Integrand size = 21, antiderivative size = 106 \[ \int \tan (c+d x) \sqrt {a+b \tan (c+d x)} \, dx=-\frac {\sqrt {a-i b} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}-\frac {\sqrt {a+i b} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 \sqrt {a+b \tan (c+d x)}}{d} \]
-arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))*(a-I*b)^(1/2)/d-arctanh((a+ b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))*(a+I*b)^(1/2)/d+2*(a+b*tan(d*x+c))^(1/2 )/d
Time = 0.12 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.94 \[ \int \tan (c+d x) \sqrt {a+b \tan (c+d x)} \, dx=-\frac {\sqrt {a-i b} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+\sqrt {a+i b} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )-2 \sqrt {a+b \tan (c+d x)}}{d} \]
-((Sqrt[a - I*b]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] + Sqrt[a + I*b]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] - 2*Sqrt[a + b*Tan[ c + d*x]])/d)
Time = 0.55 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {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.
\(\displaystyle \int \tan (c+d x) \sqrt {a+b \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x) \sqrt {a+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \frac {a \tan (c+d x)-b}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a \tan (c+d x)-b}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 4022 |
\(\displaystyle \frac {1}{2} (-b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} (-b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle -\frac {i (b+i a) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (-b+i a) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {i (b+i a) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i (-b+i a) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(-b+i a) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}-\frac {(b+i a) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {(b+i a) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {(-b+i a) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}+\frac {2 \sqrt {a+b \tan (c+d x)}}{d}\) |
-(((I*a + b)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d)) + ((I* a - b)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d) + (2*Sqrt[a + b*Tan[c + d*x]])/d
3.6.6.3.1 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]
Leaf count of result is larger than twice the leaf count of optimal. \(324\) vs. \(2(88)=176\).
Time = 1.40 (sec) , antiderivative size = 325, normalized size of antiderivative = 3.07
method | result | size |
derivativedivides | \(\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a -\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4}+\frac {\left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4}+\frac {\left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{d}\) | \(325\) |
default | \(\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a -\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4}+\frac {\left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4}+\frac {\left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{d}\) | \(325\) |
1/d*(2*(a+b*tan(d*x+c))^(1/2)+1/4*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*ln(b*tan(d *x+c)+a-(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/ 2))+(a-(a^2+b^2)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d *x+c))^(1/2)-(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)) -1/4*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2 )*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))+(a-(a^2+b^2)^(1/2))/(2*(a ^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/ 2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (84) = 168\).
Time = 0.24 (sec) , antiderivative size = 284, normalized size of antiderivative = 2.68 \[ \int \tan (c+d x) \sqrt {a+b \tan (c+d x)} \, dx=-\frac {d \sqrt {\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} + a}{d^{2}}} \log \left (d \sqrt {\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} + a}{d^{2}}} + \sqrt {b \tan \left (d x + c\right ) + a}\right ) - d \sqrt {\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} + a}{d^{2}}} \log \left (-d \sqrt {\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} + a}{d^{2}}} + \sqrt {b \tan \left (d x + c\right ) + a}\right ) + d \sqrt {-\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} - a}{d^{2}}} \log \left (d \sqrt {-\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} - a}{d^{2}}} + \sqrt {b \tan \left (d x + c\right ) + a}\right ) - d \sqrt {-\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} - a}{d^{2}}} \log \left (-d \sqrt {-\frac {d^{2} \sqrt {-\frac {b^{2}}{d^{4}}} - a}{d^{2}}} + \sqrt {b \tan \left (d x + c\right ) + a}\right ) - 4 \, \sqrt {b \tan \left (d x + c\right ) + a}}{2 \, d} \]
-1/2*(d*sqrt((d^2*sqrt(-b^2/d^4) + a)/d^2)*log(d*sqrt((d^2*sqrt(-b^2/d^4) + a)/d^2) + sqrt(b*tan(d*x + c) + a)) - d*sqrt((d^2*sqrt(-b^2/d^4) + a)/d^ 2)*log(-d*sqrt((d^2*sqrt(-b^2/d^4) + a)/d^2) + sqrt(b*tan(d*x + c) + a)) + d*sqrt(-(d^2*sqrt(-b^2/d^4) - a)/d^2)*log(d*sqrt(-(d^2*sqrt(-b^2/d^4) - a )/d^2) + sqrt(b*tan(d*x + c) + a)) - d*sqrt(-(d^2*sqrt(-b^2/d^4) - a)/d^2) *log(-d*sqrt(-(d^2*sqrt(-b^2/d^4) - a)/d^2) + sqrt(b*tan(d*x + c) + a)) - 4*sqrt(b*tan(d*x + c) + a))/d
\[ \int \tan (c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\int \sqrt {a + b \tan {\left (c + d x \right )}} \tan {\left (c + d x \right )}\, dx \]
Exception generated. \[ \int \tan (c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
Timed out. \[ \int \tan (c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\text {Timed out} \]
Time = 5.91 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.74 \[ \int \tan (c+d x) \sqrt {a+b \tan (c+d x)} \, dx=\frac {2\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d}-\mathrm {atan}\left (\frac {b^4\,\sqrt {\frac {a}{4\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {b^5\,16{}\mathrm {i}}{d}+\frac {a^2\,b^3\,16{}\mathrm {i}}{d}}+\frac {32\,a\,b^3\,\sqrt {\frac {a}{4\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {b^5\,16{}\mathrm {i}}{d}+\frac {a^2\,b^3\,16{}\mathrm {i}}{d}}\right )\,\sqrt {\frac {a-b\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {b^4\,\sqrt {\frac {a}{4\,d^2}+\frac {b\,1{}\mathrm {i}}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {b^5\,16{}\mathrm {i}}{d}+\frac {a^2\,b^3\,16{}\mathrm {i}}{d}}-\frac {32\,a\,b^3\,\sqrt {\frac {a}{4\,d^2}+\frac {b\,1{}\mathrm {i}}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {b^5\,16{}\mathrm {i}}{d}+\frac {a^2\,b^3\,16{}\mathrm {i}}{d}}\right )\,\sqrt {\frac {a+b\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i} \]
(2*(a + b*tan(c + d*x))^(1/2))/d - atan((b^4*(a/(4*d^2) - (b*1i)/(4*d^2))^ (1/2)*(a + b*tan(c + d*x))^(1/2)*32i)/((b^5*16i)/d + (a^2*b^3*16i)/d) + (3 2*a*b^3*(a/(4*d^2) - (b*1i)/(4*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((b ^5*16i)/d + (a^2*b^3*16i)/d))*((a - b*1i)/(4*d^2))^(1/2)*2i + atan((b^4*(a /(4*d^2) + (b*1i)/(4*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2)*32i)/((b^5*16i )/d + (a^2*b^3*16i)/d) - (32*a*b^3*(a/(4*d^2) + (b*1i)/(4*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((b^5*16i)/d + (a^2*b^3*16i)/d))*((a + b*1i)/(4*d^ 2))^(1/2)*2i