Integrand size = 49, antiderivative size = 251 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {(B+i (A-C)) \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a-i b} (c-i d)^{3/2} f}+\frac {(i A-B-i C) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b} (c+i d)^{3/2} f}+\frac {2 \left (c^2 C-B c d+A d^2\right ) \sqrt {a+b \tan (e+f x)}}{(b c-a d) \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}} \]
-(B+I*(A-C))*arctanh((c-I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a-I*b)^(1/2)/(c +d*tan(f*x+e))^(1/2))/(c-I*d)^(3/2)/f/(a-I*b)^(1/2)+(I*A-B-I*C)*arctanh((c +I*d)^(1/2)*(a+b*tan(f*x+e))^(1/2)/(a+I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))/( c+I*d)^(3/2)/f/(a+I*b)^(1/2)+2*(A*d^2-B*c*d+C*c^2)*(a+b*tan(f*x+e))^(1/2)/ (-a*d+b*c)/(c^2+d^2)/f/(c+d*tan(f*x+e))^(1/2)
Time = 3.53 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.10 \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {(b c-a d) \left (\frac {(i A+B-i C) (c+i d) \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+i b} \sqrt {-c+i d}}+\frac {(A+i B-C) (i c+d) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b} \sqrt {c+i d}}\right )+\frac {2 \left (c^2 C-B c d+A d^2\right ) \sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{(-b c+a d) \left (c^2+d^2\right ) f} \]
Integrate[(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2)/(Sqrt[a + b*Tan[e + f*x] ]*(c + d*Tan[e + f*x])^(3/2)),x]
-(((b*c - a*d)*(((I*A + B - I*C)*(c + I*d)*ArcTanh[(Sqrt[-c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[-a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[-a + I*b]*Sqrt[-c + I*d]) + ((A + I*B - C)*(I*c + d)*ArcTanh[(Sqrt[c + I*d]*Sqr t[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[a + I*b]*Sqrt[c + I*d])) + (2*(c^2*C - B*c*d + A*d^2)*Sqrt[a + b*Tan[e + f*x ]])/Sqrt[c + d*Tan[e + f*x]])/((-(b*c) + a*d)*(c^2 + d^2)*f))
Time = 1.38 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.20, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {3042, 4132, 27, 3042, 4099, 3042, 4098, 104, 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.
\(\displaystyle \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \tan (e+f x)+C \tan (e+f x)^2}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 4132 |
\(\displaystyle \frac {2 \int \frac {(b c-a d) (A c-C c+B d)+(b c-a d) (B c-(A-C) d) \tan (e+f x)}{2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}+\frac {2 \left (A d^2-B c d+c^2 C\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(b c-a d) (A c-C c+B d)+(b c-a d) (B c-(A-C) d) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}+\frac {2 \left (A d^2-B c d+c^2 C\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(b c-a d) (A c-C c+B d)+(b c-a d) (B c-(A-C) d) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}+\frac {2 \left (A d^2-B c d+c^2 C\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 4099 |
\(\displaystyle \frac {2 \left (A d^2-B c d+c^2 C\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {1}{2} (c-i d) (A+i B-C) (b c-a d) \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (c+i d) (A-i B-C) (b c-a d) \int \frac {i \tan (e+f x)+1}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \left (A d^2-B c d+c^2 C\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {1}{2} (c-i d) (A+i B-C) (b c-a d) \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (c+i d) (A-i B-C) (b c-a d) \int \frac {i \tan (e+f x)+1}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 4098 |
\(\displaystyle \frac {2 \left (A d^2-B c d+c^2 C\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {(c+i d) (A-i B-C) (b c-a d) \int \frac {1}{(1-i \tan (e+f x)) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{2 f}+\frac {(c-i d) (A+i B-C) (b c-a d) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{2 f}}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {2 \left (A d^2-B c d+c^2 C\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {(c-i d) (A+i B-C) (b c-a d) \int \frac {1}{-i a+b+\frac {(i c-d) (a+b \tan (e+f x))}{c+d \tan (e+f x)}}d\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{f}+\frac {(c+i d) (A-i B-C) (b c-a d) \int \frac {1}{i a+b-\frac {(i c+d) (a+b \tan (e+f x))}{c+d \tan (e+f x)}}d\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{f}}{\left (c^2+d^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 \left (A d^2-B c d+c^2 C\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {i (c-i d) (A+i B-C) (b c-a d) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a+i b} \sqrt {c+i d}}-\frac {i (c+i d) (A-i B-C) (b c-a d) \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a-i b} \sqrt {c-i d}}}{\left (c^2+d^2\right ) (b c-a d)}\) |
Int[(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2)/(Sqrt[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2)),x]
(((-I)*(A - I*B - C)*(c + I*d)*(b*c - a*d)*ArcTanh[(Sqrt[c - I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a - I*b]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[a - I*b ]*Sqrt[c - I*d]*f) + (I*(A + I*B - C)*(c - I*d)*(b*c - a*d)*ArcTanh[(Sqrt[ c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*Tan[e + f*x]] )])/(Sqrt[a + I*b]*Sqrt[c + I*d]*f))/((b*c - a*d)*(c^2 + d^2)) + (2*(c^2*C - B*c*d + A*d^2)*Sqrt[a + b*Tan[e + f*x]])/((b*c - a*d)*(c^2 + d^2)*f*Sqr t[c + d*Tan[e + f*x]])
3.2.56.3.1 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[(((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)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* (m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d )*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ [b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
Timed out.
\[\int \frac {A +B \tan \left (f x +e \right )+C \tan \left (f x +e \right )^{2}}{\sqrt {a +b \tan \left (f x +e \right )}\, \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]
Timed out. \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]
integrate((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^(1/2)/(c+d*tan( f*x+e))^(3/2),x, algorithm="fricas")
\[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}}{\sqrt {a + b \tan {\left (e + f x \right )}} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
integrate((A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(a+b*tan(f*x+e))**(1/2)/(c+d*ta n(f*x+e))**(3/2),x)
Integral((A + B*tan(e + f*x) + C*tan(e + f*x)**2)/(sqrt(a + b*tan(e + f*x) )*(c + d*tan(e + f*x))**(3/2)), x)
\[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=\int { \frac {C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A}{\sqrt {b \tan \left (f x + e\right ) + a} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integrate((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^(1/2)/(c+d*tan( f*x+e))^(3/2),x, algorithm="maxima")
integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)/(sqrt(b*tan(f*x + e) + a )*(d*tan(f*x + e) + c)^(3/2)), x)
Timed out. \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]
integrate((A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^(1/2)/(c+d*tan( f*x+e))^(3/2),x, algorithm="giac")
Timed out. \[ \int \frac {A+B \tan (e+f x)+C \tan ^2(e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Hanged} \]