Integrand size = 23, antiderivative size = 270 \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {8 a^2 b}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)} \]
1/2*(a+b)*(a^2-4*a*b+b^2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)+1/ 2*(a+b)*(a^2-4*a*b+b^2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)+1/4*( a-b)*(a^2+4*a*b+b^2)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)-1 /4*(a-b)*(a^2+4*a*b+b^2)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/ 2)+2*a*(a^2-3*b^2)/d/tan(d*x+c)^(1/2)-8/5*a^2*b/d/tan(d*x+c)^(3/2)-2/5*a^2 *(a+b*tan(d*x+c))/d/tan(d*x+c)^(5/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.55 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.38 \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \left (3 a \left (a^2-3 b^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},1,-\frac {1}{4},-\tan ^2(c+d x)\right )+b \left (5 \left (3 a^2-b^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},-\tan ^2(c+d x)\right ) \tan (c+d x)+b (9 a+5 b \tan (c+d x))\right )\right )}{15 d \tan ^{\frac {5}{2}}(c+d x)} \]
(-2*(3*a*(a^2 - 3*b^2)*Hypergeometric2F1[-5/4, 1, -1/4, -Tan[c + d*x]^2] + b*(5*(3*a^2 - b^2)*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[c + d*x]^2]*Tan[c + d*x] + b*(9*a + 5*b*Tan[c + d*x]))))/(15*d*Tan[c + d*x]^(5/2))
Time = 0.90 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.93, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.783, Rules used = {3042, 4048, 27, 3042, 4111, 27, 3042, 4012, 3042, 4017, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^3}{\tan (c+d x)^{7/2}}dx\) |
| \(\Big \downarrow \) 4048 |
| \(\displaystyle \frac {2}{5} \int \frac {12 b a^2-5 \left (a^2-3 b^2\right ) \tan (c+d x) a-b \left (3 a^2-5 b^2\right ) \tan ^2(c+d x)}{2 \tan ^{\frac {5}{2}}(c+d x)}dx-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {12 b a^2-5 \left (a^2-3 b^2\right ) \tan (c+d x) a-b \left (3 a^2-5 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x)}dx-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {12 b a^2-5 \left (a^2-3 b^2\right ) \tan (c+d x) a-b \left (3 a^2-5 b^2\right ) \tan (c+d x)^2}{\tan (c+d x)^{5/2}}dx-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4111 |
| \(\displaystyle \frac {1}{5} \left (\int -\frac {5 \left (a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx-\frac {8 a^2 b}{d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (-5 \int \frac {a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)}dx-\frac {8 a^2 b}{d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-5 \int \frac {a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)}{\tan (c+d x)^{3/2}}dx-\frac {8 a^2 b}{d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\int \frac {b \left (3 a^2-b^2\right )-a \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx-\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}\right )-\frac {8 a^2 b}{d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\int \frac {b \left (3 a^2-b^2\right )-a \left (a^2-3 b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx-\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}\right )-\frac {8 a^2 b}{d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\frac {2 \int \frac {b \left (3 a^2-b^2\right )-a \left (a^2-3 b^2\right ) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}-\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}\right )-\frac {8 a^2 b}{d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\frac {2 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}\right )-\frac {8 a^2 b}{d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\frac {2 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d}-\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}\right )-\frac {8 a^2 b}{d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\frac {2 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}\right )-\frac {8 a^2 b}{d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\frac {2 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}\right )-\frac {8 a^2 b}{d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\frac {2 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}\right )-\frac {8 a^2 b}{d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\frac {2 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}\right )-\frac {8 a^2 b}{d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\frac {2 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}\right )-\frac {8 a^2 b}{d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{5} \left (-5 \left (\frac {2 \left (\frac {1}{2} (a-b) \left (a^2+4 a b+b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\frac {1}{2} (a+b) \left (a^2-4 a b+b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}-\frac {2 a \left (a^2-3 b^2\right )}{d \sqrt {\tan (c+d x)}}\right )-\frac {8 a^2 b}{d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 a^2 (a+b \tan (c+d x))}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
(-5*((2*(-1/2*((a + b)*(a^2 - 4*a*b + b^2)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[ c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2])) + ( (a - b)*(a^2 + 4*a*b + b^2)*(-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan [c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2 *Sqrt[2])))/2))/d - (2*a*(a^2 - 3*b^2))/(d*Sqrt[Tan[c + d*x]])) - (8*a^2*b )/(d*Tan[c + d*x]^(3/2)))/5 - (2*a^2*(a + b*Tan[c + d*x]))/(5*d*Tan[c + d* x]^(5/2))
3.6.75.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ a*c, 2]}, Simp[(d*q + a*e)/(2*a*c) Int[(q + c*x^2)/(a + c*x^4), x], x] + Simp[(d*q - a*e)/(2*a*c) Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a , c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- a)*c]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[2/f Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & & NeQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1 /(d*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c *(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3*a*b^2*d) *Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*( n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 2] && LtQ [n, -1] && IntegerQ[2*m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2))), x ] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B , C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0 ]
Time = 0.06 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.90
| method | result | size |
| derivativedivides | \(\frac {\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (a^{3}-3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2 a^{3}}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}+\frac {2 a \left (a^{2}-3 b^{2}\right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {2 a^{2} b}{\tan \left (d x +c \right )^{\frac {3}{2}}}}{d}\) | \(243\) |
| default | \(\frac {\frac {\left (-3 a^{2} b +b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (a^{3}-3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2 a^{3}}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}+\frac {2 a \left (a^{2}-3 b^{2}\right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {2 a^{2} b}{\tan \left (d x +c \right )^{\frac {3}{2}}}}{d}\) | \(243\) |
| parts | \(\frac {a^{3} \left (\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2}{5 \tan \left (d x +c \right )^{\frac {5}{2}}}+\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {b^{3} \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {3 a \,b^{2} \left (-\frac {2}{\sqrt {\tan \left (d x +c \right )}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {3 a^{2} b \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\) | \(416\) |
1/d*(1/4*(-3*a^2*b+b^3)*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c) )/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^( 1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*(a^3-3*a*b^2)*2^(1/2)*(ln ((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d *x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c )^(1/2)))-2/5*a^3/tan(d*x+c)^(5/2)+2*a*(a^2-3*b^2)/tan(d*x+c)^(1/2)-2*a^2* b/tan(d*x+c)^(3/2))
Leaf count of result is larger than twice the leaf count of optimal. 1428 vs. \(2 (232) = 464\).
Time = 0.29 (sec) , antiderivative size = 1428, normalized size of antiderivative = 5.29 \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \]
-1/10*(5*d*sqrt((6*a^5*b - 20*a^3*b^3 + 6*a*b^5 + d^2*sqrt(-(a^12 - 30*a^1 0*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4) )/d^2)*log(((a^3 - 3*a*b^2)*d^3*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4) - (3*a^8*b - 46*a^6*b ^3 + 60*a^4*b^5 - 18*a^2*b^7 + b^9)*d)*sqrt((6*a^5*b - 20*a^3*b^3 + 6*a*b^ 5 + d^2*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^ 8 - 30*a^2*b^10 + b^12)/d^4))/d^2) - (a^12 - 12*a^10*b^2 - 27*a^8*b^4 + 27 *a^4*b^8 + 12*a^2*b^10 - b^12)*sqrt(tan(d*x + c)))*tan(d*x + c)^3 - 5*d*sq rt((6*a^5*b - 20*a^3*b^3 + 6*a*b^5 + d^2*sqrt(-(a^12 - 30*a^10*b^2 + 255*a ^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4))/d^2)*log(-( (a^3 - 3*a*b^2)*d^3*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4) - (3*a^8*b - 46*a^6*b^3 + 60*a^4* b^5 - 18*a^2*b^7 + b^9)*d)*sqrt((6*a^5*b - 20*a^3*b^3 + 6*a*b^5 + d^2*sqrt (-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b ^10 + b^12)/d^4))/d^2) - (a^12 - 12*a^10*b^2 - 27*a^8*b^4 + 27*a^4*b^8 + 1 2*a^2*b^10 - b^12)*sqrt(tan(d*x + c)))*tan(d*x + c)^3 - 5*d*sqrt((6*a^5*b - 20*a^3*b^3 + 6*a*b^5 - d^2*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452 *a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4))/d^2)*log(((a^3 - 3*a*b^ 2)*d^3*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4) + (3*a^8*b - 46*a^6*b^3 + 60*a^4*b^5 - 18*a...
\[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {\left (a + b \tan {\left (c + d x \right )}\right )^{3}}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {10 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 10 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 5 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 5 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \frac {8 \, {\left (5 \, a^{2} b \tan \left (d x + c\right ) + a^{3} - 5 \, {\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (d x + c\right )^{2}\right )}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{20 \, d} \]
1/20*(10*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(1/2*sqrt(2)*(sqrt( 2) + 2*sqrt(tan(d*x + c)))) + 10*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*a rctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - 5*sqrt(2)*(a^3 + 3* a^2*b - 3*a*b^2 - b^3)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 5*sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*log(-sqrt(2)*sqrt(tan(d*x + c) ) + tan(d*x + c) + 1) - 8*(5*a^2*b*tan(d*x + c) + a^3 - 5*(a^3 - 3*a*b^2)* tan(d*x + c)^2)/tan(d*x + c)^(5/2))/d
Timed out. \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
Time = 7.33 (sec) , antiderivative size = 1777, normalized size of antiderivative = 6.58 \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \]
- 2*atanh((32*a^6*d^3*tan(c + d*x)^(1/2)*((b^6*1i)/(4*d^2) - (a^6*1i)/(4*d ^2) + (3*a*b^5)/(2*d^2) + (3*a^5*b)/(2*d^2) - (a^2*b^4*15i)/(4*d^2) - (5*a ^3*b^3)/d^2 + (a^4*b^2*15i)/(4*d^2))^(1/2))/(16*a^9*d^2 + b^9*d^2*16i + 48 *a*b^8*d^2 + a^8*b*d^2*48i - a^2*b^7*d^2*288i - 736*a^3*b^6*d^2 + a^4*b^5* d^2*960i + 960*a^5*b^4*d^2 - a^6*b^3*d^2*736i - 288*a^7*b^2*d^2) - (32*b^6 *d^3*tan(c + d*x)^(1/2)*((b^6*1i)/(4*d^2) - (a^6*1i)/(4*d^2) + (3*a*b^5)/( 2*d^2) + (3*a^5*b)/(2*d^2) - (a^2*b^4*15i)/(4*d^2) - (5*a^3*b^3)/d^2 + (a^ 4*b^2*15i)/(4*d^2))^(1/2))/(16*a^9*d^2 + b^9*d^2*16i + 48*a*b^8*d^2 + a^8* b*d^2*48i - a^2*b^7*d^2*288i - 736*a^3*b^6*d^2 + a^4*b^5*d^2*960i + 960*a^ 5*b^4*d^2 - a^6*b^3*d^2*736i - 288*a^7*b^2*d^2) + (480*a^2*b^4*d^3*tan(c + d*x)^(1/2)*((b^6*1i)/(4*d^2) - (a^6*1i)/(4*d^2) + (3*a*b^5)/(2*d^2) + (3* a^5*b)/(2*d^2) - (a^2*b^4*15i)/(4*d^2) - (5*a^3*b^3)/d^2 + (a^4*b^2*15i)/( 4*d^2))^(1/2))/(16*a^9*d^2 + b^9*d^2*16i + 48*a*b^8*d^2 + a^8*b*d^2*48i - a^2*b^7*d^2*288i - 736*a^3*b^6*d^2 + a^4*b^5*d^2*960i + 960*a^5*b^4*d^2 - a^6*b^3*d^2*736i - 288*a^7*b^2*d^2) - (480*a^4*b^2*d^3*tan(c + d*x)^(1/2)* ((b^6*1i)/(4*d^2) - (a^6*1i)/(4*d^2) + (3*a*b^5)/(2*d^2) + (3*a^5*b)/(2*d^ 2) - (a^2*b^4*15i)/(4*d^2) - (5*a^3*b^3)/d^2 + (a^4*b^2*15i)/(4*d^2))^(1/2 ))/(16*a^9*d^2 + b^9*d^2*16i + 48*a*b^8*d^2 + a^8*b*d^2*48i - a^2*b^7*d^2* 288i - 736*a^3*b^6*d^2 + a^4*b^5*d^2*960i + 960*a^5*b^4*d^2 - a^6*b^3*d^2* 736i - 288*a^7*b^2*d^2))*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^...