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

Optimal. Leaf size=165 \[ -\frac{b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}-\frac{a b}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}-\frac{b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^3}+\frac{4 a b \left (a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^4}+\frac{x \left (-6 a^2 b^2+a^4+b^4\right )}{\left (a^2+b^2\right )^4} \]

[Out]

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

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Rubi [A]  time = 0.30323, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3086, 3483, 3529, 3531, 3530} \[ -\frac{b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}-\frac{a b}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}-\frac{b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^3}+\frac{4 a b \left (a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^4}+\frac{x \left (-6 a^2 b^2+a^4+b^4\right )}{\left (a^2+b^2\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3086

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

Rule 3483

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n + 1))/(d*(n + 1)
*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ
[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rubi steps

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

Mathematica [C]  time = 6.21098, size = 419, normalized size = 2.54 \[ \frac{\left (a^2-2 a b-b^2\right ) \left (a^2+2 a b-b^2\right ) (c+d x)}{d (a-i b)^4 (a+i b)^4}+\frac{4 \left (a^9 b^2+2 i a^8 b^3+2 a^7 b^4-2 i a^4 b^7-2 a^3 b^8-i a^2 b^9+i a^{10} b-a b^{10}\right ) (c+d x)}{d (a-i b)^8 (a+i b)^7}-\frac{4 i \left (a^3 b-a b^3\right ) \tan ^{-1}(\tan (c+d x))}{d \left (a^2+b^2\right )^4}-\frac{b^3 \left (6 a^2+b^2\right )}{3 a d (a-i b)^3 (a+i b)^3 (a \cos (c+d x)+b \sin (c+d x))^2}+\frac{2 \left (9 a^2 b^2 \sin (c+d x)-2 b^4 \sin (c+d x)\right )}{3 a d (a-i b)^3 (a+i b)^3 (a \cos (c+d x)+b \sin (c+d x))}+\frac{2 \left (a^3 b-a b^3\right ) \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )}{d \left (a^2+b^2\right )^4}+\frac{b^4 \sin (c+d x)}{3 a d (a-i b)^2 (a+i b)^2 (a \cos (c+d x)+b \sin (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2 - 2*a*b - b^2)*(a^2 + 2*a*b - b^2)*(c + d*x))/((a - I*b)^4*(a + I*b)^4*d) + (4*(I*a^10*b + a^9*b^2 + (2*
I)*a^8*b^3 + 2*a^7*b^4 - (2*I)*a^4*b^7 - 2*a^3*b^8 - I*a^2*b^9 - a*b^10)*(c + d*x))/((a - I*b)^8*(a + I*b)^7*d
) - ((4*I)*(a^3*b - a*b^3)*ArcTan[Tan[c + d*x]])/((a^2 + b^2)^4*d) + (2*(a^3*b - a*b^3)*Log[(a*Cos[c + d*x] +
b*Sin[c + d*x])^2])/((a^2 + b^2)^4*d) + (b^4*Sin[c + d*x])/(3*a*(a - I*b)^2*(a + I*b)^2*d*(a*Cos[c + d*x] + b*
Sin[c + d*x])^3) - (b^3*(6*a^2 + b^2))/(3*a*(a - I*b)^3*(a + I*b)^3*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^2) + (
2*(9*a^2*b^2*Sin[c + d*x] - 2*b^4*Sin[c + d*x]))/(3*a*(a - I*b)^3*(a + I*b)^3*d*(a*Cos[c + d*x] + b*Sin[c + d*
x]))

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Maple [A]  time = 0.212, size = 304, normalized size = 1.8 \begin{align*} -{\frac{b}{ \left ( 3\,{a}^{2}+3\,{b}^{2} \right ) d \left ( a+b\tan \left ( dx+c \right ) \right ) ^{3}}}-3\,{\frac{{a}^{2}b}{d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( a+b\tan \left ( dx+c \right ) \right ) }}+{\frac{{b}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{ab}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}d \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}+4\,{\frac{{a}^{3}b\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{4}}}-4\,{\frac{a{b}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{4}}}-2\,{\frac{\ln \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ){a}^{3}b}{d \left ({a}^{2}+{b}^{2} \right ) ^{4}}}+2\,{\frac{\ln \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) a{b}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{4}}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{4}}{d \left ({a}^{2}+{b}^{2} \right ) ^{4}}}-6\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{4}}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{4}}{d \left ({a}^{2}+{b}^{2} \right ) ^{4}}} \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))^4,x)

[Out]

-1/3*b/(a^2+b^2)/d/(a+b*tan(d*x+c))^3-3/d*b/(a^2+b^2)^3/(a+b*tan(d*x+c))*a^2+1/d*b^3/(a^2+b^2)^3/(a+b*tan(d*x+
c))-a*b/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^2+4/d*a^3*b/(a^2+b^2)^4*ln(a+b*tan(d*x+c))-4/d*a*b^3/(a^2+b^2)^4*ln(a+b
*tan(d*x+c))-2/d/(a^2+b^2)^4*ln(tan(d*x+c)^2+1)*a^3*b+2/d/(a^2+b^2)^4*ln(tan(d*x+c)^2+1)*a*b^3+1/d/(a^2+b^2)^4
*arctan(tan(d*x+c))*a^4-6/d/(a^2+b^2)^4*arctan(tan(d*x+c))*a^2*b^2+1/d/(a^2+b^2)^4*arctan(tan(d*x+c))*b^4

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Maxima [B]  time = 1.82541, size = 520, normalized size = 3.15 \begin{align*} \frac{\frac{3 \,{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )}{\left (d x + c\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac{12 \,{\left (a^{3} b - a b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac{6 \,{\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac{13 \, a^{4} b + 2 \, a^{2} b^{3} + b^{5} + 3 \,{\left (3 \, a^{2} b^{3} - b^{5}\right )} \tan \left (d x + c\right )^{2} + 3 \,{\left (7 \, a^{3} b^{2} - a b^{4}\right )} \tan \left (d x + c\right )}{a^{9} + 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} + a^{3} b^{6} +{\left (a^{6} b^{3} + 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} + b^{9}\right )} \tan \left (d x + c\right )^{3} + 3 \,{\left (a^{7} b^{2} + 3 \, a^{5} b^{4} + 3 \, a^{3} b^{6} + a b^{8}\right )} \tan \left (d x + c\right )^{2} + 3 \,{\left (a^{8} b + 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} + a^{2} b^{7}\right )} \tan \left (d x + c\right )}}{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))^4,x, algorithm="maxima")

[Out]

1/3*(3*(a^4 - 6*a^2*b^2 + b^4)*(d*x + c)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 12*(a^3*b - a*b^3)*
log(b*tan(d*x + c) + a)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - 6*(a^3*b - a*b^3)*log(tan(d*x + c)^2
+ 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - (13*a^4*b + 2*a^2*b^3 + b^5 + 3*(3*a^2*b^3 - b^5)*tan(
d*x + c)^2 + 3*(7*a^3*b^2 - a*b^4)*tan(d*x + c))/(a^9 + 3*a^7*b^2 + 3*a^5*b^4 + a^3*b^6 + (a^6*b^3 + 3*a^4*b^5
+ 3*a^2*b^7 + b^9)*tan(d*x + c)^3 + 3*(a^7*b^2 + 3*a^5*b^4 + 3*a^3*b^6 + a*b^8)*tan(d*x + c)^2 + 3*(a^8*b + 3
*a^6*b^3 + 3*a^4*b^5 + a^2*b^7)*tan(d*x + c)))/d

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Fricas [B]  time = 0.610633, size = 1256, normalized size = 7.61 \begin{align*} -\frac{{\left (54 \, a^{4} b^{3} - 30 \, a^{2} b^{5} + 4 \, b^{7} - 3 \,{\left (a^{7} - 9 \, a^{5} b^{2} + 19 \, a^{3} b^{4} - 3 \, a b^{6}\right )} d x\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (10 \, a^{4} b^{3} - 11 \, a^{2} b^{5} + b^{7} + 3 \,{\left (a^{5} b^{2} - 6 \, a^{3} b^{4} + a b^{6}\right )} d x\right )} \cos \left (d x + c\right ) - 6 \,{\left ({\left (a^{6} b - 4 \, a^{4} b^{3} + 3 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (a^{4} b^{3} - a^{2} b^{5}\right )} \cos \left (d x + c\right ) +{\left (a^{3} b^{4} - a b^{6} +{\left (3 \, a^{5} b^{2} - 4 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \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 (13 \, a^{3} b^{4} - 9 \, a b^{6} + 3 \,{\left (a^{4} b^{3} - 6 \, a^{2} b^{5} + b^{7}\right )} d x +{\left (18 \, a^{5} b^{2} - 58 \, a^{3} b^{4} + 12 \, a b^{6} + 3 \,{\left (3 \, a^{6} b - 19 \, a^{4} b^{3} + 9 \, a^{2} b^{5} - b^{7}\right )} d x\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \,{\left ({\left (a^{11} + a^{9} b^{2} - 6 \, a^{7} b^{4} - 14 \, a^{5} b^{6} - 11 \, a^{3} b^{8} - 3 \, a b^{10}\right )} d \cos \left (d x + c\right )^{3} + 3 \,{\left (a^{9} b^{2} + 4 \, a^{7} b^{4} + 6 \, a^{5} b^{6} + 4 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right ) +{\left ({\left (3 \, a^{10} b + 11 \, a^{8} b^{3} + 14 \, a^{6} b^{5} + 6 \, a^{4} b^{7} - a^{2} b^{9} - b^{11}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{8} b^{3} + 4 \, a^{6} b^{5} + 6 \, a^{4} b^{7} + 4 \, a^{2} b^{9} + b^{11}\right )} d\right )} \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))^4,x, algorithm="fricas")

[Out]

-1/3*((54*a^4*b^3 - 30*a^2*b^5 + 4*b^7 - 3*(a^7 - 9*a^5*b^2 + 19*a^3*b^4 - 3*a*b^6)*d*x)*cos(d*x + c)^3 - 3*(1
0*a^4*b^3 - 11*a^2*b^5 + b^7 + 3*(a^5*b^2 - 6*a^3*b^4 + a*b^6)*d*x)*cos(d*x + c) - 6*((a^6*b - 4*a^4*b^3 + 3*a
^2*b^5)*cos(d*x + c)^3 + 3*(a^4*b^3 - a^2*b^5)*cos(d*x + c) + (a^3*b^4 - a*b^6 + (3*a^5*b^2 - 4*a^3*b^4 + a*b^
6)*cos(d*x + c)^2)*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) - (13
*a^3*b^4 - 9*a*b^6 + 3*(a^4*b^3 - 6*a^2*b^5 + b^7)*d*x + (18*a^5*b^2 - 58*a^3*b^4 + 12*a*b^6 + 3*(3*a^6*b - 19
*a^4*b^3 + 9*a^2*b^5 - b^7)*d*x)*cos(d*x + c)^2)*sin(d*x + c))/((a^11 + a^9*b^2 - 6*a^7*b^4 - 14*a^5*b^6 - 11*
a^3*b^8 - 3*a*b^10)*d*cos(d*x + c)^3 + 3*(a^9*b^2 + 4*a^7*b^4 + 6*a^5*b^6 + 4*a^3*b^8 + a*b^10)*d*cos(d*x + c)
+ ((3*a^10*b + 11*a^8*b^3 + 14*a^6*b^5 + 6*a^4*b^7 - a^2*b^9 - b^11)*d*cos(d*x + c)^2 + (a^8*b^3 + 4*a^6*b^5
+ 6*a^4*b^7 + 4*a^2*b^9 + b^11)*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))**4,x)

[Out]

Timed out

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Giac [B]  time = 1.17555, size = 500, normalized size = 3.03 \begin{align*} \frac{\frac{3 \,{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )}{\left (d x + c\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac{6 \,{\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac{12 \,{\left (a^{3} b^{2} - a b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}} - \frac{22 \, a^{3} b^{4} \tan \left (d x + c\right )^{3} - 22 \, a b^{6} \tan \left (d x + c\right )^{3} + 75 \, a^{4} b^{3} \tan \left (d x + c\right )^{2} - 60 \, a^{2} b^{5} \tan \left (d x + c\right )^{2} - 3 \, b^{7} \tan \left (d x + c\right )^{2} + 87 \, a^{5} b^{2} \tan \left (d x + c\right ) - 48 \, a^{3} b^{4} \tan \left (d x + c\right ) - 3 \, a b^{6} \tan \left (d x + c\right ) + 35 \, a^{6} b - 7 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}}{{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )}{\left (b \tan \left (d x + c\right ) + a\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))^4,x, algorithm="giac")

[Out]

1/3*(3*(a^4 - 6*a^2*b^2 + b^4)*(d*x + c)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - 6*(a^3*b - a*b^3)*l
og(tan(d*x + c)^2 + 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 12*(a^3*b^2 - a*b^4)*log(abs(b*tan(d*
x + c) + a))/(a^8*b + 4*a^6*b^3 + 6*a^4*b^5 + 4*a^2*b^7 + b^9) - (22*a^3*b^4*tan(d*x + c)^3 - 22*a*b^6*tan(d*x
+ c)^3 + 75*a^4*b^3*tan(d*x + c)^2 - 60*a^2*b^5*tan(d*x + c)^2 - 3*b^7*tan(d*x + c)^2 + 87*a^5*b^2*tan(d*x +
c) - 48*a^3*b^4*tan(d*x + c) - 3*a*b^6*tan(d*x + c) + 35*a^6*b - 7*a^4*b^3 + 3*a^2*b^5 + b^7)/((a^8 + 4*a^6*b^
2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*(b*tan(d*x + c) + a)^3))/d