Integrand size = 25, antiderivative size = 195 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{3/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \]
arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a-b)^(3/2 )/d+2*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/b^(3/2)/d-a rctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a+b)^(3/2 )/d-2*a^2*tan(d*x+c)^(1/2)/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(1/2)
Time = 1.35 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.65 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=-\frac {2 \sqrt {a} \sqrt {-a+i b} \sqrt {a+i b} \left (a^2+b^2\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}+\sqrt {b} \left (\sqrt [4]{-1} (a+i b)^{3/2} b \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+\sqrt {-a+i b} \left (-2 a^2 \sqrt {a+i b} \sqrt {\tan (c+d x)}-\sqrt [4]{-1} (a-i b) b \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}\right )\right )}{(-a+i b)^{3/2} (a+i b)^{3/2} b^{3/2} d \sqrt {a+b \tan (c+d x)}} \]
-((2*Sqrt[a]*Sqrt[-a + I*b]*Sqrt[a + I*b]*(a^2 + b^2)*ArcSinh[(Sqrt[b]*Sqr t[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a] + Sqrt[b]*((-1)^(1/ 4)*(a + I*b)^(3/2)*b*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]]) /Sqrt[a + b*Tan[c + d*x]]]*Sqrt[a + b*Tan[c + d*x]] + Sqrt[-a + I*b]*(-2*a ^2*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]] - (-1)^(1/4)*(a - I*b)*b*ArcTan[((-1)^ (1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[a + b*Tan[c + d*x]])))/((-a + I*b)^(3/2)*(a + I*b)^(3/2)*b^(3/2)*d*Sqrt[a + b *Tan[c + d*x]]))
Time = 0.93 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.19, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4048, 27, 3042, 4138, 2035, 2257, 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.
\(\displaystyle \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^{5/2}}{(a+b \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 4048 |
\(\displaystyle \frac {2 \int \frac {a^2-b \tan (c+d x) a+\left (a^2+b^2\right ) \tan ^2(c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a^2-b \tan (c+d x) a+\left (a^2+b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a^2-b \tan (c+d x) a+\left (a^2+b^2\right ) \tan (c+d x)^2}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 4138 |
\(\displaystyle \frac {\int \frac {a^2-b \tan (c+d x) a+\left (a^2+b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{b d \left (a^2+b^2\right )}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {2 \int \frac {a^2-b \tan (c+d x) a+\left (a^2+b^2\right ) \tan ^2(c+d x)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right )}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 2257 |
\(\displaystyle \frac {2 \int \left (\frac {a^2+b^2}{\sqrt {a+b \tan (c+d x)}}-\frac {b^2+a \tan (c+d x) b}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right )}-\frac {2 a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^2 \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {2 \left (\frac {\left (a^2+b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}-\frac {b (b+i a) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 \sqrt {-b+i a}}+\frac {b (-b+i a) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 \sqrt {b+i a}}\right )}{b d \left (a^2+b^2\right )}\) |
(2*(-1/2*(b*(I*a + b)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b *Tan[c + d*x]]])/Sqrt[I*a - b] + ((a^2 + b^2)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[b] + ((I*a - b)*b*ArcTanh[(Sqrt[I *a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(2*Sqrt[I*a + b]))) /(b*(a^2 + b^2)*d) - (2*a^2*Sqrt[Tan[c + d*x]])/(b*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]])
3.7.41.3.1 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[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol ] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a , c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
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]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 0.91 (sec) , antiderivative size = 798946, normalized size of antiderivative = 4097.16
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 7689 vs. \(2 (159) = 318\).
Time = 2.28 (sec) , antiderivative size = 15380, normalized size of antiderivative = 78.87 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {\tan ^{\frac {5}{2}}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{\frac {5}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]