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

Optimal. Leaf size=219 \[ \frac {2 b^2 \left (3 a^2-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d (a-b)^{5/2} (a+b)^{5/2}}-\frac {b^3 \sin (c+d x)}{d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac {2 b^4 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d (a-b)^{5/2} (a+b)^{5/2}}-\frac {\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}-\frac {\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)} \]

[Out]

2*b^4*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a/(a-b)^(5/2)/(a+b)^(5/2)/d+2*b^2*(3*a^2-b^2)*arctan
h((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a/(a-b)^(5/2)/(a+b)^(5/2)/d-1/2*sin(d*x+c)/(a+b)^2/d/(1-cos(d*x+
c))-1/2*sin(d*x+c)/(a-b)^2/d/(1+cos(d*x+c))-b^3*sin(d*x+c)/(a^2-b^2)^2/d/(b+a*cos(d*x+c))

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Rubi [A]  time = 0.38, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {4397, 2897, 2648, 2659, 208, 2664, 12} \[ -\frac {b^3 \sin (c+d x)}{d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac {2 b^2 \left (3 a^2-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d (a-b)^{5/2} (a+b)^{5/2}}+\frac {2 b^4 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d (a-b)^{5/2} (a+b)^{5/2}}-\frac {\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}-\frac {\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^4*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*(a - b)^(5/2)*(a + b)^(5/2)*d) + (2*b^2*(3*a^2
- b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*(a - b)^(5/2)*(a + b)^(5/2)*d) - Sin[c + d*x]/(
2*(a + b)^2*d*(1 - Cos[c + d*x])) - Sin[c + d*x]/(2*(a - b)^2*d*(1 + Cos[c + d*x])) - (b^3*Sin[c + d*x])/((a^2
- b^2)^2*d*(b + a*Cos[c + 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 208

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

Rule 2648

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

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*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 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 2897

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rubi steps

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

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Mathematica [A]  time = 1.29, size = 131, normalized size = 0.60 \[ \frac {\frac {\csc (c+d x) \left (\left (2 a^2 b+b^3\right ) \cos (2 (c+d x))-2 a \left (a^2-b^2\right ) \cos (c+d x)-3 b^3\right )}{a \cos (c+d x)+b}-\frac {12 a b^2 \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}}{2 d (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-12*a*b^2*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + ((-3*b^3 - 2*a*(a^2 - b^2)
*Cos[c + d*x] + (2*a^2*b + b^3)*Cos[2*(c + d*x)])*Csc[c + d*x])/(b + a*Cos[c + d*x]))/(2*(a - b)^2*(a + b)^2*d
)

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fricas [A]  time = 0.54, size = 518, normalized size = 2.37 \[ \left [-\frac {2 \, a^{4} b + 2 \, a^{2} b^{3} - 4 \, b^{5} - 3 \, {\left (a^{2} b^{2} \cos \left (d x + c\right ) + a b^{3}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 2 \, {\left (2 \, a^{4} b - a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d\right )} \sin \left (d x + c\right )}, -\frac {a^{4} b + a^{2} b^{3} - 2 \, b^{5} - 3 \, {\left (a^{2} b^{2} \cos \left (d x + c\right ) + a b^{3}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - {\left (2 \, a^{4} b - a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )}{{\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d\right )} \sin \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.67, size = 282, normalized size = 1.29 \[ -\frac {\frac {12 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 5 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{3} + a^{2} b + a b^{2} - b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(12*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1
/2*c))/sqrt(-a^2 + b^2)))*a*b^2/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) + tan(1/2*d*x + 1/2*c)/(a^2 - 2*a*b
+ b^2) + (a^3*tan(1/2*d*x + 1/2*c)^2 - 3*a^2*b*tan(1/2*d*x + 1/2*c)^2 + 3*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 5*b^
3*tan(1/2*d*x + 1/2*c)^2 - a^3 + a^2*b + a*b^2 - b^3)/((a^4 - 2*a^2*b^2 + b^4)*(a*tan(1/2*d*x + 1/2*c)^3 - b*t
an(1/2*d*x + 1/2*c)^3 - a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))))/d

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maple [A]  time = 0.18, size = 155, normalized size = 0.71 \[ \frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 b^{2} \left (-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b}-\frac {3 a \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {1}{2 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a*sin(d*x+c)+b*tan(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(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.13, size = 213, normalized size = 0.97 \[ \frac {\frac {a^2-2\,a\,b+b^2}{a+b}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^3-3\,a^2\,b+3\,a\,b^2-5\,b^3\right )}{{\left (a+b\right )}^2}}{d\,\left (\left (2\,a^3-6\,a^2\,b+6\,a\,b^2-2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-2\,a^3+2\,a^2\,b+2\,a\,b^2-2\,b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,{\left (a-b\right )}^2}+\frac {6\,a\,b^2\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{3/2}}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos {\left (c + d x \right )}}{\left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(c + d*x)/(a*sin(c + d*x) + b*tan(c + d*x))**2, x)

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