x+1/2*c)^2)^3)

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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)^4/(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.558174, size = 598, normalized size = 3.6 \begin{align*} \frac{3 \, \sqrt{a^{2} + b^{2}} b^{4} \log \left (-\frac{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} - 2 \, a^{2} - b^{2} + 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{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 ) + 2 \,{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) + 2 \,{\left (2 \, a^{5} + 7 \, a^{3} b^{2} + 5 \, a b^{4} +{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(3*sqrt(a^2 + b^2)*b^4*log(-(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 - b^2 +
2*sqrt(a^2 + b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x +
c)^2 + b^2)) + 2*(a^4*b + 2*a^2*b^3 + b^5)*cos(d*x + c)^3 + 6*(a^2*b^3 + b^5)*cos(d*x + c) + 2*(2*a^5 + 7*a^3*
b^2 + 5*a*b^4 + (a^5 + 2*a^3*b^2 + a*b^4)*cos(d*x + c)^2)*sin(d*x + c))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*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)**4/(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.31919, size = 386, normalized size = 2.33 \begin{align*} -\frac{\frac{3 \, b^{4} \log \left (\frac{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} + b^{2}}} - \frac{2 \,{\left (3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 6 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 8 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2} b + 4 \, b^{3}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/3*(3*b^4*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b + 2
*sqrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) - 2*(3*a^3*tan(1/2*d*x + 1/2*c)^5 + 6*a*b^2*tan(1
/2*d*x + 1/2*c)^5 + 3*a^2*b*tan(1/2*d*x + 1/2*c)^4 + 6*b^3*tan(1/2*d*x + 1/2*c)^4 + 2*a^3*tan(1/2*d*x + 1/2*c)
^3 + 8*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*b^3*tan(1/2*d*x + 1/2*c)^2 + 3*a^3*tan(1/2*d*x + 1/2*c) + 6*a*b^2*tan(
1/2*d*x + 1/2*c) + a^2*b + 4*b^3)/((a^4 + 2*a^2*b^2 + b^4)*(tan(1/2*d*x + 1/2*c)^2 + 1)^3))/d