Integrand size = 35, antiderivative size = 201 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {i a-b} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(2 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b} d}-\frac {\sqrt {i a+b} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d} \]
(A+I*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*A*b+B*a)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^ (1/2))/d/b^(1/2)-(A-I*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+B*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/d
Time = 2.97 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.19 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt [4]{-1} \sqrt {-a+i b} (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\sqrt [4]{-1} \sqrt {a+i b} (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+\frac {(2 A b+a B) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {a+b \tan (c+d x)}}{\sqrt {a} \sqrt {b} \sqrt {1+\frac {b \tan (c+d x)}{a}}}}{d} \]
((-1)^(1/4)*Sqrt[-a + I*b]*(A - I*B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqr t[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + (-1)^(1/4)*Sqrt[a + I*b]*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[ c + d*x]]] + B*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + ((2*A*b + a*B )*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[a + b*Tan[c + d*x]])/ (Sqrt[a]*Sqrt[b]*Sqrt[1 + (b*Tan[c + d*x])/a]))/d
Time = 0.95 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3042, 4093, 27, 3042, 4138, 2035, 2257, 2009}
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.
| \(\displaystyle \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))dx\) |
| \(\Big \downarrow \) 4093 |
| \(\displaystyle \int -\frac {-\left ((2 A b+a B) \tan ^2(c+d x)\right )-2 (a A-b B) \tan (c+d x)+a B}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {1}{2} \int \frac {-\left ((2 A b+a B) \tan ^2(c+d x)\right )-2 (a A-b B) \tan (c+d x)+a B}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {1}{2} \int \frac {-\left ((2 A b+a B) \tan (c+d x)^2\right )-2 (a A-b B) \tan (c+d x)+a B}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4138 |
| \(\displaystyle \frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \frac {-\left ((2 A b+a B) \tan ^2(c+d x)\right )-2 (a A-b B) \tan (c+d x)+a B}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \frac {-\left ((2 A b+a B) \tan ^2(c+d x)\right )-2 (a A-b B) \tan (c+d x)+a B}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 2257 |
| \(\displaystyle \frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \left (\frac {-2 A b-a B}{\sqrt {a+b \tan (c+d x)}}+\frac {2 (A b+a B-(a A-b B) \tan (c+d x))}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {-\sqrt {-b+i a} (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {(a B+2 A b) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}+\sqrt {b+i a} (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\) |
-((-(Sqrt[I*a - b]*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqr t[a + b*Tan[c + d*x]]]) - ((2*A*b + a*B)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x ]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[b] + Sqrt[I*a + b]*(A - I*B)*ArcTanh[( Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d) + (B*Sqrt[ Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/d
3.5.28.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol ] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a , c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
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[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp [1/(m + n) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n - 1)* Simp[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (A*b*c + a*B*c + a*A*d - b*B*d)*(m + n)*Tan[e + f*x] + (A*b*d*(m + n) + B*(a*d*m + b*c*n))*Tan[e + f*x]^2, 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] && LtQ[0, m, 1] && LtQ[0, n, 1]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), 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*((A + B*ff*x + C*ff^2*x^ 2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f , A, B, C, 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 = 1.43 (sec) , antiderivative size = 2180698, normalized size of antiderivative = 10849.24
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 7955 vs. \(2 (161) = 322\).
Time = 3.37 (sec) , antiderivative size = 15912, normalized size of antiderivative = 79.16 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
\[ \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {a + b \tan {\left (c + d x \right )}} \sqrt {\tan {\left (c + d x \right )}}\, dx \]
\[ \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {b \tan \left (d x + c\right ) + a} \sqrt {\tan \left (d x + c\right )} \,d x } \]
\[ \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {b \tan \left (d x + c\right ) + a} \sqrt {\tan \left (d x + c\right )} \,d x } \]
Time = 132.15 (sec) , antiderivative size = 61200, normalized size of antiderivative = 304.48 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]
((2*B*a*tan(c + d*x)^(1/2))/((a + b*tan(c + d*x))^(1/2) - a^(1/2)) + (2*B* a*b*tan(c + d*x)^(3/2))/((a + b*tan(c + d*x))^(1/2) - a^(1/2))^3)/(d + (b^ 2*d*tan(c + d*x)^2)/((a + b*tan(c + d*x))^(1/2) - a^(1/2))^4 - (2*b*d*tan( c + d*x))/((a + b*tan(c + d*x))^(1/2) - a^(1/2))^2) - atan(((((A^2*b - A^2 *a*1i + B^2*a*1i - B^2*b + 2*A*B*a + A*B*b*2i)/(4*d^2))^(1/2)*((((A^2*b - A^2*a*1i + B^2*a*1i - B^2*b + 2*A*B*a + A*B*b*2i)/(4*d^2))^(1/2)*((((A^2*b - A^2*a*1i + B^2*a*1i - B^2*b + 2*A*B*a + A*B*b*2i)/(4*d^2))^(1/2)*((((27 4877906944*(1600*a^12*b^34*d^8 - 16640*a^14*b^32*d^8 + 22784*a^16*b^30*d^8 + 106496*a^18*b^28*d^8 + 65536*a^20*b^26*d^8))/d^8 - (274877906944*tan(c + d*x)*(1600*a^12*b^35*d^8 - 48000*a^14*b^33*d^8 + 155136*a^16*b^31*d^8 + 466944*a^18*b^29*d^8 + 262144*a^20*b^27*d^8))/(d^8*((a + b*tan(c + d*x))^( 1/2) - a^(1/2))^2))*((A^2*b - A^2*a*1i + B^2*a*1i - B^2*b + 2*A*B*a + A*B* b*2i)/(4*d^2))^(1/2) - (2199023255552*tan(c + d*x)^(1/2)*(2048*A*a^20*b^27 *d^6 - 12536*A*a^16*b^31*d^6 - 3328*A*a^18*b^29*d^6 - 7160*A*a^14*b^33*d^6 + 240*B*a^13*b^34*d^6 - 720*B*a^15*b^32*d^6 + 5696*B*a^17*b^30*d^6 + 1484 8*B*a^19*b^28*d^6 + 8192*B*a^21*b^26*d^6))/(d^7*((a + b*tan(c + d*x))^(1/2 ) - a^(1/2))))*((A^2*b - A^2*a*1i + B^2*a*1i - B^2*b + 2*A*B*a + A*B*b*2i) /(4*d^2))^(1/2) - (274877906944*(1200*A^2*a^12*b^35*d^6 - 1600*A^2*a^14*b^ 33*d^6 + 272464*A^2*a^16*b^31*d^6 + 573952*A^2*a^18*b^29*d^6 + 299008*A^2* a^20*b^27*d^6 - 1440*B^2*a^12*b^35*d^6 + 8352*B^2*a^14*b^33*d^6 - 320*B...