Integrand size = 33, antiderivative size = 326 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 \left (2 a A b+a^2 B-b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b (5 A b+7 a B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d} \]
1/2*(a^2*(A-B)-b^2*(A-B)-2*a*b*(A+B))*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/ d*2^(1/2)+1/2*(a^2*(A-B)-b^2*(A-B)-2*a*b*(A+B))*arctan(1+2^(1/2)*tan(d*x+c )^(1/2))/d*2^(1/2)+1/4*(2*a*b*(A-B)+a^2*(A+B)-b^2*(A+B))*ln(1-2^(1/2)*tan( d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)-1/4*(2*a*b*(A-B)+a^2*(A+B)-b^2*(A+B))*l n(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)+2*(2*A*a*b+B*a^2-B*b^2) *tan(d*x+c)^(1/2)/d+2/15*b*(5*A*b+7*B*a)*tan(d*x+c)^(3/2)/d+2/5*b*B*tan(d* x+c)^(3/2)*(a+b*tan(d*x+c))/d
Result contains complex when optimal does not.
Time = 1.31 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.46 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {15 \sqrt [4]{-1} (a-i b)^2 (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-15 (-1)^{3/4} (a+i b)^2 (A+i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 \sqrt {\tan (c+d x)} \left (15 \left (2 a A b+a^2 B-b^2 B\right )+5 b (A b+2 a B) \tan (c+d x)+3 b^2 B \tan ^2(c+d x)\right )}{15 d} \]
(15*(-1)^(1/4)*(a - I*b)^2*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] - 15*(-1)^(3/4)*(a + I*b)^2*(A + I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x ]]] + 2*Sqrt[Tan[c + d*x]]*(15*(2*a*A*b + a^2*B - b^2*B) + 5*b*(A*b + 2*a* B)*Tan[c + d*x] + 3*b^2*B*Tan[c + d*x]^2))/(15*d)
Time = 0.97 (sec) , antiderivative size = 283, normalized size of antiderivative = 0.87, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 4090, 27, 3042, 4113, 3042, 4011, 3042, 4017, 27, 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.
| \(\displaystyle \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 (A+B \tan (c+d x))dx\) |
| \(\Big \downarrow \) 4090 |
| \(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {\tan (c+d x)} \left (b (5 A b+7 a B) \tan ^2(c+d x)+5 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (5 a A-3 b B)\right )dx+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\tan (c+d x)} \left (b (5 A b+7 a B) \tan ^2(c+d x)+5 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (5 a A-3 b B)\right )dx+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\tan (c+d x)} \left (b (5 A b+7 a B) \tan (c+d x)^2+5 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (5 a A-3 b B)\right )dx+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 4113 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\tan (c+d x)} \left (5 \left (A a^2-2 b B a-A b^2\right )+5 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)\right )dx+\frac {2 b (7 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\int \sqrt {\tan (c+d x)} \left (5 \left (A a^2-2 b B a-A b^2\right )+5 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)\right )dx+\frac {2 b (7 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)-5 \left (B a^2+2 A b a-b^2 B\right )}{\sqrt {\tan (c+d x)}}dx+\frac {10 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b (7 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 \left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)-5 \left (B a^2+2 A b a-b^2 B\right )}{\sqrt {\tan (c+d x)}}dx+\frac {10 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b (7 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 \int -\frac {5 \left (B a^2+2 A b a-b^2 B-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)\right )}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {10 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b (7 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \int \frac {B a^2+2 A b a-b^2 B-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {10 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b (7 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} \left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} \left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}+\frac {10 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b (7 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} \left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} \left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d}+\frac {10 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b (7 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} \left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} \left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b (7 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} \left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} \left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b (7 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} \left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} \left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b (7 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} \left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} \left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b (7 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} \left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )-\frac {1}{2} \left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b (7 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{5} \left (-\frac {10 \left (\frac {1}{2} \left (a^2 (A+B)+2 a b (A-B)-b^2 (A+B)\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\frac {1}{2} \left (a^2 (A-B)-2 a b (A+B)-b^2 (A-B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {10 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b (7 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))}{5 d}\) |
(2*b*B*Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x]))/(5*d) + ((-10*(-1/2*((a^2* (A - B) - b^2*(A - B) - 2*a*b*(A + B))*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2])) + ((2*a *b*(A - B) + a^2*(A + B) - b^2*(A + B))*(-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 + (10*(2*a*A*b + a^2*B - b^2*B)*Sqrt[Tan[c + d*x]])/d + (2*b*(5*A*b + 7*a*B)*Tan[c + d*x]^(3/2))/(3*d))/5
3.4.87.3.1 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[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
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_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Ta n[e + f*x])^n*Simp[a^2*A*d*(m + n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m - 1) - b *(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2 , 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && !LeQ[m, -1]
Time = 0.04 (sec) , antiderivative size = 292, normalized size of antiderivative = 0.90
| method | result | size |
| derivativedivides | \(\frac {\frac {2 B \,b^{2} \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {2 A \,b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+\frac {4 B a b \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+4 A a b \left (\sqrt {\tan }\left (d x +c \right )\right )+2 B \,a^{2} \left (\sqrt {\tan }\left (d x +c \right )\right )-2 \left (\sqrt {\tan }\left (d x +c \right )\right ) B \,b^{2}+\frac {\left (-2 A a b -B \,a^{2}+B \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A \,a^{2}-A \,b^{2}-2 B a b \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(292\) |
| default | \(\frac {\frac {2 B \,b^{2} \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {2 A \,b^{2} \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+\frac {4 B a b \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+4 A a b \left (\sqrt {\tan }\left (d x +c \right )\right )+2 B \,a^{2} \left (\sqrt {\tan }\left (d x +c \right )\right )-2 \left (\sqrt {\tan }\left (d x +c \right )\right ) B \,b^{2}+\frac {\left (-2 A a b -B \,a^{2}+B \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (A \,a^{2}-A \,b^{2}-2 B a b \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\) | \(292\) |
| parts | \(\frac {\left (A \,b^{2}+2 B a b \right ) \left (\frac {2 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-\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 {\left (2 A a b +B \,a^{2}\right ) \left (2 \left (\sqrt {\tan }\left (d x +c \right )\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 {A \,a^{2} \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 {B \,b^{2} \left (\frac {2 \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-2 \left (\sqrt {\tan }\left (d x +c \right )\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}\) | \(430\) |
1/d*(2/5*B*b^2*tan(d*x+c)^(5/2)+2/3*A*b^2*tan(d*x+c)^(3/2)+4/3*B*a*b*tan(d *x+c)^(3/2)+4*A*a*b*tan(d*x+c)^(1/2)+2*B*a^2*tan(d*x+c)^(1/2)-2*tan(d*x+c) ^(1/2)*B*b^2+1/4*(-2*A*a*b-B*a^2+B*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)))+1/4*(A*a^2-A* b^2-2*B*a*b)*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*arc tan(-1+2^(1/2)*tan(d*x+c)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 4283 vs. \(2 (288) = 576\).
Time = 0.73 (sec) , antiderivative size = 4283, normalized size of antiderivative = 13.14 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
-1/30*(15*d*sqrt((2*A*B*a^4 - 12*A*B*a^2*b^2 + 2*A*B*b^4 + 4*(A^2 - B^2)*a ^3*b - 4*(A^2 - B^2)*a*b^3 + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*( A^3*B - A*B^3)*a^7*b - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*b^2 + 112*(A^3*B - A*B^3)*a^5*b^3 + 2*(19*A^4 - 102*A^2*B^2 + 19*B^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5 - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A *B^3)*a*b^7 + (A^4 - 2*A^2*B^2 + B^4)*b^8)/d^4))/d^2)*log(((A*a^2 - 2*B*a* b - A*b^2)*d^3*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*(A^3*B - A*B^3)*a^7 *b - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*b^2 + 112*(A^3*B - A*B^3)*a^5*b^3 + 2*(19*A^4 - 102*A^2*B^2 + 19*B^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5 - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A*B^3)*a*b^7 + (A^4 - 2*A^2*B^2 + B^4)*b^8)/d^4) - ((A^2*B - B^3)*a^6 + 2*(A^3 - 5*A*B^2)*a^5 *b - (23*A^2*B - 7*B^3)*a^4*b^2 - 4*(3*A^3 - 7*A*B^2)*a^3*b^3 + (23*A^2*B - 7*B^3)*a^2*b^4 + 2*(A^3 - 5*A*B^2)*a*b^5 - (A^2*B - B^3)*b^6)*d)*sqrt((2 *A*B*a^4 - 12*A*B*a^2*b^2 + 2*A*B*b^4 + 4*(A^2 - B^2)*a^3*b - 4*(A^2 - B^2 )*a*b^3 + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*(A^3*B - A*B^3)*a^7* b - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*b^2 + 112*(A^3*B - A*B^3)*a^5*b^3 + 2*(19*A^4 - 102*A^2*B^2 + 19*B^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5 - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A*B^3)*a*b^7 + (A^4 - 2*A^2*B^2 + B^4)*b^8)/d^4))/d^2) - ((A^4 - B^4)*a^8 - 8*(A^3*B + A*B^3)* a^7*b - 4*(A^4 - B^4)*a^6*b^2 - 8*(A^3*B + A*B^3)*a^5*b^3 - 10*(A^4 - B...
\[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{2} \sqrt {\tan {\left (c + d x \right )}}\, dx \]
Time = 0.37 (sec) , antiderivative size = 275, normalized size of antiderivative = 0.84 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {24 \, B b^{2} \tan \left (d x + c\right )^{\frac {5}{2}} + 30 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} - 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 30 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} - 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 15 \, \sqrt {2} {\left ({\left (A + B\right )} a^{2} + 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 15 \, \sqrt {2} {\left ({\left (A + B\right )} a^{2} + 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 40 \, {\left (2 \, B a b + A b^{2}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + 120 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \sqrt {\tan \left (d x + c\right )}}{60 \, d} \]
1/60*(24*B*b^2*tan(d*x + c)^(5/2) + 30*sqrt(2)*((A - B)*a^2 - 2*(A + B)*a* b - (A - B)*b^2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 30 *sqrt(2)*((A - B)*a^2 - 2*(A + B)*a*b - (A - B)*b^2)*arctan(-1/2*sqrt(2)*( sqrt(2) - 2*sqrt(tan(d*x + c)))) - 15*sqrt(2)*((A + B)*a^2 + 2*(A - B)*a*b - (A + B)*b^2)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 15*sq rt(2)*((A + B)*a^2 + 2*(A - B)*a*b - (A + B)*b^2)*log(-sqrt(2)*sqrt(tan(d* x + c)) + tan(d*x + c) + 1) + 40*(2*B*a*b + A*b^2)*tan(d*x + c)^(3/2) + 12 0*(B*a^2 + 2*A*a*b - B*b^2)*sqrt(tan(d*x + c)))/d
Timed out. \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\text {Timed out} \]
Time = 15.10 (sec) , antiderivative size = 3825, normalized size of antiderivative = 11.73 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
atan((A^2*a^4*tan(c + d*x)^(1/2)*((A^2*a^3*b)/d^2 - (A^2*a*b^3)/d^2 - (12* A^4*a^2*b^6*d^4 - A^4*b^8*d^4 - A^4*a^8*d^4 - 38*A^4*a^4*b^4*d^4 + 12*A^4* a^6*b^2*d^4)^(1/2)/(4*d^4))^(1/2)*32i)/((16*A^3*a^6)/d - (16*A^3*b^6)/d + (112*A^3*a^2*b^4)/d - (112*A^3*a^4*b^2)/d + (32*A*a*b*(12*A^4*a^2*b^6*d^4 - A^4*b^8*d^4 - A^4*a^8*d^4 - 38*A^4*a^4*b^4*d^4 + 12*A^4*a^6*b^2*d^4)^(1/ 2))/d^3) + (A^2*b^4*tan(c + d*x)^(1/2)*((A^2*a^3*b)/d^2 - (A^2*a*b^3)/d^2 - (12*A^4*a^2*b^6*d^4 - A^4*b^8*d^4 - A^4*a^8*d^4 - 38*A^4*a^4*b^4*d^4 + 1 2*A^4*a^6*b^2*d^4)^(1/2)/(4*d^4))^(1/2)*32i)/((16*A^3*a^6)/d - (16*A^3*b^6 )/d + (112*A^3*a^2*b^4)/d - (112*A^3*a^4*b^2)/d + (32*A*a*b*(12*A^4*a^2*b^ 6*d^4 - A^4*b^8*d^4 - A^4*a^8*d^4 - 38*A^4*a^4*b^4*d^4 + 12*A^4*a^6*b^2*d^ 4)^(1/2))/d^3) - (A^2*a^2*b^2*tan(c + d*x)^(1/2)*((A^2*a^3*b)/d^2 - (A^2*a *b^3)/d^2 - (12*A^4*a^2*b^6*d^4 - A^4*b^8*d^4 - A^4*a^8*d^4 - 38*A^4*a^4*b ^4*d^4 + 12*A^4*a^6*b^2*d^4)^(1/2)/(4*d^4))^(1/2)*192i)/((16*A^3*a^6)/d - (16*A^3*b^6)/d + (112*A^3*a^2*b^4)/d - (112*A^3*a^4*b^2)/d + (32*A*a*b*(12 *A^4*a^2*b^6*d^4 - A^4*b^8*d^4 - A^4*a^8*d^4 - 38*A^4*a^4*b^4*d^4 + 12*A^4 *a^6*b^2*d^4)^(1/2))/d^3))*((A^2*a^3*b)/d^2 - (A^2*a*b^3)/d^2 - (12*A^4*a^ 2*b^6*d^4 - A^4*b^8*d^4 - A^4*a^8*d^4 - 38*A^4*a^4*b^4*d^4 + 12*A^4*a^6*b^ 2*d^4)^(1/2)/(4*d^4))^(1/2)*2i - atan((B^2*a^4*tan(c + d*x)^(1/2)*((12*B^4 *a^2*b^6*d^4 - B^4*b^8*d^4 - B^4*a^8*d^4 - 38*B^4*a^4*b^4*d^4 + 12*B^4*a^6 *b^2*d^4)^(1/2)/(4*d^4) + (B^2*a*b^3)/d^2 - (B^2*a^3*b)/d^2)^(1/2)*32i)...