3.571 \(\int \frac{1}{(a+b \cos (c+d x) \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=95 \[ \frac{8 a \tan ^{-1}\left (\frac{2 a \tan (c+d x)+b}{\sqrt{4 a^2-b^2}}\right )}{d \left (4 a^2-b^2\right )^{3/2}}+\frac{2 b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))} \]

[Out]

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

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Rubi [A]  time = 0.108601, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2666, 2664, 12, 2660, 618, 204} \[ \frac{8 a \tan ^{-1}\left (\frac{2 a \tan (c+d x)+b}{\sqrt{4 a^2-b^2}}\right )}{d \left (4 a^2-b^2\right )^{3/2}}+\frac{2 b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2666

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

Rule 2664

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.413638, size = 94, normalized size = 0.99 \[ \frac{2 \left (\frac{4 a \tan ^{-1}\left (\frac{2 a \tan (c+d x)+b}{\sqrt{4 a^2-b^2}}\right )}{\left (4 a^2-b^2\right )^{3/2}}+\frac{b \cos (2 (c+d x))}{(2 a-b) (2 a+b) (2 a+b \sin (2 (c+d x)))}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.113, size = 139, normalized size = 1.5 \begin{align*}{\frac{{b}^{2}\tan \left ( dx+c \right ) }{d \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}a+b\tan \left ( dx+c \right ) +a \right ) a \left ( 4\,{a}^{2}-{b}^{2} \right ) }}+2\,{\frac{b}{d \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}a+b\tan \left ( dx+c \right ) +a \right ) \left ( 4\,{a}^{2}-{b}^{2} \right ) }}+8\,{\frac{a}{ \left ( 4\,{a}^{2}-{b}^{2} \right ) ^{3/2}d}\arctan \left ({\frac{b+2\,a\tan \left ( dx+c \right ) }{\sqrt{4\,{a}^{2}-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(d*x+c)*sin(d*x+c))^2,x)

[Out]

1/d/(tan(d*x+c)^2*a+b*tan(d*x+c)+a)*b^2/a/(4*a^2-b^2)*tan(d*x+c)+2/d/(tan(d*x+c)^2*a+b*tan(d*x+c)+a)*b/(4*a^2-
b^2)+8*a*arctan((b+2*a*tan(d*x+c))/(4*a^2-b^2)^(1/2))/(4*a^2-b^2)^(3/2)/d

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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+b*cos(d*x+c)*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.72315, size = 1116, normalized size = 11.75 \begin{align*} \left [-\frac{4 \, a^{2} b - b^{3} - 2 \,{\left (4 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a^{2}\right )} \sqrt{-4 \, a^{2} + b^{2}} \log \left (\frac{2 \,{\left (8 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \,{\left (8 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a^{2} - b^{2} -{\left (2 \, b \cos \left (d x + c\right )^{2} + 4 \,{\left (2 \, a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - b\right )} \sqrt{-4 \, a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{4} - b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a^{2}}\right )}{{\left (16 \, a^{4} b - 8 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (16 \, a^{5} - 8 \, a^{3} b^{2} + a b^{4}\right )} d}, -\frac{4 \, a^{2} b - b^{3} - 2 \,{\left (4 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a^{2}\right )} \sqrt{4 \, a^{2} - b^{2}} \arctan \left (-\frac{{\left (4 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b\right )} \sqrt{4 \, a^{2} - b^{2}}}{2 \,{\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, a^{2} + b^{2}}\right )}{{\left (16 \, a^{4} b - 8 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (16 \, a^{5} - 8 \, a^{3} b^{2} + a b^{4}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[-(4*a^2*b - b^3 - 2*(4*a^2*b - b^3)*cos(d*x + c)^2 - 2*(a*b*cos(d*x + c)*sin(d*x + c) + a^2)*sqrt(-4*a^2 + b^
2)*log((2*(8*a^2 - b^2)*cos(d*x + c)^4 - 4*a*b*cos(d*x + c)*sin(d*x + c) - 2*(8*a^2 - b^2)*cos(d*x + c)^2 + 2*
a^2 - b^2 - (2*b*cos(d*x + c)^2 + 4*(2*a*cos(d*x + c)^3 - a*cos(d*x + c))*sin(d*x + c) - b)*sqrt(-4*a^2 + b^2)
)/(b^2*cos(d*x + c)^4 - b^2*cos(d*x + c)^2 - 2*a*b*cos(d*x + c)*sin(d*x + c) - a^2)))/((16*a^4*b - 8*a^2*b^3 +
b^5)*d*cos(d*x + c)*sin(d*x + c) + (16*a^5 - 8*a^3*b^2 + a*b^4)*d), -(4*a^2*b - b^3 - 2*(4*a^2*b - b^3)*cos(d
*x + c)^2 + 4*(a*b*cos(d*x + c)*sin(d*x + c) + a^2)*sqrt(4*a^2 - b^2)*arctan(-(4*a*cos(d*x + c)*sin(d*x + c) +
b)*sqrt(4*a^2 - b^2)/(2*(4*a^2 - b^2)*cos(d*x + c)^2 - 4*a^2 + b^2)))/((16*a^4*b - 8*a^2*b^3 + b^5)*d*cos(d*x
+ c)*sin(d*x + c) + (16*a^5 - 8*a^3*b^2 + a*b^4)*d)]

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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+b*cos(d*x+c)*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.1771, size = 157, normalized size = 1.65 \begin{align*} \frac{\frac{8 \,{\left (\pi \left \lfloor \frac{d x + c}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{2 \, a \tan \left (d x + c\right ) + b}{\sqrt{4 \, a^{2} - b^{2}}}\right )\right )} a}{{\left (4 \, a^{2} - b^{2}\right )}^{\frac{3}{2}}} + \frac{b^{2} \tan \left (d x + c\right ) + 2 \, a b}{{\left (4 \, a^{3} - a b^{2}\right )}{\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right ) + a\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))^2,x, algorithm="giac")

[Out]

(8*(pi*floor((d*x + c)/pi + 1/2)*sgn(a) + arctan((2*a*tan(d*x + c) + b)/sqrt(4*a^2 - b^2)))*a/(4*a^2 - b^2)^(3
/2) + (b^2*tan(d*x + c) + 2*a*b)/((4*a^3 - a*b^2)*(a*tan(d*x + c)^2 + b*tan(d*x + c) + a)))/d