Integrand size = 27, antiderivative size = 209 \[ \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\frac {(i a+b)^3 \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(i a-b)^3 \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f} \]
(I*a+b)^3*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))*(c-I*d)^(1/2)/f-(I *a-b)^3*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))*(c+I*d)^(1/2)/f+2*b* (3*a^2-b^2)*(c+d*tan(f*x+e))^(1/2)/f-4/15*b^2*(-6*a*d+b*c)*(c+d*tan(f*x+e) )^(3/2)/d^2/f+2/5*b^2*(a+b*tan(f*x+e))*(c+d*tan(f*x+e))^(3/2)/d/f
Time = 2.37 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.93 \[ \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\frac {-15 i (a-i b)^3 \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+15 i (a+i b)^3 \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\frac {2 b \sqrt {c+d \tan (e+f x)} \left (15 a b c d+45 a^2 d^2-b^2 \left (2 c^2+15 d^2\right )+b d (b c+15 a d) \tan (e+f x)+3 b^2 d^2 \tan ^2(e+f x)\right )}{d^2}}{15 f} \]
((-15*I)*(a - I*b)^3*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + (15*I)*(a + I*b)^3*Sqrt[c + I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x ]]/Sqrt[c + I*d]] + (2*b*Sqrt[c + d*Tan[e + f*x]]*(15*a*b*c*d + 45*a^2*d^2 - b^2*(2*c^2 + 15*d^2) + b*d*(b*c + 15*a*d)*Tan[e + f*x] + 3*b^2*d^2*Tan[ e + f*x]^2))/d^2)/(15*f)
Time = 1.23 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.97, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.519, Rules used = {3042, 4049, 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.
| \(\displaystyle \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)}dx\) |
| \(\Big \downarrow \) 4049 |
| \(\displaystyle \frac {2 \int -\frac {1}{2} \sqrt {c+d \tan (e+f x)} \left (-5 d a^3+3 b^2 d a+2 b^2 (b c-6 a d) \tan ^2(e+f x)+2 b^3 c-5 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )dx}{5 d}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} \left (-5 d a^3+3 b^2 d a+2 b^2 (b c-6 a d) \tan ^2(e+f x)+2 b^3 c-5 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )dx}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} \left (-5 d a^3+3 b^2 d a+2 b^2 (b c-6 a d) \tan (e+f x)^2+2 b^3 c-5 b \left (3 a^2-b^2\right ) d \tan (e+f x)\right )dx}{5 d}\) |
| \(\Big \downarrow \) 4113 |
| \(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} \left (-5 a \left (a^2-3 b^2\right ) d-5 b \left (3 a^2-b^2\right ) \tan (e+f x) d\right )dx+\frac {4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\int \sqrt {c+d \tan (e+f x)} \left (-5 a \left (a^2-3 b^2\right ) d-5 b \left (3 a^2-b^2\right ) \tan (e+f x) d\right )dx+\frac {4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\int \frac {-5 d \left (c a^3-3 b d a^2-3 b^2 c a+b^3 d\right )-5 d \left (d a^3+3 b c a^2-3 b^2 d a-b^3 c\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {10 b d \left (3 a^2-b^2\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\int \frac {-5 d \left (c a^3-3 b d a^2-3 b^2 c a+b^3 d\right )-5 d \left (d a^3+3 b c a^2-3 b^2 d a-b^3 c\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {10 b d \left (3 a^2-b^2\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {-\frac {5}{2} d (a-i b)^3 (c-i d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {5}{2} d (a+i b)^3 (c+i d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {10 b d \left (3 a^2-b^2\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {-\frac {5}{2} d (a-i b)^3 (c-i d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx-\frac {5}{2} d (a+i b)^3 (c+i d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {10 b d \left (3 a^2-b^2\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {-\frac {5 i d (a-i b)^3 (c-i d) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {5 i d (a+i b)^3 (c+i d) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {10 b d \left (3 a^2-b^2\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {\frac {5 i d (a-i b)^3 (c-i d) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {5 i d (a+i b)^3 (c+i d) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {10 b d \left (3 a^2-b^2\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {-\frac {5 (a-i b)^3 (c-i d) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {5 (a+i b)^3 (c+i d) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{f}-\frac {10 b d \left (3 a^2-b^2\right ) \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {-\frac {10 b d \left (3 a^2-b^2\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {5 d (a-i b)^3 \sqrt {c-i d} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}-\frac {5 d (a+i b)^3 \sqrt {c+i d} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}+\frac {4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
(2*b^2*(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(3/2))/(5*d*f) - ((-5*(a - I*b)^3*Sqrt[c - I*d]*d*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f - (5*(a + I *b)^3*Sqrt[c + I*d]*d*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/f - (10*b*(3*a^2 - b^2)*d*Sqrt[c + d*Tan[e + f*x]])/f + (4*b^2*(b*c - 6*a*d)*(c + d*Tan[e + f*x])^(3/2))/(3*d*f))/(5*d)
3.13.29.3.1 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_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((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 - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 , 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I ntegerQ[m]) && !(IGtQ[n, 2] && ( !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]
Leaf count of result is larger than twice the leaf count of optimal. \(1704\) vs. \(2(181)=362\).
Time = 0.99 (sec) , antiderivative size = 1705, normalized size of antiderivative = 8.16
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1705\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(2073\) |
| default | \(\text {Expression too large to display}\) | \(2073\) |
-1/4/f/d*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))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a^3+1 /f*d/(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))*a^3+1/4/f/d*ln(d*t an(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))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^3*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))*(2*(c^2+ d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a^3-1/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))*a^3-1/4/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))*(2*(c^2+d^2)^(1/2)+2*c) ^(1/2)*a^3*c+2*b^3/f/d^2*(1/5*(c+d*tan(f*x+e))^(5/2)-1/3*(c+d*tan(f*x+e))^ (3/2)*c-d^2*(c+d*tan(f*x+e))^(1/2)-d^2*(-1/8*(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))+1/2*(-(c^2+d^2)^(1/2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arct an((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))+1/8*(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))+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))))+3*a...
Leaf count of result is larger than twice the leaf count of optimal. 3264 vs. \(2 (175) = 350\).
Time = 0.48 (sec) , antiderivative size = 3264, normalized size of antiderivative = 15.62 \[ \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\text {Too large to display} \]
1/30*(15*d^2*f*sqrt(-(f^2*sqrt(-(4*(9*a^10*b^2 - 60*a^8*b^4 + 118*a^6*b^6 - 60*a^4*b^8 + 9*a^2*b^10)*c^2 + 4*(3*a^11*b - 55*a^9*b^3 + 198*a^7*b^5 - 198*a^5*b^7 + 55*a^3*b^9 - 3*a*b^11)*c*d + (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^2)/f^4) + (a^6 - 15 *a^4*b^2 + 15*a^2*b^4 - b^6)*c - 2*(3*a^5*b - 10*a^3*b^3 + 3*a*b^5)*d)/f^2 )*log(-(2*(3*a^11*b - a^9*b^3 - 18*a^7*b^5 - 18*a^5*b^7 - a^3*b^9 + 3*a*b^ 11)*c + (a^12 - 12*a^10*b^2 - 27*a^8*b^4 + 27*a^4*b^8 + 12*a^2*b^10 - b^12 )*d)*sqrt(d*tan(f*x + e) + c) + ((a^3 - 3*a*b^2)*f^3*sqrt(-(4*(9*a^10*b^2 - 60*a^8*b^4 + 118*a^6*b^6 - 60*a^4*b^8 + 9*a^2*b^10)*c^2 + 4*(3*a^11*b - 55*a^9*b^3 + 198*a^7*b^5 - 198*a^5*b^7 + 55*a^3*b^9 - 3*a*b^11)*c*d + (a^1 2 - 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^2)/f^4) + (2*(9*a^7*b^2 - 33*a^5*b^4 + 19*a^3*b^6 - 3*a*b^8)*c + ( 3*a^8*b - 46*a^6*b^3 + 60*a^4*b^5 - 18*a^2*b^7 + b^9)*d)*f)*sqrt(-(f^2*sqr t(-(4*(9*a^10*b^2 - 60*a^8*b^4 + 118*a^6*b^6 - 60*a^4*b^8 + 9*a^2*b^10)*c^ 2 + 4*(3*a^11*b - 55*a^9*b^3 + 198*a^7*b^5 - 198*a^5*b^7 + 55*a^3*b^9 - 3* a*b^11)*c*d + (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^2)/f^4) + (a^6 - 15*a^4*b^2 + 15*a^2*b^4 - b^6)* c - 2*(3*a^5*b - 10*a^3*b^3 + 3*a*b^5)*d)/f^2)) - 15*d^2*f*sqrt(-(f^2*sqrt (-(4*(9*a^10*b^2 - 60*a^8*b^4 + 118*a^6*b^6 - 60*a^4*b^8 + 9*a^2*b^10)*c^2 + 4*(3*a^11*b - 55*a^9*b^3 + 198*a^7*b^5 - 198*a^5*b^7 + 55*a^3*b^9 - ...
\[ \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\int \left (a + b \tan {\left (e + f x \right )}\right )^{3} \sqrt {c + d \tan {\left (e + f x \right )}}\, dx \]
\[ \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{3} \sqrt {d \tan \left (f x + e\right ) + c} \,d x } \]
Timed out. \[ \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\text {Timed out} \]
Time = 25.51 (sec) , antiderivative size = 10306, normalized size of antiderivative = 49.31 \[ \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\text {Too large to display} \]
atan(((((8*(4*b^3*d^4*f^2 - 12*a^2*b*d^4*f^2 + 4*b^3*c^2*d^2*f^2 - 12*a^2* b*c^2*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(((8*b^6*c*f^2 - 8*a^6*c*f^2 - 120*a^2*b^4*c*f^2 + 120*a^4*b^2*c*f^2 - 160*a^3*b^3*d*f^2 + 48*a*b^5*d*f^2 + 48*a^5*b*d*f^2)^2/64 - f^4*(a^12*c^2 + a^12*d^2 + b^12 *c^2 + b^12*d^2 + 6*a^2*b^10*c^2 + 15*a^4*b^8*c^2 + 20*a^6*b^6*c^2 + 15*a^ 8*b^4*c^2 + 6*a^10*b^2*c^2 + 6*a^2*b^10*d^2 + 15*a^4*b^8*d^2 + 20*a^6*b^6* d^2 + 15*a^8*b^4*d^2 + 6*a^10*b^2*d^2))^(1/2) + a^6*c*f^2 - b^6*c*f^2 + 15 *a^2*b^4*c*f^2 - 15*a^4*b^2*c*f^2 + 20*a^3*b^3*d*f^2 - 6*a*b^5*d*f^2 - 6*a ^5*b*d*f^2)/(4*f^4))^(1/2))*(-(((8*b^6*c*f^2 - 8*a^6*c*f^2 - 120*a^2*b^4*c *f^2 + 120*a^4*b^2*c*f^2 - 160*a^3*b^3*d*f^2 + 48*a*b^5*d*f^2 + 48*a^5*b*d *f^2)^2/64 - f^4*(a^12*c^2 + a^12*d^2 + b^12*c^2 + b^12*d^2 + 6*a^2*b^10*c ^2 + 15*a^4*b^8*c^2 + 20*a^6*b^6*c^2 + 15*a^8*b^4*c^2 + 6*a^10*b^2*c^2 + 6 *a^2*b^10*d^2 + 15*a^4*b^8*d^2 + 20*a^6*b^6*d^2 + 15*a^8*b^4*d^2 + 6*a^10* b^2*d^2))^(1/2) + a^6*c*f^2 - b^6*c*f^2 + 15*a^2*b^4*c*f^2 - 15*a^4*b^2*c* f^2 + 20*a^3*b^3*d*f^2 - 6*a*b^5*d*f^2 - 6*a^5*b*d*f^2)/(4*f^4))^(1/2) + ( 16*(c + d*tan(e + f*x))^(1/2)*(a^6*d^4 - b^6*d^4 + 15*a^2*b^4*d^4 - 15*a^4 *b^2*d^4 - a^6*c^2*d^2 + b^6*c^2*d^2 - 40*a^3*b^3*c*d^3 - 15*a^2*b^4*c^2*d ^2 + 15*a^4*b^2*c^2*d^2 + 12*a*b^5*c*d^3 + 12*a^5*b*c*d^3))/f^2)*(-(((8*b^ 6*c*f^2 - 8*a^6*c*f^2 - 120*a^2*b^4*c*f^2 + 120*a^4*b^2*c*f^2 - 160*a^3*b^ 3*d*f^2 + 48*a*b^5*d*f^2 + 48*a^5*b*d*f^2)^2/64 - f^4*(a^12*c^2 + a^12*...