\(\int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^2} \, dx\) [571]

Optimal result

Integrand size = 18, antiderivative size = 95 \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^2} \, dx=\frac {8 a \arctan \left (\frac {b+2 a \tan (c+d x)}{\sqrt {4 a^2-b^2}}\right )}{\left (4 a^2-b^2\right )^{3/2} d}+\frac {2 b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right ) d (2 a+b \sin (2 c+2 d x))} \]

[Out]

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+2*b*cos(2*d*x+2*c)/(4*a^2-b^2)/d/(2*a+b*s
in(2*d*x+2*c))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2745, 2743, 12, 2739, 632, 210} \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^2} \, dx=\frac {8 a \arctan \left (\frac {2 a \tan (c+d x)+b}{\sqrt {4 a^2-b^2}}\right )}{d \left (4 a^2-b^2\right )^{3/2}}+\frac {2 b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))} \]

[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 12

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

Rule 210

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

Rule 632

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 2739

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 2743

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] && Integ
erQ[2*n]

Rule 2745

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]

Rubi steps \begin{align*} \text {integral}& = \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) \text {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) \text {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 \arctan \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] (verified)

Time = 0.55 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^2} \, dx=\frac {2 \left (\frac {4 a \arctan \left (\frac {b+2 a \tan (c+d x)}{\sqrt {4 a^2-b^2}}\right )}{\left (4 a^2-b^2\right )^{3/2}}+\frac {b \cos (2 (c+d x))}{(2 a-b) (2 a+b) (2 a+b \sin (2 (c+d x)))}\right )}{d} \]

[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

Maple [A] (verified)

Time = 1.36 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.20

method	result	size
derivativedivides	\(\frac {\frac {\frac {b^{2} \tan \left (d x +c \right )}{a \left (4 a^{2}-b^{2}\right )}+\frac {2 b}{4 a^{2}-b^{2}}}{a \tan \left (d x +c \right )^{2}+b \tan \left (d x +c \right )+a}+\frac {8 a \arctan \left (\frac {b +2 a \tan \left (d x +c \right )}{\sqrt {4 a^{2}-b^{2}}}\right )}{\left (4 a^{2}-b^{2}\right )^{\frac {3}{2}}}}{d}\)	\(114\)
default	\(\frac {\frac {\frac {b^{2} \tan \left (d x +c \right )}{a \left (4 a^{2}-b^{2}\right )}+\frac {2 b}{4 a^{2}-b^{2}}}{a \tan \left (d x +c \right )^{2}+b \tan \left (d x +c \right )+a}+\frac {8 a \arctan \left (\frac {b +2 a \tan \left (d x +c \right )}{\sqrt {4 a^{2}-b^{2}}}\right )}{\left (4 a^{2}-b^{2}\right )^{\frac {3}{2}}}}{d}\)	\(114\)
risch	\(\frac {8 \,{\mathrm e}^{2 i \left (d x +c \right )} a +4 i b}{\left (4 a^{2}-b^{2}\right ) d \left (b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )}-\frac {4 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \sqrt {-4 a^{2}+b^{2}}-4 a^{2}+b^{2}}{b \sqrt {-4 a^{2}+b^{2}}}\right )}{\sqrt {-4 a^{2}+b^{2}}\, \left (2 a +b \right ) \left (2 a -b \right ) d}+\frac {4 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \sqrt {-4 a^{2}+b^{2}}+4 a^{2}-b^{2}}{b \sqrt {-4 a^{2}+b^{2}}}\right )}{\sqrt {-4 a^{2}+b^{2}}\, \left (2 a +b \right ) \left (2 a -b \right ) d}\)	\(235\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (91) = 182\).

Time = 0.28 (sec) , antiderivative size = 493, normalized size of antiderivative = 5.19 \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^2} \, dx=\left [-\frac {4 \, a^{2} b - b^{3} - 2 \, {\left (4 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a^{2}\right )} \sqrt {-4 \, a^{2} + b^{2}} \log \left (\frac {2 \, {\left (8 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, {\left (8 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a^{2} - b^{2} - {\left (2 \, b \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - b\right )} \sqrt {-4 \, a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{4} - b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a^{2}}\right )}{{\left (16 \, a^{4} b - 8 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (16 \, a^{5} - 8 \, a^{3} b^{2} + a b^{4}\right )} d}, -\frac {4 \, a^{2} b - b^{3} - 2 \, {\left (4 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a^{2}\right )} \sqrt {4 \, a^{2} - b^{2}} \arctan \left (-\frac {{\left (4 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b\right )} \sqrt {4 \, a^{2} - b^{2}}}{2 \, {\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, a^{2} + b^{2}}\right )}{{\left (16 \, a^{4} b - 8 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (16 \, a^{5} - 8 \, a^{3} b^{2} + a b^{4}\right )} d}\right ] \]

[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)]

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b^2-4*a^2>0)', see `assume?` f
or more deta

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.22 \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^2} \, dx=\frac {\frac {8 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {2 \, a \tan \left (d x + c\right ) + b}{\sqrt {4 \, a^{2} - b^{2}}}\right )\right )} a}{{\left (4 \, a^{2} - b^{2}\right )}^{\frac {3}{2}}} + \frac {b^{2} \tan \left (d x + c\right ) + 2 \, a b}{{\left (4 \, a^{3} - a b^{2}\right )} {\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right ) + a\right )}}}{d} \]

[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

Mupad [B] (verification not implemented)

Time = 27.40 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.91 \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^2} \, dx=\frac {\frac {2\,b}{4\,a^2-b^2}+\frac {b^2\,\mathrm {tan}\left (c+d\,x\right )}{a\,\left (4\,a^2-b^2\right )}}{d\,\left (a\,{\mathrm {tan}\left (c+d\,x\right )}^2+b\,\mathrm {tan}\left (c+d\,x\right )+a\right )}+\frac {8\,a\,\mathrm {atan}\left (\frac {\left (4\,a^2-b^2\right )\,\left (\frac {8\,a^2\,\mathrm {tan}\left (c+d\,x\right )}{{\left (2\,a+b\right )}^{3/2}\,{\left (2\,a-b\right )}^{3/2}}+\frac {4\,a\,\left (4\,a^2\,b-b^3\right )}{{\left (2\,a+b\right )}^{3/2}\,\left (4\,a^2-b^2\right )\,{\left (2\,a-b\right )}^{3/2}}\right )}{4\,a}\right )}{d\,{\left (2\,a+b\right )}^{3/2}\,{\left (2\,a-b\right )}^{3/2}} \]

[In]

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

[Out]

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