Integrand size = 21, antiderivative size = 126 \[ \int \frac {\tan ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {x}{a}+\frac {\left (2 a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))}{2 b^3 d}-\frac {2 (a-b)^{3/2} (a+b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a b^3 d}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\sec (c+d x) \tan (c+d x)}{2 b d} \]
x/a+1/2*(2*a^2-3*b^2)*arctanh(sin(d*x+c))/b^3/d-2*(a-b)^(3/2)*(a+b)^(3/2)* arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a/b^3/d-a*tan(d*x+c)/b ^2/d+1/2*sec(d*x+c)*tan(d*x+c)/b/d
Leaf count is larger than twice the leaf count of optimal. \(287\) vs. \(2(126)=252\).
Time = 1.91 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.28 \[ \int \frac {\tan ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {(b+a \cos (c+d x)) \sec (c+d x) \left (\frac {4 c}{a}+\frac {4 d x}{a}+\frac {8 \left (a^2-b^2\right )^{3/2} \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a b^3}-\frac {4 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{b^3}+\frac {6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{b}+\frac {4 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{b^3}-\frac {6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{b}+\frac {1}{b \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {1}{b \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 a \tan (c+d x)}{b^2}\right )}{4 d (a+b \sec (c+d x))} \]
((b + a*Cos[c + d*x])*Sec[c + d*x]*((4*c)/a + (4*d*x)/a + (8*(a^2 - b^2)^( 3/2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*b^3) - (4*a^ 2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/b^3 + (6*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/b + (4*a^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/ b^3 - (6*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/b + 1/(b*(Cos[(c + d*x) /2] - Sin[(c + d*x)/2])^2) - 1/(b*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) - (4*a*Tan[c + d*x])/b^2))/(4*d*(a + b*Sec[c + d*x]))
Time = 0.79 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.16, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 4386, 3042, 3372, 25, 3042, 3536, 3042, 3138, 221, 4257}
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.
| \(\displaystyle \int \frac {\tan ^4(c+d x)}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cot \left (c+d x+\frac {\pi }{2}\right )^4}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4386 |
| \(\displaystyle \int \frac {\sin (c+d x) \tan ^3(c+d x)}{a \cos (c+d x)+b}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos \left (c+d x+\frac {\pi }{2}\right )^4}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )}dx\) |
| \(\Big \downarrow \) 3372 |
| \(\displaystyle -\frac {\int -\frac {\left (2 a^2+b \cos (c+d x) a-3 b^2+2 b^2 \cos ^2(c+d x)\right ) \sec (c+d x)}{b+a \cos (c+d x)}dx}{2 b^2}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\left (2 a^2+b \cos (c+d x) a-3 b^2+2 b^2 \cos ^2(c+d x)\right ) \sec (c+d x)}{b+a \cos (c+d x)}dx}{2 b^2}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 a^2+b \sin \left (c+d x+\frac {\pi }{2}\right ) a-3 b^2+2 b^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 b^2}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3536 |
| \(\displaystyle \frac {-\frac {2 \left (a^2-b^2\right )^2 \int \frac {1}{b+a \cos (c+d x)}dx}{a b}+\frac {\left (2 a^2-3 b^2\right ) \int \sec (c+d x)dx}{b}+\frac {2 b^2 x}{a}}{2 b^2}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {2 \left (a^2-b^2\right )^2 \int \frac {1}{b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a b}+\frac {\left (2 a^2-3 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}+\frac {2 b^2 x}{a}}{2 b^2}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {-\frac {4 \left (a^2-b^2\right )^2 \int \frac {1}{-\left ((a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a b d}+\frac {\left (2 a^2-3 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}+\frac {2 b^2 x}{a}}{2 b^2}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\left (2 a^2-3 b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}-\frac {4 \left (a^2-b^2\right )^2 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a b d \sqrt {a-b} \sqrt {a+b}}+\frac {2 b^2 x}{a}}{2 b^2}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 b d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\left (2 a^2-3 b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {4 \left (a^2-b^2\right )^2 \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a b d \sqrt {a-b} \sqrt {a+b}}+\frac {2 b^2 x}{a}}{2 b^2}-\frac {a \tan (c+d x)}{b^2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 b d}\) |
((2*b^2*x)/a + ((2*a^2 - 3*b^2)*ArcTanh[Sin[c + d*x]])/(b*d) - (4*(a^2 - b ^2)^2*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]* b*Sqrt[a + b]*d))/(2*b^2) - (a*Tan[c + d*x])/(b^2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*b*d)
3.3.95.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[Cos[e + f*x]*(a + b* Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x])^(n + 1)/(a*d*f*(n + 1))), x] + (-Si mp[b*(m + n + 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((d*Sin[e + f*x] )^(n + 2)/(a^2*d^2*f*(n + 1)*(n + 2))), x] - Simp[1/(a^2*d^2*(n + 1)*(n + 2 )) Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 2)*Simp[a^2*n*(n + 2) - b^2*(m + n + 2)*(m + n + 3) + a*b*m*Sin[e + f*x] - (a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 4))*Sin[e + f*x]^2, x], x], x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n]) && !m < -1 && LtQ[n, -1] && (LtQ[n, -2] || EqQ[m + n + 4, 0])
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)])), x_Symbol] :> Simp[C*(x/(b*d)), x] + (Simp[(A*b^2 - a*b*B + a^2*C) /(b*(b*c - a*d)) Int[1/(a + b*Sin[e + f*x]), x], x] - Simp[(c^2*C - B*c*d + A*d^2)/(d*(b*c - a*d)) Int[1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a , b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Int[Cos[c + d*x]^m*((b + a*Sin[c + d*x])^n/Sin[c + d*x]^(m + n)), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && IntegerQ[m] && (IntegerQ[m/2] || LeQ[m, 1])
Leaf count of result is larger than twice the leaf count of optimal. \(233\) vs. \(2(113)=226\).
Time = 0.88 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.86
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \left (a -b \right ) \left (a^{3}+a^{2} b -a \,b^{2}-b^{3}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} a \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-2 a -b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (2 a^{2}-3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {1}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-2 a -b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-2 a^{2}+3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{3}}}{d}\) | \(234\) |
| default | \(\frac {-\frac {2 \left (a -b \right ) \left (a^{3}+a^{2} b -a \,b^{2}-b^{3}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{3} a \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-2 a -b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (2 a^{2}-3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {1}{2 b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-2 a -b}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-2 a^{2}+3 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{3}}}{d}\) | \(234\) |
| risch | \(\frac {x}{a}-\frac {i \left (b \,{\mathrm e}^{3 i \left (d x +c \right )}+2 a \,{\mathrm e}^{2 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}+2 a \right )}{d \,b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{d \,b^{3}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{d \,b^{3}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d b}+\frac {\sqrt {a^{2}-b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \sqrt {a^{2}-b^{2}}-b}{a}\right )}{d \,b^{3}}-\frac {\sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \sqrt {a^{2}-b^{2}}-b}{a}\right )}{d b a}-\frac {\sqrt {a^{2}-b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {b +i \sqrt {a^{2}-b^{2}}}{a}\right )}{d \,b^{3}}+\frac {\sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {b +i \sqrt {a^{2}-b^{2}}}{a}\right )}{d b a}\) | \(369\) |
1/d*(-2/b^3*(a-b)*(a^3+a^2*b-a*b^2-b^3)/a/((a-b)*(a+b))^(1/2)*arctanh((a-b )*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2))-1/2/b/(tan(1/2*d*x+1/2*c)+1)^2-1 /2*(-2*a-b)/b^2/(tan(1/2*d*x+1/2*c)+1)+1/2*(2*a^2-3*b^2)/b^3*ln(tan(1/2*d* x+1/2*c)+1)+2/a*arctan(tan(1/2*d*x+1/2*c))+1/2/b/(tan(1/2*d*x+1/2*c)-1)^2- 1/2*(-2*a-b)/b^2/(tan(1/2*d*x+1/2*c)-1)+1/2/b^3*(-2*a^2+3*b^2)*ln(tan(1/2* d*x+1/2*c)-1))
Time = 0.41 (sec) , antiderivative size = 444, normalized size of antiderivative = 3.52 \[ \int \frac {\tan ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\left [\frac {4 \, b^{3} d x \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{2} - b^{2}\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, a^{2} b \cos \left (d x + c\right ) - a b^{2}\right )} \sin \left (d x + c\right )}{4 \, a b^{3} d \cos \left (d x + c\right )^{2}}, \frac {4 \, b^{3} d x \cos \left (d x + c\right )^{2} - 4 \, {\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{2} + {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, a^{2} b \cos \left (d x + c\right ) - a b^{2}\right )} \sin \left (d x + c\right )}{4 \, a b^{3} d \cos \left (d x + c\right )^{2}}\right ] \]
[1/4*(4*b^3*d*x*cos(d*x + c)^2 - 2*(a^2 - b^2)^(3/2)*cos(d*x + c)^2*log((2 *a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*co s(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*co s(d*x + c) + b^2)) + (2*a^3 - 3*a*b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1 ) - (2*a^3 - 3*a*b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*(2*a^2*b*c os(d*x + c) - a*b^2)*sin(d*x + c))/(a*b^3*d*cos(d*x + c)^2), 1/4*(4*b^3*d* x*cos(d*x + c)^2 - 4*(a^2 - b^2)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2) *(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^2 + (2*a^3 - 3*a*b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*a^3 - 3*a*b^2)*cos(d* x + c)^2*log(-sin(d*x + c) + 1) - 2*(2*a^2*b*cos(d*x + c) - a*b^2)*sin(d*x + c))/(a*b^3*d*cos(d*x + c)^2)]
\[ \int \frac {\tan ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\int \frac {\tan ^{4}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\tan ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 476 vs. \(2 (113) = 226\).
Time = 0.90 (sec) , antiderivative size = 476, normalized size of antiderivative = 3.78 \[ \int \frac {\tan ^4(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\frac {2 \, {\left ({\left (a^{2} + a b - b^{2}\right )} \sqrt {-a^{2} + b^{2}} {\left | a \right |} {\left | -a + b \right |} {\left | b \right |} + {\left (a^{3} b + a^{2} b^{2} - a b^{3} - 2 \, b^{4}\right )} \sqrt {-a^{2} + b^{2}} {\left | -a + b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b^{4} + \sqrt {b^{8} + {\left (a b^{3} + b^{4}\right )} {\left (a b^{3} - b^{4}\right )}}}{a b^{3} - b^{4}}}}\right )\right )}}{{\left (a b^{2} - b^{3}\right )} a^{2} b^{2} + {\left (a b^{4} - b^{5}\right )} {\left | a \right |} {\left | b \right |}} + \frac {2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} - a b^{4} + 2 \, b^{5} - a^{3} {\left | a \right |} {\left | b \right |} + 2 \, a b^{2} {\left | a \right |} {\left | b \right |} - b^{3} {\left | a \right |} {\left | b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b^{4} - \sqrt {b^{8} + {\left (a b^{3} + b^{4}\right )} {\left (a b^{3} - b^{4}\right )}}}{a b^{3} - b^{4}}}}\right )\right )}}{a^{2} b^{4} - b^{4} {\left | a \right |} {\left | b \right |}} - \frac {{\left (2 \, a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} + \frac {{\left (2 \, a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}} - \frac {2 \, {\left (2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} b^{2}}}{2 \, d} \]
-1/2*(2*((a^2 + a*b - b^2)*sqrt(-a^2 + b^2)*abs(a)*abs(-a + b)*abs(b) + (a ^3*b + a^2*b^2 - a*b^3 - 2*b^4)*sqrt(-a^2 + b^2)*abs(-a + b))*(pi*floor(1/ 2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt(-(b^4 + sqrt(b^8 + (a*b^3 + b^4)*(a*b^3 - b^4)))/(a*b^3 - b^4))))/((a*b^2 - b^3)*a^2*b^2 + (a*b^4 - b^5)*abs(a)*abs(b)) + 2*(a^4*b - 2*a^2*b^3 - a*b^4 + 2*b^5 - a^3* abs(a)*abs(b) + 2*a*b^2*abs(a)*abs(b) - b^3*abs(a)*abs(b))*(pi*floor(1/2*( d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt(-(b^4 - sqrt(b^8 + ( a*b^3 + b^4)*(a*b^3 - b^4)))/(a*b^3 - b^4))))/(a^2*b^4 - b^4*abs(a)*abs(b) ) - (2*a^2 - 3*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 + (2*a^2 - 3*b^ 2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^3 - 2*(2*a*tan(1/2*d*x + 1/2*c)^3 + b*tan(1/2*d*x + 1/2*c)^3 - 2*a*tan(1/2*d*x + 1/2*c) + b*tan(1/2*d*x + 1/ 2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*b^2))/d
Time = 16.06 (sec) , antiderivative size = 6062, normalized size of antiderivative = 48.11 \[ \int \frac {\tan ^4(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \]
(atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*cos(c/2 + (d*x)/2)^4*3i) /(b*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2)^4 - 2*cos(c/2 + (d*x)/2)^ 2*sin(c/2 + (d*x)/2)^2)) + (atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2 ))*sin(c/2 + (d*x)/2)^4*3i)/(b*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2 )^4 - 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2)) + (cos(c/2 + (d*x)/2)* sin(c/2 + (d*x)/2)^3)/(b*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2)^4 - 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2)) + (cos(c/2 + (d*x)/2)^3*sin( c/2 + (d*x)/2))/(b*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2)^4 - 2*cos( c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2)) + (2*atan((4*a^13*sin(c/2 + (d*x)/ 2) + 4*a^12*b*sin(c/2 + (d*x)/2) + 12*a^2*b^11*sin(c/2 + (d*x)/2) + 12*a^3 *b^10*sin(c/2 + (d*x)/2) + 15*a^4*b^9*sin(c/2 + (d*x)/2) + 15*a^5*b^8*sin( c/2 + (d*x)/2) - 59*a^6*b^7*sin(c/2 + (d*x)/2) - 59*a^7*b^6*sin(c/2 + (d*x )/2) + 57*a^8*b^5*sin(c/2 + (d*x)/2) + 57*a^9*b^4*sin(c/2 + (d*x)/2) - 24* a^10*b^3*sin(c/2 + (d*x)/2) - 24*a^11*b^2*sin(c/2 + (d*x)/2))/(a*cos(c/2 + (d*x)/2)*(12*a*b^11 + 4*a^11*b + 4*a^12 + 12*a^2*b^10 + 15*a^3*b^9 + 15*a ^4*b^8 - 59*a^5*b^7 - 59*a^6*b^6 + 57*a^7*b^5 + 57*a^8*b^4 - 24*a^9*b^3 - 24*a^10*b^2)))*cos(c/2 + (d*x)/2)^4)/(a*d*(cos(c/2 + (d*x)/2)^4 + sin(c/2 + (d*x)/2)^4 - 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2)) + (2*atan((4* a^13*sin(c/2 + (d*x)/2) + 4*a^12*b*sin(c/2 + (d*x)/2) + 12*a^2*b^11*sin(c/ 2 + (d*x)/2) + 12*a^3*b^10*sin(c/2 + (d*x)/2) + 15*a^4*b^9*sin(c/2 + (d...