Integrand size = 25, antiderivative size = 102 \[ \int \frac {a+b \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {(i a+b) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d} f}+\frac {(i a-b) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d} f} \] Output:
-(I*a+b)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(1/2)/f+(I* a-b)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/(c+I*d)^(1/2)/f
Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.99 \[ \int \frac {a+b \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {i \left (-\frac {(a-i b) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {(a+i b) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}\right )}{f} \] Input:
Integrate[(a + b*Tan[e + f*x])/Sqrt[c + d*Tan[e + f*x]],x]
Output:
(I*(-(((a - I*b)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/Sqrt[c - I*d]) + ((a + I*b)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/Sqrt[ c + I*d]))/f
Time = 0.44 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {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.
\(\displaystyle \int \frac {a+b \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {1}{2} (a+i b) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} (a+i b) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a-i b) \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {i (a-i b) \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}-\frac {i (a+i b) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {i (a+i b) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{2 f}-\frac {i (a-i b) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a+i b) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {(a-i b) \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(a-i b) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {(a+i b) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f \sqrt {c+i d}}\) |
Input:
Int[(a + b*Tan[e + f*x])/Sqrt[c + d*Tan[e + f*x]],x]
Output:
((a - I*b)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/(Sqrt[c - I*d]*f) + ((a + I *b)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/(Sqrt[c + I*d]*f)
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]
Leaf count of result is larger than twice the leaf count of optimal. \(1891\) vs. \(2(84)=168\).
Time = 0.50 (sec) , antiderivative size = 1892, normalized size of antiderivative = 18.55
method | result | size |
parts | \(\text {Expression too large to display}\) | \(1892\) |
derivativedivides | \(\text {Expression too large to display}\) | \(3976\) |
default | \(\text {Expression too large to display}\) | \(3976\) |
Input:
int((a+b*tan(f*x+e))/(c+d*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
Output:
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)*l...
Leaf count of result is larger than twice the leaf count of optimal. 1669 vs. \(2 (79) = 158\).
Time = 0.12 (sec) , antiderivative size = 1669, normalized size of antiderivative = 16.36 \[ \int \frac {a+b \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:
integrate((a+b*tan(f*x+e))/(c+d*tan(f*x+e))^(1/2),x, algorithm="fricas")
Output:
-1/2*sqrt(-((c^2 + d^2)*f^2*sqrt(-(4*a^2*b^2*c^2 - 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2)/((c^4 + 2*c^2*d^2 + d^4)*f^4)) + 2*a*b*d + ( a^2 - b^2)*c)/((c^2 + d^2)*f^2))*log((2*(a^3*b + a*b^3)*c - (a^4 - b^4)*d) *sqrt(d*tan(f*x + e) + c) + ((a*c^3 + b*c^2*d + a*c*d^2 + b*d^3)*f^3*sqrt( -(4*a^2*b^2*c^2 - 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2)/((c ^4 + 2*c^2*d^2 + d^4)*f^4)) + (2*a*b^2*c^2 - (3*a^2*b - b^3)*c*d + (a^3 - a*b^2)*d^2)*f)*sqrt(-((c^2 + d^2)*f^2*sqrt(-(4*a^2*b^2*c^2 - 4*(a^3*b - a* b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2)/((c^4 + 2*c^2*d^2 + d^4)*f^4)) + 2 *a*b*d + (a^2 - b^2)*c)/((c^2 + d^2)*f^2))) + 1/2*sqrt(-((c^2 + d^2)*f^2*s qrt(-(4*a^2*b^2*c^2 - 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2) /((c^4 + 2*c^2*d^2 + d^4)*f^4)) + 2*a*b*d + (a^2 - b^2)*c)/((c^2 + d^2)*f^ 2))*log((2*(a^3*b + a*b^3)*c - (a^4 - b^4)*d)*sqrt(d*tan(f*x + e) + c) - ( (a*c^3 + b*c^2*d + a*c*d^2 + b*d^3)*f^3*sqrt(-(4*a^2*b^2*c^2 - 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2)/((c^4 + 2*c^2*d^2 + d^4)*f^4)) + (2*a*b^2*c^2 - (3*a^2*b - b^3)*c*d + (a^3 - a*b^2)*d^2)*f)*sqrt(-((c^2 + d^2)*f^2*sqrt(-(4*a^2*b^2*c^2 - 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2)/((c^4 + 2*c^2*d^2 + d^4)*f^4)) + 2*a*b*d + (a^2 - b^2)*c)/((c^2 + d^2)*f^2))) + 1/2*sqrt(((c^2 + d^2)*f^2*sqrt(-(4*a^2*b^2*c^2 - 4*(a^3*b - a*b^3)*c*d + (a^4 - 2*a^2*b^2 + b^4)*d^2)/((c^4 + 2*c^2*d^2 + d^4)*f^4) ) - 2*a*b*d - (a^2 - b^2)*c)/((c^2 + d^2)*f^2))*log((2*(a^3*b + a*b^3)*...
\[ \int \frac {a+b \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {a + b \tan {\left (e + f x \right )}}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \] Input:
integrate((a+b*tan(f*x+e))/(c+d*tan(f*x+e))**(1/2),x)
Output:
Integral((a + b*tan(e + f*x))/sqrt(c + d*tan(e + f*x)), x)
Exception generated. \[ \int \frac {a+b \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((a+b*tan(f*x+e))/(c+d*tan(f*x+e))^(1/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(d-c>0)', see `assume?` for more details)Is
Timed out. \[ \int \frac {a+b \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Timed out} \] Input:
integrate((a+b*tan(f*x+e))/(c+d*tan(f*x+e))^(1/2),x, algorithm="giac")
Output:
Timed out
Time = 4.24 (sec) , antiderivative size = 2909, normalized size of antiderivative = 28.52 \[ \int \frac {a+b \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:
int((a + b*tan(e + f*x))/(c + d*tan(e + f*x))^(1/2),x)
Output:
2*atanh((8*c*d^2*(- (-16*a^4*d^2*f^4)^(1/2)/(16*(c^2*f^4 + d^2*f^4)) - (a^ 2*c*f^2)/(4*(c^2*f^4 + d^2*f^4)))^(1/2)*(c + d*tan(e + f*x))^(1/2)*(-16*a^ 4*d^2*f^4)^(1/2))/((16*a^3*c*d^5*f^5)/(c^2*f^4 + d^2*f^4) + (4*a*d^5*f^4*( -16*a^4*d^2*f^4)^(1/2))/(c^2*f^5 + d^2*f^5) + (16*a^3*c^3*d^3*f^5)/(c^2*f^ 4 + d^2*f^4) + (4*a*c^2*d^3*f^4*(-16*a^4*d^2*f^4)^(1/2))/(c^2*f^5 + d^2*f^ 5)) - (32*a^2*d^2*(- (-16*a^4*d^2*f^4)^(1/2)/(16*(c^2*f^4 + d^2*f^4)) - (a ^2*c*f^2)/(4*(c^2*f^4 + d^2*f^4)))^(1/2)*(c + d*tan(e + f*x))^(1/2))/((16* a^3*c*d^3*f^3)/(c^2*f^4 + d^2*f^4) + (4*a*d^3*f^2*(-16*a^4*d^2*f^4)^(1/2)) /(c^2*f^5 + d^2*f^5)) + (32*a^2*c^2*d^2*f^2*(- (-16*a^4*d^2*f^4)^(1/2)/(16 *(c^2*f^4 + d^2*f^4)) - (a^2*c*f^2)/(4*(c^2*f^4 + d^2*f^4)))^(1/2)*(c + d* tan(e + f*x))^(1/2))/((16*a^3*c*d^5*f^5)/(c^2*f^4 + d^2*f^4) + (4*a*d^5*f^ 4*(-16*a^4*d^2*f^4)^(1/2))/(c^2*f^5 + d^2*f^5) + (16*a^3*c^3*d^3*f^5)/(c^2 *f^4 + d^2*f^4) + (4*a*c^2*d^3*f^4*(-16*a^4*d^2*f^4)^(1/2))/(c^2*f^5 + d^2 *f^5)))*(- (-16*a^4*d^2*f^4)^(1/2)/(16*(c^2*f^4 + d^2*f^4)) - (a^2*c*f^2)/ (4*(c^2*f^4 + d^2*f^4)))^(1/2) - 2*atanh((32*a^2*d^2*((-16*a^4*d^2*f^4)^(1 /2)/(16*(c^2*f^4 + d^2*f^4)) - (a^2*c*f^2)/(4*(c^2*f^4 + d^2*f^4)))^(1/2)* (c + d*tan(e + f*x))^(1/2))/((16*a^3*c*d^3*f^3)/(c^2*f^4 + d^2*f^4) - (4*a *d^3*f^2*(-16*a^4*d^2*f^4)^(1/2))/(c^2*f^5 + d^2*f^5)) + (8*c*d^2*((-16*a^ 4*d^2*f^4)^(1/2)/(16*(c^2*f^4 + d^2*f^4)) - (a^2*c*f^2)/(4*(c^2*f^4 + d^2* f^4)))^(1/2)*(c + d*tan(e + f*x))^(1/2)*(-16*a^4*d^2*f^4)^(1/2))/((16*a...
\[ \int \frac {a+b \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {2 \sqrt {d \tan \left (f x +e \right )+c}\, a -\left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}}{d \tan \left (f x +e \right )+c}d x \right ) a d f +\left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )}{d \tan \left (f x +e \right )+c}d x \right ) b d f}{d f} \] Input:
int((a+b*tan(f*x+e))/(c+d*tan(f*x+e))^(1/2),x)
Output:
(2*sqrt(tan(e + f*x)*d + c)*a - int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x) **2)/(tan(e + f*x)*d + c),x)*a*d*f + int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)*d + c),x)*b*d*f)/(d*f)