Integrand size = 21, antiderivative size = 152 \[ \int \frac {\cos ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {a \left (3 a^4+10 a^2 b^2+15 b^4\right ) x}{8 \left (a^2+b^2\right )^3}+\frac {b^5 \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {\cos ^4(c+d x) (b+a \tan (c+d x))}{4 \left (a^2+b^2\right ) d}+\frac {\cos ^2(c+d x) \left (4 b^3+a \left (3 a^2+7 b^2\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d} \] Output:
1/8*a*(3*a^4+10*a^2*b^2+15*b^4)*x/(a^2+b^2)^3+b^5*ln(a*cos(d*x+c)+b*sin(d* x+c))/(a^2+b^2)^3/d+1/4*cos(d*x+c)^4*(b+a*tan(d*x+c))/(a^2+b^2)/d+1/8*cos( d*x+c)^2*(4*b^3+a*(3*a^2+7*b^2)*tan(d*x+c))/(a^2+b^2)^2/d
Time = 0.82 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.48 \[ \int \frac {\cos ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {-\left (\left (8 b^6+\sqrt {-b^2} \left (3 a^5+10 a^3 b^2+15 a b^4\right )\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )\right )+16 b^6 \log (a+b \tan (c+d x))-\left (8 b^6-\sqrt {-b^2} \left (3 a^5+10 a^3 b^2+15 a b^4\right )\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+4 b \left (a^2+b^2\right )^2 \cos ^4(c+d x) (b+a \tan (c+d x))+2 \left (a^2+b^2\right ) \cos ^2(c+d x) \left (4 b^4+a b \left (3 a^2+7 b^2\right ) \tan (c+d x)\right )}{16 b \left (a^2+b^2\right )^3 d} \] Input:
Integrate[Cos[c + d*x]^4/(a + b*Tan[c + d*x]),x]
Output:
(-((8*b^6 + Sqrt[-b^2]*(3*a^5 + 10*a^3*b^2 + 15*a*b^4))*Log[Sqrt[-b^2] - b *Tan[c + d*x]]) + 16*b^6*Log[a + b*Tan[c + d*x]] - (8*b^6 - Sqrt[-b^2]*(3* a^5 + 10*a^3*b^2 + 15*a*b^4))*Log[Sqrt[-b^2] + b*Tan[c + d*x]] + 4*b*(a^2 + b^2)^2*Cos[c + d*x]^4*(b + a*Tan[c + d*x]) + 2*(a^2 + b^2)*Cos[c + d*x]^ 2*(4*b^4 + a*b*(3*a^2 + 7*b^2)*Tan[c + d*x]))/(16*b*(a^2 + b^2)^3*d)
Time = 0.48 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.61, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3987, 27, 496, 25, 686, 25, 657, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\cos ^4(c+d x)}{a+b \tan (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (c+d x)^4 (a+b \tan (c+d x))}dx\) |
| \(\Big \downarrow \) 3987 |
| \(\displaystyle \frac {\int \frac {b^6}{(a+b \tan (c+d x)) \left (\tan ^2(c+d x) b^2+b^2\right )^3}d(b \tan (c+d x))}{b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b^5 \int \frac {1}{(a+b \tan (c+d x)) \left (\tan ^2(c+d x) b^2+b^2\right )^3}d(b \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 496 |
| \(\displaystyle \frac {b^5 \left (\frac {a b \tan (c+d x)+b^2}{4 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}-\frac {\int -\frac {3 a^2+3 b \tan (c+d x) a+4 b^2}{(a+b \tan (c+d x)) \left (\tan ^2(c+d x) b^2+b^2\right )^2}d(b \tan (c+d x))}{4 b^2 \left (a^2+b^2\right )}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^5 \left (\frac {\int \frac {3 a^2+3 b \tan (c+d x) a+4 b^2}{(a+b \tan (c+d x)) \left (\tan ^2(c+d x) b^2+b^2\right )^2}d(b \tan (c+d x))}{4 b^2 \left (a^2+b^2\right )}+\frac {a b \tan (c+d x)+b^2}{4 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 686 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {a b \left (3 a^2+7 b^2\right ) \tan (c+d x)+4 b^4}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )}-\frac {\int -\frac {3 a^4+7 b^2 a^2+b \left (3 a^2+7 b^2\right ) \tan (c+d x) a+8 b^4}{(a+b \tan (c+d x)) \left (\tan ^2(c+d x) b^2+b^2\right )}d(b \tan (c+d x))}{2 b^2 \left (a^2+b^2\right )}}{4 b^2 \left (a^2+b^2\right )}+\frac {a b \tan (c+d x)+b^2}{4 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {\int \frac {3 a^4+7 b^2 a^2+b \left (3 a^2+7 b^2\right ) \tan (c+d x) a+8 b^4}{(a+b \tan (c+d x)) \left (\tan ^2(c+d x) b^2+b^2\right )}d(b \tan (c+d x))}{2 b^2 \left (a^2+b^2\right )}+\frac {a b \left (3 a^2+7 b^2\right ) \tan (c+d x)+4 b^4}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )}}{4 b^2 \left (a^2+b^2\right )}+\frac {a b \tan (c+d x)+b^2}{4 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle \frac {b^5 \left (\frac {\frac {\int \left (\frac {8 b^4}{\left (a^2+b^2\right ) (a+b \tan (c+d x))}+\frac {3 a^5+10 b^2 a^3+15 b^4 a-8 b^5 \tan (c+d x)}{\left (a^2+b^2\right ) \left (\tan ^2(c+d x) b^2+b^2\right )}\right )d(b \tan (c+d x))}{2 b^2 \left (a^2+b^2\right )}+\frac {a b \left (3 a^2+7 b^2\right ) \tan (c+d x)+4 b^4}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )}}{4 b^2 \left (a^2+b^2\right )}+\frac {a b \tan (c+d x)+b^2}{4 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^5 \left (\frac {a b \tan (c+d x)+b^2}{4 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}+\frac {\frac {a b \left (3 a^2+7 b^2\right ) \tan (c+d x)+4 b^4}{2 b^2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )}+\frac {-\frac {4 b^4 \log \left (b^2 \tan ^2(c+d x)+b^2\right )}{a^2+b^2}+\frac {8 b^4 \log (a+b \tan (c+d x))}{a^2+b^2}+\frac {a \left (3 a^4+10 a^2 b^2+15 b^4\right ) \arctan (\tan (c+d x))}{b \left (a^2+b^2\right )}}{2 b^2 \left (a^2+b^2\right )}}{4 b^2 \left (a^2+b^2\right )}\right )}{d}\) |
Input:
Int[Cos[c + d*x]^4/(a + b*Tan[c + d*x]),x]
Output:
(b^5*((b^2 + a*b*Tan[c + d*x])/(4*b^2*(a^2 + b^2)*(b^2 + b^2*Tan[c + d*x]^ 2)^2) + (((a*(3*a^4 + 10*a^2*b^2 + 15*b^4)*ArcTan[Tan[c + d*x]])/(b*(a^2 + b^2)) + (8*b^4*Log[a + b*Tan[c + d*x]])/(a^2 + b^2) - (4*b^4*Log[b^2 + b^ 2*Tan[c + d*x]^2])/(a^2 + b^2))/(2*b^2*(a^2 + b^2)) + (4*b^4 + a*b*(3*a^2 + 7*b^2)*Tan[c + d*x])/(2*b^2*(a^2 + b^2)*(b^2 + b^2*Tan[c + d*x]^2)))/(4* b^2*(a^2 + b^2))))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-(a*d + b*c*x))*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1)*(b*c^2 + a*d^2))), x] + Simp[1/(2*a*(p + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*Simp[b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3) + b*c*d*(n + 2 *p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[p, -1] && IntQuad raticQ[a, 0, b, c, d, n, p, x]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
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] + Simp[ 1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)) Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Sim p[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] && LtQ [p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[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]
Time = 8.10 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.30
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{5} \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}+\frac {\frac {\left (\frac {3}{8} a^{5}+\frac {5}{4} a^{3} b^{2}+\frac {7}{8} a \,b^{4}\right ) \tan \left (d x +c \right )^{3}+\left (\frac {1}{2} a^{2} b^{3}+\frac {1}{2} b^{5}\right ) \tan \left (d x +c \right )^{2}+\left (\frac {7}{4} a^{3} b^{2}+\frac {9}{8} a \,b^{4}+\frac {5}{8} a^{5}\right ) \tan \left (d x +c \right )+\frac {a^{4} b}{4}+a^{2} b^{3}+\frac {3 b^{5}}{4}}{\left (1+\tan \left (d x +c \right )^{2}\right )^{2}}-\frac {b^{5} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\frac {\left (3 a^{5}+10 a^{3} b^{2}+15 a \,b^{4}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{\left (a^{2}+b^{2}\right )^{3}}}{d}\) | \(197\) |
| default | \(\frac {\frac {b^{5} \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right )^{3}}+\frac {\frac {\left (\frac {3}{8} a^{5}+\frac {5}{4} a^{3} b^{2}+\frac {7}{8} a \,b^{4}\right ) \tan \left (d x +c \right )^{3}+\left (\frac {1}{2} a^{2} b^{3}+\frac {1}{2} b^{5}\right ) \tan \left (d x +c \right )^{2}+\left (\frac {7}{4} a^{3} b^{2}+\frac {9}{8} a \,b^{4}+\frac {5}{8} a^{5}\right ) \tan \left (d x +c \right )+\frac {a^{4} b}{4}+a^{2} b^{3}+\frac {3 b^{5}}{4}}{\left (1+\tan \left (d x +c \right )^{2}\right )^{2}}-\frac {b^{5} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\frac {\left (3 a^{5}+10 a^{3} b^{2}+15 a \,b^{4}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{8}}{\left (a^{2}+b^{2}\right )^{3}}}{d}\) | \(197\) |
| risch | \(\frac {9 x a b}{8 i a^{3}-24 i a \,b^{2}+24 a^{2} b -8 b^{3}}+\frac {3 i x \,a^{2}}{8 i a^{3}-24 i a \,b^{2}+24 a^{2} b -8 b^{3}}-\frac {8 i x \,b^{2}}{8 i a^{3}-24 i a \,b^{2}+24 a^{2} b -8 b^{3}}-\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )} b}{16 \left (-2 i a b +a^{2}-b^{2}\right ) d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a}{8 \left (-2 i a b +a^{2}-b^{2}\right ) d}-\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )} b}{16 \left (i b +a \right )^{2} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a}{8 \left (i b +a \right )^{2} d}-\frac {2 i b^{5} x}{a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}}-\frac {2 i b^{5} c}{d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}+\frac {b^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )}-\frac {b \cos \left (4 d x +4 c \right )}{32 d \left (-a^{2}-b^{2}\right )}-\frac {a \sin \left (4 d x +4 c \right )}{32 d \left (-a^{2}-b^{2}\right )}\) | \(396\) |
Input:
int(cos(d*x+c)^4/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(b^5/(a^2+b^2)^3*ln(a+b*tan(d*x+c))+1/(a^2+b^2)^3*(((3/8*a^5+5/4*a^3*b ^2+7/8*a*b^4)*tan(d*x+c)^3+(1/2*a^2*b^3+1/2*b^5)*tan(d*x+c)^2+(7/4*a^3*b^2 +9/8*a*b^4+5/8*a^5)*tan(d*x+c)+1/4*a^4*b+a^2*b^3+3/4*b^5)/(1+tan(d*x+c)^2) ^2-1/2*b^5*ln(1+tan(d*x+c)^2)+1/8*(3*a^5+10*a^3*b^2+15*a*b^4)*arctan(tan(d *x+c))))
Time = 0.11 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {4 \, b^{5} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{4} + {\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} d x + 4 \, {\left (a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 7 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} d} \] Input:
integrate(cos(d*x+c)^4/(a+b*tan(d*x+c)),x, algorithm="fricas")
Output:
1/8*(4*b^5*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^ 2 + b^2) + 2*(a^4*b + 2*a^2*b^3 + b^5)*cos(d*x + c)^4 + (3*a^5 + 10*a^3*b^ 2 + 15*a*b^4)*d*x + 4*(a^2*b^3 + b^5)*cos(d*x + c)^2 + (2*(a^5 + 2*a^3*b^2 + a*b^4)*cos(d*x + c)^3 + (3*a^5 + 10*a^3*b^2 + 7*a*b^4)*cos(d*x + c))*si n(d*x + c))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d)
\[ \int \frac {\cos ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\int \frac {\cos ^{4}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \] Input:
integrate(cos(d*x+c)**4/(a+b*tan(d*x+c)),x)
Output:
Integral(cos(c + d*x)**4/(a + b*tan(c + d*x)), x)
Time = 0.13 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.78 \[ \int \frac {\cos ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {8 \, b^{5} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {4 \, b^{5} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {4 \, b^{3} \tan \left (d x + c\right )^{2} + {\left (3 \, a^{3} + 7 \, a b^{2}\right )} \tan \left (d x + c\right )^{3} + 2 \, a^{2} b + 6 \, b^{3} + {\left (5 \, a^{3} + 9 \, a b^{2}\right )} \tan \left (d x + c\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{4} + a^{4} + 2 \, a^{2} b^{2} + b^{4} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{2}}}{8 \, d} \] Input:
integrate(cos(d*x+c)^4/(a+b*tan(d*x+c)),x, algorithm="maxima")
Output:
1/8*(8*b^5*log(b*tan(d*x + c) + a)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 4 *b^5*log(tan(d*x + c)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (3*a^5 + 10*a^3*b^2 + 15*a*b^4)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + ( 4*b^3*tan(d*x + c)^2 + (3*a^3 + 7*a*b^2)*tan(d*x + c)^3 + 2*a^2*b + 6*b^3 + (5*a^3 + 9*a*b^2)*tan(d*x + c))/((a^4 + 2*a^2*b^2 + b^4)*tan(d*x + c)^4 + a^4 + 2*a^2*b^2 + b^4 + 2*(a^4 + 2*a^2*b^2 + b^4)*tan(d*x + c)^2))/d
Time = 0.15 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.82 \[ \int \frac {\cos ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {b^{6} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b d + 3 \, a^{4} b^{3} d + 3 \, a^{2} b^{5} d + b^{7} d} - \frac {b^{5} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, {\left (a^{6} d + 3 \, a^{4} b^{2} d + 3 \, a^{2} b^{4} d + b^{6} d\right )}} + \frac {{\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} {\left (d x + c\right )}}{8 \, {\left (a^{6} d + 3 \, a^{4} b^{2} d + 3 \, a^{2} b^{4} d + b^{6} d\right )}} + \frac {2 \, a^{4} b + 8 \, a^{2} b^{3} + 6 \, b^{5} + {\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 7 \, a b^{4}\right )} \tan \left (d x + c\right )^{3} + 4 \, {\left (a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )^{2} + {\left (5 \, a^{5} + 14 \, a^{3} b^{2} + 9 \, a b^{4}\right )} \tan \left (d x + c\right )}{8 \, {\left (a^{2} + b^{2}\right )}^{3} {\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2} d} \] Input:
integrate(cos(d*x+c)^4/(a+b*tan(d*x+c)),x, algorithm="giac")
Output:
b^6*log(abs(b*tan(d*x + c) + a))/(a^6*b*d + 3*a^4*b^3*d + 3*a^2*b^5*d + b^ 7*d) - 1/2*b^5*log(tan(d*x + c)^2 + 1)/(a^6*d + 3*a^4*b^2*d + 3*a^2*b^4*d + b^6*d) + 1/8*(3*a^5 + 10*a^3*b^2 + 15*a*b^4)*(d*x + c)/(a^6*d + 3*a^4*b^ 2*d + 3*a^2*b^4*d + b^6*d) + 1/8*(2*a^4*b + 8*a^2*b^3 + 6*b^5 + (3*a^5 + 1 0*a^3*b^2 + 7*a*b^4)*tan(d*x + c)^3 + 4*(a^2*b^3 + b^5)*tan(d*x + c)^2 + ( 5*a^5 + 14*a^3*b^2 + 9*a*b^4)*tan(d*x + c))/((a^2 + b^2)^3*(tan(d*x + c)^2 + 1)^2*d)
Time = 1.25 (sec) , antiderivative size = 318, normalized size of antiderivative = 2.09 \[ \int \frac {\cos ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {\frac {a^2\,b+3\,b^3}{4\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (3\,a^3+7\,a\,b^2\right )}{8\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (5\,a^3+9\,a\,b^2\right )}{8\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4+2\,{\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-a^2\,3{}\mathrm {i}+9\,a\,b+b^2\,8{}\mathrm {i}\right )}{16\,d\,\left (-a^3-a^2\,b\,3{}\mathrm {i}+3\,a\,b^2+b^3\,1{}\mathrm {i}\right )}+\frac {b^5\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d\,{\left (a^2+b^2\right )}^3}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (-3\,a^2+a\,b\,9{}\mathrm {i}+8\,b^2\right )}{16\,d\,\left (-a^3\,1{}\mathrm {i}-3\,a^2\,b+a\,b^2\,3{}\mathrm {i}+b^3\right )} \] Input:
int(cos(c + d*x)^4/(a + b*tan(c + d*x)),x)
Output:
((a^2*b + 3*b^3)/(4*(a^4 + b^4 + 2*a^2*b^2)) + (b^3*tan(c + d*x)^2)/(2*(a^ 4 + b^4 + 2*a^2*b^2)) + (tan(c + d*x)^3*(7*a*b^2 + 3*a^3))/(8*(a^4 + b^4 + 2*a^2*b^2)) + (tan(c + d*x)*(9*a*b^2 + 5*a^3))/(8*(a^4 + b^4 + 2*a^2*b^2) ))/(d*(2*tan(c + d*x)^2 + tan(c + d*x)^4 + 1)) - (log(tan(c + d*x) - 1i)*( 9*a*b - a^2*3i + b^2*8i))/(16*d*(3*a*b^2 - a^2*b*3i - a^3 + b^3*1i)) + (b^ 5*log(a + b*tan(c + d*x)))/(d*(a^2 + b^2)^3) - (log(tan(c + d*x) + 1i)*(a* b*9i - 3*a^2 + 8*b^2))/(16*d*(a*b^2*3i - 3*a^2*b - a^3*1i + b^3))
Time = 0.16 (sec) , antiderivative size = 353, normalized size of antiderivative = 2.32 \[ \int \frac {\cos ^4(c+d x)}{a+b \tan (c+d x)} \, dx=\frac {-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{5}-4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{3} b^{2}-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a \,b^{4}+5 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{5}+14 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b^{2}+9 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{4}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b^{5}+8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -a \right ) b^{5}+2 \sin \left (d x +c \right )^{4} a^{4} b +4 \sin \left (d x +c \right )^{4} a^{2} b^{3}+2 \sin \left (d x +c \right )^{4} b^{5}-4 \sin \left (d x +c \right )^{2} a^{4} b -12 \sin \left (d x +c \right )^{2} a^{2} b^{3}-8 \sin \left (d x +c \right )^{2} b^{5}+3 a^{5} c +3 a^{5} d x +4 a^{4} b +10 a^{3} b^{2} c +10 a^{3} b^{2} d x +12 a^{2} b^{3}+15 a \,b^{4} c +15 a \,b^{4} d x +8 b^{5}}{8 d \left (a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}\right )} \] Input:
int(cos(d*x+c)^4/(a+b*tan(d*x+c)),x)
Output:
( - 2*cos(c + d*x)*sin(c + d*x)**3*a**5 - 4*cos(c + d*x)*sin(c + d*x)**3*a **3*b**2 - 2*cos(c + d*x)*sin(c + d*x)**3*a*b**4 + 5*cos(c + d*x)*sin(c + d*x)*a**5 + 14*cos(c + d*x)*sin(c + d*x)*a**3*b**2 + 9*cos(c + d*x)*sin(c + d*x)*a*b**4 - 8*log(tan((c + d*x)/2)**2 + 1)*b**5 + 8*log(tan((c + d*x)/ 2)**2*a - 2*tan((c + d*x)/2)*b - a)*b**5 + 2*sin(c + d*x)**4*a**4*b + 4*si n(c + d*x)**4*a**2*b**3 + 2*sin(c + d*x)**4*b**5 - 4*sin(c + d*x)**2*a**4* b - 12*sin(c + d*x)**2*a**2*b**3 - 8*sin(c + d*x)**2*b**5 + 3*a**5*c + 3*a **5*d*x + 4*a**4*b + 10*a**3*b**2*c + 10*a**3*b**2*d*x + 12*a**2*b**3 + 15 *a*b**4*c + 15*a*b**4*d*x + 8*b**5)/(8*d*(a**6 + 3*a**4*b**2 + 3*a**2*b**4 + b**6))