Integrand size = 27, antiderivative size = 157 \[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=-\frac {i (a-i b)^2 \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {i (a+i b)^2 \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {4 a b \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f} \]
-I*(a-I*b)^2*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))*(c-I*d)^(1/2)/f +I*(a+I*b)^2*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))*(c+I*d)^(1/2)/f +4*a*b*(c+d*tan(f*x+e))^(1/2)/f+2/3*b^2*(c+d*tan(f*x+e))^(3/2)/d/f
Time = 0.62 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.95 \[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=\frac {-3 i (a-i b)^2 \sqrt {c-i d} d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+3 i (a+i b)^2 \sqrt {c+i d} d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+2 b \sqrt {c+d \tan (e+f x)} (b c+6 a d+b d \tan (e+f x))}{3 d f} \]
((-3*I)*(a - I*b)^2*Sqrt[c - I*d]*d*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[ c - I*d]] + (3*I)*(a + I*b)^2*Sqrt[c + I*d]*d*ArcTanh[Sqrt[c + d*Tan[e + f *x]]/Sqrt[c + I*d]] + 2*b*Sqrt[c + d*Tan[e + f*x]]*(b*c + 6*a*d + b*d*Tan[ e + f*x]))/(3*d*f)
Time = 0.76 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.86, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {3042, 4026, 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.
| \(\displaystyle \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}dx\) |
| \(\Big \downarrow \) 4026 |
| \(\displaystyle \int \left (a^2+2 b \tan (e+f x) a-b^2\right ) \sqrt {c+d \tan (e+f x)}dx+\frac {2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a^2+2 b \tan (e+f x) a-b^2\right ) \sqrt {c+d \tan (e+f x)}dx+\frac {2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \frac {c a^2-2 b d a-b^2 c+\left (d a^2+2 b c a-b^2 d\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {4 a b \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {c a^2-2 b d a-b^2 c+\left (d a^2+2 b c a-b^2 d\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {4 a b \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {1}{2} (a-i b)^2 (c-i d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b)^2 (c+i d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {4 a b \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} (a-i b)^2 (c-i d) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b)^2 (c+i d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {4 a b \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {i (a-i b)^2 (c-i d) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (a+i b)^2 (c+i d) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {4 a b \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {i (a-i b)^2 (c-i d) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {i (a+i b)^2 (c+i d) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {4 a b \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a-i b)^2 (c-i d) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(a+i b)^2 (c+i d) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {4 a b \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a-i b)^2 \sqrt {c-i d} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {(a+i b)^2 \sqrt {c+i d} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}+\frac {4 a b \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f}\) |
((a - I*b)^2*Sqrt[c - I*d]*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f + ((a + I *b)^2*Sqrt[c + I*d]*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/f + (4*a*b*Sqrt[c + d*Tan[e + f*x]])/f + (2*b^2*(c + d*Tan[e + f*x])^(3/2))/(3*d*f)
3.13.30.3.1 Defintions of rubi rules used
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_)])^2, x_Symbol] :> Simp[d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*( m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e + f* x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && !LeQ [m, -1] && !(EqQ[m, 2] && EqQ[a, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(1313\) vs. \(2(131)=262\).
Time = 0.92 (sec) , antiderivative size = 1314, normalized size of antiderivative = 8.37
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1314\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(1507\) |
| default | \(\text {Expression too large to display}\) | \(1507\) |
-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^2+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^2+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^2*c+1/4*a^2/f/d*(2*(c^2+d^2)^(1/2) +2*c)^(1/2)*(c^2+d^2)^(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))+a^2/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))-1/4*a^2/f/d*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*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))+b^2*(2/3/d/f*(c+d*tan(f*x+e))^(3/2)+1/4/f/d*(2*(c^2+d^2)^(1/2)+ 2*c)^(1/2)*(c^2+d^2)^(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/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))-1/4/f/d*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*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))-1/4/f/d*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(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/ f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2*(...
Leaf count of result is larger than twice the leaf count of optimal. 2203 vs. \(2 (123) = 246\).
Time = 0.34 (sec) , antiderivative size = 2203, normalized size of antiderivative = 14.03 \[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=\text {Too large to display} \]
-1/6*(3*d*f*sqrt(-(f^2*sqrt(-(16*(a^6*b^2 - 2*a^4*b^4 + a^2*b^6)*c^2 + 8*( a^7*b - 7*a^5*b^3 + 7*a^3*b^5 - a*b^7)*c*d + (a^8 - 12*a^6*b^2 + 38*a^4*b^ 4 - 12*a^2*b^6 + b^8)*d^2)/f^4) + (a^4 - 6*a^2*b^2 + b^4)*c - 4*(a^3*b - a *b^3)*d)/f^2)*log((4*(a^7*b + a^5*b^3 - a^3*b^5 - a*b^7)*c + (a^8 - 4*a^6* b^2 - 10*a^4*b^4 - 4*a^2*b^6 + b^8)*d)*sqrt(d*tan(f*x + e) + c) + ((a^2 - b^2)*f^3*sqrt(-(16*(a^6*b^2 - 2*a^4*b^4 + a^2*b^6)*c^2 + 8*(a^7*b - 7*a^5* b^3 + 7*a^3*b^5 - a*b^7)*c*d + (a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)*d^2)/f^4) + 2*(4*(a^4*b^2 - a^2*b^4)*c + (a^5*b - 6*a^3*b^3 + a*b^ 5)*d)*f)*sqrt(-(f^2*sqrt(-(16*(a^6*b^2 - 2*a^4*b^4 + a^2*b^6)*c^2 + 8*(a^7 *b - 7*a^5*b^3 + 7*a^3*b^5 - a*b^7)*c*d + (a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)*d^2)/f^4) + (a^4 - 6*a^2*b^2 + b^4)*c - 4*(a^3*b - a*b^ 3)*d)/f^2)) - 3*d*f*sqrt(-(f^2*sqrt(-(16*(a^6*b^2 - 2*a^4*b^4 + a^2*b^6)*c ^2 + 8*(a^7*b - 7*a^5*b^3 + 7*a^3*b^5 - a*b^7)*c*d + (a^8 - 12*a^6*b^2 + 3 8*a^4*b^4 - 12*a^2*b^6 + b^8)*d^2)/f^4) + (a^4 - 6*a^2*b^2 + b^4)*c - 4*(a ^3*b - a*b^3)*d)/f^2)*log((4*(a^7*b + a^5*b^3 - a^3*b^5 - a*b^7)*c + (a^8 - 4*a^6*b^2 - 10*a^4*b^4 - 4*a^2*b^6 + b^8)*d)*sqrt(d*tan(f*x + e) + c) - ((a^2 - b^2)*f^3*sqrt(-(16*(a^6*b^2 - 2*a^4*b^4 + a^2*b^6)*c^2 + 8*(a^7*b - 7*a^5*b^3 + 7*a^3*b^5 - a*b^7)*c*d + (a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12 *a^2*b^6 + b^8)*d^2)/f^4) + 2*(4*(a^4*b^2 - a^2*b^4)*c + (a^5*b - 6*a^3*b^ 3 + a*b^5)*d)*f)*sqrt(-(f^2*sqrt(-(16*(a^6*b^2 - 2*a^4*b^4 + a^2*b^6)*c...
\[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=\int \left (a + b \tan {\left (e + f x \right )}\right )^{2} \sqrt {c + d \tan {\left (e + f x \right )}}\, dx \]
\[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=\int { {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \sqrt {d \tan \left (f x + e\right ) + c} \,d x } \]
Timed out. \[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=\text {Timed out} \]
Time = 11.78 (sec) , antiderivative size = 3722, normalized size of antiderivative = 23.71 \[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=\text {Too large to display} \]
(2*b^2*(c + d*tan(e + f*x))^(3/2))/(3*d*f) - atan(((((8*(8*a*b*d^4*f^2 + 8 *a*b*c^2*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(a^4*c - a^ 4*d*1i + b^4*c - b^4*d*1i - 6*a^2*b^2*c + a^2*b^2*d*6i + a*b^3*c*4i - a^3* b*c*4i + 4*a*b^3*d - 4*a^3*b*d)/(4*f^2))^(1/2))*(-(a^4*c - a^4*d*1i + b^4* c - b^4*d*1i - 6*a^2*b^2*c + a^2*b^2*d*6i + a*b^3*c*4i - a^3*b*c*4i + 4*a* b^3*d - 4*a^3*b*d)/(4*f^2))^(1/2) + (16*(c + d*tan(e + f*x))^(1/2)*(a^4*d^ 4 + b^4*d^4 - 6*a^2*b^2*d^4 - a^4*c^2*d^2 - b^4*c^2*d^2 + 6*a^2*b^2*c^2*d^ 2 - 8*a*b^3*c*d^3 + 8*a^3*b*c*d^3))/f^2)*(-(a^4*c - a^4*d*1i + b^4*c - b^4 *d*1i - 6*a^2*b^2*c + a^2*b^2*d*6i + a*b^3*c*4i - a^3*b*c*4i + 4*a*b^3*d - 4*a^3*b*d)/(4*f^2))^(1/2)*1i - (((8*(8*a*b*d^4*f^2 + 8*a*b*c^2*d^2*f^2))/ f^3 + 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(a^4*c - a^4*d*1i + b^4*c - b^ 4*d*1i - 6*a^2*b^2*c + a^2*b^2*d*6i + a*b^3*c*4i - a^3*b*c*4i + 4*a*b^3*d - 4*a^3*b*d)/(4*f^2))^(1/2))*(-(a^4*c - a^4*d*1i + b^4*c - b^4*d*1i - 6*a^ 2*b^2*c + a^2*b^2*d*6i + a*b^3*c*4i - a^3*b*c*4i + 4*a*b^3*d - 4*a^3*b*d)/ (4*f^2))^(1/2) - (16*(c + d*tan(e + f*x))^(1/2)*(a^4*d^4 + b^4*d^4 - 6*a^2 *b^2*d^4 - a^4*c^2*d^2 - b^4*c^2*d^2 + 6*a^2*b^2*c^2*d^2 - 8*a*b^3*c*d^3 + 8*a^3*b*c*d^3))/f^2)*(-(a^4*c - a^4*d*1i + b^4*c - b^4*d*1i - 6*a^2*b^2*c + a^2*b^2*d*6i + a*b^3*c*4i - a^3*b*c*4i + 4*a*b^3*d - 4*a^3*b*d)/(4*f^2) )^(1/2)*1i)/((((8*(8*a*b*d^4*f^2 + 8*a*b*c^2*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(a^4*c - a^4*d*1i + b^4*c - b^4*d*1i - 6*a^2*b...