Integrand size = 29, antiderivative size = 301 \[ \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{3/2} (c-i d)^{3/2} f}+\frac {i \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2} (c+i d)^{3/2} f}-\frac {2 b^2}{\left (a^2+b^2\right ) (b c-a d) f \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{\left (a^2+b^2\right ) (b c-a d)^2 \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}} \]
-I*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))/(a-I*b)^(3/2)/(c-I*d)^(3/2)/f+I*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))/(a+I*b)^(3/2)/(c+I*d) ^(3/2)/f-2*b^2/(a^2+b^2)/(-a*d+b*c)/f/(a+b*tan(f*x+e))^(1/2)/(c+d*tan(f*x+ e))^(1/2)-2*d*(a^2*d^2+b^2*(c^2+2*d^2))*(a+b*tan(f*x+e))^(1/2)/(a^2+b^2)/( -a*d+b*c)^2/(c^2+d^2)/f/(c+d*tan(f*x+e))^(1/2)
Time = 5.11 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {\frac {(b c-a d)^2 \left (\frac {i (a+i b) (c+i d) \text {arctanh}\left (\frac {\sqrt {-c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+i b} \sqrt {-c+i d}}+\frac {(i a+b) (c-i d) \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b} \sqrt {c+i d}}\right )}{(-b c+a d) \left (c^2+d^2\right )}+\frac {2 b^2}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{(b c-a d) \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{\left (a^2+b^2\right ) (b c-a d) f} \]
-((((b*c - a*d)^2*((I*(a + I*b)*(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]) + ((I*a + b)*(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])))/((-(b*c) + a*d)*(c^2 + d^2)) + (2*b^2)/(Sqrt[a + b*Tan[ e + f*x]]*Sqrt[c + d*Tan[e + f*x]]) + (2*d*(a^2*d^2 + b^2*(c^2 + 2*d^2))*S qrt[a + b*Tan[e + f*x]])/((b*c - a*d)*(c^2 + d^2)*Sqrt[c + d*Tan[e + f*x]] ))/((a^2 + b^2)*(b*c - a*d)*f))
Time = 1.80 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.27, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 4052, 27, 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.
\(\displaystyle \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 4052 |
\(\displaystyle -\frac {2 \int -\frac {-d a^2+b c a-2 b^2 d \tan ^2(e+f x)-2 b^2 d-b (b c-a d) \tan (e+f x)}{2 \sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}dx}{\left (a^2+b^2\right ) (b c-a d)}-\frac {2 b^2}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {-d a^2+b c a-2 b^2 d \tan ^2(e+f x)-2 b^2 d-b (b c-a d) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}dx}{\left (a^2+b^2\right ) (b c-a d)}-\frac {2 b^2}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {-d a^2+b c a-2 b^2 d \tan (e+f x)^2-2 b^2 d-b (b c-a d) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}dx}{\left (a^2+b^2\right ) (b c-a d)}-\frac {2 b^2}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 4132 |
\(\displaystyle \frac {\frac {2 \int \frac {(b c-a d)^2 (a c-b d)-(b c-a d)^2 (b c+a d) \tan (e+f x)}{2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}}{\left (a^2+b^2\right ) (b c-a d)}-\frac {2 b^2}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {(b c-a d)^2 (a c-b d)-(b c-a d)^2 (b c+a d) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}}{\left (a^2+b^2\right ) (b c-a d)}-\frac {2 b^2}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {(b c-a d)^2 (a c-b d)-(b c-a d)^2 (b c+a d) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}}{\left (a^2+b^2\right ) (b c-a d)}-\frac {2 b^2}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 4099 |
\(\displaystyle -\frac {2 b^2}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {1}{2} (a-i b) (c-i d) (b c-a d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b) (c+i d) (b c-a d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}}{\left (a^2+b^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 b^2}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {1}{2} (a-i b) (c-i d) (b c-a d)^2 \int \frac {1-i \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx+\frac {1}{2} (a+i b) (c+i d) (b c-a d)^2 \int \frac {i \tan (e+f x)+1}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{\left (c^2+d^2\right ) (b c-a d)}}{\left (a^2+b^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 4098 |
\(\displaystyle -\frac {2 b^2}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {(a+i b) (c+i d) (b c-a d)^2 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{2 f}+\frac {(a-i b) (c-i d) (b c-a d)^2 \int \frac {1}{(i \tan (e+f x)+1) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}d\tan (e+f x)}{2 f}}{\left (c^2+d^2\right ) (b c-a d)}}{\left (a^2+b^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle -\frac {2 b^2}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {(a-i b) (c-i d) (b c-a d)^2 \int \frac {1}{-i a+b+\frac {(i c-d) (a+b \tan (e+f x))}{c+d \tan (e+f x)}}d\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{f}+\frac {(a+i b) (c+i d) (b c-a d)^2 \int \frac {1}{i a+b-\frac {(i c+d) (a+b \tan (e+f x))}{c+d \tan (e+f x)}}d\frac {\sqrt {a+b \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{f}}{\left (c^2+d^2\right ) (b c-a d)}}{\left (a^2+b^2\right ) (b c-a d)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 b^2}{f \left (a^2+b^2\right ) (b c-a d) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {2 d \left (a^2 d^2+b^2 \left (c^2+2 d^2\right )\right ) \sqrt {a+b \tan (e+f x)}}{f \left (c^2+d^2\right ) (b c-a d) \sqrt {c+d \tan (e+f x)}}+\frac {\frac {i (a-i b) (c-i d) (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a+i b} \sqrt {c+i d}}-\frac {i (a+i b) (c+i d) (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{f \sqrt {a-i b} \sqrt {c-i d}}}{\left (c^2+d^2\right ) (b c-a d)}}{\left (a^2+b^2\right ) (b c-a d)}\) |
(-2*b^2)/((a^2 + b^2)*(b*c - a*d)*f*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Ta n[e + f*x]]) + ((((-I)*(a + I*b)*(c + I*d)*(b*c - a*d)^2*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 - I*d)*(b*c - a*d)^2*Ar cTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + I*b]*Sqrt[c + d*T an[e + f*x]])])/(Sqrt[a + I*b]*Sqrt[c + I*d]*f))/((b*c - a*d)*(c^2 + d^2)) - (2*d*(a^2*d^2 + b^2*(c^2 + 2*d^2))*Sqrt[a + b*Tan[e + f*x]])/((b*c - a* d)*(c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]))/((a^2 + b^2)*(b*c - a*d))
3.13.94.3.1 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_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + Simp[1 /((m + 1)*(a^2 + b^2)*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x], x], x] / ; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || Integ erQ[m]) && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 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_.) + (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.
\[\int \frac {1}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {3}{2}} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 32905 vs. \(2 (249) = 498\).
Time = 40.78 (sec) , antiderivative size = 32905, normalized size of antiderivative = 109.32 \[ \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(((2*b*d+2*a*c)^2>0)', see `assum e?` for mo
Timed out. \[ \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(a+b \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]