3.229 \(\int \frac{1}{(a \cos (c+d x)+b \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=98 \[ \frac{2 \sin (c+d x)}{3 a d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))}-\frac{b \cos (c+d x)-a \sin (c+d x)}{3 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^3} \]

[Out]

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

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Rubi [A]  time = 0.041788, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3076, 3075} \[ \frac{2 \sin (c+d x)}{3 a d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))}-\frac{b \cos (c+d x)-a \sin (c+d x)}{3 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Cos[c + d*x] - a*Sin[c + d*x])/(3*(a^2 + b^2)*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (2*Sin[c + d*x])/(3
*a*(a^2 + b^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))^4} \, dx &=-\frac{b \cos (c+d x)-a \sin (c+d x)}{3 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{2 \int \frac{1}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx}{3 \left (a^2+b^2\right )}\\ &=-\frac{b \cos (c+d x)-a \sin (c+d x)}{3 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{2 \sin (c+d x)}{3 a \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.290424, size = 85, normalized size = 0.87 \[ \frac{\sin (c+d x) \left (\left (a^2-b^2\right ) \cos (2 (c+d x))+2 a^2+b^2\right )-a b \cos (3 (c+d x))}{3 a d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*b*Cos[3*(c + d*x)]) + (2*a^2 + b^2 + (a^2 - b^2)*Cos[2*(c + d*x)])*Sin[c + d*x])/(3*a*(a^2 + b^2)*d*(a*Co
s[c + d*x] + b*Sin[c + d*x])^3)

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Maple [A]  time = 0.181, size = 64, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ( -{\frac{1}{{b}^{3} \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{{a}^{2}+{b}^{2}}{3\,{b}^{3} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{3}}}+{\frac{a}{{b}^{3} \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))^4,x)

[Out]

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

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Maxima [A]  time = 1.01743, size = 115, normalized size = 1.17 \begin{align*} -\frac{3 \, b^{2} \tan \left (d x + c\right )^{2} + 3 \, a b \tan \left (d x + c\right ) + a^{2} + b^{2}}{3 \,{\left (b^{6} \tan \left (d x + c\right )^{3} + 3 \, a b^{5} \tan \left (d x + c\right )^{2} + 3 \, a^{2} b^{4} \tan \left (d x + c\right ) + a^{3} b^{3}\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))^4,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.21456, size = 470, normalized size = 4.8 \begin{align*} -\frac{2 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right ) -{\left (a^{3} + 3 \, a b^{2} + 2 \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \,{\left ({\left (a^{7} - a^{5} b^{2} - 5 \, a^{3} b^{4} - 3 \, a b^{6}\right )} d \cos \left (d x + c\right )^{3} + 3 \,{\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) +{\left ({\left (3 \, a^{6} b + 5 \, a^{4} b^{3} + a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{2} +{\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\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))^4,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.17313, size = 68, normalized size = 0.69 \begin{align*} -\frac{3 \, b^{2} \tan \left (d x + c\right )^{2} + 3 \, a b \tan \left (d x + c\right ) + a^{2} + b^{2}}{3 \,{\left (b \tan \left (d x + c\right ) + a\right )}^{3} b^{3} 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))^4,x, algorithm="giac")

[Out]

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