Integrand size = 29, antiderivative size = 264 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+b \tan (e+f x)}} \, dx=-\frac {i (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a-i b} f}+\frac {i (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b} f}+\frac {d^{3/2} (5 b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{b^{3/2} f}+\frac {d^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f} \]
d^(3/2)*(-a*d+5*b*c)*arctanh(d^(1/2)*(a+b*tan(f*x+e))^(1/2)/b^(1/2)/(c+d*t an(f*x+e))^(1/2))/b^(3/2)/f-I*(c-I*d)^(5/2)*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))/f/(a-I*b)^(1/2)+I*(c+ I*d)^(5/2)*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))/f/(a+I*b)^(1/2)+d^2*(a+b*tan(f*x+e))^(1/2)*(c+d*tan(f* x+e))^(1/2)/b/f
Time = 2.97 (sec) , antiderivative size = 432, normalized size of antiderivative = 1.64 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+b \tan (e+f x)}} \, dx=\frac {\frac {b \left (\sqrt {-b^2} c \left (c^2-3 d^2\right )-b d \left (-3 c^2+d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {-c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+\sqrt {-b^2}} \sqrt {-c+\frac {\sqrt {-b^2} d}{b}}}+\frac {b \left (\sqrt {-b^2} c \left (c^2-3 d^2\right )+b d \left (-3 c^2+d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+\frac {\sqrt {-b^2} d}{b}}}+b d^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}+\frac {\sqrt {b} d^{3/2} (5 b c-a d) \sqrt {c-\frac {a d}{b}} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c-\frac {a d}{b}}}\right ) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}{\sqrt {c+d \tan (e+f x)}}}{b^2 f} \]
((b*(Sqrt[-b^2]*c*(c^2 - 3*d^2) - b*d*(-3*c^2 + d^2))*ArcTanh[(Sqrt[-c + ( Sqrt[-b^2]*d)/b]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[-a + Sqrt[-b^2]]*Sqrt[c + d*Tan[e + f*x]])])/(Sqrt[-a + Sqrt[-b^2]]*Sqrt[-c + (Sqrt[-b^2]*d)/b]) + (b*(Sqrt[-b^2]*c*(c^2 - 3*d^2) + b*d*(-3*c^2 + d^2))*ArcTanh[(Sqrt[c + (Sq rt[-b^2]*d)/b]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[a + Sqrt[-b^2]]*Sqrt[c + d* Tan[e + f*x]])])/(Sqrt[a + Sqrt[-b^2]]*Sqrt[c + (Sqrt[-b^2]*d)/b]) + b*d^2 *Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]] + (Sqrt[b]*d^(3/2)*(5*b *c - a*d)*Sqrt[c - (a*d)/b]*ArcSinh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sq rt[b]*Sqrt[c - (a*d)/b])]*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])/Sqrt [c + d*Tan[e + f*x]])/(b^2*f)
Time = 1.16 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 4049, 27, 3042, 4138, 2348, 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.
| \(\displaystyle \int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+b \tan (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+b \tan (e+f x)}}dx\) |
| \(\Big \downarrow \) 4049 |
| \(\displaystyle \frac {\int \frac {2 b c^3+d^2 (5 b c-a d) \tan ^2(e+f x)-d^2 (b c+a d)+2 b d \left (3 c^2-d^2\right ) \tan (e+f x)}{2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{b}+\frac {d^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {2 b c^3+d^2 (5 b c-a d) \tan ^2(e+f x)-d^2 (b c+a d)+2 b d \left (3 c^2-d^2\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{2 b}+\frac {d^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 b c^3+d^2 (5 b c-a d) \tan (e+f x)^2-d^2 (b c+a d)+2 b d \left (3 c^2-d^2\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{2 b}+\frac {d^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}\) |
| \(\Big \downarrow \) 4138 |
| \(\displaystyle \frac {\int \frac {2 b c^3+d^2 (5 b c-a d) \tan ^2(e+f x)-d^2 (b c+a d)+2 b d \left (3 c^2-d^2\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left (\tan ^2(e+f x)+1\right )}d\tan (e+f x)}{2 b f}+\frac {d^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}\) |
| \(\Big \downarrow \) 2348 |
| \(\displaystyle \frac {d^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\int \left (-\frac {(a d-5 b c) d^2}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {2 b d^3-6 b c^2 d+i \left (2 b c^3-6 b c d^2\right )}{2 (i-\tan (e+f x)) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {-2 b d^3+6 b c^2 d+i \left (2 b c^3-6 b c d^2\right )}{2 (\tan (e+f x)+i) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\right )d\tan (e+f x)}{2 b f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{b f}+\frac {\frac {2 d^{3/2} (5 b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {b}}-\frac {2 i b (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a-i b}}+\frac {2 i b (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b}}}{2 b f}\) |
(((-2*I)*b*(c - I*d)^(5/2)*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] + ((2*I)*b*(c + I*d)^(5/2)*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] + (2*d^(3/2)*(5*b*c - a*d)*Arc Tanh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]]) ])/Sqrt[b])/(2*b*f) + (d^2*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x ]])/(b*f)
3.13.79.3.1 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[(Px_)*((c_) + (d_.)*(x_))^(m_.)*((e_) + (f_.)*(x_))^(n_.)*((a_.) + (b_. )*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(c + d*x)^m*(e + f*x)^ n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[P x, x] && (IntegerQ[p] || (IntegerQ[2*p] && IntegerQ[m] && ILtQ[n, 0])) && !(IGtQ[m, 0] && IGtQ[n, 0])
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 - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*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] && GtQ[m, 2] && (GeQ[n, -1] || I ntegerQ[m]) && !(IGtQ[n, 2] && ( !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]
Timed out.
\[\int \frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{\sqrt {a +b \tan \left (f x +e \right )}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 12777 vs. \(2 (204) = 408\).
Time = 13.39 (sec) , antiderivative size = 25581, normalized size of antiderivative = 96.90 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+b \tan (e+f x)}} \, dx=\text {Too large to display} \]
\[ \int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+b \tan (e+f x)}} \, dx=\int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\sqrt {a + b \tan {\left (e + f x \right )}}}\, dx \]
\[ \int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+b \tan (e+f x)}} \, dx=\int { \frac {{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{\sqrt {b \tan \left (f x + e\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+b \tan (e+f x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{\sqrt {a+b \tan (e+f x)}} \, dx=\int \frac {{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{\sqrt {a+b\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \]