3.257 \(\int \frac{\cos ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=243 \[ -\frac{b^5 \sin (c+d x)}{a^2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac{2 b^6 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{2 b^4 \left (5 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{2 b x}{a^3}-\frac{\sin (c+d x)}{a^2 d}-\frac{\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}-\frac{\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)} \]

[Out]

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

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Rubi [A]  time = 0.610712, antiderivative size = 243, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4397, 2897, 2648, 2637, 2664, 12, 2659, 208} \[ -\frac{b^5 \sin (c+d x)}{a^2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac{2 b^6 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{2 b^4 \left (5 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{2 b x}{a^3}-\frac{\sin (c+d x)}{a^2 d}-\frac{\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}-\frac{\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4397

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

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 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, 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 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 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]

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx &=\int \frac{\cos ^3(c+d x) \cot ^2(c+d x)}{(b+a \cos (c+d x))^2} \, dx\\ &=-\int \left (-\frac{2 b}{a^3}-\frac{1}{2 (a-b)^2 (-1-\cos (c+d x))}-\frac{1}{2 (a+b)^2 (1-\cos (c+d x))}+\frac{\cos (c+d x)}{a^2}+\frac{b^5}{a^3 \left (a^2-b^2\right ) (-b-a \cos (c+d x))^2}+\frac{b^4 \left (5 a^2-3 b^2\right )}{a^3 \left (a^2-b^2\right )^2 (-b-a \cos (c+d x))}\right ) \, dx\\ &=\frac{2 b x}{a^3}-\frac{\int \cos (c+d x) \, dx}{a^2}+\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{\left (b^4 \left (5 a^2-3 b^2\right )\right ) \int \frac{1}{-b-a \cos (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )^2}-\frac{b^5 \int \frac{1}{(-b-a \cos (c+d x))^2} \, dx}{a^3 \left (a^2-b^2\right )}\\ &=\frac{2 b x}{a^3}-\frac{\sin (c+d x)}{a^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^5 \sin (c+d x)}{a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac{b^5 \int \frac{b}{-b-a \cos (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )^2}-\frac{\left (2 b^4 \left (5 a^2-3 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^3 \left (a^2-b^2\right )^2 d}\\ &=\frac{2 b x}{a^3}+\frac{2 b^4 \left (5 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}-\frac{\sin (c+d x)}{a^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^5 \sin (c+d x)}{a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac{b^6 \int \frac{1}{-b-a \cos (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )^2}\\ &=\frac{2 b x}{a^3}+\frac{2 b^4 \left (5 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}-\frac{\sin (c+d x)}{a^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^5 \sin (c+d x)}{a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac{\left (2 b^6\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^3 \left (a^2-b^2\right )^2 d}\\ &=\frac{2 b x}{a^3}+\frac{2 b^6 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}+\frac{2 b^4 \left (5 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}-\frac{\sin (c+d x)}{a^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^5 \sin (c+d x)}{a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 3.06031, size = 164, normalized size = 0.67 \[ -\frac{\frac{2 b^5 \sin (c+d x)}{a^2 (a-b)^2 (a+b)^2 (a \cos (c+d x)+b)}+\frac{4 b^4 \left (5 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{5/2}}-\frac{4 b (c+d x)}{a^3}+\frac{2 \sin (c+d x)}{a^2}+\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{(a-b)^2}+\frac{\cot \left (\frac{1}{2} (c+d x)\right )}{(a+b)^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.186, size = 291, normalized size = 1.2 \begin{align*} -{\frac{1}{2\,d \left ({a}^{2}-2\,ab+{b}^{2} \right ) }\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{{a}^{2}d \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+4\,{\frac{b\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{{a}^{3}d}}+2\,{\frac{{b}^{5}\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}+10\,{\frac{{b}^{4}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}a\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-4\,{\frac{{b}^{6}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}{a}^{3}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{1}{2\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^2,x)

[Out]

-1/2/d/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)-2/d/a^2*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)+4/d/a^3*b*arctan
(tan(1/2*d*x+1/2*c))+2/d*b^5/(a+b)^2/(a-b)^2/a^2*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)
^2*b-a-b)+10/d*b^4/(a+b)^2/(a-b)^2/a/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))
-4/d*b^6/(a+b)^2/(a-b)^2/a^3/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-1/2/d/(
a+b)^2/tan(1/2*d*x+1/2*c)

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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(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.70098, size = 1839, normalized size = 7.57 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.29026, size = 639, normalized size = 2.63 \begin{align*} -\frac{\frac{4 \,{\left (5 \, a^{2} b^{4} - 2 \, b^{6}\right )}{\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 )}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} 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{5 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 7 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 5 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 7 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 4 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 8 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 4 \, a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 6 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 6 \, a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{5} + a^{4} b + a^{3} b^{2} - a^{2} b^{3}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 2 \, 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 )}} - \frac{4 \,{\left (d x + c\right )} b}{a^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/2*(4*(5*a^2*b^4 - 2*b^6)*(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^7 - 2*a^5*b^2 + a^3*b^4)*sqrt(-a^2 + b^2)) + tan(1/2*d*x + 1/
2*c)/(a^2 - 2*a*b + b^2) + (5*a^5*tan(1/2*d*x + 1/2*c)^4 - 7*a^4*b*tan(1/2*d*x + 1/2*c)^4 - 5*a^3*b^2*tan(1/2*
d*x + 1/2*c)^4 + 7*a^2*b^3*tan(1/2*d*x + 1/2*c)^4 + 4*a*b^4*tan(1/2*d*x + 1/2*c)^4 - 8*b^5*tan(1/2*d*x + 1/2*c
)^4 - 4*a^5*tan(1/2*d*x + 1/2*c)^2 - 6*a^4*b*tan(1/2*d*x + 1/2*c)^2 + 12*a^3*b^2*tan(1/2*d*x + 1/2*c)^2 + 6*a^
2*b^3*tan(1/2*d*x + 1/2*c)^2 - 4*a*b^4*tan(1/2*d*x + 1/2*c)^2 - 8*b^5*tan(1/2*d*x + 1/2*c)^2 - a^5 + a^4*b + a
^3*b^2 - a^2*b^3)/((a^6 - 2*a^4*b^2 + a^2*b^4)*(a*tan(1/2*d*x + 1/2*c)^5 - b*tan(1/2*d*x + 1/2*c)^5 - 2*b*tan(
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))) - 4*(d*x + c)*b/a^3)/d