\(\int \frac {\cos ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx\) [559]

Optimal result

Integrand size = 21, antiderivative size = 235 \[ \int \frac {\cos ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {3 \left (a^6+5 a^4 b^2+15 a^2 b^4-5 b^6\right ) x}{8 \left (a^2+b^2\right )^4}+\frac {6 a b^5 \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^4 d}+\frac {3 b \left (a^2-b^2\right ) \left (a^2+5 b^2\right )}{8 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac {\cos ^4(c+d x) (b+a \tan (c+d x))}{4 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\cos ^2(c+d x) \left (b \left (a^2-5 b^2\right )-3 a \left (a^2+3 b^2\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \]

[Out]

3/8*(a^6+5*a^4*b^2+15*a^2*b^4-5*b^6)*x/(a^2+b^2)^4+6*a*b^5*ln(a*cos(d*x+c)+b*sin(d*x+c))/(a^2+b^2)^4/d+3/8*b*(
a^2-b^2)*(a^2+5*b^2)/(a^2+b^2)^3/d/(a+b*tan(d*x+c))+1/4*cos(d*x+c)^4*(b+a*tan(d*x+c))/(a^2+b^2)/d/(a+b*tan(d*x
+c))-1/8*cos(d*x+c)^2*(b*(a^2-5*b^2)-3*a*(a^2+3*b^2)*tan(d*x+c))/(a^2+b^2)^2/d/(a+b*tan(d*x+c))

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3587, 755, 837, 815, 649, 209, 266} \[ \int \frac {\cos ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {3 b \left (a^2-b^2\right ) \left (a^2+5 b^2\right )}{8 d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}+\frac {\cos ^4(c+d x) (a \tan (c+d x)+b)}{4 d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\cos ^2(c+d x) \left (b \left (a^2-5 b^2\right )-3 a \left (a^2+3 b^2\right ) \tan (c+d x)\right )}{8 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac {6 a b^5 \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^4}+\frac {3 x \left (a^6+5 a^4 b^2+15 a^2 b^4-5 b^6\right )}{8 \left (a^2+b^2\right )^4} \]

[In]

Int[Cos[c + d*x]^4/(a + b*Tan[c + d*x])^2,x]

[Out]

(3*(a^6 + 5*a^4*b^2 + 15*a^2*b^4 - 5*b^6)*x)/(8*(a^2 + b^2)^4) + (6*a*b^5*Log[a*Cos[c + d*x] + b*Sin[c + d*x]]
)/((a^2 + b^2)^4*d) + (3*b*(a^2 - b^2)*(a^2 + 5*b^2))/(8*(a^2 + b^2)^3*d*(a + b*Tan[c + d*x])) + (Cos[c + d*x]
^4*(b + a*Tan[c + d*x]))/(4*(a^2 + b^2)*d*(a + b*Tan[c + d*x])) - (Cos[c + d*x]^2*(b*(a^2 - 5*b^2) - 3*a*(a^2
+ 3*b^2)*Tan[c + d*x]))/(8*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

Rule 209

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

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

Rule 755

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 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 837

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] +
Dist[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 3587

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{(a+x)^2 \left (1+\frac {x^2}{b^2}\right )^3} \, dx,x,b \tan (c+d x)\right )}{b d} \\ & = \frac {\cos ^4(c+d x) (b+a \tan (c+d x))}{4 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {b \text {Subst}\left (\int \frac {-5-\frac {3 a^2}{b^2}-\frac {4 a x}{b^2}}{(a+x)^2 \left (1+\frac {x^2}{b^2}\right )^2} \, dx,x,b \tan (c+d x)\right )}{4 \left (a^2+b^2\right ) d} \\ & = \frac {\cos ^4(c+d x) (b+a \tan (c+d x))}{4 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\cos ^2(c+d x) \left (b \left (a^2-5 b^2\right )-3 a \left (a^2+3 b^2\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {b^5 \text {Subst}\left (\int \frac {\frac {3 \left (a^4+2 a^2 b^2+5 b^4\right )}{b^6}+\frac {6 a \left (a^2+3 b^2\right ) x}{b^6}}{(a+x)^2 \left (1+\frac {x^2}{b^2}\right )} \, dx,x,b \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d} \\ & = \frac {\cos ^4(c+d x) (b+a \tan (c+d x))}{4 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\cos ^2(c+d x) \left (b \left (a^2-5 b^2\right )-3 a \left (a^2+3 b^2\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {b^5 \text {Subst}\left (\int \left (\frac {3 \left (-a^4-4 a^2 b^2+5 b^4\right )}{b^4 \left (a^2+b^2\right ) (a+x)^2}+\frac {48 a}{\left (a^2+b^2\right )^2 (a+x)}-\frac {3 \left (-a^6-5 a^4 b^2-15 a^2 b^4+5 b^6+16 a b^4 x\right )}{b^4 \left (a^2+b^2\right )^2 \left (b^2+x^2\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d} \\ & = \frac {6 a b^5 \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^4 d}+\frac {3 b \left (a^2-b^2\right ) \left (a^2+5 b^2\right )}{8 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac {\cos ^4(c+d x) (b+a \tan (c+d x))}{4 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\cos ^2(c+d x) \left (b \left (a^2-5 b^2\right )-3 a \left (a^2+3 b^2\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {(3 b) \text {Subst}\left (\int \frac {-a^6-5 a^4 b^2-15 a^2 b^4+5 b^6+16 a b^4 x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^4 d} \\ & = \frac {6 a b^5 \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^4 d}+\frac {3 b \left (a^2-b^2\right ) \left (a^2+5 b^2\right )}{8 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac {\cos ^4(c+d x) (b+a \tan (c+d x))}{4 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\cos ^2(c+d x) \left (b \left (a^2-5 b^2\right )-3 a \left (a^2+3 b^2\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left (6 a b^5\right ) \text {Subst}\left (\int \frac {x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{\left (a^2+b^2\right )^4 d}+\frac {\left (3 b \left (a^6+5 a^4 b^2+15 a^2 b^4-5 b^6\right )\right ) \text {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^4 d} \\ & = \frac {3 \left (a^6+5 a^4 b^2+15 a^2 b^4-5 b^6\right ) x}{8 \left (a^2+b^2\right )^4}+\frac {6 a b^5 \log (\cos (c+d x))}{\left (a^2+b^2\right )^4 d}+\frac {6 a b^5 \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^4 d}+\frac {3 b \left (a^2-b^2\right ) \left (a^2+5 b^2\right )}{8 \left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}+\frac {\cos ^4(c+d x) (b+a \tan (c+d x))}{4 \left (a^2+b^2\right ) d (a+b \tan (c+d x))}-\frac {\cos ^2(c+d x) \left (b \left (a^2-5 b^2\right )-3 a \left (a^2+3 b^2\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.60 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.77 \[ \int \frac {\cos ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {4 b \cos ^4(c+d x) (b+a \tan (c+d x))+\frac {2 b \cos ^2(c+d x) \left (-a^2 b+5 b^3+3 a \left (a^2+3 b^2\right ) \tan (c+d x)\right )}{a^2+b^2}-\frac {\sqrt {-b^2} \left (6 a \left (a^2+b^2\right ) \left (a^2+3 b^2\right ) \left (\left (a-\sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+2 \sqrt {-b^2} \log (a+b \tan (c+d x))-\left (a+\sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )\right ) (a+b \tan (c+d x))+3 \left (a^4+4 a^2 b^2-5 b^4\right ) \left (2 \sqrt {-b^2} \left (a^2+b^2\right )+\left (-a^2+b^2+2 a \sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right ) (a+b \tan (c+d x))-4 a \sqrt {-b^2} \log (a+b \tan (c+d x)) (a+b \tan (c+d x))+\left (a^2-b^2+2 a \sqrt {-b^2}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right ) (a+b \tan (c+d x))\right )\right )}{\left (a^2+b^2\right )^3}}{16 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))} \]

[In]

Integrate[Cos[c + d*x]^4/(a + b*Tan[c + d*x])^2,x]

[Out]

(4*b*Cos[c + d*x]^4*(b + a*Tan[c + d*x]) + (2*b*Cos[c + d*x]^2*(-(a^2*b) + 5*b^3 + 3*a*(a^2 + 3*b^2)*Tan[c + d
*x]))/(a^2 + b^2) - (Sqrt[-b^2]*(6*a*(a^2 + b^2)*(a^2 + 3*b^2)*((a - Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*
x]] + 2*Sqrt[-b^2]*Log[a + b*Tan[c + d*x]] - (a + Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]])*(a + b*Tan[c +
d*x]) + 3*(a^4 + 4*a^2*b^2 - 5*b^4)*(2*Sqrt[-b^2]*(a^2 + b^2) + (-a^2 + b^2 + 2*a*Sqrt[-b^2])*Log[Sqrt[-b^2]
- b*Tan[c + d*x]]*(a + b*Tan[c + d*x]) - 4*a*Sqrt[-b^2]*Log[a + b*Tan[c + d*x]]*(a + b*Tan[c + d*x]) + (a^2 -
b^2 + 2*a*Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]]*(a + b*Tan[c + d*x]))))/(a^2 + b^2)^3)/(16*b*(a^2 + b^2
)*d*(a + b*Tan[c + d*x]))

Maple [A] (verified)

Time = 16.90 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (\frac {3}{8} a^{6}+\frac {15}{8} a^{4} b^{2}+\frac {5}{8} a^{2} b^{4}-\frac {7}{8} b^{6}\right ) \left (\tan ^{3}\left (d x +c \right )\right )+\left (2 a^{3} b^{3}+2 a \,b^{5}\right ) \left (\tan ^{2}\left (d x +c \right )\right )+\left (\frac {17}{8} a^{4} b^{2}+\frac {3}{8} a^{2} b^{4}-\frac {9}{8} b^{6}+\frac {5}{8} a^{6}\right ) \tan \left (d x +c \right )+\frac {a^{5} b}{2}+3 a^{3} b^{3}+\frac {5 a \,b^{5}}{2}}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}-3 a \,b^{5} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+\frac {3 \left (a^{6}+5 a^{4} b^{2}+15 a^{2} b^{4}-5 b^{6}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{\left (a^{2}+b^{2}\right )^{4}}-\frac {b^{5}}{\left (a^{2}+b^{2}\right )^{3} \left (a +b \tan \left (d x +c \right )\right )}+\frac {6 b^{5} a \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{4}}}{d}\)	\(248\)
default	\(\frac {\frac {\frac {\left (\frac {3}{8} a^{6}+\frac {15}{8} a^{4} b^{2}+\frac {5}{8} a^{2} b^{4}-\frac {7}{8} b^{6}\right ) \left (\tan ^{3}\left (d x +c \right )\right )+\left (2 a^{3} b^{3}+2 a \,b^{5}\right ) \left (\tan ^{2}\left (d x +c \right )\right )+\left (\frac {17}{8} a^{4} b^{2}+\frac {3}{8} a^{2} b^{4}-\frac {9}{8} b^{6}+\frac {5}{8} a^{6}\right ) \tan \left (d x +c \right )+\frac {a^{5} b}{2}+3 a^{3} b^{3}+\frac {5 a \,b^{5}}{2}}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{2}}-3 a \,b^{5} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+\frac {3 \left (a^{6}+5 a^{4} b^{2}+15 a^{2} b^{4}-5 b^{6}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{\left (a^{2}+b^{2}\right )^{4}}-\frac {b^{5}}{\left (a^{2}+b^{2}\right )^{3} \left (a +b \tan \left (d x +c \right )\right )}+\frac {6 b^{5} a \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{4}}}{d}\)	\(248\)
risch	\(\frac {12 i x a b}{32 i a^{3} b -32 i a \,b^{3}-8 a^{4}+48 a^{2} b^{2}-8 b^{4}}-\frac {3 x \,a^{2}}{32 i a^{3} b -32 i a \,b^{3}-8 a^{4}+48 a^{2} b^{2}-8 b^{4}}+\frac {15 x \,b^{2}}{32 i a^{3} b -32 i a \,b^{3}-8 a^{4}+48 a^{2} b^{2}-8 b^{4}}-\frac {i {\mathrm e}^{4 i \left (d x +c \right )}}{64 \left (-2 i a b +a^{2}-b^{2}\right ) d}-\frac {{\mathrm e}^{2 i \left (d x +c \right )} b}{4 \left (-3 i b \,a^{2}+i b^{3}+a^{3}-3 a \,b^{2}\right ) d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a}{8 \left (-3 i b \,a^{2}+i b^{3}+a^{3}-3 a \,b^{2}\right ) d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} b}{4 \left (2 i a b +a^{2}-b^{2}\right ) \left (i b +a \right ) d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a}{8 \left (2 i a b +a^{2}-b^{2}\right ) \left (i b +a \right ) d}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{64 \left (2 i a b +a^{2}-b^{2}\right ) d}-\frac {12 i a \,b^{5} x}{a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 b^{6} a^{2}+b^{8}}-\frac {12 i a \,b^{5} c}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 b^{6} a^{2}+b^{8}\right )}-\frac {2 i b^{6}}{\left (-i a +b \right )^{3} d \left (i a +b \right )^{4} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )}+\frac {6 a \,b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{d \left (a^{8}+4 a^{6} b^{2}+6 a^{4} b^{4}+4 b^{6} a^{2}+b^{8}\right )}\)	\(562\)

[In]

int(cos(d*x+c)^4/(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(1/(a^2+b^2)^4*(((3/8*a^6+15/8*a^4*b^2+5/8*a^2*b^4-7/8*b^6)*tan(d*x+c)^3+(2*a^3*b^3+2*a*b^5)*tan(d*x+c)^2+
(17/8*a^4*b^2+3/8*a^2*b^4-9/8*b^6+5/8*a^6)*tan(d*x+c)+1/2*a^5*b+3*a^3*b^3+5/2*a*b^5)/(1+tan(d*x+c)^2)^2-3*a*b^
5*ln(1+tan(d*x+c)^2)+3/8*(a^6+5*a^4*b^2+15*a^2*b^4-5*b^6)*arctan(tan(d*x+c)))-b^5/(a^2+b^2)^3/(a+b*tan(d*x+c))
+6*b^5/(a^2+b^2)^4*a*ln(a+b*tan(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.80 \[ \int \frac {\cos ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {4 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{5} - 2 \, {\left (a^{6} b - 3 \, a^{4} b^{3} - 9 \, a^{2} b^{5} - 5 \, b^{7}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{6} b + 8 \, a^{4} b^{3} - 9 \, a^{2} b^{5} - 30 \, b^{7} + 6 \, {\left (a^{7} + 5 \, a^{5} b^{2} + 15 \, a^{3} b^{4} - 5 \, a b^{6}\right )} d x\right )} \cos \left (d x + c\right ) + 48 \, {\left (a^{2} b^{5} \cos \left (d x + c\right ) + a b^{6} \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - {\left (3 \, a^{5} b^{2} + 22 \, a^{3} b^{4} + 3 \, a b^{6} - 4 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{4} - 6 \, {\left (a^{6} b + 5 \, a^{4} b^{3} + 15 \, a^{2} b^{5} - 5 \, b^{7}\right )} d x - 6 \, {\left (a^{7} + 5 \, a^{5} b^{2} + 7 \, a^{3} b^{4} + 3 \, a b^{6}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left ({\left (a^{9} + 4 \, a^{7} b^{2} + 6 \, a^{5} b^{4} + 4 \, a^{3} b^{6} + a b^{8}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}\right )} d \sin \left (d x + c\right )\right )}} \]

[In]

integrate(cos(d*x+c)^4/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*(4*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cos(d*x + c)^5 - 2*(a^6*b - 3*a^4*b^3 - 9*a^2*b^5 - 5*b^7)*cos(d
*x + c)^3 + (3*a^6*b + 8*a^4*b^3 - 9*a^2*b^5 - 30*b^7 + 6*(a^7 + 5*a^5*b^2 + 15*a^3*b^4 - 5*a*b^6)*d*x)*cos(d*
x + c) + 48*(a^2*b^5*cos(d*x + c) + a*b^6*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(
d*x + c)^2 + b^2) - (3*a^5*b^2 + 22*a^3*b^4 + 3*a*b^6 - 4*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cos(d*x + c)^4
- 6*(a^6*b + 5*a^4*b^3 + 15*a^2*b^5 - 5*b^7)*d*x - 6*(a^7 + 5*a^5*b^2 + 7*a^3*b^4 + 3*a*b^6)*cos(d*x + c)^2)*
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)]

Timed out. \[ \int \frac {\cos ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4/(a+b*tan(d*x+c))**2,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 502 vs. \(2 (228) = 456\).

Time = 0.48 (sec) , antiderivative size = 502, normalized size of antiderivative = 2.14 \[ \int \frac {\cos ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {48 \, a b^{5} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {24 \, a b^{5} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {3 \, {\left (a^{6} + 5 \, a^{4} b^{2} + 15 \, a^{2} b^{4} - 5 \, b^{6}\right )} {\left (d x + c\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {4 \, a^{4} b + 20 \, a^{2} b^{3} - 8 \, b^{5} + 3 \, {\left (a^{4} b + 4 \, a^{2} b^{3} - 5 \, b^{5}\right )} \tan \left (d x + c\right )^{4} + 3 \, {\left (a^{5} + 4 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \tan \left (d x + c\right )^{3} + {\left (5 \, a^{4} b + 28 \, a^{2} b^{3} - 25 \, b^{5}\right )} \tan \left (d x + c\right )^{2} + {\left (5 \, a^{5} + 16 \, a^{3} b^{2} + 11 \, a b^{4}\right )} \tan \left (d x + c\right )}{a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6} + {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \tan \left (d x + c\right )^{5} + {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \tan \left (d x + c\right )^{3} + 2 \, {\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \tan \left (d x + c\right )^{2} + {\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \tan \left (d x + c\right )}}{8 \, d} \]

[In]

integrate(cos(d*x+c)^4/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/8*(48*a*b^5*log(b*tan(d*x + c) + a)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - 24*a*b^5*log(tan(d*x +
c)^2 + 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 3*(a^6 + 5*a^4*b^2 + 15*a^2*b^4 - 5*b^6)*(d*x + c
)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + (4*a^4*b + 20*a^2*b^3 - 8*b^5 + 3*(a^4*b + 4*a^2*b^3 - 5*b
^5)*tan(d*x + c)^4 + 3*(a^5 + 4*a^3*b^2 + 3*a*b^4)*tan(d*x + c)^3 + (5*a^4*b + 28*a^2*b^3 - 25*b^5)*tan(d*x +
c)^2 + (5*a^5 + 16*a^3*b^2 + 11*a*b^4)*tan(d*x + c))/(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6 + (a^6*b + 3*a^4*b^3
+ 3*a^2*b^5 + b^7)*tan(d*x + c)^5 + (a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*tan(d*x + c)^4 + 2*(a^6*b + 3*a^4*b
^3 + 3*a^2*b^5 + b^7)*tan(d*x + c)^3 + 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*tan(d*x + c)^2 + (a^6*b + 3*a^4
*b^3 + 3*a^2*b^5 + b^7)*tan(d*x + c)))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 464 vs. \(2 (228) = 456\).

Time = 0.47 (sec) , antiderivative size = 464, normalized size of antiderivative = 1.97 \[ \int \frac {\cos ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {48 \, a b^{6} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}} - \frac {24 \, a b^{5} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac {3 \, {\left (a^{6} + 5 \, a^{4} b^{2} + 15 \, a^{2} b^{4} - 5 \, b^{6}\right )} {\left (d x + c\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac {8 \, {\left (6 \, a b^{6} \tan \left (d x + c\right ) + 7 \, a^{2} b^{5} + b^{7}\right )}}{{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}} + \frac {36 \, a b^{5} \tan \left (d x + c\right )^{4} + 3 \, a^{6} \tan \left (d x + c\right )^{3} + 15 \, a^{4} b^{2} \tan \left (d x + c\right )^{3} + 5 \, a^{2} b^{4} \tan \left (d x + c\right )^{3} - 7 \, b^{6} \tan \left (d x + c\right )^{3} + 16 \, a^{3} b^{3} \tan \left (d x + c\right )^{2} + 88 \, a b^{5} \tan \left (d x + c\right )^{2} + 5 \, a^{6} \tan \left (d x + c\right ) + 17 \, a^{4} b^{2} \tan \left (d x + c\right ) + 3 \, a^{2} b^{4} \tan \left (d x + c\right ) - 9 \, b^{6} \tan \left (d x + c\right ) + 4 \, a^{5} b + 24 \, a^{3} b^{3} + 56 \, a b^{5}}{{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} {\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2}}}{8 \, d} \]

[In]

integrate(cos(d*x+c)^4/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/8*(48*a*b^6*log(abs(b*tan(d*x + c) + a))/(a^8*b + 4*a^6*b^3 + 6*a^4*b^5 + 4*a^2*b^7 + b^9) - 24*a*b^5*log(ta
n(d*x + c)^2 + 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 3*(a^6 + 5*a^4*b^2 + 15*a^2*b^4 - 5*b^6)*(
d*x + c)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - 8*(6*a*b^6*tan(d*x + c) + 7*a^2*b^5 + b^7)/((a^8 +
4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*(b*tan(d*x + c) + a)) + (36*a*b^5*tan(d*x + c)^4 + 3*a^6*tan(d*x + c)
^3 + 15*a^4*b^2*tan(d*x + c)^3 + 5*a^2*b^4*tan(d*x + c)^3 - 7*b^6*tan(d*x + c)^3 + 16*a^3*b^3*tan(d*x + c)^2 +
88*a*b^5*tan(d*x + c)^2 + 5*a^6*tan(d*x + c) + 17*a^4*b^2*tan(d*x + c) + 3*a^2*b^4*tan(d*x + c) - 9*b^6*tan(d
*x + c) + 4*a^5*b + 24*a^3*b^3 + 56*a*b^5)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*(tan(d*x + c)^2 +
1)^2))/d

Mupad [B] (verification not implemented)

Time = 5.33 (sec) , antiderivative size = 463, normalized size of antiderivative = 1.97 \[ \int \frac {\cos ^4(c+d x)}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {3\,{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^4\,b+4\,a^2\,b^3-5\,b^5\right )}{8\,\left (a^6+3\,a^4\,b^2+3\,a^2\,b^4+b^6\right )}+\frac {a^4\,b+5\,a^2\,b^3-2\,b^5}{2\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (5\,a^3+11\,a\,b^2\right )}{8\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {3\,{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (a^3+3\,a\,b^2\right )}{8\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (5\,a^4\,b+28\,a^2\,b^3-25\,b^5\right )}{8\,\left (a^2+b^2\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^5+a\,{\mathrm {tan}\left (c+d\,x\right )}^4+2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3+2\,a\,{\mathrm {tan}\left (c+d\,x\right )}^2+b\,\mathrm {tan}\left (c+d\,x\right )+a\right )}+\frac {3\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (-a^2+a\,b\,4{}\mathrm {i}+5\,b^2\right )}{16\,d\,\left (a^4\,1{}\mathrm {i}+4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}-4\,a\,b^3+b^4\,1{}\mathrm {i}\right )}+\frac {3\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (a^2+a\,b\,4{}\mathrm {i}-5\,b^2\right )}{16\,d\,\left (a^4\,1{}\mathrm {i}-4\,a^3\,b-a^2\,b^2\,6{}\mathrm {i}+4\,a\,b^3+b^4\,1{}\mathrm {i}\right )}+\frac {6\,a\,b^5\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d\,{\left (a^2+b^2\right )}^4} \]

[In]

int(cos(c + d*x)^4/(a + b*tan(c + d*x))^2,x)

[Out]

((3*tan(c + d*x)^4*(a^4*b - 5*b^5 + 4*a^2*b^3))/(8*(a^6 + b^6 + 3*a^2*b^4 + 3*a^4*b^2)) + (a^4*b - 2*b^5 + 5*a
^2*b^3)/(2*(a^2 + b^2)*(a^4 + b^4 + 2*a^2*b^2)) + (tan(c + d*x)*(11*a*b^2 + 5*a^3))/(8*(a^4 + b^4 + 2*a^2*b^2)
) + (3*tan(c + d*x)^3*(3*a*b^2 + a^3))/(8*(a^4 + b^4 + 2*a^2*b^2)) + (tan(c + d*x)^2*(5*a^4*b - 25*b^5 + 28*a^
2*b^3))/(8*(a^2 + b^2)*(a^4 + b^4 + 2*a^2*b^2)))/(d*(a + b*tan(c + d*x) + 2*a*tan(c + d*x)^2 + a*tan(c + d*x)^
4 + 2*b*tan(c + d*x)^3 + b*tan(c + d*x)^5)) + (3*log(tan(c + d*x) + 1i)*(a*b*4i - a^2 + 5*b^2))/(16*d*(4*a^3*b
- 4*a*b^3 + a^4*1i + b^4*1i - a^2*b^2*6i)) + (3*log(tan(c + d*x) - 1i)*(a*b*4i + a^2 - 5*b^2))/(16*d*(4*a*b^3
- 4*a^3*b + a^4*1i + b^4*1i - a^2*b^2*6i)) + (6*a*b^5*log(a + b*tan(c + d*x)))/(d*(a^2 + b^2)^4)