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

Optimal. Leaf size=241 \[ \frac{\cos ^3(c+d x) (a \tan (c+d x)+b)}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{\cos (c+d x) \left (b \left (a^2-4 b^2\right )-a \left (2 a^2+7 b^2\right ) \tan (c+d x)\right )}{3 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{b \left (9 a^2 b^2+2 a^4-8 b^4\right ) \sec (c+d x)}{3 d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}-\frac{5 a b^4 \cos (c+d x) \sqrt{\sec ^2(c+d x)} \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{d \left (a^2+b^2\right )^{7/2}} \]

[Out]

(-5*a*b^4*ArcTanh[(b - a*Tan[c + d*x])/(Sqrt[a^2 + b^2]*Sqrt[Sec[c + d*x]^2])]*Cos[c + d*x]*Sqrt[Sec[c + d*x]^
2])/((a^2 + b^2)^(7/2)*d) + (b*(2*a^4 + 9*a^2*b^2 - 8*b^4)*Sec[c + d*x])/(3*(a^2 + b^2)^3*d*(a + b*Tan[c + d*x
])) + (Cos[c + d*x]^3*(b + a*Tan[c + d*x]))/(3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])) - (Cos[c + d*x]*(b*(a^2 - 4
*b^2) - a*(2*a^2 + 7*b^2)*Tan[c + d*x]))/(3*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.257687, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3512, 741, 823, 807, 725, 206} \[ \frac{\cos ^3(c+d x) (a \tan (c+d x)+b)}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{\cos (c+d x) \left (b \left (a^2-4 b^2\right )-a \left (2 a^2+7 b^2\right ) \tan (c+d x)\right )}{3 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac{b \left (9 a^2 b^2+2 a^4-8 b^4\right ) \sec (c+d x)}{3 d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}-\frac{5 a b^4 \cos (c+d x) \sqrt{\sec ^2(c+d x)} \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{d \left (a^2+b^2\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a*b^4*ArcTanh[(b - a*Tan[c + d*x])/(Sqrt[a^2 + b^2]*Sqrt[Sec[c + d*x]^2])]*Cos[c + d*x]*Sqrt[Sec[c + d*x]^
2])/((a^2 + b^2)^(7/2)*d) + (b*(2*a^4 + 9*a^2*b^2 - 8*b^4)*Sec[c + d*x])/(3*(a^2 + b^2)^3*d*(a + b*Tan[c + d*x
])) + (Cos[c + d*x]^3*(b + a*Tan[c + d*x]))/(3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])) - (Cos[c + d*x]*(b*(a^2 - 4
*b^2) - a*(2*a^2 + 7*b^2)*Tan[c + d*x]))/(3*(a^2 + b^2)^2*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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[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 ^3(c+d x)}{(a+b \tan (c+d x))^2} \, dx &=\frac{\left (\cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(a+x)^2 \left (1+\frac{x^2}{b^2}\right )^{5/2}} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{\cos ^3(c+d x) (b+a \tan (c+d x))}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac{\left (b \cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{-2 \left (2+\frac{a^2}{b^2}\right )-\frac{3 a x}{b^2}}{(a+x)^2 \left (1+\frac{x^2}{b^2}\right )^{3/2}} \, dx,x,b \tan (c+d x)\right )}{3 \left (a^2+b^2\right ) d}\\ &=\frac{\cos ^3(c+d x) (b+a \tan (c+d x))}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac{\cos (c+d x) \left (b \left (a^2-4 b^2\right )-a \left (2 a^2+7 b^2\right ) \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\left (b^5 \cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{-\frac{2 \left (a^2-4 b^2\right )}{b^4}+\frac{a \left (2 a^2+7 b^2\right ) x}{b^6}}{(a+x)^2 \sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d}\\ &=\frac{b \left (2 a^4+9 a^2 b^2-8 b^4\right ) \sec (c+d x)}{3 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac{\cos ^3(c+d x) (b+a \tan (c+d x))}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac{\cos (c+d x) \left (b \left (a^2-4 b^2\right )-a \left (2 a^2+7 b^2\right ) \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac{\left (5 a b^3 \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 )}{\left (a^2+b^2\right )^3 d}\\ &=\frac{b \left (2 a^4+9 a^2 b^2-8 b^4\right ) \sec (c+d x)}{3 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac{\cos ^3(c+d x) (b+a \tan (c+d x))}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac{\cos (c+d x) \left (b \left (a^2-4 b^2\right )-a \left (2 a^2+7 b^2\right ) \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\left (5 a b^3 \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 )}{\left (a^2+b^2\right )^3 d}\\ &=-\frac{5 a b^4 \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)}}{\left (a^2+b^2\right )^{7/2} d}+\frac{b \left (2 a^4+9 a^2 b^2-8 b^4\right ) \sec (c+d x)}{3 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac{\cos ^3(c+d x) (b+a \tan (c+d x))}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac{\cos (c+d x) \left (b \left (a^2-4 b^2\right )-a \left (2 a^2+7 b^2\right ) \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.16903, size = 249, normalized size = 1.03 \[ \frac{\sec (c+d x) \left (\left (a^2+b^2\right ) \left (40 a^3 b^2 \sin (2 (c+d x))+2 a^3 b^2 \sin (4 (c+d x))+20 b^3 \left (a^2+b^2\right ) \cos (2 (c+d x))+b \left (a^2+b^2\right )^2 \cos (4 (c+d x))+90 a^2 b^3+15 a^4 b+10 a^5 \sin (2 (c+d x))+a^5 \sin (4 (c+d x))+30 a b^4 \sin (2 (c+d x))+a b^4 \sin (4 (c+d x))-45 b^5\right )+240 a b^4 \sqrt{a^2+b^2} (a \cos (c+d x)+b \sin (c+d x)) \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )\right )}{24 d \left (a^2+b^2\right )^4 (a+b \tan (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]*(240*a*b^4*Sqrt[a^2 + b^2]*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]]*(a*Cos[c + d*x] +
b*Sin[c + d*x]) + (a^2 + b^2)*(15*a^4*b + 90*a^2*b^3 - 45*b^5 + 20*b^3*(a^2 + b^2)*Cos[2*(c + d*x)] + b*(a^2 +
b^2)^2*Cos[4*(c + d*x)] + 10*a^5*Sin[2*(c + d*x)] + 40*a^3*b^2*Sin[2*(c + d*x)] + 30*a*b^4*Sin[2*(c + d*x)] +
a^5*Sin[4*(c + d*x)] + 2*a^3*b^2*Sin[4*(c + d*x)] + a*b^4*Sin[4*(c + d*x)])))/(24*(a^2 + b^2)^4*d*(a + b*Tan[
c + d*x]))

________________________________________________________________________________________

Maple [A]  time = 0.116, size = 320, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{ \left ( -{a}^{4}-3\,{a}^{2}{b}^{2}+2\,{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+ \left ( -2\,b{a}^{3}-6\,{b}^{3}a \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( -2/3\,{a}^{4}-6\,{a}^{2}{b}^{2}+8/3\,{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-8\,{b}^{3}a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+ \left ( -{a}^{4}-3\,{a}^{2}{b}^{2}+2\,{b}^{4} \right ) \tan \left ( 1/2\,dx+c/2 \right ) -2/3\,b{a}^{3}-14/3\,{b}^{3}a}{ \left ({a}^{2}+{b}^{2} \right ) \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-2\,{\frac{{b}^{4}}{ \left ({a}^{2}+{b}^{2} \right ) \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) } \left ({\frac{1}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tan \left ( 1/2\,dx+c/2 \right ) b-a} \left ( -{\frac{{b}^{2}\tan \left ( 1/2\,dx+c/2 \right ) }{a}}-b \right ) }-5\,{\frac{a}{\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)^3/(a+b*tan(d*x+c))^2,x)

[Out]

1/d*(-2/(a^2+b^2)/(a^4+2*a^2*b^2+b^4)*((-a^4-3*a^2*b^2+2*b^4)*tan(1/2*d*x+1/2*c)^5+(-2*a^3*b-6*a*b^3)*tan(1/2*
d*x+1/2*c)^4+(-2/3*a^4-6*a^2*b^2+8/3*b^4)*tan(1/2*d*x+1/2*c)^3-8*b^3*a*tan(1/2*d*x+1/2*c)^2+(-a^4-3*a^2*b^2+2*
b^4)*tan(1/2*d*x+1/2*c)-2/3*b*a^3-14/3*b^3*a)/(1+tan(1/2*d*x+1/2*c)^2)^3-2*b^4/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)*(
(-b^2/a*tan(1/2*d*x+1/2*c)-b)/(tan(1/2*d*x+1/2*c)^2*a-2*tan(1/2*d*x+1/2*c)*b-a)-5*a/(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))))

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.2147, size = 946, normalized size = 3.93 \begin{align*} \frac{4 \, a^{6} b + 22 \, a^{4} b^{3} + 2 \, a^{2} b^{5} - 16 \, b^{7} + 2 \,{\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (a^{6} b - 2 \, a^{4} b^{3} - 7 \, a^{2} b^{5} - 4 \, b^{7}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (a^{2} b^{4} \cos \left (d x + c\right ) + a b^{5} \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 ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} +{\left (2 \, a^{7} + 11 \, a^{5} b^{2} + 16 \, a^{3} b^{4} + 7 \, a b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left ({\left (a^{9} + 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} + 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right ) +{\left (a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}\right )} d \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(4*a^6*b + 22*a^4*b^3 + 2*a^2*b^5 - 16*b^7 + 2*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cos(d*x + c)^4 - 2*(a
^6*b - 2*a^4*b^3 - 7*a^2*b^5 - 4*b^7)*cos(d*x + c)^2 + 15*(a^2*b^4*cos(d*x + c) + a*b^5*sin(d*x + c))*sqrt(a^2
+ b^2)*log(-(2*a*b*cos(d*x + c)*sin(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*((
a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cos(d*x + c)^3 + (2*a^7 + 11*a^5*b^2 + 16*a^3*b^4 + 7*a*b^6)*cos(d*x + c)
)*sin(d*x + c))/((a^9 + 4*a^7*b^2 + 6*a^5*b^4 + 4*a^3*b^6 + a*b^8)*d*cos(d*x + c) + (a^8*b + 4*a^6*b^3 + 6*a^4
*b^5 + 4*a^2*b^7 + b^9)*d*sin(d*x + c))

________________________________________________________________________________________

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+b*tan(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.6133, size = 591, normalized size = 2.45 \begin{align*} -\frac{\frac{15 \, a b^{4} \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{6 \,{\left (b^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a b^{5}\right )}}{{\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a 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 )}} - \frac{2 \,{\left (3 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 18 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, a^{3} b + 14 \, a 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 )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(15*a*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)) - 6*(b^6*tan(1/2*d*x + 1/2*c) + a*
b^5)/((a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)) - 2*(3*
a^4*tan(1/2*d*x + 1/2*c)^5 + 9*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*a^3*b*tan(1/2
*d*x + 1/2*c)^4 + 18*a*b^3*tan(1/2*d*x + 1/2*c)^4 + 2*a^4*tan(1/2*d*x + 1/2*c)^3 + 18*a^2*b^2*tan(1/2*d*x + 1/
2*c)^3 - 8*b^4*tan(1/2*d*x + 1/2*c)^3 + 24*a*b^3*tan(1/2*d*x + 1/2*c)^2 + 3*a^4*tan(1/2*d*x + 1/2*c) + 9*a^2*b
^2*tan(1/2*d*x + 1/2*c) - 6*b^4*tan(1/2*d*x + 1/2*c) + 2*a^3*b + 14*a*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)^3))/d