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

Optimal. Leaf size=221 \[ \frac{\cos (c+d x) (a \tan (c+d x)+b)}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{a b \left (2 a^2-13 b^2\right ) \sec (c+d x)}{2 d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}+\frac{b \left (2 a^2-3 b^2\right ) \sec (c+d x)}{2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}-\frac{3 b^2 \left (4 a^2-b^2\right ) \cos (c+d x) \sqrt{\sec ^2(c+d x)} \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{2 d \left (a^2+b^2\right )^{7/2}} \]

[Out]

(-3*b^2*(4*a^2 - b^2)*ArcTanh[(b - a*Tan[c + d*x])/(Sqrt[a^2 + b^2]*Sqrt[Sec[c + d*x]^2])]*Cos[c + d*x]*Sqrt[S
ec[c + d*x]^2])/(2*(a^2 + b^2)^(7/2)*d) + (b*(2*a^2 - 3*b^2)*Sec[c + d*x])/(2*(a^2 + b^2)^2*d*(a + b*Tan[c + d
*x])^2) + (Cos[c + d*x]*(b + a*Tan[c + d*x]))/((a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (a*b*(2*a^2 - 13*b^2)*S
ec[c + d*x])/(2*(a^2 + b^2)^3*d*(a + b*Tan[c + d*x]))

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Rubi [A]  time = 0.213629, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3512, 741, 835, 807, 725, 206} \[ \frac{\cos (c+d x) (a \tan (c+d x)+b)}{d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac{a b \left (2 a^2-13 b^2\right ) \sec (c+d x)}{2 d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}+\frac{b \left (2 a^2-3 b^2\right ) \sec (c+d x)}{2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}-\frac{3 b^2 \left (4 a^2-b^2\right ) \cos (c+d x) \sqrt{\sec ^2(c+d x)} \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{2 d \left (a^2+b^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b^2*(4*a^2 - b^2)*ArcTanh[(b - a*Tan[c + d*x])/(Sqrt[a^2 + b^2]*Sqrt[Sec[c + d*x]^2])]*Cos[c + d*x]*Sqrt[S
ec[c + d*x]^2])/(2*(a^2 + b^2)^(7/2)*d) + (b*(2*a^2 - 3*b^2)*Sec[c + d*x])/(2*(a^2 + b^2)^2*d*(a + b*Tan[c + d
*x])^2) + (Cos[c + d*x]*(b + a*Tan[c + d*x]))/((a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (a*b*(2*a^2 - 13*b^2)*S
ec[c + d*x])/(2*(a^2 + b^2)^3*d*(a + b*Tan[c + d*x]))

Rule 3512

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(d^(2
*IntPart[m/2])*(d*Sec[e + f*x])^(2*FracPart[m/2]))/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]), Subst[Int[(a + x)^n*(
1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] &&
!IntegerQ[m/2]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\cos (c+d x)}{(a+b \tan (c+d x))^3} \, dx &=\frac{\left (\cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(a+x)^3 \left (1+\frac{x^2}{b^2}\right )^{3/2}} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{\cos (c+d x) (b+a \tan (c+d x))}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac{\left (b \cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{-3-\frac{2 a x}{b^2}}{(a+x)^3 \sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{\left (a^2+b^2\right ) d}\\ &=\frac{b \left (2 a^2-3 b^2\right ) \sec (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}+\frac{\cos (c+d x) (b+a \tan (c+d x))}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{\left (b^3 \cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{\frac{10 a}{b^2}+\frac{\left (2 a^2-3 b^2\right ) x}{b^4}}{(a+x)^2 \sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d}\\ &=\frac{b \left (2 a^2-3 b^2\right ) \sec (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}+\frac{\cos (c+d x) (b+a \tan (c+d x))}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a b \left (2 a^2-13 b^2\right ) \sec (c+d x)}{2 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac{\left (3 b \left (4 a^2-b^2\right ) \cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(a+x) \sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3 d}\\ &=\frac{b \left (2 a^2-3 b^2\right ) \sec (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}+\frac{\cos (c+d x) (b+a \tan (c+d x))}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a b \left (2 a^2-13 b^2\right ) \sec (c+d x)}{2 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}-\frac{\left (3 b \left (4 a^2-b^2\right ) \cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a^2}{b^2}-x^2} \, dx,x,\frac{1-\frac{a \tan (c+d x)}{b}}{\sqrt{\sec ^2(c+d x)}}\right )}{2 \left (a^2+b^2\right )^3 d}\\ &=-\frac{3 b^2 \left (4 a^2-b^2\right ) \tanh ^{-1}\left (\frac{b \left (1-\frac{a \tan (c+d x)}{b}\right )}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right ) \cos (c+d x) \sqrt{\sec ^2(c+d x)}}{2 \left (a^2+b^2\right )^{7/2} d}+\frac{b \left (2 a^2-3 b^2\right ) \sec (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}+\frac{\cos (c+d x) (b+a \tan (c+d x))}{\left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac{a b \left (2 a^2-13 b^2\right ) \sec (c+d x)}{2 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.84247, size = 183, normalized size = 0.83 \[ \frac{\frac{\sec ^2(c+d x) \left (b \left (a^2+b^2\right )^2 \cos (3 (c+d x))+b \left (-22 a^2 b^2+11 a^4-3 b^4\right ) \cos (c+d x)+2 a \sin (c+d x) \left (\left (a^2+b^2\right )^2 \cos (2 (c+d x))+4 a^2 b^2+a^4-12 b^4\right )\right )}{\left (a^2+b^2\right )^3 (a+b \tan (c+d x))^2}-\frac{12 b^2 \left (b^2-4 a^2\right ) \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{7/2}}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-12*b^2*(-4*a^2 + b^2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(7/2) + (Sec[c + d*x]
^2*(b*(11*a^4 - 22*a^2*b^2 - 3*b^4)*Cos[c + d*x] + b*(a^2 + b^2)^2*Cos[3*(c + d*x)] + 2*a*(a^4 + 4*a^2*b^2 - 1
2*b^4 + (a^2 + b^2)^2*Cos[2*(c + d*x)])*Sin[c + d*x]))/((a^2 + b^2)^3*(a + b*Tan[c + d*x])^2))/(4*d)

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Maple [A]  time = 0.129, size = 283, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{ \left ( -{a}^{3}+3\,a{b}^{2} \right ) \tan \left ( 1/2\,dx+c/2 \right ) -3\,b{a}^{2}+{b}^{3}}{ \left ({a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6} \right ) \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-2\,{\frac{{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{3}} \left ({\frac{1}{ \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a \right ) ^{2}} \left ( -1/2\,{\frac{{b}^{2} \left ( 9\,{a}^{2}+2\,{b}^{2} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{a}}-1/2\,{\frac{b \left ( 8\,{a}^{4}-15\,{a}^{2}{b}^{2}-2\,{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{{a}^{2}}}+1/2\,{\frac{{b}^{2} \left ( 23\,{a}^{2}+2\,{b}^{2} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{a}}+4\,b{a}^{2}+1/2\,{b}^{3} \right ) }-3/2\,{\frac{4\,{a}^{2}-{b}^{2}}{\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-2/(a^6+3*a^4*b^2+3*a^2*b^4+b^6)*((-a^3+3*a*b^2)*tan(1/2*d*x+1/2*c)-3*b*a^2+b^3)/(1+tan(1/2*d*x+1/2*c)^2)
-2*b^2/(a^2+b^2)^3*((-1/2*b^2*(9*a^2+2*b^2)/a*tan(1/2*d*x+1/2*c)^3-1/2*b*(8*a^4-15*a^2*b^2-2*b^4)/a^2*tan(1/2*
d*x+1/2*c)^2+1/2*b^2*(23*a^2+2*b^2)/a*tan(1/2*d*x+1/2*c)+4*b*a^2+1/2*b^3)/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*
x+1/2*c)*b-a)^2-3/2*(4*a^2-b^2)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2))))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.30406, size = 1073, normalized size = 4.86 \begin{align*} \frac{4 \,{\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (4 \, a^{2} b^{4} - b^{6} +{\left (4 \, a^{4} b^{2} - 5 \, a^{2} b^{4} + b^{6}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (4 \, a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 2 \,{\left (4 \, a^{6} b - 10 \, a^{4} b^{3} - 17 \, a^{2} b^{5} - 3 \, b^{7}\right )} \cos \left (d x + c\right ) + 2 \,{\left (2 \, a^{5} b^{2} - 11 \, a^{3} b^{4} - 13 \, a b^{6} + 2 \,{\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \,{\left ({\left (a^{10} + 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} - 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} - b^{10}\right )} d \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{9} b + 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} + 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{8} b^{2} + 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} + 4 \, a^{2} b^{8} + b^{10}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cos(d*x + c)^3 - 3*(4*a^2*b^4 - b^6 + (4*a^4*b^2 - 5*a^2*b^4 + b^
6)*cos(d*x + c)^2 + 2*(4*a^3*b^3 - a*b^5)*cos(d*x + c)*sin(d*x + c))*sqrt(a^2 + b^2)*log((2*a*b*cos(d*x + c)*s
in(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 - b^2 - 2*sqrt(a^2 + b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/
(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2)) + 2*(4*a^6*b - 10*a^4*b^3 - 17*a^2*b^5 -
3*b^7)*cos(d*x + c) + 2*(2*a^5*b^2 - 11*a^3*b^4 - 13*a*b^6 + 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cos(d*x
+ c)^2)*sin(d*x + c))/((a^10 + 3*a^8*b^2 + 2*a^6*b^4 - 2*a^4*b^6 - 3*a^2*b^8 - b^10)*d*cos(d*x + c)^2 + 2*(a^9
*b + 4*a^7*b^3 + 6*a^5*b^5 + 4*a^3*b^7 + a*b^9)*d*cos(d*x + c)*sin(d*x + c) + (a^8*b^2 + 4*a^6*b^4 + 6*a^4*b^6
+ 4*a^2*b^8 + b^10)*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos{\left (c + d x \right )}}{\left (a + b \tan{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+b*tan(d*x+c))**3,x)

[Out]

Integral(cos(c + d*x)/(a + b*tan(c + d*x))**3, x)

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Giac [A]  time = 2.27872, size = 539, normalized size = 2.44 \begin{align*} -\frac{\frac{3 \,{\left (4 \, a^{2} b^{2} - b^{4}\right )} \log \left (\frac{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} \sqrt{a^{2} + b^{2}}} - \frac{4 \,{\left (a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a^{2} b - b^{3}\right )}}{{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}} - \frac{2 \,{\left (9 \, a^{3} b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, a b^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 8 \, a^{4} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 15 \, a^{2} b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, b^{7} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 23 \, a^{3} b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a b^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8 \, a^{4} b^{3} - a^{2} b^{5}\right )}}{{\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*(4*a^2*b^2 - b^4)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/
2*c) - 2*b + 2*sqrt(a^2 + b^2)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*sqrt(a^2 + b^2)) - 4*(a^3*tan(1/2*d*x +
1/2*c) - 3*a*b^2*tan(1/2*d*x + 1/2*c) + 3*a^2*b - b^3)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*(tan(1/2*d*x + 1/2
*c)^2 + 1)) - 2*(9*a^3*b^4*tan(1/2*d*x + 1/2*c)^3 + 2*a*b^6*tan(1/2*d*x + 1/2*c)^3 + 8*a^4*b^3*tan(1/2*d*x + 1
/2*c)^2 - 15*a^2*b^5*tan(1/2*d*x + 1/2*c)^2 - 2*b^7*tan(1/2*d*x + 1/2*c)^2 - 23*a^3*b^4*tan(1/2*d*x + 1/2*c) -
2*a*b^6*tan(1/2*d*x + 1/2*c) - 8*a^4*b^3 - a^2*b^5)/((a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6)*(a*tan(1/2*d*x +
1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)^2))/d