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

Optimal. Leaf size=310 \[ \frac{\cos ^3(c+d x) (a \tan (c+d x)+b)}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{\cos (c+d x) \left (b \left (2 a^2-5 b^2\right )-a \left (2 a^2+9 b^2\right ) \tan (c+d x)\right )}{3 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}+\frac{a b \left (28 a^2 b^2+4 a^4-81 b^4\right ) \sec (c+d x)}{6 d \left (a^2+b^2\right )^4 (a+b \tan (c+d x))}+\frac{b \left (24 a^2 b^2+4 a^4-15 b^4\right ) \sec (c+d x)}{6 d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))^2}-\frac{5 b^4 \left (6 a^2-b^2\right ) \cos (c+d x) \sqrt{\sec ^2(c+d x)} \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{2 d \left (a^2+b^2\right )^{9/2}} \]

[Out]

(-5*b^4*(6*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)^(9/2)*d) + (b*(4*a^4 + 24*a^2*b^2 - 15*b^4)*Sec[c + d*x])/(6*(a^2 + b^2)^3*d*(a
+ b*Tan[c + d*x])^2) + (Cos[c + d*x]^3*(b + a*Tan[c + d*x]))/(3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (a*b*
(4*a^4 + 28*a^2*b^2 - 81*b^4)*Sec[c + d*x])/(6*(a^2 + b^2)^4*d*(a + b*Tan[c + d*x])) - (Cos[c + d*x]*(b*(2*a^2
- 5*b^2) - a*(2*a^2 + 9*b^2)*Tan[c + d*x]))/(3*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x])^2)

________________________________________________________________________________________

Rubi [A]  time = 0.384532, antiderivative size = 310, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3512, 741, 823, 835, 807, 725, 206} \[ \frac{\cos ^3(c+d x) (a \tan (c+d x)+b)}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac{\cos (c+d x) \left (b \left (2 a^2-5 b^2\right )-a \left (2 a^2+9 b^2\right ) \tan (c+d x)\right )}{3 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}+\frac{a b \left (28 a^2 b^2+4 a^4-81 b^4\right ) \sec (c+d x)}{6 d \left (a^2+b^2\right )^4 (a+b \tan (c+d x))}+\frac{b \left (24 a^2 b^2+4 a^4-15 b^4\right ) \sec (c+d x)}{6 d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))^2}-\frac{5 b^4 \left (6 a^2-b^2\right ) \cos (c+d x) \sqrt{\sec ^2(c+d x)} \tanh ^{-1}\left (\frac{b-a \tan (c+d x)}{\sqrt{a^2+b^2} \sqrt{\sec ^2(c+d x)}}\right )}{2 d \left (a^2+b^2\right )^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b^4*(6*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)^(9/2)*d) + (b*(4*a^4 + 24*a^2*b^2 - 15*b^4)*Sec[c + d*x])/(6*(a^2 + b^2)^3*d*(a
+ b*Tan[c + d*x])^2) + (Cos[c + d*x]^3*(b + a*Tan[c + d*x]))/(3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) + (a*b*
(4*a^4 + 28*a^2*b^2 - 81*b^4)*Sec[c + d*x])/(6*(a^2 + b^2)^4*d*(a + b*Tan[c + d*x])) - (Cos[c + d*x]*(b*(2*a^2
- 5*b^2) - a*(2*a^2 + 9*b^2)*Tan[c + d*x]))/(3*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x])^2)

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 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 ^3(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 )^{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))^2}-\frac{\left (b \cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{-5-\frac{2 a^2}{b^2}-\frac{4 a x}{b^2}}{(a+x)^3 \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))^2}-\frac{\cos (c+d x) \left (b \left (2 a^2-5 b^2\right )-a \left (2 a^2+9 b^2\right ) \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}+\frac{\left (b^5 \cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{-\frac{3 \left (2 a^2-5 b^2\right )}{b^4}+\frac{2 a \left (2 a^2+9 b^2\right ) x}{b^6}}{(a+x)^3 \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 (4 a^4+24 a^2 b^2-15 b^4\right ) \sec (c+d x)}{6 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))^2}+\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))^2}-\frac{\cos (c+d x) \left (b \left (2 a^2-5 b^2\right )-a \left (2 a^2+9 b^2\right ) \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}-\frac{\left (b^7 \cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{\frac{2 a \left (2 a^2-33 b^2\right )}{b^6}-\frac{\left (4 a^4+24 a^2 b^2-15 b^4\right ) x}{b^8}}{(a+x)^2 \sqrt{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (c+d x)\right )}{6 \left (a^2+b^2\right )^3 d}\\ &=\frac{b \left (4 a^4+24 a^2 b^2-15 b^4\right ) \sec (c+d x)}{6 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))^2}+\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))^2}+\frac{a b \left (4 a^4+28 a^2 b^2-81 b^4\right ) \sec (c+d x)}{6 \left (a^2+b^2\right )^4 d (a+b \tan (c+d x))}-\frac{\cos (c+d x) \left (b \left (2 a^2-5 b^2\right )-a \left (2 a^2+9 b^2\right ) \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}+\frac{\left (5 b^3 \left (6 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 )^4 d}\\ &=\frac{b \left (4 a^4+24 a^2 b^2-15 b^4\right ) \sec (c+d x)}{6 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))^2}+\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))^2}+\frac{a b \left (4 a^4+28 a^2 b^2-81 b^4\right ) \sec (c+d x)}{6 \left (a^2+b^2\right )^4 d (a+b \tan (c+d x))}-\frac{\cos (c+d x) \left (b \left (2 a^2-5 b^2\right )-a \left (2 a^2+9 b^2\right ) \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}-\frac{\left (5 b^3 \left (6 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 )^4 d}\\ &=-\frac{5 b^4 \left (6 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 )^{9/2} d}+\frac{b \left (4 a^4+24 a^2 b^2-15 b^4\right ) \sec (c+d x)}{6 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))^2}+\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))^2}+\frac{a b \left (4 a^4+28 a^2 b^2-81 b^4\right ) \sec (c+d x)}{6 \left (a^2+b^2\right )^4 d (a+b \tan (c+d x))}-\frac{\cos (c+d x) \left (b \left (2 a^2-5 b^2\right )-a \left (2 a^2+9 b^2\right ) \tan (c+d x)\right )}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}\\ \end{align*}

Mathematica [A]  time = 1.83266, size = 371, normalized size = 1.2 \[ \frac{\sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \left (\frac{6 b^6 \tan (c+d x)}{a \left (a^2+b^2\right )^3}-\frac{6 b^5 \left (12 a^2+b^2\right ) (a+b \tan (c+d x))}{a \left (a^2+b^2\right )^4}+\frac{9 b \left (14 a^2 b^2+a^4-3 b^4\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (a^2+b^2\right )^4}-\frac{b \left (b^2-3 a^2\right ) \cos (c+d x) \cos (3 (c+d x)) (a+b \tan (c+d x))^2}{\left (a^2+b^2\right )^3}+\frac{a \left (a^2-3 b^2\right ) \sin (3 (c+d x)) \cos (c+d x) (a+b \tan (c+d x))^2}{\left (a^2+b^2\right )^3}+\frac{9 a \left (6 a^2 b^2+a^4-11 b^4\right ) \tan (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (a^2+b^2\right )^4}-\frac{60 b^4 \left (b^2-6 a^2\right ) \cos (c+d x) (a+b \tan (c+d x))^2 \tanh ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )-b}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{9/2}}\right )}{12 d (a+b \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])*((9*b*(a^4 + 14*a^2*b^2 - 3*b^4)*(a*Cos[c + d*x] + b*Sin[c +
d*x])^2)/(a^2 + b^2)^4 + (6*b^6*Tan[c + d*x])/(a*(a^2 + b^2)^3) + (9*a*(a^4 + 6*a^2*b^2 - 11*b^4)*(a*Cos[c +
d*x] + b*Sin[c + d*x])^2*Tan[c + d*x])/(a^2 + b^2)^4 - (6*b^5*(12*a^2 + b^2)*(a + b*Tan[c + d*x]))/(a*(a^2 + b
^2)^4) - (60*b^4*(-6*a^2 + b^2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]]*Cos[c + d*x]*(a + b*Tan[c +
d*x])^2)/(a^2 + b^2)^(9/2) - (b*(-3*a^2 + b^2)*Cos[c + d*x]*Cos[3*(c + d*x)]*(a + b*Tan[c + d*x])^2)/(a^2 + b
^2)^3 + (a*(a^2 - 3*b^2)*Cos[c + d*x]*Sin[3*(c + d*x)]*(a + b*Tan[c + d*x])^2)/(a^2 + b^2)^3))/(12*d*(a + b*Ta
n[c + d*x])^3)

________________________________________________________________________________________

Maple [A]  time = 0.155, size = 457, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{ \left ({a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6} \right ) \left ({a}^{2}+{b}^{2} \right ) \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}} \left ( \left ( -{a}^{5}-4\,{a}^{3}{b}^{2}+9\,a{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+ \left ( -3\,{a}^{4}b-12\,{a}^{2}{b}^{3}+3\,{b}^{5} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( -2/3\,{a}^{5}-{\frac{32\,{a}^{3}{b}^{2}}{3}}+14\,a{b}^{4} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+ \left ( -20\,{a}^{2}{b}^{3}+4\,{b}^{5} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+ \left ( -{a}^{5}-4\,{a}^{3}{b}^{2}+9\,a{b}^{4} \right ) \tan \left ( 1/2\,dx+c/2 \right ) -{a}^{4}b-{\frac{32\,{a}^{2}{b}^{3}}{3}}+7/3\,{b}^{5} \right ) }-2\,{\frac{{b}^{4}}{ \left ({a}^{6}+3\,{a}^{4}{b}^{2}+3\,{a}^{2}{b}^{4}+{b}^{6} \right ) \left ({a}^{2}+{b}^{2} \right ) } \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 ( 13\,{a}^{2}+2\,{b}^{2} \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{a}}-1/2\,{\frac{b \left ( 12\,{a}^{4}-23\,{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 ( 35\,{a}^{2}+2\,{b}^{2} \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{a}}+6\,b{a}^{2}+1/2\,{b}^{3} \right ) }-5/2\,{\frac{6\,{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)^3/(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^2+b^2)*((-a^5-4*a^3*b^2+9*a*b^4)*tan(1/2*d*x+1/2*c)^5+(-3*a^4*b-12*a^
2*b^3+3*b^5)*tan(1/2*d*x+1/2*c)^4+(-2/3*a^5-32/3*a^3*b^2+14*a*b^4)*tan(1/2*d*x+1/2*c)^3+(-20*a^2*b^3+4*b^5)*ta
n(1/2*d*x+1/2*c)^2+(-a^5-4*a^3*b^2+9*a*b^4)*tan(1/2*d*x+1/2*c)-a^4*b-32/3*a^2*b^3+7/3*b^5)/(1+tan(1/2*d*x+1/2*
c)^2)^3-2*b^4/(a^2+b^2)/(a^6+3*a^4*b^2+3*a^2*b^4+b^6)*((-1/2*b^2*(13*a^2+2*b^2)/a*tan(1/2*d*x+1/2*c)^3-1/2*b*(
12*a^4-23*a^2*b^2-2*b^4)/a^2*tan(1/2*d*x+1/2*c)^2+1/2*b^2*(35*a^2+2*b^2)/a*tan(1/2*d*x+1/2*c)+6*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)^2-5/2*(6*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))))

________________________________________________________________________________________

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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.79722, size = 1391, normalized size = 4.49 \begin{align*} \frac{4 \,{\left (a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}\right )} \cos \left (d x + c\right )^{5} - 4 \,{\left (2 \, a^{8} b + a^{6} b^{3} - 9 \, a^{4} b^{5} - 13 \, a^{2} b^{7} - 5 \, b^{9}\right )} \cos \left (d x + c\right )^{3} - 15 \,{\left (6 \, a^{2} b^{6} - b^{8} +{\left (6 \, a^{4} b^{4} - 7 \, a^{2} b^{6} + b^{8}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (6 \, a^{3} b^{5} - a b^{7}\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 (8 \, a^{8} b + 64 \, a^{6} b^{3} - 16 \, a^{4} b^{5} - 87 \, a^{2} b^{7} - 15 \, b^{9}\right )} \cos \left (d x + c\right ) + 2 \,{\left (4 \, a^{7} b^{2} + 32 \, a^{5} b^{4} - 53 \, a^{3} b^{6} - 81 \, a b^{8} + 2 \,{\left (a^{9} + 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} + 4 \, a^{3} b^{6} + a b^{8}\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (2 \, a^{9} + 15 \, a^{7} b^{2} + 33 \, a^{5} b^{4} + 29 \, a^{3} b^{6} + 9 \, a b^{8}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \,{\left ({\left (a^{12} + 4 \, a^{10} b^{2} + 5 \, a^{8} b^{4} - 5 \, a^{4} b^{8} - 4 \, a^{2} b^{10} - b^{12}\right )} d \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{11} b + 5 \, a^{9} b^{3} + 10 \, a^{7} b^{5} + 10 \, a^{5} b^{7} + 5 \, a^{3} b^{9} + a b^{11}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{10} b^{2} + 5 \, a^{8} b^{4} + 10 \, a^{6} b^{6} + 10 \, a^{4} b^{8} + 5 \, a^{2} b^{10} + b^{12}\right )} d\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))^3,x, algorithm="fricas")

[Out]

1/12*(4*(a^8*b + 4*a^6*b^3 + 6*a^4*b^5 + 4*a^2*b^7 + b^9)*cos(d*x + c)^5 - 4*(2*a^8*b + a^6*b^3 - 9*a^4*b^5 -
13*a^2*b^7 - 5*b^9)*cos(d*x + c)^3 - 15*(6*a^2*b^6 - b^8 + (6*a^4*b^4 - 7*a^2*b^6 + b^8)*cos(d*x + c)^2 + 2*(6
*a^3*b^5 - a*b^7)*cos(d*x + c)*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*(8*a^8*b + 64*a^6*b^3 - 16*a^4*b^5 - 87*a^2*b^7 - 15*b^9)*co
s(d*x + c) + 2*(4*a^7*b^2 + 32*a^5*b^4 - 53*a^3*b^6 - 81*a*b^8 + 2*(a^9 + 4*a^7*b^2 + 6*a^5*b^4 + 4*a^3*b^6 +
a*b^8)*cos(d*x + c)^4 + 2*(2*a^9 + 15*a^7*b^2 + 33*a^5*b^4 + 29*a^3*b^6 + 9*a*b^8)*cos(d*x + c)^2)*sin(d*x + c
))/((a^12 + 4*a^10*b^2 + 5*a^8*b^4 - 5*a^4*b^8 - 4*a^2*b^10 - b^12)*d*cos(d*x + c)^2 + 2*(a^11*b + 5*a^9*b^3 +
10*a^7*b^5 + 10*a^5*b^7 + 5*a^3*b^9 + a*b^11)*d*cos(d*x + c)*sin(d*x + c) + (a^10*b^2 + 5*a^8*b^4 + 10*a^6*b^
6 + 10*a^4*b^8 + 5*a^2*b^10 + b^12)*d)

________________________________________________________________________________________

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))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 2.19951, size = 864, normalized size = 2.79 \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+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/6*(15*(6*a^2*b^4 - b^6)*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^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*sqrt(a^2 + b^2)) - 6*(13*a^
3*b^6*tan(1/2*d*x + 1/2*c)^3 + 2*a*b^8*tan(1/2*d*x + 1/2*c)^3 + 12*a^4*b^5*tan(1/2*d*x + 1/2*c)^2 - 23*a^2*b^7
*tan(1/2*d*x + 1/2*c)^2 - 2*b^9*tan(1/2*d*x + 1/2*c)^2 - 35*a^3*b^6*tan(1/2*d*x + 1/2*c) - 2*a*b^8*tan(1/2*d*x
+ 1/2*c) - 12*a^4*b^5 - a^2*b^7)/((a^10 + 4*a^8*b^2 + 6*a^6*b^4 + 4*a^4*b^6 + a^2*b^8)*(a*tan(1/2*d*x + 1/2*c
)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)^2) - 4*(3*a^5*tan(1/2*d*x + 1/2*c)^5 + 12*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 -
27*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 9*a^4*b*tan(1/2*d*x + 1/2*c)^4 + 36*a^2*b^3*tan(1/2*d*x + 1/2*c)^4 - 9*b^5*
tan(1/2*d*x + 1/2*c)^4 + 2*a^5*tan(1/2*d*x + 1/2*c)^3 + 32*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 42*a*b^4*tan(1/2*d
*x + 1/2*c)^3 + 60*a^2*b^3*tan(1/2*d*x + 1/2*c)^2 - 12*b^5*tan(1/2*d*x + 1/2*c)^2 + 3*a^5*tan(1/2*d*x + 1/2*c)
+ 12*a^3*b^2*tan(1/2*d*x + 1/2*c) - 27*a*b^4*tan(1/2*d*x + 1/2*c) + 3*a^4*b + 32*a^2*b^3 - 7*b^5)/((a^8 + 4*a
^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*(tan(1/2*d*x + 1/2*c)^2 + 1)^3))/d