Integrand size = 27, antiderivative size = 209 \[ \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\frac {(i a+b)^3 \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(i a-b)^3 \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f} \] Output:
(I*a+b)^3*(c-I*d)^(1/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f-(I *a-b)^3*(c+I*d)^(1/2)*arctanh((c+d*tan(f*x+e))^(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 = 1.53 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.93 \[ \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\frac {-15 i (a-i b)^3 \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+15 i (a+i b)^3 \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\frac {2 b \sqrt {c+d \tan (e+f x)} \left (15 a b c d+45 a^2 d^2-b^2 \left (2 c^2+15 d^2\right )+b d (b c+15 a d) \tan (e+f x)+3 b^2 d^2 \tan ^2(e+f x)\right )}{d^2}}{15 f} \] Input:
Integrate[(a + b*Tan[e + f*x])^3*Sqrt[c + d*Tan[e + f*x]],x]
Output:
((-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 |
\(\displaystyle \frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {-\frac {10 b d \left (3 a^2-b^2\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {5 d (a-i b)^3 \sqrt {c-i d} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}-\frac {5 d (a+i b)^3 \sqrt {c+i d} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}+\frac {4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{3 d f}}{5 d}\) |
Input:
Int[(a + b*Tan[e + f*x])^3*Sqrt[c + d*Tan[e + f*x]],x]
Output:
(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)
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 = 0.49 (sec) , antiderivative size = 1705, normalized size of antiderivative = 8.16
method | result | size |
parts | \(\text {Expression too large to display}\) | \(1705\) |
derivativedivides | \(\text {Expression too large to display}\) | \(2060\) |
default | \(\text {Expression too large to display}\) | \(2060\) |
Input:
int((a+b*tan(f*x+e))^3*(c+d*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
Output:
-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))*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(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*a^3/f/d*(2*(c^2+d^2)^(1/2) +2*c)^(1/2)*(c^2+d^2)^(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))-a^3/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))-1/4*a^3/f/d*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*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*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-(c+d*tan(f*x+e))^(1/2)*d^2-d^2*(-1/8*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*l n(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)*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))+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*b...
Leaf count of result is larger than twice the leaf count of optimal. 3264 vs. \(2 (175) = 350\).
Time = 0.31 (sec) , antiderivative size = 3264, normalized size of antiderivative = 15.62 \[ \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\text {Too large to display} \] Input:
integrate((a+b*tan(f*x+e))^3*(c+d*tan(f*x+e))^(1/2),x, algorithm="fricas")
Output:
Too large to include
\[ \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\int \left (a + b \tan {\left (e + f x \right )}\right )^{3} \sqrt {c + d \tan {\left (e + f x \right )}}\, dx \] Input:
integrate((a+b*tan(f*x+e))**3*(c+d*tan(f*x+e))**(1/2),x)
Output:
Integral((a + b*tan(e + f*x))**3*sqrt(c + d*tan(e + f*x)), x)
\[ \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{3} \sqrt {d \tan \left (f x + e\right ) + c} \,d x } \] Input:
integrate((a+b*tan(f*x+e))^3*(c+d*tan(f*x+e))^(1/2),x, algorithm="maxima")
Output:
integrate((b*tan(f*x + e) + a)^3*sqrt(d*tan(f*x + e) + c), x)
Exception generated. \[ \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+b*tan(f*x+e))^3*(c+d*tan(f*x+e))^(1/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{%%%{1,[0,19,7]%%%}+%%%{8,[0,17,7]%%%}+%%%{28,[0,15,7]%%%}+ %%%{56,[0
Time = 19.26 (sec) , antiderivative size = 10306, normalized size of antiderivative = 49.31 \[ \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\text {Too large to display} \] Input:
int((a + b*tan(e + f*x))^3*(c + d*tan(e + f*x))^(1/2),x)
Output:
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*...
\[ \int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\left (\int \sqrt {d \tan \left (f x +e \right )+c}d x \right ) a^{3}+\left (\int \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{3}d x \right ) b^{3}+3 \left (\int \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}d x \right ) a \,b^{2}+3 \left (\int \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )d x \right ) a^{2} b \] Input:
int((a+b*tan(f*x+e))^3*(c+d*tan(f*x+e))^(1/2),x)
Output:
int(sqrt(tan(e + f*x)*d + c),x)*a**3 + int(sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**3,x)*b**3 + 3*int(sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2,x)*a*b* *2 + 3*int(sqrt(tan(e + f*x)*d + c)*tan(e + f*x),x)*a**2*b