Integrand size = 47, antiderivative size = 287 \[ \int \frac {(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {(a-i b)^2 (B+i (A-C)) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d} f}+\frac {(a+i b)^2 (i A-B-i C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d} f}+\frac {2 \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{15 d^3 f}-\frac {2 b (4 b c C-5 b B d-4 a C d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{15 d^2 f}+\frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f} \]
-(a-I*b)^2*(B+I*(A-C))*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f/(c- I*d)^(1/2)+(a+I*b)^2*(I*A-B-I*C)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1 /2))/f/(c+I*d)^(1/2)+2/15*(12*a^2*C*d^2-10*a*b*d*(-3*B*d+2*C*c)+b^2*(8*c^2 *C-10*B*c*d+15*(A-C)*d^2))*(c+d*tan(f*x+e))^(1/2)/d^3/f-2/15*b*(-5*B*b*d-4 *C*a*d+4*C*b*c)*(c+d*tan(f*x+e))^(1/2)*tan(f*x+e)/d^2/f+2/5*C*(c+d*tan(f*x +e))^(1/2)*(a+b*tan(f*x+e))^2/d/f
Time = 6.36 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.44 \[ \int \frac {(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}+\frac {2 \left (\frac {b (-4 b c C+5 b B d+4 a C d) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {2 \left (\frac {i \sqrt {c-i d} \left (\frac {15}{4} i \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2+\frac {15}{4} \left (2 a b B-a^2 (A-C)+b^2 (A-C)\right ) d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(-c+i d) f}-\frac {i \sqrt {c+i d} \left (-\frac {15}{4} i \left (a^2 B-b^2 B+2 a b (A-C)\right ) d^2+\frac {15}{4} \left (2 a b B-a^2 (A-C)+b^2 (A-C)\right ) d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(-c-i d) f}+\frac {\left (-12 a^2 C d^2+10 a b d (2 c C-3 B d)-b^2 \left (8 c^2 C-10 B c d+15 (A-C) d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{2 d f}\right )}{3 d}\right )}{5 d} \]
Integrate[((a + b*Tan[e + f*x])^2*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2)) /Sqrt[c + d*Tan[e + f*x]],x]
(2*C*(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]])/(5*d*f) + (2*((b*(-4 *b*c*C + 5*b*B*d + 4*a*C*d)*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(3*d*f) - (2*((I*Sqrt[c - I*d]*(((15*I)/4)*(a^2*B - b^2*B + 2*a*b*(A - C))*d^2 + (15*(2*a*b*B - a^2*(A - C) + b^2*(A - C))*d^2)/4)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((-c + I*d)*f) - (I*Sqrt[c + I*d]*(((-15*I)/4)*(a ^2*B - b^2*B + 2*a*b*(A - C))*d^2 + (15*(2*a*b*B - a^2*(A - C) + b^2*(A - C))*d^2)/4)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((-c - I*d)*f ) + ((-12*a^2*C*d^2 + 10*a*b*d*(2*c*C - 3*B*d) - b^2*(8*c^2*C - 10*B*c*d + 15*(A - C)*d^2))*Sqrt[c + d*Tan[e + f*x]])/(2*d*f)))/(3*d)))/(5*d)
Time = 1.82 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.319, Rules used = {3042, 4130, 27, 3042, 4120, 27, 3042, 4113, 3042, 4022, 3042, 4020, 25, 73, 221}
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.
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan (e+f x)^2\right )}{\sqrt {c+d \tan (e+f x)}}dx\) |
| \(\Big \downarrow \) 4130 |
| \(\displaystyle \frac {2 \int -\frac {(a+b \tan (e+f x)) \left ((4 b c C-4 a d C-5 b B d) \tan ^2(e+f x)-5 (A b-C b+a B) d \tan (e+f x)+4 b c C-a (5 A-C) d\right )}{2 \sqrt {c+d \tan (e+f x)}}dx}{5 d}+\frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\int \frac {(a+b \tan (e+f x)) \left ((4 b c C-4 a d C-5 b B d) \tan ^2(e+f x)-5 (A b-C b+a B) d \tan (e+f x)+4 b c C-a (5 A-C) d\right )}{\sqrt {c+d \tan (e+f x)}}dx}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\int \frac {(a+b \tan (e+f x)) \left ((4 b c C-4 a d C-5 b B d) \tan (e+f x)^2-5 (A b-C b+a B) d \tan (e+f x)+4 b c C-a (5 A-C) d\right )}{\sqrt {c+d \tan (e+f x)}}dx}{5 d}\) |
| \(\Big \downarrow \) 4120 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}-\frac {2 \int -\frac {-2 c (4 c C-5 B d) b^2+20 a c C d b-3 a^2 (5 A-C) d^2-\left (\left (8 C c^2-10 B d c+15 (A-C) d^2\right ) b^2-10 a d (2 c C-3 B d) b+12 a^2 C d^2\right ) \tan ^2(e+f x)-15 \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2 \tan (e+f x)}{2 \sqrt {c+d \tan (e+f x)}}dx}{3 d}}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {\int \frac {-2 c (4 c C-5 B d) b^2+20 a c C d b-3 a^2 (5 A-C) d^2-\left (\left (8 C c^2-10 B d c+15 (A-C) d^2\right ) b^2-10 a d (2 c C-3 B d) b+12 a^2 C d^2\right ) \tan ^2(e+f x)-15 \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2 \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}+\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {\int \frac {-2 c (4 c C-5 B d) b^2+20 a c C d b-3 a^2 (5 A-C) d^2-\left (\left (8 C c^2-10 B d c+15 (A-C) d^2\right ) b^2-10 a d (2 c C-3 B d) b+12 a^2 C d^2\right ) \tan (e+f x)^2-15 \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2 \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{3 d}+\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 4113 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {\int \frac {15 \left (-\left ((A-C) a^2\right )+2 b B a+b^2 (A-C)\right ) d^2-15 \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2 \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}}{3 d}+\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {\int \frac {15 \left (-\left ((A-C) a^2\right )+2 b B a+b^2 (A-C)\right ) d^2-15 \left (B a^2+2 b (A-C) a-b^2 B\right ) d^2 \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}}{3 d}+\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {-\frac {15}{2} d^2 (a+i b)^2 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {15}{2} d^2 (a-i b)^2 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}}{3 d}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {-\frac {15}{2} d^2 (a+i b)^2 (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {15}{2} d^2 (a-i b)^2 (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}}{3 d}}{5 d}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {-\frac {15 i d^2 (a-i b)^2 (A-i B-C) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {15 i d^2 (a+i b)^2 (A+i B-C) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}}{3 d}}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {\frac {15 i d^2 (a-i b)^2 (A-i B-C) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {15 i d^2 (a+i b)^2 (A+i B-C) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}}{3 d}}{5 d}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {-\frac {15 d (a-i b)^2 (A-i B-C) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {15 d (a+i b)^2 (A+i B-C) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}}{3 d}}{5 d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 C (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}}{5 d f}-\frac {\frac {2 b \tan (e+f x) (-4 a C d-5 b B d+4 b c C) \sqrt {c+d \tan (e+f x)}}{3 d f}+\frac {-\frac {2 \sqrt {c+d \tan (e+f x)} \left (12 a^2 C d^2-10 a b d (2 c C-3 B d)+b^2 \left (15 d^2 (A-C)-10 B c d+8 c^2 C\right )\right )}{d f}-\frac {15 d^2 (a-i b)^2 (A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}-\frac {15 d^2 (a+i b)^2 (A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}}{3 d}}{5 d}\) |
Int[((a + b*Tan[e + f*x])^2*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2))/Sqrt[ c + d*Tan[e + f*x]],x]
(2*C*(a + b*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]])/(5*d*f) - ((2*b*(4*b *c*C - 5*b*B*d - 4*a*C*d)*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(3*d*f) + ((-15*(a - I*b)^2*(A - I*B - C)*d^2*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/( Sqrt[c - I*d]*f) - (15*(a + I*b)^2*(A + I*B - C)*d^2*ArcTan[Tan[e + f*x]/S qrt[c + I*d]])/(Sqrt[c + I*d]*f) - (2*(12*a^2*C*d^2 - 10*a*b*d*(2*c*C - 3* B*d) + b^2*(8*c^2*C - 10*B*c*d + 15*(A - C)*d^2))*Sqrt[c + d*Tan[e + f*x]] )/(d*f))/(3*d))/(5*d)
3.2.11.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
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]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) *(x_)])^(n_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[b*C*Tan[e + f*x]*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 2))), x] - Simp[1/(d*(n + 2)) Int[(c + d*Tan[e + f*x])^n*Si mp[b*c*C - a*A*d*(n + 2) - (A*b + a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C* d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && !LtQ[n, -1]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_. ) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C *(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ c, 0] && NeQ[a, 0])))
Leaf count of result is larger than twice the leaf count of optimal. \(5512\) vs. \(2(254)=508\).
Time = 0.16 (sec) , antiderivative size = 5513, normalized size of antiderivative = 19.21
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(5513\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(18289\) |
| default | \(\text {Expression too large to display}\) | \(18289\) |
int((a+b*tan(f*x+e))^2*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+e))^(1 /2),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 25627 vs. \(2 (244) = 488\).
Time = 4.81 (sec) , antiderivative size = 25627, normalized size of antiderivative = 89.29 \[ \int \frac {(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]
integrate((a+b*tan(f*x+e))^2*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+ e))^(1/2),x, algorithm="fricas")
\[ \int \frac {(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{2} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \]
Integral((a + b*tan(e + f*x))**2*(A + B*tan(e + f*x) + C*tan(e + f*x)**2)/ sqrt(c + d*tan(e + f*x)), x)
Timed out. \[ \int \frac {(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Timed out} \]
integrate((a+b*tan(f*x+e))^2*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+ e))^(1/2),x, algorithm="maxima")
Timed out. \[ \int \frac {(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Timed out} \]
integrate((a+b*tan(f*x+e))^2*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+ e))^(1/2),x, algorithm="giac")
Time = 43.42 (sec) , antiderivative size = 21254, normalized size of antiderivative = 74.06 \[ \int \frac {(a+b \tan (e+f x))^2 \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]
int(((a + b*tan(e + f*x))^2*(A + B*tan(e + f*x) + C*tan(e + f*x)^2))/(c + d*tan(e + f*x))^(1/2),x)
atan(((((16*(2*C*b^2*d^3*f^2 - 2*C*a^2*d^3*f^2 + 4*C*a*b*c*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*((((8*C^2*a^4*c*f^2 + 8*C^2*b^4*c*f^2 - 32*C^2*a*b^3*d*f^2 + 32*C^2*a^3*b*d*f^2 - 48*C^2*a^2*b^2*c*f^2)^2/4 - ( 16*c^2*f^4 + 16*d^2*f^4)*(C^4*a^8 + C^4*b^8 + 4*C^4*a^2*b^6 + 6*C^4*a^4*b^ 4 + 4*C^4*a^6*b^2))^(1/2) - 4*C^2*a^4*c*f^2 - 4*C^2*b^4*c*f^2 + 16*C^2*a*b ^3*d*f^2 - 16*C^2*a^3*b*d*f^2 + 24*C^2*a^2*b^2*c*f^2)/(16*(c^2*f^4 + d^2*f ^4)))^(1/2))*((((8*C^2*a^4*c*f^2 + 8*C^2*b^4*c*f^2 - 32*C^2*a*b^3*d*f^2 + 32*C^2*a^3*b*d*f^2 - 48*C^2*a^2*b^2*c*f^2)^2/4 - (16*c^2*f^4 + 16*d^2*f^4) *(C^4*a^8 + C^4*b^8 + 4*C^4*a^2*b^6 + 6*C^4*a^4*b^4 + 4*C^4*a^6*b^2))^(1/2 ) - 4*C^2*a^4*c*f^2 - 4*C^2*b^4*c*f^2 + 16*C^2*a*b^3*d*f^2 - 16*C^2*a^3*b* d*f^2 + 24*C^2*a^2*b^2*c*f^2)/(16*(c^2*f^4 + d^2*f^4)))^(1/2) - (16*(c + d *tan(e + f*x))^(1/2)*(C^2*a^4*d^2 + C^2*b^4*d^2 - 6*C^2*a^2*b^2*d^2))/f^2) *((((8*C^2*a^4*c*f^2 + 8*C^2*b^4*c*f^2 - 32*C^2*a*b^3*d*f^2 + 32*C^2*a^3*b *d*f^2 - 48*C^2*a^2*b^2*c*f^2)^2/4 - (16*c^2*f^4 + 16*d^2*f^4)*(C^4*a^8 + C^4*b^8 + 4*C^4*a^2*b^6 + 6*C^4*a^4*b^4 + 4*C^4*a^6*b^2))^(1/2) - 4*C^2*a^ 4*c*f^2 - 4*C^2*b^4*c*f^2 + 16*C^2*a*b^3*d*f^2 - 16*C^2*a^3*b*d*f^2 + 24*C ^2*a^2*b^2*c*f^2)/(16*(c^2*f^4 + d^2*f^4)))^(1/2)*1i - (((16*(2*C*b^2*d^3* f^2 - 2*C*a^2*d^3*f^2 + 4*C*a*b*c*d^2*f^2))/f^3 + 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*((((8*C^2*a^4*c*f^2 + 8*C^2*b^4*c*f^2 - 32*C^2*a*b^3*d*f^2 + 3 2*C^2*a^3*b*d*f^2 - 48*C^2*a^2*b^2*c*f^2)^2/4 - (16*c^2*f^4 + 16*d^2*f^...