3.231 \(\int \frac{1}{(a \cos (c+d x)+b \sin (c+d x))^6} \, dx\)

Optimal. Leaf size=151 \[ \frac{8 \sin (c+d x)}{15 a d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))}-\frac{4 (b \cos (c+d x)-a \sin (c+d x))}{15 d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^3}-\frac{b \cos (c+d x)-a \sin (c+d x)}{5 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^5} \]

[Out]

-(b*Cos[c + d*x] - a*Sin[c + d*x])/(5*(a^2 + b^2)*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) - (4*(b*Cos[c + d*x]
- a*Sin[c + d*x]))/(15*(a^2 + b^2)^2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (8*Sin[c + d*x])/(15*a*(a^2 + b^
2)^2*d*(a*Cos[c + d*x] + b*Sin[c + d*x]))

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Rubi [A]  time = 0.0693007, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3076, 3075} \[ \frac{8 \sin (c+d x)}{15 a d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))}-\frac{4 (b \cos (c+d x)-a \sin (c+d x))}{15 d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^3}-\frac{b \cos (c+d x)-a \sin (c+d x)}{5 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^5} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(-6),x]

[Out]

-(b*Cos[c + d*x] - a*Sin[c + d*x])/(5*(a^2 + b^2)*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^5) - (4*(b*Cos[c + d*x]
- a*Sin[c + d*x]))/(15*(a^2 + b^2)^2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (8*Sin[c + d*x])/(15*a*(a^2 + b^
2)^2*d*(a*Cos[c + d*x] + b*Sin[c + d*x]))

Rule 3076

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

Rule 3075

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

Rubi steps

\begin{align*} \int \frac{1}{(a \cos (c+d x)+b \sin (c+d x))^6} \, dx &=-\frac{b \cos (c+d x)-a \sin (c+d x)}{5 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^5}+\frac{4 \int \frac{1}{(a \cos (c+d x)+b \sin (c+d x))^4} \, dx}{5 \left (a^2+b^2\right )}\\ &=-\frac{b \cos (c+d x)-a \sin (c+d x)}{5 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac{4 (b \cos (c+d x)-a \sin (c+d x))}{15 \left (a^2+b^2\right )^2 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{8 \int \frac{1}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx}{15 \left (a^2+b^2\right )^2}\\ &=-\frac{b \cos (c+d x)-a \sin (c+d x)}{5 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^5}-\frac{4 (b \cos (c+d x)-a \sin (c+d x))}{15 \left (a^2+b^2\right )^2 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{8 \sin (c+d x)}{15 a \left (a^2+b^2\right )^2 d (a \cos (c+d x)+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.531786, size = 182, normalized size = 1.21 \[ \frac{20 a^2 b^2 \sin (c+d x)-6 a^2 b^2 \sin (5 (c+d x))-10 a b \left (a^2+b^2\right ) \cos (3 (c+d x))+\left (4 a b^3-4 a^3 b\right ) \cos (5 (c+d x))+10 a^4 \sin (c+d x)+5 a^4 \sin (3 (c+d x))+a^4 \sin (5 (c+d x))+10 b^4 \sin (c+d x)-5 b^4 \sin (3 (c+d x))+b^4 \sin (5 (c+d x))}{30 a d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cos[c + d*x] + b*Sin[c + d*x])^(-6),x]

[Out]

(-10*a*b*(a^2 + b^2)*Cos[3*(c + d*x)] + (-4*a^3*b + 4*a*b^3)*Cos[5*(c + d*x)] + 10*a^4*Sin[c + d*x] + 20*a^2*b
^2*Sin[c + d*x] + 10*b^4*Sin[c + d*x] + 5*a^4*Sin[3*(c + d*x)] - 5*b^4*Sin[3*(c + d*x)] + a^4*Sin[5*(c + d*x)]
- 6*a^2*b^2*Sin[5*(c + d*x)] + b^4*Sin[5*(c + d*x)])/(30*a*(a^2 + b^2)^2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^
5)

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Maple [A]  time = 0.225, size = 125, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{a \left ({a}^{2}+{b}^{2} \right ) }{{b}^{5} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{4}}}-{\frac{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}{5\,{b}^{5} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{5}}}-{\frac{1}{{b}^{5} \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{6\,{a}^{2}+2\,{b}^{2}}{3\,{b}^{5} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{3}}}+2\,{\frac{a}{{b}^{5} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cos(d*x+c)+b*sin(d*x+c))^6,x)

[Out]

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

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Maxima [A]  time = 1.04956, size = 235, normalized size = 1.56 \begin{align*} -\frac{15 \, b^{4} \tan \left (d x + c\right )^{4} + 30 \, a b^{3} \tan \left (d x + c\right )^{3} + 3 \, a^{4} + a^{2} b^{2} + 3 \, b^{4} + 10 \,{\left (3 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{2} + 5 \,{\left (3 \, a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )}{15 \,{\left (b^{10} \tan \left (d x + c\right )^{5} + 5 \, a b^{9} \tan \left (d x + c\right )^{4} + 10 \, a^{2} b^{8} \tan \left (d x + c\right )^{3} + 10 \, a^{3} b^{7} \tan \left (d x + c\right )^{2} + 5 \, a^{4} b^{6} \tan \left (d x + c\right ) + a^{5} b^{5}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))^6,x, algorithm="maxima")

[Out]

-1/15*(15*b^4*tan(d*x + c)^4 + 30*a*b^3*tan(d*x + c)^3 + 3*a^4 + a^2*b^2 + 3*b^4 + 10*(3*a^2*b^2 + b^4)*tan(d*
x + c)^2 + 5*(3*a^3*b + a*b^3)*tan(d*x + c))/((b^10*tan(d*x + c)^5 + 5*a*b^9*tan(d*x + c)^4 + 10*a^2*b^8*tan(d
*x + c)^3 + 10*a^3*b^7*tan(d*x + c)^2 + 5*a^4*b^6*tan(d*x + c) + a^5*b^5)*d)

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Fricas [B]  time = 2.77556, size = 984, normalized size = 6.52 \begin{align*} -\frac{8 \,{\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{5} - 20 \,{\left (a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} - 5 \,{\left (a^{4} b + 6 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right ) -{\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4} + 8 \,{\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} + 4 \,{\left (a^{5} + 10 \, a^{3} b^{2} - 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \,{\left ({\left (a^{11} - 7 \, a^{9} b^{2} - 22 \, a^{7} b^{4} - 14 \, a^{5} b^{6} + 5 \, a^{3} b^{8} + 5 \, a b^{10}\right )} d \cos \left (d x + c\right )^{5} + 10 \,{\left (a^{9} b^{2} + 2 \, a^{7} b^{4} - 2 \, a^{3} b^{8} - a b^{10}\right )} d \cos \left (d x + c\right )^{3} + 5 \,{\left (a^{7} b^{4} + 3 \, a^{5} b^{6} + 3 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right ) +{\left ({\left (5 \, a^{10} b + 5 \, a^{8} b^{3} - 14 \, a^{6} b^{5} - 22 \, a^{4} b^{7} - 7 \, a^{2} b^{9} + b^{11}\right )} d \cos \left (d x + c\right )^{4} + 2 \,{\left (5 \, a^{8} b^{3} + 14 \, a^{6} b^{5} + 12 \, a^{4} b^{7} + 2 \, a^{2} b^{9} - b^{11}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{6} b^{5} + 3 \, a^{4} b^{7} + 3 \, 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(1/(a*cos(d*x+c)+b*sin(d*x+c))^6,x, algorithm="fricas")

[Out]

-1/15*(8*(5*a^4*b - 10*a^2*b^3 + b^5)*cos(d*x + c)^5 - 20*(a^4*b - 6*a^2*b^3 + b^5)*cos(d*x + c)^3 - 5*(a^4*b
+ 6*a^2*b^3 - 3*b^5)*cos(d*x + c) - (3*a^5 + 10*a^3*b^2 + 15*a*b^4 + 8*(a^5 - 10*a^3*b^2 + 5*a*b^4)*cos(d*x +
c)^4 + 4*(a^5 + 10*a^3*b^2 - 15*a*b^4)*cos(d*x + c)^2)*sin(d*x + c))/((a^11 - 7*a^9*b^2 - 22*a^7*b^4 - 14*a^5*
b^6 + 5*a^3*b^8 + 5*a*b^10)*d*cos(d*x + c)^5 + 10*(a^9*b^2 + 2*a^7*b^4 - 2*a^3*b^8 - a*b^10)*d*cos(d*x + c)^3
+ 5*(a^7*b^4 + 3*a^5*b^6 + 3*a^3*b^8 + a*b^10)*d*cos(d*x + c) + ((5*a^10*b + 5*a^8*b^3 - 14*a^6*b^5 - 22*a^4*b
^7 - 7*a^2*b^9 + b^11)*d*cos(d*x + c)^4 + 2*(5*a^8*b^3 + 14*a^6*b^5 + 12*a^4*b^7 + 2*a^2*b^9 - b^11)*d*cos(d*x
+ c)^2 + (a^6*b^5 + 3*a^4*b^7 + 3*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(1/(a*cos(d*x+c)+b*sin(d*x+c))**6,x)

[Out]

Timed out

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Giac [A]  time = 1.12876, size = 159, normalized size = 1.05 \begin{align*} -\frac{15 \, b^{4} \tan \left (d x + c\right )^{4} + 30 \, a b^{3} \tan \left (d x + c\right )^{3} + 30 \, a^{2} b^{2} \tan \left (d x + c\right )^{2} + 10 \, b^{4} \tan \left (d x + c\right )^{2} + 15 \, a^{3} b \tan \left (d x + c\right ) + 5 \, a b^{3} \tan \left (d x + c\right ) + 3 \, a^{4} + a^{2} b^{2} + 3 \, b^{4}}{15 \,{\left (b \tan \left (d x + c\right ) + a\right )}^{5} b^{5} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))^6,x, algorithm="giac")

[Out]

-1/15*(15*b^4*tan(d*x + c)^4 + 30*a*b^3*tan(d*x + c)^3 + 30*a^2*b^2*tan(d*x + c)^2 + 10*b^4*tan(d*x + c)^2 + 1
5*a^3*b*tan(d*x + c) + 5*a*b^3*tan(d*x + c) + 3*a^4 + a^2*b^2 + 3*b^4)/((b*tan(d*x + c) + a)^5*b^5*d)