Integrand size = 45, antiderivative size = 224 \[ \int (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=-\frac {(i a+b) (A-i B-C) \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {(i a-b) (A+i B-C) \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 (A b+a B-b C) \sqrt {c+d \tan (e+f x)}}{f}-\frac {2 (2 b c C-5 b B d-5 a C d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{5 d f} \]
-(I*a+b)*(A-I*B-C)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))*(c-I*d)^( 1/2)/f+(I*a-b)*(A+I*B-C)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))*(c+ I*d)^(1/2)/f+2*(A*b+B*a-C*b)*(c+d*tan(f*x+e))^(1/2)/f-2/15*(-5*B*b*d-5*C*a *d+2*C*b*c)*(c+d*tan(f*x+e))^(3/2)/d^2/f+2/5*b*C*tan(f*x+e)*(c+d*tan(f*x+e ))^(3/2)/d/f
Time = 2.16 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.98 \[ \int (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\frac {\frac {2 (-2 b c C+5 b B d+5 a C d) (c+d \tan (e+f x))^{3/2}}{d}+6 b C \tan (e+f x) (c+d \tan (e+f x))^{3/2}+15 (i a+b) (A-i B-C) d \left (-\sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+d \tan (e+f x)}\right )+15 (-i a+b) (A+i B-C) d \left (-\sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\sqrt {c+d \tan (e+f x)}\right )}{15 d f} \]
Integrate[(a + b*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]]*(A + B*Tan[e + f*x ] + C*Tan[e + f*x]^2),x]
((2*(-2*b*c*C + 5*b*B*d + 5*a*C*d)*(c + d*Tan[e + f*x])^(3/2))/d + 6*b*C*T an[e + f*x]*(c + d*Tan[e + f*x])^(3/2) + 15*(I*a + b)*(A - I*B - C)*d*(-(S qrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]]) + Sqrt[c + d *Tan[e + f*x]]) + 15*((-I)*a + b)*(A + I*B - C)*d*(-(Sqrt[c + I*d]*ArcTanh [Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]]) + Sqrt[c + d*Tan[e + f*x]]))/(15 *d*f)
Time = 1.23 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.97, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.311, Rules used = {3042, 4120, 27, 3042, 4113, 3042, 4011, 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.
| \(\displaystyle \int (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan (e+f x)^2\right )dx\) |
| \(\Big \downarrow \) 4120 |
| \(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {2 \int \frac {1}{2} \sqrt {c+d \tan (e+f x)} \left ((2 b c C-5 a d C-5 b B d) \tan ^2(e+f x)-5 (A b-C b+a B) d \tan (e+f x)+2 b c C-5 a A d\right )dx}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} \left ((2 b c C-5 a d C-5 b B d) \tan ^2(e+f x)-5 (A b-C b+a B) d \tan (e+f x)+2 b c C-5 a A d\right )dx}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} \left ((2 b c C-5 a d C-5 b B d) \tan (e+f x)^2-5 (A b-C b+a B) d \tan (e+f x)+2 b c C-5 a A d\right )dx}{5 d}\) |
| \(\Big \downarrow \) 4113 |
| \(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} (5 (b B-a (A-C)) d-5 (A b-C b+a B) d \tan (e+f x))dx+\frac {2 (-5 a C d-5 b B d+2 b c C) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} (5 (b B-a (A-C)) d-5 (A b-C b+a B) d \tan (e+f x))dx+\frac {2 (-5 a C d-5 b B d+2 b c C) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\int \frac {5 d (b B c+b (A-C) d-a (A c-C c-B d))-5 d (A b c+a B c-b C c+a A d-b B d-a C d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {10 d (a B+A b-b C) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (-5 a C d-5 b B d+2 b c C) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\int \frac {5 d (b B c+b (A-C) d-a (A c-C c-B d))-5 d (A b c+a B c-b C c+a A d-b B d-a C d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {10 d (a B+A b-b C) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (-5 a C d-5 b B d+2 b c C) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {-\frac {5}{2} d (a+i b) (c+i d) (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {5}{2} d (a-i b) (c-i d) (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {10 d (a B+A b-b C) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (-5 a C d-5 b B d+2 b c C) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {-\frac {5}{2} d (a+i b) (c+i d) (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {5}{2} d (a-i b) (c-i d) (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {10 d (a B+A b-b C) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (-5 a C d-5 b B d+2 b c C) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {-\frac {5 i d (a-i b) (c-i d) (A-i B-C) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {5 i d (a+i b) (c+i d) (A+i B-C) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {10 d (a B+A b-b C) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (-5 a C d-5 b B d+2 b c C) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\frac {5 i d (a-i b) (c-i d) (A-i B-C) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {5 i d (a+i b) (c+i d) (A+i B-C) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {10 d (a B+A b-b C) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (-5 a C d-5 b B d+2 b c C) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {-\frac {5 (a+i b) (c+i d) (A+i B-C) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {5 (a-i b) (c-i d) (A-i B-C) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {10 d (a B+A b-b C) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (-5 a C d-5 b B d+2 b c C) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 b C \tan (e+f x) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {-\frac {5 d (a-i b) \sqrt {c-i d} (A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}-\frac {5 d (a+i b) \sqrt {c+i d} (A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}-\frac {10 d (a B+A b-b C) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 (-5 a C d-5 b B d+2 b c C) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
(2*b*C*Tan[e + f*x]*(c + d*Tan[e + f*x])^(3/2))/(5*d*f) - ((-5*(a - I*b)*( A - I*B - C)*Sqrt[c - I*d]*d*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f - (5*(a + I*b)*(A + I*B - C)*Sqrt[c + I*d]*d*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/ f - (10*(A*b + a*B - b*C)*d*Sqrt[c + d*Tan[e + f*x]])/f + (2*(2*b*c*C - 5* b*B*d - 5*a*C*d)*(c + d*Tan[e + f*x])^(3/2))/(3*d*f))/(5*d)
3.1.92.3.1 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[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[((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]
Leaf count of result is larger than twice the leaf count of optimal. \(2217\) vs. \(2(194)=388\).
Time = 0.18 (sec) , antiderivative size = 2218, normalized size of antiderivative = 9.90
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(2218\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(3028\) |
| default | \(\text {Expression too large to display}\) | \(3028\) |
int((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f*x+e)^2 ),x,method=_RETURNVERBOSE)
1/f*(A*b+B*a)*(2*(c+d*tan(f*x+e))^(1/2)+1/4*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)* ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^ 2+d^2)^(1/2))+((c^2+d^2)^(1/2)-c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2 *(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2 *c)^(1/2))-1/4*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f* x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))+(-(c^2+d^2)^(1/ 2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c ^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)))+2/3/f/d*B*b*(c+d *tan(f*x+e))^(3/2)+2/3/f/d*C*a*(c+d*tan(f*x+e))^(3/2)-1/4/f/d*ln((c+d*tan( f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2) )*B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*b-1/4/f/d*ln((c+d*tan(f* x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))* C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a+1/f*d/(2*(c^2+d^2)^(1/2) -2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2) )/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*B*b+1/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*a rctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2 )^(1/2)-2*c)^(1/2))*C*a+1/4/f/d*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/ 2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*B*(2*(c^2+d^2)^(1/2)+2*c)^(1 /2)*b*c+1/4/f/d*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d* tan(f*x+e)-c-(c^2+d^2)^(1/2))*C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*c+1/4/f...
Leaf count of result is larger than twice the leaf count of optimal. 12410 vs. \(2 (187) = 374\).
Time = 1.66 (sec) , antiderivative size = 12410, normalized size of antiderivative = 55.40 \[ \int (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Too large to display} \]
integrate((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f* x+e)^2),x, algorithm="fricas")
\[ \int (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\int \left (a + b \tan {\left (e + f x \right )}\right ) \sqrt {c + d \tan {\left (e + f x \right )}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )\, dx \]
Integral((a + b*tan(e + f*x))*sqrt(c + d*tan(e + f*x))*(A + B*tan(e + f*x) + C*tan(e + f*x)**2), x)
\[ \int (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\int { {\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (b \tan \left (f x + e\right ) + a\right )} \sqrt {d \tan \left (f x + e\right ) + c} \,d x } \]
integrate((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f* x+e)^2),x, algorithm="maxima")
integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(b*tan(f*x + e) + a)*sqr t(d*tan(f*x + e) + c), x)
Timed out. \[ \int (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Timed out} \]
integrate((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))*(A+B*tan(f*x+e)+C*tan(f* x+e)^2),x, algorithm="giac")
Time = 57.65 (sec) , antiderivative size = 22955, normalized size of antiderivative = 102.48 \[ \int (a+b \tan (e+f x)) \sqrt {c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx=\text {Too large to display} \]
int((a + b*tan(e + f*x))*(c + d*tan(e + f*x))^(1/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)^2),x)
((2*B*a*d - 4*C*a*c)/(d*f) + (4*C*a*c)/(d*f))*(c + d*tan(e + f*x))^(1/2) + ((2*B*b*d - 6*C*b*c)/(3*d^2*f) + (4*C*b*c)/(3*d^2*f))*(c + d*tan(e + f*x) )^(3/2) + (c + d*tan(e + f*x))^(1/2)*(2*c*((2*B*b*d - 6*C*b*c)/(d^2*f) + ( 4*C*b*c)/(d^2*f)) + (2*A*b*d^2 + 6*C*b*c^2 - 4*B*b*c*d)/(d^2*f) - (2*C*b*( d^4*f + c^2*d^2*f))/(d^4*f^2)) - atan(((((8*(4*A*b*d^4*f^2 - 4*C*b*d^4*f^2 + 4*A*b*c^2*d^2*f^2 - 4*C*b*c^2*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan(e + f *x))^(1/2)*((A^2*b^2*c)/(4*f^2) - (4*A*C^3*b^4*d^2*f^4 - B^4*b^4*d^2*f^4 - C^4*b^4*d^2*f^4 - A^4*b^4*d^2*f^4 + 4*A^3*C*b^4*d^2*f^4 - 4*A^2*B^2*b^4*c ^2*f^4 + 2*A^2*B^2*b^4*d^2*f^4 - 6*A^2*C^2*b^4*d^2*f^4 - 4*B^2*C^2*b^4*c^2 *f^4 + 2*B^2*C^2*b^4*d^2*f^4 + 4*A*B^3*b^4*c*d*f^4 - 4*A^3*B*b^4*c*d*f^4 + 4*B*C^3*b^4*c*d*f^4 - 4*B^3*C*b^4*c*d*f^4 + 8*A*B^2*C*b^4*c^2*f^4 - 4*A*B ^2*C*b^4*d^2*f^4 - 12*A*B*C^2*b^4*c*d*f^4 + 12*A^2*B*C*b^4*c*d*f^4)^(1/2)/ (4*f^4) - (B^2*b^2*c)/(4*f^2) + (C^2*b^2*c)/(4*f^2) - (A*B*b^2*d)/(2*f^2) - (A*C*b^2*c)/(2*f^2) + (B*C*b^2*d)/(2*f^2))^(1/2))*((A^2*b^2*c)/(4*f^2) - (4*A*C^3*b^4*d^2*f^4 - B^4*b^4*d^2*f^4 - C^4*b^4*d^2*f^4 - A^4*b^4*d^2*f^ 4 + 4*A^3*C*b^4*d^2*f^4 - 4*A^2*B^2*b^4*c^2*f^4 + 2*A^2*B^2*b^4*d^2*f^4 - 6*A^2*C^2*b^4*d^2*f^4 - 4*B^2*C^2*b^4*c^2*f^4 + 2*B^2*C^2*b^4*d^2*f^4 + 4* A*B^3*b^4*c*d*f^4 - 4*A^3*B*b^4*c*d*f^4 + 4*B*C^3*b^4*c*d*f^4 - 4*B^3*C*b^ 4*c*d*f^4 + 8*A*B^2*C*b^4*c^2*f^4 - 4*A*B^2*C*b^4*d^2*f^4 - 12*A*B*C^2*b^4 *c*d*f^4 + 12*A^2*B*C*b^4*c*d*f^4)^(1/2)/(4*f^4) - (B^2*b^2*c)/(4*f^2) ...