Integrand size = 29, antiderivative size = 264 \[ \int \frac {(a+b \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i (a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c-i d} f}+\frac {i (a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c+i d} f}-\frac {b^{3/2} (b c-5 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f} \]
-b^(3/2)*(-5*a*d+b*c)*arctanh(d^(1/2)*(a+b*tan(f*x+e))^(1/2)/b^(1/2)/(c+d* tan(f*x+e))^(1/2))/d^(3/2)/f-I*(a-I*b)^(5/2)*arctanh((c-I*d)^(1/2)*(a+b*ta n(f*x+e))^(1/2)/(a-I*b)^(1/2)/(c+d*tan(f*x+e))^(1/2))/f/(c-I*d)^(1/2)+I*(a +I*b)^(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/(c+I*d)^(1/2)+b^2*(a+b*tan(f*x+e))^(1/2)*(c+d*tan(f *x+e))^(1/2)/d/f
Time = 3.51 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.66 \[ \int \frac {(a+b \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {\frac {\left (3 a^2 b^2-b^4+a \sqrt {-b^2} \left (a^2-3 b^2\right )\right ) d \text {arctanh}\left (\frac {\sqrt {-c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {-a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a+\sqrt {-b^2}} \sqrt {-c+\frac {\sqrt {-b^2} d}{b}}}+\frac {\left (-3 a^2 b^2+b^4+a^3 \sqrt {-b^2}+3 a \left (-b^2\right )^{3/2}\right ) d \text {arctanh}\left (\frac {\sqrt {c+\frac {\sqrt {-b^2} d}{b}} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+\sqrt {-b^2}} \sqrt {c+\frac {\sqrt {-b^2} d}{b}}}+b^3 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}-\frac {b^{5/2} (b c-5 a d) \sqrt {c-\frac {a d}{b}} \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c-\frac {a d}{b}}}\right ) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}{\sqrt {d} \sqrt {c+d \tan (e+f x)}}}{b d f} \]
(((3*a^2*b^2 - b^4 + a*Sqrt[-b^2]*(a^2 - 3*b^2))*d*ArcTanh[(Sqrt[-c + (Sqr t[-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]) + ((- 3*a^2*b^2 + b^4 + a^3*Sqrt[-b^2] + 3*a*(-b^2)^(3/2))*d*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^3 *Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]] - (b^(5/2)*(b*c - 5*a*d )*Sqrt[c - (a*d)/b]*ArcSinh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sq rt[c - (a*d)/b])]*Sqrt[(b*(c + d*Tan[e + f*x]))/(b*c - a*d)])/(Sqrt[d]*Sqr t[c + d*Tan[e + f*x]]))/(b*d*f)
Time = 1.16 (sec) , antiderivative size = 268, 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.
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}}dx\) |
| \(\Big \downarrow \) 4049 |
| \(\displaystyle \frac {\int \frac {2 d a^3-b^2 (b c-5 a d) \tan ^2(e+f x)-b^2 (b c+a d)+2 b \left (3 a^2-b^2\right ) d \tan (e+f x)}{2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{d}+\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {2 d a^3-b^2 (b c-5 a d) \tan ^2(e+f x)-b^2 (b c+a d)+2 b \left (3 a^2-b^2\right ) d \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{2 d}+\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 d a^3-b^2 (b c-5 a d) \tan (e+f x)^2-b^2 (b c+a d)+2 b \left (3 a^2-b^2\right ) d \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{2 d}+\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\) |
| \(\Big \downarrow \) 4138 |
| \(\displaystyle \frac {\int \frac {2 d a^3-b^2 (b c-5 a d) \tan ^2(e+f x)-b^2 (b c+a d)+2 b \left (3 a^2-b^2\right ) d \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left (\tan ^2(e+f x)+1\right )}d\tan (e+f x)}{2 d f}+\frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}\) |
| \(\Big \downarrow \) 2348 |
| \(\displaystyle \frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\frac {\int \left (-\frac {(b c-5 a d) b^2}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {2 d b^3-6 a^2 d b+i \left (2 a^3 d-6 a b^2 d\right )}{2 (i-\tan (e+f x)) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {-2 d b^3+6 a^2 d b+i \left (2 a^3 d-6 a b^2 d\right )}{2 (\tan (e+f x)+i) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\right )d\tan (e+f x)}{2 d f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{d f}+\frac {-\frac {2 b^{3/2} (b c-5 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d}}-\frac {2 i d (a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c-i d}}+\frac {2 i d (a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {c+i d}}}{2 d f}\) |
(((-2*I)*(a - I*b)^(5/2)*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[c - I*d] + ((2*I)*(a + I *b)^(5/2)*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[c + I*d] - (2*b^(3/2)*(b*c - 5*a*d)*Arc Tanh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]]) ])/Sqrt[d])/(2*d*f) + (b^2*Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x ]])/(d*f)
3.13.83.3.1 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.
\[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{\sqrt {c +d \tan \left (f x +e \right )}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 12228 vs. \(2 (204) = 408\).
Time = 18.45 (sec) , antiderivative size = 24483, normalized size of antiderivative = 92.74 \[ \int \frac {(a+b \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Too large to display} \]
\[ \int \frac {(a+b \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \]
\[ \int \frac {(a+b \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\int { \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{\sqrt {d \tan \left (f x + e\right ) + c}} \,d x } \]
Timed out. \[ \int \frac {(a+b \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+b \tan (e+f x))^{5/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \]