Integrand size = 25, antiderivative size = 115 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {i \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}-\frac {i \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d} \]
I*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/(I*a-b)^ (1/2)-I*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/( I*a+b)^(1/2)
Time = 0.10 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {\sqrt [4]{-1} \left (-\frac {\arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {\arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )}{d} \]
((-1)^(1/4)*(-(ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[ a + b*Tan[c + d*x]]]/Sqrt[-a + I*b]) + ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sq rt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/Sqrt[a + I*b]))/d
Time = 0.30 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4058, 613, 104, 218, 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.
\(\displaystyle \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4058 |
\(\displaystyle \frac {\int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 613 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {1}{\sqrt {\tan (c+d x)} (\tan (c+d x)+i) \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)-\frac {1}{2} \int \frac {1}{(i-\tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {\int \frac {1}{\frac {(a-i b) \tan (c+d x)}{a+b \tan (c+d x)}+i}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}-\int \frac {1}{i-\frac {(a+i b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\int \frac {1}{\frac {(a-i b) \tan (c+d x)}{a+b \tan (c+d x)}+i}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}+\frac {i \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {i \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}-\frac {i \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b+i a}}}{d}\) |
((I*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/S qrt[I*a - b] - (I*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Ta n[c + d*x]]])/Sqrt[I*a + b])/d
3.7.35.3.1 Defintions of rubi rules used
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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[Sqrt[(e_.)*(x_)]/(Sqrt[(c_) + (d_.)*(x_)]*((a_) + (b_.)*(x_)^2)), x_Sym bol] :> Simp[e/(2*b) Int[1/(Sqrt[e*x]*Sqrt[c + d*x]*(Rt[-a/b, 2] + x)), x ], x] - Simp[e/(2*b) Int[1/(Sqrt[e*x]*Sqrt[c + d*x]*(Rt[-a/b, 2] - x)), x ], x] /; FreeQ[{a, b, c, d, e}, x]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S imp[ff/f Subst[Int[(a + b*ff*x)^m*((c + d*ff*x)^n/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a *d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 0.61 (sec) , antiderivative size = 939566, normalized size of antiderivative = 8170.14
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 3933 vs. \(2 (87) = 174\).
Time = 0.69 (sec) , antiderivative size = 3933, normalized size of antiderivative = 34.20 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]
-1/8*sqrt(((a^2 + b^2)*d^2*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - b)/( (a^2 + b^2)*d^2))*log((2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a*b^4 )*tan(d*x + c) + (2*(a^5*b + 3*a^3*b^3 + 2*a*b^5)*d^2*tan(d*x + c) - (a^6 + 4*a^4*b^2 + 7*a^2*b^4 + 4*b^6)*d^2)*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d ^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + ((a^6 + 7*a^4*b^2 + 12 *a^2*b^4)*d*tan(d*x + c)^2 + 2*(a^5*b + a^3*b^3 - 4*a*b^5)*d*tan(d*x + c) - (a^6 + 3*a^4*b^2 + 4*a^2*b^4)*d - 2*((a^4*b^3 + 5*a^2*b^5 + 4*b^7)*d^3*t an(d*x + c)^2 + (a^7 + 6*a^5*b^2 + 13*a^3*b^4 + 8*a*b^6)*d^3*tan(d*x + c) + (a^6*b + 3*a^4*b^3 + 2*a^2*b^5)*d^3)*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)* d^4)))*sqrt(((a^2 + b^2)*d^2*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - b) /((a^2 + b^2)*d^2)))/(tan(d*x + c)^2 + 1)) - 1/8*sqrt(((a^2 + b^2)*d^2*sqr t(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - b)/((a^2 + b^2)*d^2))*log(-(2*(2*a ^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x + c) + (2*(a^5*b + 3*a^3*b^3 + 2*a*b^5)*d^2*tan(d*x + c) - (a^6 + 4*a^4*b^2 + 7*a^2*b^4 + 4*b ^6)*d^2)*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(b*tan(d*x + c) + a )*sqrt(tan(d*x + c)) + ((a^6 + 7*a^4*b^2 + 12*a^2*b^4)*d*tan(d*x + c)^2 + 2*(a^5*b + a^3*b^3 - 4*a*b^5)*d*tan(d*x + c) - (a^6 + 3*a^4*b^2 + 4*a^2*b^ 4)*d - 2*((a^4*b^3 + 5*a^2*b^5 + 4*b^7)*d^3*tan(d*x + c)^2 + (a^7 + 6*a^5* b^2 + 13*a^3*b^4 + 8*a*b^6)*d^3*tan(d*x + c) + (a^6*b + 3*a^4*b^3 + 2*a^2* b^5)*d^3)*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))*sqrt(((a^2 + b^2)*d...
\[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {\sqrt {\tan {\left (c + d x \right )}}}{\sqrt {a + b \tan {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {\sqrt {\tan \left (d x + c\right )}}{\sqrt {b \tan \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \]
Time = 51.58 (sec) , antiderivative size = 4102, normalized size of antiderivative = 35.67 \[ \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]
(log(((-1/(d^2*(a*1i + b)))^(1/2)*(((-1/(d^2*(a*1i + b)))^(1/2)*(((-1/(d^2 *(a*1i + b)))^(1/2)*(((-1/(d^2*(a*1i + b)))^(1/2)*(((-1/(d^2*(a*1i + b)))^ (1/2)*(536870912*a^8*b^16*(-1/(d^2*(a*1i + b)))^(1/2)*(8*a^2 + 7*b^2)*(8*a ^2 + 7*b^2 - (17*b^3*tan(c + d*x))/((a + b*tan(c + d*x))^(1/2) - a^(1/2))^ 2 - (16*a^2*b*tan(c + d*x))/((a + b*tan(c + d*x))^(1/2) - a^(1/2))^2) + (1 073741824*a^7*b^17*tan(c + d*x)^(1/2)*(48*a^4 + 5*b^4 + 52*a^2*b^2))/(d*(( a + b*tan(c + d*x))^(1/2) - a^(1/2)))))/2 + (268435456*a^6*b^17*(144*a^4 - b^4 + 112*a^2*b^2))/d^2 + (268435456*a^6*b^16*tan(c + d*x)*(256*a^6 + b^6 - 270*a^2*b^4 - 32*a^4*b^2))/(d^2*((a + b*tan(c + d*x))^(1/2) - a^(1/2))^ 2)))/2 + (536870912*a^7*b^18*tan(c + d*x)^(1/2)*(48*a^2 + 5*b^2))/(d^3*((a + b*tan(c + d*x))^(1/2) - a^(1/2)))))/2 + (134217728*a^6*b^16*(32*a^4 - b ^4 + 40*a^2*b^2))/d^4 + (134217728*a^6*b^17*tan(c + d*x)*(192*a^4 + b^4 - 92*a^2*b^2))/(d^4*((a + b*tan(c + d*x))^(1/2) - a^(1/2))^2)))/2 + (6710886 4*a^7*b^17*tan(c + d*x)^(1/2)*(48*a^2 + 5*b^2))/(d^5*((a + b*tan(c + d*x)) ^(1/2) - a^(1/2)))))/2 + (16777216*a^6*b^17*(16*a^2 - b^2))/d^6 + (1677721 6*a^6*b^16*tan(c + d*x)*(16*a^2 - b^2)^2)/(d^6*((a + b*tan(c + d*x))^(1/2) - a^(1/2))^2))*(-1/(a*d^2*1i + b*d^2))^(1/2))/2 - log(((-1/(d^2*(a*1i + b )))^(1/2)*(((-1/(d^2*(a*1i + b)))^(1/2)*(((-1/(d^2*(a*1i + b)))^(1/2)*(((- 1/(d^2*(a*1i + b)))^(1/2)*(((-1/(d^2*(a*1i + b)))^(1/2)*(536870912*a^8*b^1 6*(-1/(d^2*(a*1i + b)))^(1/2)*(8*a^2 + 7*b^2)*(8*a^2 + 7*b^2 - (17*b^3*...