Integrand size = 35, antiderivative size = 169 \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {\sqrt {i a-b} (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {2 \sqrt {b} B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\sqrt {i a+b} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]
-(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*(I* a-b)^(1/2)/d+2*B*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))* b^(1/2)/d-(I*A+B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^ (1/2))*(I*a+b)^(1/2)/d
Time = 1.10 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {\sqrt [4]{-1} \left (\sqrt {-a+i b} (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\sqrt {a+i b} (-i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )+\frac {2 \sqrt {a} \sqrt {b} B \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}}{d} \]
((-1)^(1/4)*(Sqrt[-a + I*b]*(I*A + 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]*((-I)*A + B)*A rcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x ]]]) + (2*Sqrt[a]*Sqrt[b]*B*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]* Sqrt[1 + (b*Tan[c + d*x])/a])/Sqrt[a + b*Tan[c + d*x]])/d
Time = 1.18 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 4097, 3042, 4099, 3042, 4098, 104, 216, 219, 4117, 65, 219}
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.
| \(\displaystyle \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4097 |
| \(\displaystyle \int \frac {a A-b B+(A b+a B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+b B \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a A-b B+(A b+a B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+b B \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4099 |
| \(\displaystyle \frac {1}{2} (a+i b) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a-i b) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+b B \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} (a+i b) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a-i b) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+b B \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4098 |
| \(\displaystyle \frac {(a-i b) (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}+\frac {(a+i b) (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}+b B \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {(a+i b) (A+i B) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {(a-i b) (A-i B) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+b B \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {(a-i b) (A-i B) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+b B \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {(a+i b) (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle b B \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {(a+i b) (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {(a-i b) (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}\) |
| \(\Big \downarrow \) 4117 |
| \(\displaystyle \frac {b B \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{d}+\frac {(a+i b) (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {(a-i b) (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}\) |
| \(\Big \downarrow \) 65 |
| \(\displaystyle \frac {2 b B \int \frac {1}{1-\frac {b \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {(a+i b) (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {(a-i b) (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(a+i b) (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {(a-i b) (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}+\frac {2 \sqrt {b} B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\) |
((a + I*b)*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b* Tan[c + d*x]]])/(Sqrt[I*a - b]*d) + (2*Sqrt[b]*B*ArcTanh[(Sqrt[b]*Sqrt[Tan [c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + ((a - I*b)*(A - I*B)*ArcTanh[(S qrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b] *d)
3.5.29.3.1 Defintions of rubi rules used
Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2 Sub st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d }, x] && !GtQ[c, 0]
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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(Sqrt[(a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> In t[Simp[a*A - b*B + (A*b + a*B)*Tan[e + f*x], x]/(Sqrt[a + b*Tan[e + f*x]]*S qrt[c + d*Tan[e + f*x]]), x] + Simp[b*B Int[(1 + Tan[e + f*x]^2)/(Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e , f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[A^2/f Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e + f* x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 + B^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(A + I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T an[e + f*x]), x], x] + Simp[(A - I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d *Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A , B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2 + B^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A/f Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 2.28 (sec) , antiderivative size = 2178123, normalized size of antiderivative = 12888.30
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 7969 vs. \(2 (130) = 260\).
Time = 2.38 (sec) , antiderivative size = 15940, normalized size of antiderivative = 94.32 \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\text {Too large to display} \]
\[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {a + b \tan {\left (c + d x \right )}}}{\sqrt {\tan {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {b \tan \left (d x + c\right ) + a}}{\sqrt {\tan \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\text {Timed out} \]
Time = 20.29 (sec) , antiderivative size = 1141, normalized size of antiderivative = 6.75 \[ \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\text {Too large to display} \]
atanh((a^(3/2)*d*tan(c + d*x)^(1/2)*(((-A^4*a^2*d^4)^(1/2) - A^2*b*d^2)/d^ 4)^(1/2)*(-A^4*a^2*d^4)^(1/2) - a*d*tan(c + d*x)^(1/2)*(((-A^4*a^2*d^4)^(1 /2) - A^2*b*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)*(-A^4*a^2*d^4)^(1/2 ) + A^2*a^(3/2)*b*d^3*tan(c + d*x)^(1/2)*(((-A^4*a^2*d^4)^(1/2) - A^2*b*d^ 2)/d^4)^(1/2) - A^2*a*b*d^3*tan(c + d*x)^(1/2)*(((-A^4*a^2*d^4)^(1/2) - A^ 2*b*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2))/(A^3*a^3*d^2 - A*a*b*(-A^4 *a^2*d^4)^(1/2) - A*b^2*tan(c + d*x)*(-A^4*a^2*d^4)^(1/2) - A^3*a^(5/2)*d^ 2*(a + b*tan(c + d*x))^(1/2) + A^3*a^2*b*d^2*tan(c + d*x) + A*a^(1/2)*b*(a + b*tan(c + d*x))^(1/2)*(-A^4*a^2*d^4)^(1/2)))*(((-A^4*a^2*d^4)^(1/2) - A ^2*b*d^2)/d^4)^(1/2) - atanh((a^(3/2)*d*tan(c + d*x)^(1/2)*(-((-A^4*a^2*d^ 4)^(1/2) + A^2*b*d^2)/d^4)^(1/2)*(-A^4*a^2*d^4)^(1/2) - a*d*tan(c + d*x)^( 1/2)*(-((-A^4*a^2*d^4)^(1/2) + A^2*b*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^ (1/2)*(-A^4*a^2*d^4)^(1/2) - A^2*a^(3/2)*b*d^3*tan(c + d*x)^(1/2)*(-((-A^4 *a^2*d^4)^(1/2) + A^2*b*d^2)/d^4)^(1/2) + A^2*a*b*d^3*tan(c + d*x)^(1/2)*( -((-A^4*a^2*d^4)^(1/2) + A^2*b*d^2)/d^4)^(1/2)*(a + b*tan(c + d*x))^(1/2)) /(A^3*a^3*d^2 + A*a*b*(-A^4*a^2*d^4)^(1/2) + A*b^2*tan(c + d*x)*(-A^4*a^2* d^4)^(1/2) - A^3*a^(5/2)*d^2*(a + b*tan(c + d*x))^(1/2) + A^3*a^2*b*d^2*ta n(c + d*x) - A*a^(1/2)*b*(a + b*tan(c + d*x))^(1/2)*(-A^4*a^2*d^4)^(1/2))) *(-((-A^4*a^2*d^4)^(1/2) + A^2*b*d^2)/d^4)^(1/2) + atanh((2*((a^(1/2)*d*ta n(c + d*x)^(1/2)*(((-B^4*a^2*d^4)^(1/2) + B^2*b*d^2)/d^4)^(1/2))/2 - (d...