3.122 \(\int \frac{\cos ^4(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=145 \[ \frac{b^4}{a d \left (a^2+b^2\right )^2 (a \cot (c+d x)+b)}-\frac{\sin ^2(c+d x) \left (2 a b-\left (a^2-b^2\right ) \cot (c+d x)\right )}{2 d \left (a^2+b^2\right )^2}+\frac{4 a b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{x \left (6 a^2 b^2+a^4-3 b^4\right )}{2 \left (a^2+b^2\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.292613, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3088, 1647, 1629, 635, 203, 260} \[ \frac{b^4}{a d \left (a^2+b^2\right )^2 (a \cot (c+d x)+b)}-\frac{\sin ^2(c+d x) \left (2 a b-\left (a^2-b^2\right ) \cot (c+d x)\right )}{2 d \left (a^2+b^2\right )^2}+\frac{4 a b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{x \left (6 a^2 b^2+a^4-3 b^4\right )}{2 \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4/(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

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

Rule 3088

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> -Dist[d^(-1), Subst[Int[(x^m*(b + a*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Cot[c + d*x]], x] /; FreeQ[
{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^4}{(b+a x)^2 \left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac{\left (2 a b-\left (a^2-b^2\right ) \cot (c+d x)\right ) \sin ^2(c+d x)}{2 \left (a^2+b^2\right )^2 d}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{b^2 \left (a^2-b^2\right )}{\left (a^2+b^2\right )^2}-\frac{2 a b x}{a^2+b^2}-\frac{\left (a^4+5 a^2 b^2+2 b^4\right ) x^2}{\left (a^2+b^2\right )^2}}{(b+a x)^2 \left (1+x^2\right )} \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=-\frac{\left (2 a b-\left (a^2-b^2\right ) \cot (c+d x)\right ) \sin ^2(c+d x)}{2 \left (a^2+b^2\right )^2 d}+\frac{\operatorname{Subst}\left (\int \left (-\frac{2 b^4}{\left (a^2+b^2\right )^2 (b+a x)^2}+\frac{8 a^2 b^3}{\left (a^2+b^2\right )^3 (b+a x)}+\frac{-a^4-6 a^2 b^2+3 b^4-8 a b^3 x}{\left (a^2+b^2\right )^3 \left (1+x^2\right )}\right ) \, dx,x,\cot (c+d x)\right )}{2 d}\\ &=\frac{b^4}{a \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac{4 a b^3 \log (b+a \cot (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac{\left (2 a b-\left (a^2-b^2\right ) \cot (c+d x)\right ) \sin ^2(c+d x)}{2 \left (a^2+b^2\right )^2 d}+\frac{\operatorname{Subst}\left (\int \frac{-a^4-6 a^2 b^2+3 b^4-8 a b^3 x}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 \left (a^2+b^2\right )^3 d}\\ &=\frac{b^4}{a \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac{4 a b^3 \log (b+a \cot (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac{\left (2 a b-\left (a^2-b^2\right ) \cot (c+d x)\right ) \sin ^2(c+d x)}{2 \left (a^2+b^2\right )^2 d}-\frac{\left (4 a b^3\right ) \operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,\cot (c+d x)\right )}{\left (a^2+b^2\right )^3 d}-\frac{\left (a^4+6 a^2 b^2-3 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{2 \left (a^2+b^2\right )^3 d}\\ &=\frac{\left (a^4+6 a^2 b^2-3 b^4\right ) x}{2 \left (a^2+b^2\right )^3}+\frac{b^4}{a \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac{4 a b^3 \log (b+a \cot (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac{4 a b^3 \log (\sin (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac{\left (2 a b-\left (a^2-b^2\right ) \cot (c+d x)\right ) \sin ^2(c+d x)}{2 \left (a^2+b^2\right )^2 d}\\ \end{align*}

Mathematica [A]  time = 0.896495, size = 149, normalized size = 1.03 \[ \frac{2 \left (6 a^2 b^2+a^4-3 b^4\right ) (c+d x)+\left (a^2-b^2\right ) \left (a^2+b^2\right ) \sin (2 (c+d x))+2 a b \left (a^2+b^2\right ) \cos (2 (c+d x))+\frac{4 b^4 \left (a^2+b^2\right ) \sin (c+d x)}{a (a \cos (c+d x)+b \sin (c+d x))}+16 a b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{4 d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

(2*(a^4 + 6*a^2*b^2 - 3*b^4)*(c + d*x) + 2*a*b*(a^2 + b^2)*Cos[2*(c + d*x)] + 16*a*b^3*Log[a*Cos[c + d*x] + b*
Sin[c + d*x]] + (4*b^4*(a^2 + b^2)*Sin[c + d*x])/(a*(a*Cos[c + d*x] + b*Sin[c + d*x])) + (a^2 - b^2)*(a^2 + b^
2)*Sin[2*(c + d*x)])/(4*(a^2 + b^2)^3*d)

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Maple [B]  time = 0.161, size = 292, normalized size = 2. \begin{align*} -{\frac{{b}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( a+b\tan \left ( dx+c \right ) \right ) }}+4\,{\frac{a{b}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{{a}^{4}\tan \left ( dx+c \right ) }{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }}-{\frac{\tan \left ( dx+c \right ){b}^{4}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }}+{\frac{{a}^{3}b}{d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }}+{\frac{a{b}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }}-2\,{\frac{a{b}^{3}\ln \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+3\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{3\,\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{4}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{4}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}} \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))^2,x)

[Out]

-1/d*b^3/(a^2+b^2)^2/(a+b*tan(d*x+c))+4/d*b^3/(a^2+b^2)^3*a*ln(a+b*tan(d*x+c))+1/2/d/(a^2+b^2)^3/(tan(d*x+c)^2
+1)*tan(d*x+c)*a^4-1/2/d/(a^2+b^2)^3/(tan(d*x+c)^2+1)*tan(d*x+c)*b^4+1/d/(a^2+b^2)^3/(tan(d*x+c)^2+1)*a^3*b+1/
d/(a^2+b^2)^3/(tan(d*x+c)^2+1)*a*b^3-2/d/(a^2+b^2)^3*a*b^3*ln(tan(d*x+c)^2+1)+3/d/(a^2+b^2)^3*arctan(tan(d*x+c
))*a^2*b^2-3/2/d/(a^2+b^2)^3*arctan(tan(d*x+c))*b^4+1/2/d/(a^2+b^2)^3*arctan(tan(d*x+c))*a^4

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Maxima [B]  time = 1.76085, size = 381, normalized size = 2.63 \begin{align*} \frac{\frac{8 \, a b^{3} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{4 \, a b^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{{\left (a^{4} + 6 \, a^{2} b^{2} - 3 \, b^{4}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{2 \, a^{2} b - 2 \, b^{3} +{\left (a^{2} b - 3 \, b^{3}\right )} \tan \left (d x + c\right )^{2} +{\left (a^{3} + a b^{2}\right )} \tan \left (d x + c\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4} +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )^{3} +{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \tan \left (d x + c\right )^{2} +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )}}{2 \, 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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.565004, size = 614, normalized size = 4.23 \begin{align*} \frac{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} -{\left (a^{2} b^{3} + 3 \, b^{5} -{\left (a^{5} + 6 \, a^{3} b^{2} - 3 \, a b^{4}\right )} d x\right )} \cos \left (d x + c\right ) + 4 \,{\left (a^{2} b^{3} \cos \left (d x + c\right ) + a b^{4} \sin \left (d x + c\right )\right )} \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} b^{2} - a b^{4} -{\left (a^{4} b + 6 \, a^{2} b^{3} - 3 \, b^{5}\right )} d x -{\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 )}{2 \,{\left ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} d \sin \left (d x + c\right )\right )}} \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))^2,x, algorithm="fricas")

[Out]

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

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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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.13898, size = 338, normalized size = 2.33 \begin{align*} \frac{\frac{8 \, a b^{4} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} - \frac{4 \, a b^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{{\left (a^{4} + 6 \, a^{2} b^{2} - 3 \, b^{4}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{a^{2} b \tan \left (d x + c\right )^{2} - 3 \, 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 ) + 2 \, a^{2} b - 2 \, b^{3}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b \tan \left (d x + c\right )^{3} + a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right ) + a\right )}}}{2 \, 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))^2,x, algorithm="giac")

[Out]

1/2*(8*a*b^4*log(abs(b*tan(d*x + c) + a))/(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7) - 4*a*b^3*log(tan(d*x + c)^2 +
1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (a^4 + 6*a^2*b^2 - 3*b^4)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b
^6) + (a^2*b*tan(d*x + c)^2 - 3*b^3*tan(d*x + c)^2 + a^3*tan(d*x + c) + a*b^2*tan(d*x + c) + 2*a^2*b - 2*b^3)/
((a^4 + 2*a^2*b^2 + b^4)*(b*tan(d*x + c)^3 + a*tan(d*x + c)^2 + b*tan(d*x + c) + a)))/d