Integrand size = 35, antiderivative size = 133 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {(i A+B-i C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d} f}-\frac {(B-i (A-C)) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d} f}+\frac {2 C \sqrt {c+d \tan (e+f x)}}{d f} \]
-(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) -(B-I*(A-C))*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/f/(c+I*d)^(1/2) +2*C*(c+d*tan(f*x+e))^(1/2)/d/f
Time = 0.24 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.97 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {-\frac {i (A-i B-C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {i (A+i B-C) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}+\frac {2 C \sqrt {c+d \tan (e+f x)}}{d}}{f} \]
(((-I)*(A - I*B - C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/Sqrt [c - I*d] + (I*(A + I*B - C)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d ]])/Sqrt[c + I*d] + (2*C*Sqrt[c + d*Tan[e + f*x]])/d)/f
Time = 0.59 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.84, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {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.
\(\displaystyle \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \tan (e+f x)+C \tan (e+f x)^2}{\sqrt {c+d \tan (e+f x)}}dx\) |
\(\Big \downarrow \) 4113 |
\(\displaystyle \int \frac {A-C+B \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 C \sqrt {c+d \tan (e+f x)}}{d f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A-C+B \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 C \sqrt {c+d \tan (e+f x)}}{d f}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {1}{2} (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 C \sqrt {c+d \tan (e+f x)}}{d f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} (A+i B-C) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (A-i B-C) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 C \sqrt {c+d \tan (e+f x)}}{d f}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {i (A-i B-C) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (A+i B-C) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 C \sqrt {c+d \tan (e+f x)}}{d f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {i (A-i B-C) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}+\frac {i (A+i B-C) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}+\frac {2 C \sqrt {c+d \tan (e+f x)}}{d f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(A+i B-C) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(A-i B-C) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {2 C \sqrt {c+d \tan (e+f x)}}{d f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(A-i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {(A+i B-C) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}+\frac {2 C \sqrt {c+d \tan (e+f x)}}{d f}\) |
((A - I*B - C)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + ((A + I*B - C)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f) + (2*C*S qrt[c + d*Tan[e + f*x]])/(d*f)
3.2.13.3.1 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[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]
Leaf count of result is larger than twice the leaf count of optimal. \(3462\) vs. \(2(112)=224\).
Time = 0.12 (sec) , antiderivative size = 3463, normalized size of antiderivative = 26.04
method | result | size |
parts | \(\text {Expression too large to display}\) | \(3463\) |
derivativedivides | \(\text {Expression too large to display}\) | \(5570\) |
default | \(\text {Expression too large to display}\) | \(5570\) |
A*(-1/4/f/d/(c^2+d^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))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2-1/4/ f*d/(c^2+d^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))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+1/4/f/d/(c^2+d^2 )^(3/2)*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))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^3+1/4/f*d/(c^2+d^2)^ (3/2)*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))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c-1/f/d/(c^2+d^2)^(1/2)/ (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))*c^2-1/f*d/(c^2+d^2)^(1/ 2)/(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/f/d/(c^2+d^2)^(3/2 )/(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))*c^4+3/f*d/(c^2+d^2)^( 3/2)/(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))*c^2+2/f*d^3/(c^2+d ^2)^(3/2)/(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/(c^2+ d^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))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2+1/4/f*d/(c^2+d^2)*...
Leaf count of result is larger than twice the leaf count of optimal. 3194 vs. \(2 (105) = 210\).
Time = 0.39 (sec) , antiderivative size = 3194, normalized size of antiderivative = 24.02 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]
1/2*(d*f*sqrt(-((c^2 + d^2)*f^2*sqrt(-(4*(A^2*B^2 - 2*A*B^2*C + B^2*C^2)*c ^2 - 4*(A^3*B - A*B^3 + 3*A*B*C^2 - B*C^3 - (3*A^2*B - B^3)*C)*c*d + (A^4 - 2*A^2*B^2 + B^4 - 4*A*C^3 + C^4 + 2*(3*A^2 - B^2)*C^2 - 4*(A^3 - A*B^2)* C)*d^2)/((c^4 + 2*c^2*d^2 + d^4)*f^4)) + (A^2 - B^2 - 2*A*C + C^2)*c + 2*( A*B - B*C)*d)/((c^2 + d^2)*f^2))*log(-(2*(A^3*B + A*B^3 + 3*A*B*C^2 - B*C^ 3 - (3*A^2*B + B^3)*C)*c - (A^4 - B^4 - 4*A^3*C + 6*A^2*C^2 - 4*A*C^3 + C^ 4)*d)*sqrt(d*tan(f*x + e) + c) + (((A - C)*c^3 + B*c^2*d + (A - C)*c*d^2 + B*d^3)*f^3*sqrt(-(4*(A^2*B^2 - 2*A*B^2*C + B^2*C^2)*c^2 - 4*(A^3*B - A*B^ 3 + 3*A*B*C^2 - B*C^3 - (3*A^2*B - B^3)*C)*c*d + (A^4 - 2*A^2*B^2 + B^4 - 4*A*C^3 + C^4 + 2*(3*A^2 - B^2)*C^2 - 4*(A^3 - A*B^2)*C)*d^2)/((c^4 + 2*c^ 2*d^2 + d^4)*f^4)) + (2*(A*B^2 - B^2*C)*c^2 - (3*A^2*B - B^3 - 6*A*B*C + 3 *B*C^2)*c*d + (A^3 - A*B^2 + 3*A*C^2 - C^3 - (3*A^2 - B^2)*C)*d^2)*f)*sqrt (-((c^2 + d^2)*f^2*sqrt(-(4*(A^2*B^2 - 2*A*B^2*C + B^2*C^2)*c^2 - 4*(A^3*B - A*B^3 + 3*A*B*C^2 - B*C^3 - (3*A^2*B - B^3)*C)*c*d + (A^4 - 2*A^2*B^2 + B^4 - 4*A*C^3 + C^4 + 2*(3*A^2 - B^2)*C^2 - 4*(A^3 - A*B^2)*C)*d^2)/((c^4 + 2*c^2*d^2 + d^4)*f^4)) + (A^2 - B^2 - 2*A*C + C^2)*c + 2*(A*B - B*C)*d) /((c^2 + d^2)*f^2))) - d*f*sqrt(-((c^2 + d^2)*f^2*sqrt(-(4*(A^2*B^2 - 2*A* B^2*C + B^2*C^2)*c^2 - 4*(A^3*B - A*B^3 + 3*A*B*C^2 - B*C^3 - (3*A^2*B - B ^3)*C)*c*d + (A^4 - 2*A^2*B^2 + B^4 - 4*A*C^3 + C^4 + 2*(3*A^2 - B^2)*C^2 - 4*(A^3 - A*B^2)*C)*d^2)/((c^4 + 2*c^2*d^2 + d^4)*f^4)) + (A^2 - B^2 -...
\[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \]
\[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx=\int { \frac {C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A}{\sqrt {d \tan \left (f x + e\right ) + c}} \,d x } \]
Timed out. \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Timed out} \]
Time = 12.95 (sec) , antiderivative size = 4326, normalized size of antiderivative = 32.53 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]
2*atanh((32*C^2*d^2*((-16*C^4*d^2*f^4)^(1/2)/(16*(c^2*f^4 + d^2*f^4)) - (C ^2*c*f^2)/(4*(c^2*f^4 + d^2*f^4)))^(1/2)*(c + d*tan(e + f*x))^(1/2))/((16* C^3*c*d^3*f^3)/(c^2*f^4 + d^2*f^4) - (4*C*d^3*f^2*(-16*C^4*d^2*f^4)^(1/2)) /(c^2*f^5 + d^2*f^5)) + (8*c*d^2*((-16*C^4*d^2*f^4)^(1/2)/(16*(c^2*f^4 + d ^2*f^4)) - (C^2*c*f^2)/(4*(c^2*f^4 + d^2*f^4)))^(1/2)*(c + d*tan(e + f*x)) ^(1/2)*(-16*C^4*d^2*f^4)^(1/2))/((16*C^3*c*d^5*f^5)/(c^2*f^4 + d^2*f^4) - (4*C*d^5*f^4*(-16*C^4*d^2*f^4)^(1/2))/(c^2*f^5 + d^2*f^5) + (16*C^3*c^3*d^ 3*f^5)/(c^2*f^4 + d^2*f^4) - (4*C*c^2*d^3*f^4*(-16*C^4*d^2*f^4)^(1/2))/(c^ 2*f^5 + d^2*f^5)) - (32*C^2*c^2*d^2*f^2*((-16*C^4*d^2*f^4)^(1/2)/(16*(c^2* f^4 + d^2*f^4)) - (C^2*c*f^2)/(4*(c^2*f^4 + d^2*f^4)))^(1/2)*(c + d*tan(e + f*x))^(1/2))/((16*C^3*c*d^5*f^5)/(c^2*f^4 + d^2*f^4) - (4*C*d^5*f^4*(-16 *C^4*d^2*f^4)^(1/2))/(c^2*f^5 + d^2*f^5) + (16*C^3*c^3*d^3*f^5)/(c^2*f^4 + d^2*f^4) - (4*C*c^2*d^3*f^4*(-16*C^4*d^2*f^4)^(1/2))/(c^2*f^5 + d^2*f^5)) )*((-16*C^4*d^2*f^4)^(1/2)/(16*(c^2*f^4 + d^2*f^4)) - (C^2*c*f^2)/(4*(c^2* f^4 + d^2*f^4)))^(1/2) - 2*atanh((8*c*d^2*(- (-16*C^4*d^2*f^4)^(1/2)/(16*( c^2*f^4 + d^2*f^4)) - (C^2*c*f^2)/(4*(c^2*f^4 + d^2*f^4)))^(1/2)*(c + d*ta n(e + f*x))^(1/2)*(-16*C^4*d^2*f^4)^(1/2))/((16*C^3*c*d^5*f^5)/(c^2*f^4 + d^2*f^4) + (4*C*d^5*f^4*(-16*C^4*d^2*f^4)^(1/2))/(c^2*f^5 + d^2*f^5) + (16 *C^3*c^3*d^3*f^5)/(c^2*f^4 + d^2*f^4) + (4*C*c^2*d^3*f^4*(-16*C^4*d^2*f^4) ^(1/2))/(c^2*f^5 + d^2*f^5)) - (32*C^2*d^2*(- (-16*C^4*d^2*f^4)^(1/2)/(...