3.230 \(\int \frac{1}{(a \cos (c+d x)+b \sin (c+d x))^5} \, dx\)

Optimal. Leaf size=156 \[ -\frac{3 (b \cos (c+d x)-a \sin (c+d x))}{8 d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^2}-\frac{b \cos (c+d x)-a \sin (c+d x)}{4 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^4}-\frac{3 \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{8 d \left (a^2+b^2\right )^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0860091, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3076, 3074, 206} \[ -\frac{3 (b \cos (c+d x)-a \sin (c+d x))}{8 d \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^2}-\frac{b \cos (c+d x)-a \sin (c+d x)}{4 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^4}-\frac{3 \tanh ^{-1}\left (\frac{b \cos (c+d x)-a \sin (c+d x)}{\sqrt{a^2+b^2}}\right )}{8 d \left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(-5),x]

[Out]

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

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

Mathematica [A]  time = 1.12765, size = 157, normalized size = 1.01 \[ \frac{\frac{-11 b \left (a^2+b^2\right ) \cos (c+d x)+\left (3 b^3-9 a^2 b\right ) \cos (3 (c+d x))+2 a \sin (c+d x) \left (3 \left (a^2-3 b^2\right ) \cos (2 (c+d x))+7 a^2+b^2\right )}{4 \left (a^2+b^2\right )^2 (a \cos (c+d x)+b \sin (c+d x))^4}+\frac{6 \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cos[c + d*x] + b*Sin[c + d*x])^(-5),x]

[Out]

((6*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) + (-11*b*(a^2 + b^2)*Cos[c + d*x] +
(-9*a^2*b + 3*b^3)*Cos[3*(c + d*x)] + 2*a*(7*a^2 + b^2 + 3*(a^2 - 3*b^2)*Cos[2*(c + d*x)])*Sin[c + d*x])/(4*(a
^2 + b^2)^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^4))/(8*d)

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Maple [B]  time = 0.222, size = 514, normalized size = 3.3 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{ \left ( a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-2\,b\tan \left ( 1/2\,dx+c/2 \right ) -a \right ) ^{4}} \left ( -1/8\,{\frac{ \left ( 5\,{a}^{4}+16\,{a}^{2}{b}^{2}+8\,{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}}{a \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}+3/8\,{\frac{b \left ({a}^{4}+16\,{a}^{2}{b}^{2}+8\,{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}}{{a}^{2} \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}-1/8\,{\frac{ \left ( 3\,{a}^{6}-36\,{a}^{4}{b}^{2}+56\,{a}^{2}{b}^{4}+32\,{b}^{6} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{{a}^{3} \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}+1/8\,{\frac{b \left ( 15\,{a}^{6}-114\,{a}^{4}{b}^{2}-8\,{a}^{2}{b}^{4}+16\,{b}^{6} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}}{{a}^{4} \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}-1/8\,{\frac{ \left ( 3\,{a}^{6}+84\,{a}^{4}{b}^{2}-56\,{a}^{2}{b}^{4}-32\,{b}^{6} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{{a}^{3} \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}-1/8\,{\frac{b \left ( 23\,{a}^{4}-64\,{a}^{2}{b}^{2}-24\,{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{{a}^{2} \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}-1/8\,{\frac{ \left ( 5\,{a}^{4}-24\,{a}^{2}{b}^{2}-8\,{b}^{4} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{a \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}+1/8\,{\frac{b \left ( 5\,{a}^{2}+2\,{b}^{2} \right ) }{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}} \right ) }+{\frac{3}{4\,{a}^{4}+8\,{a}^{2}{b}^{2}+4\,{b}^{4}}{\it Artanh} \left ({\frac{1}{2} \left ( 2\,\tan \left ( 1/2\,dx+c/2 \right ) a-2\,b \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{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))^5,x)

[Out]

1/d*(-2*(-1/8*(5*a^4+16*a^2*b^2+8*b^4)/a/(a^4+2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)^7+3/8*b*(a^4+16*a^2*b^2+8*b^4)
/a^2/(a^4+2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)^6-1/8/a^3*(3*a^6-36*a^4*b^2+56*a^2*b^4+32*b^6)/(a^4+2*a^2*b^2+b^4)
*tan(1/2*d*x+1/2*c)^5+1/8/a^4*b*(15*a^6-114*a^4*b^2-8*a^2*b^4+16*b^6)/(a^4+2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)^4
-1/8/a^3*(3*a^6+84*a^4*b^2-56*a^2*b^4-32*b^6)/(a^4+2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)^3-1/8*b*(23*a^4-64*a^2*b^
2-24*b^4)/a^2/(a^4+2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)^2-1/8*(5*a^4-24*a^2*b^2-8*b^4)/a/(a^4+2*a^2*b^2+b^4)*tan(
1/2*d*x+1/2*c)+1/8*b*(5*a^2+2*b^2)/(a^4+2*a^2*b^2+b^4))/(a*tan(1/2*d*x+1/2*c)^2-2*b*tan(1/2*d*x+1/2*c)-a)^4+3/
4/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*tan(1/2*d*x+1/2*c)*a-2*b)/(a^2+b^2)^(1/2)))

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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(1/(a*cos(d*x+c)+b*sin(d*x+c))^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.53538, size = 1216, normalized size = 7.79 \begin{align*} -\frac{6 \,{\left (3 \, a^{4} b + 2 \, a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left ({\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + b^{4} + 2 \,{\left (3 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (a b^{3} \cos \left (d x + c\right ) +{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{a^{2} + b^{2}} \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 (4 \, a^{4} b - a^{2} b^{3} - 5 \, b^{5}\right )} \cos \left (d x + c\right ) - 2 \,{\left (2 \, a^{5} + 7 \, a^{3} b^{2} + 5 \, a b^{4} + 3 \,{\left (a^{5} - 2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \,{\left ({\left (a^{10} - 3 \, a^{8} b^{2} - 14 \, a^{6} b^{4} - 14 \, a^{4} b^{6} - 3 \, a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )^{4} + 2 \,{\left (3 \, a^{8} b^{2} + 8 \, a^{6} b^{4} + 6 \, a^{4} b^{6} - b^{10}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{6} b^{4} + 3 \, a^{4} b^{6} + 3 \, a^{2} b^{8} + b^{10}\right )} d + 4 \,{\left ({\left (a^{9} b + 2 \, a^{7} b^{3} - 2 \, a^{3} b^{7} - a b^{9}\right )} d \cos \left (d x + c\right )^{3} +{\left (a^{7} b^{3} + 3 \, a^{5} b^{5} + 3 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right )\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))^5,x, algorithm="fricas")

[Out]

-1/16*(6*(3*a^4*b + 2*a^2*b^3 - b^5)*cos(d*x + c)^3 - 3*((a^4 - 6*a^2*b^2 + b^4)*cos(d*x + c)^4 + b^4 + 2*(3*a
^2*b^2 - b^4)*cos(d*x + c)^2 + 4*(a*b^3*cos(d*x + c) + (a^3*b - a*b^3)*cos(d*x + c)^3)*sin(d*x + c))*sqrt(a^2
+ b^2)*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*(4*
a^4*b - a^2*b^3 - 5*b^5)*cos(d*x + c) - 2*(2*a^5 + 7*a^3*b^2 + 5*a*b^4 + 3*(a^5 - 2*a^3*b^2 - 3*a*b^4)*cos(d*x
+ c)^2)*sin(d*x + c))/((a^10 - 3*a^8*b^2 - 14*a^6*b^4 - 14*a^4*b^6 - 3*a^2*b^8 + b^10)*d*cos(d*x + c)^4 + 2*(
3*a^8*b^2 + 8*a^6*b^4 + 6*a^4*b^6 - b^10)*d*cos(d*x + c)^2 + (a^6*b^4 + 3*a^4*b^6 + 3*a^2*b^8 + b^10)*d + 4*((
a^9*b + 2*a^7*b^3 - 2*a^3*b^7 - a*b^9)*d*cos(d*x + c)^3 + (a^7*b^3 + 3*a^5*b^5 + 3*a^3*b^7 + a*b^9)*d*cos(d*x
+ c))*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))**5,x)

[Out]

Timed out

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Giac [B]  time = 1.34513, size = 794, normalized size = 5.09 \begin{align*} \text{result too large to display} \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))^5,x, algorithm="giac")

[Out]

-1/8*(3*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*sqr
t(a^2 + b^2)))/((a^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) - 2*(5*a^7*tan(1/2*d*x + 1/2*c)^7 + 16*a^5*b^2*tan(1/
2*d*x + 1/2*c)^7 + 8*a^3*b^4*tan(1/2*d*x + 1/2*c)^7 - 3*a^6*b*tan(1/2*d*x + 1/2*c)^6 - 48*a^4*b^3*tan(1/2*d*x
+ 1/2*c)^6 - 24*a^2*b^5*tan(1/2*d*x + 1/2*c)^6 + 3*a^7*tan(1/2*d*x + 1/2*c)^5 - 36*a^5*b^2*tan(1/2*d*x + 1/2*c
)^5 + 56*a^3*b^4*tan(1/2*d*x + 1/2*c)^5 + 32*a*b^6*tan(1/2*d*x + 1/2*c)^5 - 15*a^6*b*tan(1/2*d*x + 1/2*c)^4 +
114*a^4*b^3*tan(1/2*d*x + 1/2*c)^4 + 8*a^2*b^5*tan(1/2*d*x + 1/2*c)^4 - 16*b^7*tan(1/2*d*x + 1/2*c)^4 + 3*a^7*
tan(1/2*d*x + 1/2*c)^3 + 84*a^5*b^2*tan(1/2*d*x + 1/2*c)^3 - 56*a^3*b^4*tan(1/2*d*x + 1/2*c)^3 - 32*a*b^6*tan(
1/2*d*x + 1/2*c)^3 + 23*a^6*b*tan(1/2*d*x + 1/2*c)^2 - 64*a^4*b^3*tan(1/2*d*x + 1/2*c)^2 - 24*a^2*b^5*tan(1/2*
d*x + 1/2*c)^2 + 5*a^7*tan(1/2*d*x + 1/2*c) - 24*a^5*b^2*tan(1/2*d*x + 1/2*c) - 8*a^3*b^4*tan(1/2*d*x + 1/2*c)
- 5*a^6*b - 2*a^4*b^3)/((a^8 + 2*a^6*b^2 + a^4*b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)
^4))/d