\(\int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\) [647]

Optimal result

Integrand size = 25, antiderivative size = 317 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {i \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}-\frac {5 a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{7/2} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b^3 \left (a^2+b^2\right )^2 d} \] Output:

-I*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a-b)^( 
5/2)/d-5*a*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/b^(7/2 
)/d+I*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a+ 
b)^(5/2)/d-2/3*a^2*tan(d*x+c)^(5/2)/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(3/2)-2 
/3*a^2*(5*a^2+11*b^2)*tan(d*x+c)^(3/2)/b^2/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^ 
(1/2)+(5*a^4+10*a^2*b^2+b^4)*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/b^3/( 
a^2+b^2)^2/d
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 4.49 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.82 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {\frac {6 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {7}{2},\frac {9}{2},-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {7}{2}}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{a^2 \sqrt {a+b \tan (c+d x)}}+7 \left (-\frac {3 \sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{5/2}}+\frac {3 \sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^{5/2}}-\frac {2 a \sqrt {\tan (c+d x)} \left (6 a b+\left (a^2+7 b^2\right ) \tan (c+d x)\right )}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^{3/2}}\right )}{21 d} \] Input:

Integrate[Tan[c + d*x]^(9/2)/(a + b*Tan[c + d*x])^(5/2),x]
 

Output:

((6*Hypergeometric2F1[5/2, 7/2, 9/2, -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^( 
7/2)*Sqrt[1 + (b*Tan[c + d*x])/a])/(a^2*Sqrt[a + b*Tan[c + d*x]]) + 7*((-3 
*(-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + 
 b*Tan[c + d*x]]])/(-a + I*b)^(5/2) + (3*(-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqr 
t[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(a + I*b)^(5/2) 
- (2*a*Sqrt[Tan[c + d*x]]*(6*a*b + (a^2 + 7*b^2)*Tan[c + d*x]))/((a^2 + b^ 
2)^2*(a + b*Tan[c + d*x])^(3/2))))/(21*d)
 

Rubi [A] (verified)

Time = 1.98 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.17, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4048, 27, 3042, 4128, 27, 3042, 4130, 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.

\(\displaystyle \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^{9/2}}{(a+b \tan (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 4048

\(\displaystyle \frac {2 \int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (5 a^2-3 b \tan (c+d x) a+\left (5 a^2+3 b^2\right ) \tan ^2(c+d x)\right )}{2 (a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (5 a^2-3 b \tan (c+d x) a+\left (5 a^2+3 b^2\right ) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\tan (c+d x)^{3/2} \left (5 a^2-3 b \tan (c+d x) a+\left (5 a^2+3 b^2\right ) \tan (c+d x)^2\right )}{(a+b \tan (c+d x))^{3/2}}dx}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4128

\(\displaystyle \frac {\frac {2 \int \frac {3 \sqrt {\tan (c+d x)} \left (-2 a \tan (c+d x) b^3+\left (5 a^4+10 b^2 a^2+b^4\right ) \tan ^2(c+d x)+a^2 \left (5 a^2+11 b^2\right )\right )}{2 \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \int \frac {\sqrt {\tan (c+d x)} \left (-2 a \tan (c+d x) b^3+\left (5 a^4+10 b^2 a^2+b^4\right ) \tan ^2(c+d x)+a^2 \left (5 a^2+11 b^2\right )\right )}{\sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \int \frac {\sqrt {\tan (c+d x)} \left (-2 a \tan (c+d x) b^3+\left (5 a^4+10 b^2 a^2+b^4\right ) \tan (c+d x)^2+a^2 \left (5 a^2+11 b^2\right )\right )}{\sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {\frac {3 \left (\frac {\int -\frac {-2 \left (a^2-b^2\right ) \tan (c+d x) b^3+5 a \left (a^2+b^2\right )^2 \tan ^2(c+d x)+a \left (5 a^4+10 b^2 a^2+b^4\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b}+\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}\right )}{b \left (a^2+b^2\right )}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \left (\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {-2 \left (a^2-b^2\right ) \tan (c+d x) b^3+5 a \left (a^2+b^2\right )^2 \tan ^2(c+d x)+a \left (5 a^4+10 b^2 a^2+b^4\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{2 b}\right )}{b \left (a^2+b^2\right )}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \left (\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {-2 \left (a^2-b^2\right ) \tan (c+d x) b^3+5 a \left (a^2+b^2\right )^2 \tan (c+d x)^2+a \left (5 a^4+10 b^2 a^2+b^4\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{2 b}\right )}{b \left (a^2+b^2\right )}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4138

\(\displaystyle \frac {\frac {3 \left (\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {-2 \left (a^2-b^2\right ) \tan (c+d x) b^3+5 a \left (a^2+b^2\right )^2 \tan ^2(c+d x)+a \left (5 a^4+10 b^2 a^2+b^4\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 b d}\right )}{b \left (a^2+b^2\right )}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 2035

\(\displaystyle \frac {\frac {3 \left (\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {-2 \left (a^2-b^2\right ) \tan (c+d x) b^3+5 a \left (a^2+b^2\right )^2 \tan ^2(c+d x)+a \left (5 a^4+10 b^2 a^2+b^4\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{b d}\right )}{b \left (a^2+b^2\right )}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 2257

\(\displaystyle \frac {\frac {3 \left (\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \left (\frac {5 a \left (a^2+b^2\right )^2}{\sqrt {a+b \tan (c+d x)}}-\frac {2 \left (2 a b^4+\left (a^2-b^2\right ) \tan (c+d x) b^3\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{b d}\right )}{b \left (a^2+b^2\right )}-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 a^2 \tan ^{\frac {5}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {2 a^2 \left (5 a^2+11 b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {3 \left (\frac {\left (5 a^4+10 a^2 b^2+b^4\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\frac {5 a \left (a^2+b^2\right )^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}-\frac {i b^3 (a-i b)^2 \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}+\frac {i b^3 (a+i b)^2 \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b+i a}}}{b d}\right )}{b \left (a^2+b^2\right )}}{3 b \left (a^2+b^2\right )}\)

Input:

Int[Tan[c + d*x]^(9/2)/(a + b*Tan[c + d*x])^(5/2),x]
 

Output:

(-2*a^2*Tan[c + d*x]^(5/2))/(3*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) 
 + ((-2*a^2*(5*a^2 + 11*b^2)*Tan[c + d*x]^(3/2))/(b*(a^2 + b^2)*d*Sqrt[a + 
 b*Tan[c + d*x]]) + (3*(-((((-I)*(a - I*b)^2*b^3*ArcTan[(Sqrt[I*a - b]*Sqr 
t[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[I*a - b] + (5*a*(a^2 + b^ 
2)^2*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[ 
b] + (I*(a + I*b)^2*b^3*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a 
+ b*Tan[c + d*x]]])/Sqrt[I*a + b])/(b*d)) + ((5*a^4 + 10*a^2*b^2 + b^4)*Sq 
rt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/(b*d)))/(b*(a^2 + b^2)))/(3*b*( 
a^2 + b^2))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

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]
 

rule 2257

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4048

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]
 

rule 4128

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*d^2 + c*(c*C - B*d))*(a + b*Tan[e + 
 f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Sim 
p[1/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e 
 + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c*m + a*d* 
(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b 
*(d*(B*c - A*d)*(m + n + 1) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], 
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ 
[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 4130

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[C*(a + b*Tan[e + f*x])^m*((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 - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C 
*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* 
m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

rule 4138

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]
 

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 1.49 (sec) , antiderivative size = 1493684, normalized size of antiderivative = 4711.94

\[\text {output too large to display}\]

Input:

int(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c))^(5/2),x)
 

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 11619 vs. \(2 (265) = 530\).

Time = 3.86 (sec) , antiderivative size = 23240, normalized size of antiderivative = 73.31 \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)**(9/2)/(a+b*tan(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{\frac {9}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate(tan(d*x + c)^(9/2)/(b*tan(d*x + c) + a)^(5/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{9/2}}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:

int(tan(c + d*x)^(9/2)/(a + b*tan(c + d*x))^(5/2),x)
 

Output:

int(tan(c + d*x)^(9/2)/(a + b*tan(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {9}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{4}}{\tan \left (d x +c \right )^{3} b^{3}+3 \tan \left (d x +c \right )^{2} a \,b^{2}+3 \tan \left (d x +c \right ) a^{2} b +a^{3}}d x \] Input:

int(tan(d*x+c)^(9/2)/(a+b*tan(d*x+c))^(5/2),x)
 

Output:

int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**4)/(tan(c + 
 d*x)**3*b**3 + 3*tan(c + d*x)**2*a*b**2 + 3*tan(c + d*x)*a**2*b + a**3),x 
)