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

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

[Out]

-5/2*b^4*(6*a^2-b^2)*arctanh((b-a*tan(d*x+c))/(a^2+b^2)^(1/2)/(sec(d*x+c)^2)^(1/2))*cos(d*x+c)*(sec(d*x+c)^2)^
(1/2)/(a^2+b^2)^(9/2)/d+1/6*b*(4*a^4+24*a^2*b^2-15*b^4)*sec(d*x+c)/(a^2+b^2)^3/d/(a+b*tan(d*x+c))^2+1/3*cos(d*
x+c)^3*(b+a*tan(d*x+c))/(a^2+b^2)/d/(a+b*tan(d*x+c))^2+1/6*a*b*(4*a^4+28*a^2*b^2-81*b^4)*sec(d*x+c)/(a^2+b^2)^
4/d/(a+b*tan(d*x+c))-1/3*cos(d*x+c)*(b*(2*a^2-5*b^2)-a*(2*a^2+9*b^2)*tan(d*x+c))/(a^2+b^2)^2/d/(a+b*tan(d*x+c)
)^2

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Rubi [A]  time = 0.38, antiderivative size = 310, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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 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]

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*}

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Mathematica [A]  time = 2.00, size = 371, normalized size = 1.20 \[ \frac {\sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \left (-\frac {b \left (b^2-3 a^2\right ) \cos (c+d x) \cos (3 (c+d x)) (a+b \tan (c+d x))^2}{\left (a^2+b^2\right )^3}+\frac {a \left (a^2-3 b^2\right ) \sin (3 (c+d x)) \cos (c+d x) (a+b \tan (c+d x))^2}{\left (a^2+b^2\right )^3}+\frac {6 b^6 \tan (c+d x)}{a \left (a^2+b^2\right )^3}-\frac {6 b^5 \left (12 a^2+b^2\right ) (a+b \tan (c+d x))}{a \left (a^2+b^2\right )^4}-\frac {60 b^4 \left (b^2-6 a^2\right ) \cos (c+d x) (a+b \tan (c+d x))^2 \tanh ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )-b}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{9/2}}+\frac {9 b \left (a^4+14 a^2 b^2-3 b^4\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (a^2+b^2\right )^4}+\frac {9 a \left (a^4+6 a^2 b^2-11 b^4\right ) \tan (c+d x) (a \cos (c+d x)+b \sin (c+d x))^2}{\left (a^2+b^2\right )^4}\right )}{12 d (a+b \tan (c+d x))^3} \]

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)

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fricas [B]  time = 0.84, size = 619, normalized size = 2.00 \[ \frac {4 \, {\left (a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}\right )} \cos \left (d x + c\right )^{5} - 4 \, {\left (2 \, a^{8} b + a^{6} b^{3} - 9 \, a^{4} b^{5} - 13 \, a^{2} b^{7} - 5 \, b^{9}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left (6 \, a^{2} b^{6} - b^{8} + {\left (6 \, a^{4} b^{4} - 7 \, a^{2} b^{6} + b^{8}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, a^{3} b^{5} - a b^{7}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 2 \, {\left (8 \, a^{8} b + 64 \, a^{6} b^{3} - 16 \, a^{4} b^{5} - 87 \, a^{2} b^{7} - 15 \, b^{9}\right )} \cos \left (d x + c\right ) + 2 \, {\left (4 \, a^{7} b^{2} + 32 \, a^{5} b^{4} - 53 \, a^{3} b^{6} - 81 \, a b^{8} + 2 \, {\left (a^{9} + 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} + 4 \, a^{3} b^{6} + a b^{8}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, a^{9} + 15 \, a^{7} b^{2} + 33 \, a^{5} b^{4} + 29 \, a^{3} b^{6} + 9 \, a b^{8}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, {\left ({\left (a^{12} + 4 \, a^{10} b^{2} + 5 \, a^{8} b^{4} - 5 \, a^{4} b^{8} - 4 \, a^{2} b^{10} - b^{12}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{11} b + 5 \, a^{9} b^{3} + 10 \, a^{7} b^{5} + 10 \, a^{5} b^{7} + 5 \, a^{3} b^{9} + a b^{11}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{10} b^{2} + 5 \, a^{8} b^{4} + 10 \, a^{6} b^{6} + 10 \, a^{4} b^{8} + 5 \, a^{2} b^{10} + b^{12}\right )} d\right )}} \]

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)

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giac [B]  time = 4.34, size = 640, normalized size = 2.06 \[ -\frac {\frac {15 \, {\left (6 \, a^{2} b^{4} - b^{6}\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^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} \sqrt {a^{2} + b^{2}}} - \frac {6 \, {\left (13 \, a^{3} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, a^{4} b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 23 \, a^{2} b^{7} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b^{9} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 35 \, a^{3} b^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a b^{8} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, a^{4} b^{5} - a^{2} b^{7}\right )}}{{\left (a^{10} + 4 \, a^{8} b^{2} + 6 \, a^{6} b^{4} + 4 \, a^{4} b^{6} + a^{2} b^{8}\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}} - \frac {4 \, {\left (3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 36 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 9 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 32 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 42 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 27 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{4} b + 32 \, a^{2} b^{3} - 7 \, b^{5}\right )}}{{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

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

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maple [A]  time = 0.56, size = 457, normalized size = 1.47 \[ \frac {-\frac {2 b^{4} \left (\frac {-\frac {b^{2} \left (13 a^{2}+2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {b \left (12 a^{4}-23 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2}}+\frac {b^{2} \left (35 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}+6 a^{2} b +\frac {b^{3}}{2}}{\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right )^{2}}-\frac {5 \left (6 a^{2}-b^{2}\right ) \arctanh \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right ) \left (a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}\right )}-\frac {2 \left (\left (-a^{5}-4 b^{2} a^{3}+9 a \,b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-3 a^{4} b -12 a^{2} b^{3}+3 b^{5}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {2}{3} a^{5}-\frac {32}{3} b^{2} a^{3}+14 a \,b^{4}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-20 a^{2} b^{3}+4 b^{5}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a^{5}-4 b^{2} a^{3}+9 a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a^{4} b -\frac {32 a^{2} b^{3}}{3}+\frac {7 b^{5}}{3}\right )}{\left (a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}\right ) \left (a^{2}+b^{2}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}}{d} \]

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*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*a^2*b+1/2*b^3)/(
a*tan(1/2*d*x+1/2*c)^2-2*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)))-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*b^2*a^3+14*a*b^4)*tan(1/2*d*x+1/2*c
)^3+(-20*a^2*b^3+4*b^5)*tan(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)

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maxima [B]  time = 0.54, size = 1229, normalized size = 3.96 \[ \text {result too large to display} \]

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]

-1/6*(15*(6*a^2*b^4 - b^6)*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt(a^2 + b^2))/(b - a*sin(d*x + c)/(
cos(d*x + c) + 1) - 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)) - 2*(6
*a^8*b + 64*a^6*b^3 - 50*a^4*b^5 - 3*a^2*b^7 + (6*a^9 + 48*a^7*b^2 + 202*a^5*b^4 - 161*a^3*b^6 - 6*a*b^8)*sin(
d*x + c)/(cos(d*x + c) + 1) + 2*(6*a^8*b + 56*a^6*b^3 - 14*a^4*b^5 - 67*a^2*b^7 - 3*b^9)*sin(d*x + c)^2/(cos(d
*x + c) + 1)^2 - 4*(2*a^9 - 4*a^7*b^2 - 86*a^5*b^4 + 133*a^3*b^6 + 3*a*b^8)*sin(d*x + c)^3/(cos(d*x + c) + 1)^
3 + 2*(8*a^8*b + 28*a^6*b^3 + 188*a^4*b^5 - 156*a^2*b^7 - 9*b^9)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 2*(2*a^
9 + 4*a^7*b^2 + 62*a^5*b^4 - 255*a^3*b^6)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 2*(14*a^8*b + 56*a^6*b^3 - 246
*a^4*b^5 + 141*a^2*b^7 + 9*b^9)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 4*(2*a^9 + 8*a^7*b^2 + 42*a^5*b^4 + 33*a
^3*b^6 - 3*a*b^8)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 3*(2*a^8*b + 8*a^6*b^3 - 78*a^4*b^5 + 23*a^2*b^7 + 2*b
^9)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 3*(2*a^9 + 8*a^7*b^2 - 18*a^5*b^4 + 13*a^3*b^6 + 2*a*b^8)*sin(d*x +
c)^9/(cos(d*x + c) + 1)^9)/(a^12 + 4*a^10*b^2 + 6*a^8*b^4 + 4*a^6*b^6 + a^4*b^8 + 4*(a^11*b + 4*a^9*b^3 + 6*a^
7*b^5 + 4*a^5*b^7 + a^3*b^9)*sin(d*x + c)/(cos(d*x + c) + 1) + (a^12 + 8*a^10*b^2 + 22*a^8*b^4 + 28*a^6*b^6 +
17*a^4*b^8 + 4*a^2*b^10)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 8*(a^11*b + 4*a^9*b^3 + 6*a^7*b^5 + 4*a^5*b^7 +
a^3*b^9)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 2*(a^12 - 2*a^10*b^2 - 18*a^8*b^4 - 32*a^6*b^6 - 23*a^4*b^8 -
6*a^2*b^10)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2*(a^12 - 2*a^10*b^2 - 18*a^8*b^4 - 32*a^6*b^6 - 23*a^4*b^8
- 6*a^2*b^10)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 8*(a^11*b + 4*a^9*b^3 + 6*a^7*b^5 + 4*a^5*b^7 + a^3*b^9)*s
in(d*x + c)^7/(cos(d*x + c) + 1)^7 + (a^12 + 8*a^10*b^2 + 22*a^8*b^4 + 28*a^6*b^6 + 17*a^4*b^8 + 4*a^2*b^10)*s
in(d*x + c)^8/(cos(d*x + c) + 1)^8 - 4*(a^11*b + 4*a^9*b^3 + 6*a^7*b^5 + 4*a^5*b^7 + a^3*b^9)*sin(d*x + c)^9/(
cos(d*x + c) + 1)^9 + (a^12 + 4*a^10*b^2 + 6*a^8*b^4 + 4*a^6*b^6 + a^4*b^8)*sin(d*x + c)^10/(cos(d*x + c) + 1)
^10))/d

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mupad [B]  time = 8.83, size = 1128, normalized size = 3.64 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*tan(c/2 + (d*x)/2)^5*(2*a^7 - 255*a*b^6 + 62*a^3*b^4 + 4*a^5*b^2))/(3*(a^8 + b^8 + 4*a^2*b^6 + 6*a^4*b^4 +
4*a^6*b^2)) + (6*a^6*b - 3*b^7 - 50*a^2*b^5 + 64*a^4*b^3)/(3*(a^2 + b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2))
+ (2*tan(c/2 + (d*x)/2)^2*(6*a^6*b - 3*b^7 - 64*a^2*b^5 + 50*a^4*b^3))/(3*a^2*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*
b^2)) + (tan(c/2 + (d*x)/2)^9*(2*a^8 + 2*b^8 + 13*a^2*b^6 - 18*a^4*b^4 + 8*a^6*b^2))/(a*(a^8 + b^8 + 4*a^2*b^6
+ 6*a^4*b^4 + 4*a^6*b^2)) - (4*tan(c/2 + (d*x)/2)^7*(2*a^6 - 3*b^6 + 36*a^2*b^4 + 6*a^4*b^2))/(3*a*(a^6 + b^6
+ 3*a^2*b^4 + 3*a^4*b^2)) - (tan(c/2 + (d*x)/2)^8*(2*a^8*b + 2*b^9 + 23*a^2*b^7 - 78*a^4*b^5 + 8*a^6*b^3))/(a
^2*(a^8 + b^8 + 4*a^2*b^6 + 6*a^4*b^4 + 4*a^6*b^2)) + (2*tan(c/2 + (d*x)/2)^4*(8*a^8*b - 9*b^9 - 156*a^2*b^7 +
188*a^4*b^5 + 28*a^6*b^3))/(3*a^2*(a^2 + b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) - (2*tan(c/2 + (d*x)/2)^6*
(14*a^8*b + 9*b^9 + 141*a^2*b^7 - 246*a^4*b^5 + 56*a^6*b^3))/(3*a^2*(a^2 + b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4
*b^2)) + (tan(c/2 + (d*x)/2)*(6*a^8 - 6*b^8 - 161*a^2*b^6 + 202*a^4*b^4 + 48*a^6*b^2))/(3*a*(a^2 + b^2)*(a^6 +
b^6 + 3*a^2*b^4 + 3*a^4*b^2)) - (4*tan(c/2 + (d*x)/2)^3*(2*a^8 + 3*b^8 + 133*a^2*b^6 - 86*a^4*b^4 - 4*a^6*b^2
))/(3*a*(a^2 + b^2)*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)))/(d*(a^2*tan(c/2 + (d*x)/2)^10 - tan(c/2 + (d*x)/2)^6
*(2*a^2 - 12*b^2) - tan(c/2 + (d*x)/2)^4*(2*a^2 - 12*b^2) + a^2 + tan(c/2 + (d*x)/2)^2*(a^2 + 4*b^2) + tan(c/2
+ (d*x)/2)^8*(a^2 + 4*b^2) + 8*a*b*tan(c/2 + (d*x)/2)^3 - 8*a*b*tan(c/2 + (d*x)/2)^7 - 4*a*b*tan(c/2 + (d*x)/
2)^9 + 4*a*b*tan(c/2 + (d*x)/2))) + (b^4*atan((a^8*b*1i + b^9*1i + a^2*b^7*4i + a^4*b^5*6i + a^6*b^3*4i - a^9*
tan(c/2 + (d*x)/2)*1i - a*b^8*tan(c/2 + (d*x)/2)*1i - a^3*b^6*tan(c/2 + (d*x)/2)*4i - a^5*b^4*tan(c/2 + (d*x)/
2)*6i - a^7*b^2*tan(c/2 + (d*x)/2)*4i)/(a^2 + b^2)^(9/2))*(6*a^2 - b^2)*5i)/(d*(a^2 + b^2)^(9/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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