Integrand size = 29, antiderivative size = 273 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^{3/2}} \, dx=-\frac {i (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a-i b)^{3/2} f}+\frac {i (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{(a+i b)^{3/2} f}+\frac {2 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{b^{3/2} f}-\frac {2 (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}{b \left (a^2+b^2\right ) f \sqrt {a+b \tan (e+f x)}} \]
-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))/(a-I*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))/(a+I*b) ^(3/2)/f+2*d^(5/2)*arctanh(d^(1/2)*(a+b*tan(f*x+e))^(1/2)/b^(1/2)/(c+d*tan (f*x+e))^(1/2))/b^(3/2)/f-2*(-a*d+b*c)^2*(c+d*tan(f*x+e))^(1/2)/b/(a^2+b^2 )/f/(a+b*tan(f*x+e))^(1/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1187\) vs. \(2(273)=546\).
Time = 6.29 (sec) , antiderivative size = 1187, normalized size of antiderivative = 4.35 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^{3/2}} \, dx=-\frac {i (-c-i d) \left (-\left ((-c-i d) \left (-\frac {2 \sqrt {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 )}{(-a-i b) \sqrt {a+i b}}+\frac {2 \sqrt {c+d \tan (e+f x)}}{(-a-i b) \sqrt {a+b \tan (e+f x)}}\right )\right )-\frac {2 d \sqrt {c+d \tan (e+f x)} \left (1+\frac {b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac {b^2 c}{b c-a d}-\frac {a b d}{b c-a d}\right )}\right )^{3/2} \left (1-\frac {\sqrt {b} \sqrt {d} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b c-a d} \sqrt {\frac {b^2 c}{b c-a d}-\frac {a b d}{b c-a d}}}\right ) \sqrt {a+b \tan (e+f x)}}{\sqrt {b c-a d} \sqrt {\frac {b^2 c}{b c-a d}-\frac {a b d}{b c-a d}} \sqrt {1+\frac {b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac {b^2 c}{b c-a d}-\frac {a b d}{b c-a d}\right )}}}\right )}{b \sqrt {\frac {b}{\frac {b^2 c}{b c-a d}-\frac {a b d}{b c-a d}}} \sqrt {a+b \tan (e+f x)} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}} \left (-1-\frac {b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac {b^2 c}{b c-a d}-\frac {a b d}{b c-a d}\right )}\right )}\right )}{2 f}-\frac {i (-c+i d) \left (-\left ((-c+i d) \left (-\frac {2 \sqrt {-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 )}{(a-i b) \sqrt {-a+i b}}+\frac {2 \sqrt {c+d \tan (e+f x)}}{(a-i b) \sqrt {a+b \tan (e+f x)}}\right )\right )+\frac {2 d \sqrt {c+d \tan (e+f x)} \left (1+\frac {b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac {b^2 c}{b c-a d}-\frac {a b d}{b c-a d}\right )}\right )^{3/2} \left (1-\frac {\sqrt {b} \sqrt {d} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b c-a d} \sqrt {\frac {b^2 c}{b c-a d}-\frac {a b d}{b c-a d}}}\right ) \sqrt {a+b \tan (e+f x)}}{\sqrt {b c-a d} \sqrt {\frac {b^2 c}{b c-a d}-\frac {a b d}{b c-a d}} \sqrt {1+\frac {b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac {b^2 c}{b c-a d}-\frac {a b d}{b c-a d}\right )}}}\right )}{b \sqrt {\frac {b}{\frac {b^2 c}{b c-a d}-\frac {a b d}{b c-a d}}} \sqrt {a+b \tan (e+f x)} \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}} \left (-1-\frac {b d (a+b \tan (e+f x))}{(b c-a d) \left (\frac {b^2 c}{b c-a d}-\frac {a b d}{b c-a d}\right )}\right )}\right )}{2 f} \]
((-1/2*I)*(-c - I*d)*(-((-c - I*d)*((-2*Sqrt[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]])])/(( -a - I*b)*Sqrt[a + I*b]) + (2*Sqrt[c + d*Tan[e + f*x]])/((-a - I*b)*Sqrt[a + b*Tan[e + f*x]]))) - (2*d*Sqrt[c + d*Tan[e + f*x]]*(1 + (b*d*(a + b*Tan [e + f*x]))/((b*c - a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))))^(3/ 2)*(1 - (Sqrt[b]*Sqrt[d]*ArcSinh[(Sqrt[b]*Sqrt[d]*Sqrt[a + b*Tan[e + f*x]] )/(Sqrt[b*c - a*d]*Sqrt[(b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)])]*Sqrt[ a + b*Tan[e + f*x]])/(Sqrt[b*c - a*d]*Sqrt[(b^2*c)/(b*c - a*d) - (a*b*d)/( b*c - a*d)]*Sqrt[1 + (b*d*(a + b*Tan[e + f*x]))/((b*c - a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)))])))/(b*Sqrt[b/((b^2*c)/(b*c - a*d) - (a*b* d)/(b*c - a*d))]*Sqrt[a + b*Tan[e + f*x]]*Sqrt[(b*(c + d*Tan[e + f*x]))/(b *c - a*d)]*(-1 - (b*d*(a + b*Tan[e + f*x]))/((b*c - a*d)*((b^2*c)/(b*c - a *d) - (a*b*d)/(b*c - a*d)))))))/f - ((I/2)*(-c + I*d)*(-((-c + I*d)*((-2*S qrt[-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]])])/((a - I*b)*Sqrt[-a + I*b]) + (2*Sqrt[c + d*Tan[e + f*x]])/((a - I*b)*Sqrt[a + b*Tan[e + f*x]]))) + (2*d*Sqrt[c + d *Tan[e + f*x]]*(1 + (b*d*(a + b*Tan[e + f*x]))/((b*c - a*d)*((b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d))))^(3/2)*(1 - (Sqrt[b]*Sqrt[d]*ArcSinh[(Sqrt[ b]*Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b*c - a*d]*Sqrt[(b^2*c)/(b*c - a*d) - (a*b*d)/(b*c - a*d)])]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b*c - a*d...
Time = 1.34 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 4048, 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.
| \(\displaystyle \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4048 |
| \(\displaystyle \frac {2 \int \frac {a^2 d^3+\left (a^2+b^2\right ) \tan ^2(e+f x) d^3+3 b^2 c^2 d+a b c \left (c^2-3 d^2\right )+b \left (a d \left (3 c^2-d^2\right )-b \left (c^3-3 c d^2\right )\right ) \tan (e+f x)}{2 \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}{b f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a^2 d^3+\left (a^2+b^2\right ) \tan ^2(e+f x) d^3+3 b^2 c^2 d+a b c \left (c^2-3 d^2\right )-b \left (b c^3-3 a d c^2-3 b d^2 c+a d^3\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}{b f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a^2 d^3+\left (a^2+b^2\right ) \tan (e+f x)^2 d^3+3 b^2 c^2 d+a b c \left (c^2-3 d^2\right )-b \left (b c^3-3 a d c^2-3 b d^2 c+a d^3\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}{b f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4138 |
| \(\displaystyle \frac {\int \frac {a^2 d^3+\left (a^2+b^2\right ) \tan ^2(e+f x) d^3+3 b^2 c^2 d+a b c \left (c^2-3 d^2\right )+b \left (a d \left (3 c^2-d^2\right )-b \left (c^3-3 c d^2\right )\right ) \tan (e+f x)}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)} \left (\tan ^2(e+f x)+1\right )}d\tan (e+f x)}{b f \left (a^2+b^2\right )}-\frac {2 (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}{b f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}\) |
| \(\Big \downarrow \) 2348 |
| \(\displaystyle -\frac {2 (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}{b f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}+\frac {\int \left (\frac {\left (a^2+b^2\right ) d^3}{\sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {b^2 c^3-3 a b d c^2-3 b^2 d^2 c+a b d^3+i \left (a b c^3+3 b^2 d c^2-3 a b d^2 c-b^2 d^3\right )}{2 (i-\tan (e+f x)) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {-b^2 c^3+3 a b d c^2+3 b^2 d^2 c-a b d^3+i \left (a b c^3+3 b^2 d c^2-3 a b d^2 c-b^2 d^3\right )}{2 (\tan (e+f x)+i) \sqrt {a+b \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}\right )d\tan (e+f x)}{b f \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 (b c-a d)^2 \sqrt {c+d \tan (e+f x)}}{b f \left (a^2+b^2\right ) \sqrt {a+b \tan (e+f x)}}+\frac {\frac {2 d^{5/2} \left (a^2+b^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b \tan (e+f x)}}{\sqrt {b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {b}}-\frac {b (-b+i a) (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c-i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a-i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a-i b}}+\frac {b (b+i a) (c+i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+i d} \sqrt {a+b \tan (e+f x)}}{\sqrt {a+i b} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a+i b}}}{b f \left (a^2+b^2\right )}\) |
(-(((I*a - b)*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]) + (b*(I*a + b)*(c + I*d)^(5/2)*ArcTanh[(Sqrt[c + I*d]*Sqrt[a + b*Tan[e + f*x]])/(Sq rt[a + I*b]*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[a + I*b] + (2*(a^2 + b^2)*d^( 5/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*Tan[e + f*x]])/(Sqrt[b]*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[b])/(b*(a^2 + b^2)*f) - (2*(b*c - a*d)^2*Sqrt[c + d*Tan[e + f*x]])/(b*(a^2 + b^2)*f*Sqrt[a + b*Tan[e + f*x]])
3.13.80.3.1 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*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1 /(d*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c *(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3*a*b^2*d) *Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*( n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 2] && LtQ [n, -1] && IntegerQ[2*m]
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.
\[\int \frac {\left (c +d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{\left (a +b \tan \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 21401 vs. \(2 (213) = 426\).
Time = 24.06 (sec) , antiderivative size = 42829, normalized size of antiderivative = 156.88 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^{3/2}} \, dx=\int \frac {\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^{3/2}} \, dx=\int { \frac {{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{(a+b \tan (e+f x))^{3/2}} \, dx=\int \frac {{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]