Integrand size = 35, antiderivative size = 206 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=-\frac {(A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}+\frac {(2 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{3/2} d}-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d} \]
(2*A*b-B*a)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/b^(3/ 2)/d-(A+I*B)*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)-(A-I*B)*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)+B*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/b/ d
Time = 2.01 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.19 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {\frac {\sqrt [4]{-1} b (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {(-1)^{3/4} b (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+\frac {\sqrt {a} (2 A b-a B) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}}{b d} \]
(((-1)^(1/4)*b*(I*A + B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d* x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] + ((-1)^(3/4)*b*(A + I*B)*A rcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x ]]])/Sqrt[a + I*b] + B*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + (Sqrt [a]*(2*A*b - a*B)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + ( b*Tan[c + d*x])/a])/(Sqrt[b]*Sqrt[a + b*Tan[c + d*x]]))/(b*d)
Time = 0.95 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3042, 4090, 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.
\(\displaystyle \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^{3/2} (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4090 |
\(\displaystyle \frac {\int -\frac {-\left ((2 A b-a B) \tan ^2(c+d x)\right )+2 b B \tan (c+d x)+a B}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {-\left ((2 A b-a B) \tan ^2(c+d x)\right )+2 b B \tan (c+d x)+a B}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{2 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {-\left ((2 A b-a B) \tan (c+d x)^2\right )+2 b B \tan (c+d x)+a B}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{2 b}\) |
\(\Big \downarrow \) 4138 |
\(\displaystyle \frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {-\left ((2 A b-a B) \tan ^2(c+d x)\right )+2 b B \tan (c+d x)+a B}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {-\left ((2 A b-a B) \tan ^2(c+d x)\right )+2 b B \tan (c+d x)+a B}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{b d}\) |
\(\Big \downarrow \) 2257 |
\(\displaystyle \frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \left (\frac {a B-2 A b}{\sqrt {a+b \tan (c+d x)}}+\frac {2 (A b+B \tan (c+d x) b)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{b d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {B \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\frac {b (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}-\frac {(2 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}+\frac {b (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b+i a}}}{b d}\) |
-(((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] - ((2*A*b - a*B)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d* x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[b] + (b*(A - I*B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[I*a + b])/(b*d)) + (B*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/(b*d)
3.5.52.3.1 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*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Ta n[e + f*x])^n*Simp[a^2*A*d*(m + n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m - 1) - b *(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2 , 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
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 = 0.95 (sec) , antiderivative size = 1889462, normalized size of antiderivative = 9172.15
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 10163 vs. \(2 (166) = 332\).
Time = 4.85 (sec) , antiderivative size = 20328, normalized size of antiderivative = 98.68 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]
\[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\sqrt {a + b \tan {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{\frac {3}{2}}}{\sqrt {b \tan \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \]