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]

1/5*(-b*cos(d*x+c)+a*sin(d*x+c))/(a^2+b^2)/d/(a*cos(d*x+c)+b*sin(d*x+c))^5-4/15*(b*cos(d*x+c)-a*sin(d*x+c))/(a
^2+b^2)^2/d/(a*cos(d*x+c)+b*sin(d*x+c))^3+8/15*sin(d*x+c)/a/(a^2+b^2)^2/d/(a*cos(d*x+c)+b*sin(d*x+c))

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Rubi [A]  time = 0.07, antiderivative size = 151, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.52, size = 182, normalized size = 1.21 \[ \frac {10 a^4 \sin (c+d x)+5 a^4 \sin (3 (c+d x))+a^4 \sin (5 (c+d x))+\left (4 a b^3-4 a^3 b\right ) \cos (5 (c+d x))+20 a^2 b^2 \sin (c+d x)-6 a^2 b^2 \sin (5 (c+d x))-10 a b \left (a^2+b^2\right ) \cos (3 (c+d x))+10 b^4 \sin (c+d x)-5 b^4 \sin (3 (c+d x))+b^4 \sin (5 (c+d x))}{30 a d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^5} \]

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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fricas [B]  time = 2.05, size = 441, normalized size = 2.92 \[ -\frac {8 \, {\left (5 \, a^{4} b - 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{5} - 20 \, {\left (a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} - 5 \, {\left (a^{4} b + 6 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right ) - {\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4} + 8 \, {\left (a^{5} - 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (a^{5} + 10 \, a^{3} b^{2} - 15 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, {\left ({\left (a^{11} - 7 \, a^{9} b^{2} - 22 \, a^{7} b^{4} - 14 \, a^{5} b^{6} + 5 \, a^{3} b^{8} + 5 \, a b^{10}\right )} d \cos \left (d x + c\right )^{5} + 10 \, {\left (a^{9} b^{2} + 2 \, a^{7} b^{4} - 2 \, a^{3} b^{8} - a b^{10}\right )} d \cos \left (d x + c\right )^{3} + 5 \, {\left (a^{7} b^{4} + 3 \, a^{5} b^{6} + 3 \, a^{3} b^{8} + a b^{10}\right )} d \cos \left (d x + c\right ) + {\left ({\left (5 \, a^{10} b + 5 \, a^{8} b^{3} - 14 \, a^{6} b^{5} - 22 \, a^{4} b^{7} - 7 \, a^{2} b^{9} + b^{11}\right )} d \cos \left (d x + c\right )^{4} + 2 \, {\left (5 \, a^{8} b^{3} + 14 \, a^{6} b^{5} + 12 \, a^{4} b^{7} + 2 \, a^{2} b^{9} - b^{11}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{6} b^{5} + 3 \, a^{4} b^{7} + 3 \, a^{2} b^{9} + b^{11}\right )} d\right )} \sin \left (d x + c\right )\right )}} \]

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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giac [A]  time = 0.22, size = 118, normalized size = 0.78 \[ -\frac {15 \, b^{4} \tan \left (d x + c\right )^{4} + 30 \, a b^{3} \tan \left (d x + c\right )^{3} + 30 \, a^{2} b^{2} \tan \left (d x + c\right )^{2} + 10 \, b^{4} \tan \left (d x + c\right )^{2} + 15 \, a^{3} b \tan \left (d x + c\right ) + 5 \, a b^{3} \tan \left (d x + c\right ) + 3 \, a^{4} + a^{2} b^{2} + 3 \, b^{4}}{15 \, {\left (b \tan \left (d x + c\right ) + a\right )}^{5} b^{5} d} \]

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)

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maple [A]  time = 0.64, size = 125, normalized size = 0.83 \[ \frac {-\frac {a^{4}+2 a^{2} b^{2}+b^{4}}{5 b^{5} \left (a +b \tan \left (d x +c \right )\right )^{5}}+\frac {a \left (a^{2}+b^{2}\right )}{b^{5} \left (a +b \tan \left (d x +c \right )\right )^{4}}-\frac {6 a^{2}+2 b^{2}}{3 b^{5} \left (a +b \tan \left (d x +c \right )\right )^{3}}+\frac {2 a}{b^{5} \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {1}{b^{5} \left (a +b \tan \left (d x +c \right )\right )}}{d} \]

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

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maxima [A]  time = 0.79, size = 174, normalized size = 1.15 \[ -\frac {15 \, b^{4} \tan \left (d x + c\right )^{4} + 30 \, a b^{3} \tan \left (d x + c\right )^{3} + 3 \, a^{4} + a^{2} b^{2} + 3 \, b^{4} + 10 \, {\left (3 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{2} + 5 \, {\left (3 \, a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )}{15 \, {\left (b^{10} \tan \left (d x + c\right )^{5} + 5 \, a b^{9} \tan \left (d x + c\right )^{4} + 10 \, a^{2} b^{8} \tan \left (d x + c\right )^{3} + 10 \, a^{3} b^{7} \tan \left (d x + c\right )^{2} + 5 \, a^{4} b^{6} \tan \left (d x + c\right ) + a^{5} b^{5}\right )} d} \]

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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mupad [B]  time = 5.15, size = 470, normalized size = 3.11 \[ \frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{a}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a}-\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (7\,a^2\,b-6\,b^3\right )}{3\,a^4}+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (7\,a^2\,b-6\,b^3\right )}{3\,a^4}-\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^2-6\,b^2\right )}{3\,a^3}-\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (a^2-6\,b^2\right )}{3\,a^3}+\frac {8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^2}-\frac {8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{a^2}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (29\,a^4-112\,a^2\,b^2+24\,b^4\right )}{15\,a^5}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (10\,a^5-120\,a^3\,b^2+80\,a\,b^4\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (40\,a^4\,b-80\,a^2\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (40\,a^4\,b-80\,a^2\,b^3\right )-a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (10\,a^5-120\,a^3\,b^2+80\,a\,b^4\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (60\,a^4\,b-160\,a^2\,b^3+32\,b^5\right )+a^5-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (5\,a^5-40\,a^3\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (5\,a^5-40\,a^3\,b^2\right )+10\,a^4\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+10\,a^4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*tan(c/2 + (d*x)/2)^9)/a + (2*tan(c/2 + (d*x)/2))/a - (8*tan(c/2 + (d*x)/2)^4*(7*a^2*b - 6*b^3))/(3*a^4) +
(8*tan(c/2 + (d*x)/2)^6*(7*a^2*b - 6*b^3))/(3*a^4) - (8*tan(c/2 + (d*x)/2)^3*(a^2 - 6*b^2))/(3*a^3) - (8*tan(c
/2 + (d*x)/2)^7*(a^2 - 6*b^2))/(3*a^3) + (8*b*tan(c/2 + (d*x)/2)^2)/a^2 - (8*b*tan(c/2 + (d*x)/2)^8)/a^2 + (4*
tan(c/2 + (d*x)/2)^5*(29*a^4 + 24*b^4 - 112*a^2*b^2))/(15*a^5))/(d*(tan(c/2 + (d*x)/2)^4*(80*a*b^4 + 10*a^5 -
120*a^3*b^2) - tan(c/2 + (d*x)/2)^3*(40*a^4*b - 80*a^2*b^3) - tan(c/2 + (d*x)/2)^7*(40*a^4*b - 80*a^2*b^3) - a
^5*tan(c/2 + (d*x)/2)^10 - tan(c/2 + (d*x)/2)^6*(80*a*b^4 + 10*a^5 - 120*a^3*b^2) + tan(c/2 + (d*x)/2)^5*(60*a
^4*b + 32*b^5 - 160*a^2*b^3) + a^5 - tan(c/2 + (d*x)/2)^2*(5*a^5 - 40*a^3*b^2) + tan(c/2 + (d*x)/2)^8*(5*a^5 -
40*a^3*b^2) + 10*a^4*b*tan(c/2 + (d*x)/2) + 10*a^4*b*tan(c/2 + (d*x)/2)^9))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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