\(\int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx\) [537]

Optimal result

Integrand size = 23, antiderivative size = 282 \[ \int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{3/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{3/2} d}-\frac {2 a^2 \tan ^3(c+d x)}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {2 \left (16 a^4+6 a^2 b^2-5 b^4\right ) \sqrt {a+b \tan (c+d x)}}{5 b^4 \left (a^2+b^2\right ) d}-\frac {2 a \left (8 a^2+3 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{5 b^3 \left (a^2+b^2\right ) d}+\frac {2 \left (6 a^2+b^2\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b^2 \left (a^2+b^2\right ) d} \] Output:

-arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(3/2)/d-arctanh((a+ 
b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(3/2)/d-2*a^2*tan(d*x+c)^3/b/(a 
^2+b^2)/d/(a+b*tan(d*x+c))^(1/2)+2/5*(16*a^4+6*a^2*b^2-5*b^4)*(a+b*tan(d*x 
+c))^(1/2)/b^4/(a^2+b^2)/d-2/5*a*(8*a^2+3*b^2)*tan(d*x+c)*(a+b*tan(d*x+c)) 
^(1/2)/b^3/(a^2+b^2)/d+2/5*(6*a^2+b^2)*tan(d*x+c)^2*(a+b*tan(d*x+c))^(1/2) 
/b^2/(a^2+b^2)/d
 

Mathematica [A] (verified)

Time = 4.08 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.04 \[ \int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {\frac {5 b \left (b^2-a \sqrt {-b^2}\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {-b^2}}}\right )}{\sqrt {-b^2} \left (a^2+b^2\right ) \sqrt {a-\sqrt {-b^2}}}+\frac {5 b \left (-a+\sqrt {-b^2}\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+\sqrt {-b^2}}}\right )}{\left (a^2+b^2\right ) \sqrt {a+\sqrt {-b^2}}}+\frac {2 a^3 \left (8 a^2+3 b^2\right )}{b^3 \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {4 a \tan ^2(c+d x)}{b \sqrt {a+b \tan (c+d x)}}+\frac {2 \tan ^3(c+d x)}{\sqrt {a+b \tan (c+d x)}}+\frac {2 \left (8 a^2-5 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b^3}}{5 b d} \] Input:

Integrate[Tan[c + d*x]^5/(a + b*Tan[c + d*x])^(3/2),x]
 

Output:

((5*b*(b^2 - a*Sqrt[-b^2])*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - Sqrt[ 
-b^2]]])/(Sqrt[-b^2]*(a^2 + b^2)*Sqrt[a - Sqrt[-b^2]]) + (5*b*(-a + Sqrt[- 
b^2])*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + Sqrt[-b^2]]])/((a^2 + b^2) 
*Sqrt[a + Sqrt[-b^2]]) + (2*a^3*(8*a^2 + 3*b^2))/(b^3*(a^2 + b^2)*Sqrt[a + 
 b*Tan[c + d*x]]) - (4*a*Tan[c + d*x]^2)/(b*Sqrt[a + b*Tan[c + d*x]]) + (2 
*Tan[c + d*x]^3)/Sqrt[a + b*Tan[c + d*x]] + (2*(8*a^2 - 5*b^2)*Sqrt[a + b* 
Tan[c + d*x]])/b^3)/(5*b*d)
 

Rubi [A] (warning: unable to verify)

Time = 1.68 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.01, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.783, Rules used = {3042, 4048, 27, 3042, 4130, 27, 3042, 4130, 27, 3042, 4113, 3042, 4022, 3042, 4020, 25, 73, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4048

\(\displaystyle \frac {2 \int \frac {\tan ^2(c+d x) \left (6 a^2-b \tan (c+d x) a+\left (6 a^2+b^2\right ) \tan ^2(c+d x)\right )}{2 \sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\tan ^2(c+d x) \left (6 a^2-b \tan (c+d x) a+\left (6 a^2+b^2\right ) \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\tan (c+d x)^2 \left (6 a^2-b \tan (c+d x) a+\left (6 a^2+b^2\right ) \tan (c+d x)^2\right )}{\sqrt {a+b \tan (c+d x)}}dx}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {\frac {2 \int -\frac {\tan (c+d x) \left (5 \tan (c+d x) b^3+3 a \left (8 a^2+3 b^2\right ) \tan ^2(c+d x)+4 a \left (6 a^2+b^2\right )\right )}{2 \sqrt {a+b \tan (c+d x)}}dx}{5 b}+\frac {2 \left (6 a^2+b^2\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 \left (6 a^2+b^2\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\int \frac {\tan (c+d x) \left (5 \tan (c+d x) b^3+3 a \left (8 a^2+3 b^2\right ) \tan ^2(c+d x)+4 a \left (6 a^2+b^2\right )\right )}{\sqrt {a+b \tan (c+d x)}}dx}{5 b}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 \left (6 a^2+b^2\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\int \frac {\tan (c+d x) \left (5 \tan (c+d x) b^3+3 a \left (8 a^2+3 b^2\right ) \tan (c+d x)^2+4 a \left (6 a^2+b^2\right )\right )}{\sqrt {a+b \tan (c+d x)}}dx}{5 b}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {\frac {2 \left (6 a^2+b^2\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 \int -\frac {3 \left (5 a \tan (c+d x) b^3+\left (16 a^4+6 b^2 a^2-5 b^4\right ) \tan ^2(c+d x)+2 a^2 \left (8 a^2+3 b^2\right )\right )}{2 \sqrt {a+b \tan (c+d x)}}dx}{3 b}+\frac {2 a \left (8 a^2+3 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{b d}}{5 b}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 \left (6 a^2+b^2\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 a \left (8 a^2+3 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {5 a \tan (c+d x) b^3+\left (16 a^4+6 b^2 a^2-5 b^4\right ) \tan ^2(c+d x)+2 a^2 \left (8 a^2+3 b^2\right )}{\sqrt {a+b \tan (c+d x)}}dx}{b}}{5 b}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 \left (6 a^2+b^2\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 a \left (8 a^2+3 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {5 a \tan (c+d x) b^3+\left (16 a^4+6 b^2 a^2-5 b^4\right ) \tan (c+d x)^2+2 a^2 \left (8 a^2+3 b^2\right )}{\sqrt {a+b \tan (c+d x)}}dx}{b}}{5 b}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {\frac {2 \left (6 a^2+b^2\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 a \left (8 a^2+3 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {5 b^4+5 a \tan (c+d x) b^3}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (16 a^4+6 a^2 b^2-5 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b}}{5 b}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 \left (6 a^2+b^2\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 a \left (8 a^2+3 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\int \frac {5 b^4+5 a \tan (c+d x) b^3}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (16 a^4+6 a^2 b^2-5 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b}}{5 b}}{b \left (a^2+b^2\right )}-\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\)

\(\Big \downarrow \) 4022

\(\displaystyle -\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {2 \left (6 a^2+b^2\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 a \left (8 a^2+3 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\frac {5}{2} b^3 (b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {5}{2} b^3 (-b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (16 a^4+6 a^2 b^2-5 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b}}{5 b}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {2 \left (6 a^2+b^2\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 a \left (8 a^2+3 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\frac {5}{2} b^3 (b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {5}{2} b^3 (-b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (16 a^4+6 a^2 b^2-5 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b}}{5 b}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 4020

\(\displaystyle -\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {2 \left (6 a^2+b^2\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 a \left (8 a^2+3 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{b d}-\frac {-\frac {5 i b^3 (-b+i a) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {5 i b^3 (b+i a) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (16 a^4+6 a^2 b^2-5 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b}}{5 b}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {2 \left (6 a^2+b^2\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 a \left (8 a^2+3 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\frac {5 i b^3 (-b+i a) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {5 i b^3 (b+i a) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (16 a^4+6 a^2 b^2-5 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b}}{5 b}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {2 \left (6 a^2+b^2\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 a \left (8 a^2+3 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\frac {5 b^2 (b+i a) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}-\frac {5 b^2 (-b+i a) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{d}+\frac {2 \left (16 a^4+6 a^2 b^2-5 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{b}}{5 b}}{b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {2 a^2 \tan ^3(c+d x)}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {2 \left (6 a^2+b^2\right ) \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 a \left (8 a^2+3 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{b d}-\frac {\frac {2 \left (16 a^4+6 a^2 b^2-5 b^4\right ) \sqrt {a+b \tan (c+d x)}}{b d}-\frac {5 b^3 (-b+i a) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {5 b^3 (b+i a) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}}{b}}{5 b}}{b \left (a^2+b^2\right )}\)

Input:

Int[Tan[c + d*x]^5/(a + b*Tan[c + d*x])^(3/2),x]
 

Output:

(-2*a^2*Tan[c + d*x]^3)/(b*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]) + ((2*( 
6*a^2 + b^2)*Tan[c + d*x]^2*Sqrt[a + b*Tan[c + d*x]])/(5*b*d) - ((2*a*(8*a 
^2 + 3*b^2)*Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(b*d) - ((-5*(I*a - b)* 
b^3*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + (5*b^3*(I*a + 
b)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d) + (2*(16*a^4 + 6* 
a^2*b^2 - 5*b^4)*Sqrt[a + b*Tan[c + d*x]])/(b*d))/b)/(5*b))/(b*(a^2 + b^2) 
)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042

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

rule 4020

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f)   Subst[Int[(a + (b/d)*x)^m/(d^2 + 
c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
 

rule 4022

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2   Int[(a + b*Tan[e + f*x])^m 
*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]
 

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 4113

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + 
 b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si 
mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && 
NeQ[A*b^2 - a*b*B + a^2*C, 0] &&  !LeQ[m, -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])))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1831\) vs. \(2(252)=504\).

Time = 0.26 (sec) , antiderivative size = 1832, normalized size of antiderivative = 6.50

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1832\)
default	\(\text {Expression too large to display}\)	\(1832\)

Input:

int(tan(d*x+c)^5/(a+b*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-2/d/b^4*a*(a+b*tan(d*x+c))^(3/2)+6/d/b^4*a^2*(a+b*tan(d*x+c))^(1/2)-1/2/d 
*b^2/(a^2+b^2)^(5/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2) 
^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a+2/d*b^2 
/(a^2+b^2)^(5/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^ 
(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^2+1/ 
d*b^2/(a^2+b^2)^2/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c)) 
^(1/2)-(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+1/2 
/d*b^2/(a^2+b^2)^(5/2)*ln(b*tan(d*x+c)+a-(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^ 
2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a+1/d*b 
^2/(a^2+b^2)^2/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1 
/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+2/d*b^ 
2/(a^2+b^2)^(5/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c)) 
^(1/2)-(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^2-2 
/d/b^2*(a+b*tan(d*x+c))^(1/2)+2/5/d/b^4*(a+b*tan(d*x+c))^(5/2)+2/d*b^4/(a^ 
2+b^2)^(5/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2 
)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))-1/d*b^2/(a 
^2+b^2)^(3/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/ 
2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))+2/d/b^4*a 
^5/(a^2+b^2)/(a+b*tan(d*x+c))^(1/2)+1/4/d/(a^2+b^2)^2*ln(b*tan(d*x+c)+a+(a 
+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2076 vs. \(2 (248) = 496\).

Time = 0.14 (sec) , antiderivative size = 2076, normalized size of antiderivative = 7.36 \[ \int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-1/10*(5*((a^2*b^5 + b^7)*d*tan(d*x + c) + (a^3*b^4 + a*b^6)*d)*sqrt(((a^6 
 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/(( 
a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^ 
12)*d^4)) + a^3 - 3*a*b^2)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))*log( 
-(3*a^2 - b^2)*sqrt(b*tan(d*x + c) + a) + (2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 
+ a*b^6)*d^3*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15* 
a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) - (3*a^4 - 4* 
a^2*b^2 + b^4)*d)*sqrt(((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(9*a 
^4*b^2 - 6*a^2*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 
15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + a^3 - 3*a*b^2)/((a^6 + 3*a^4*b^2 + 
 3*a^2*b^4 + b^6)*d^2))) - 5*((a^2*b^5 + b^7)*d*tan(d*x + c) + (a^3*b^4 + 
a*b^6)*d)*sqrt(((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(9*a^4*b^2 - 
 6*a^2*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b 
^8 + 6*a^2*b^10 + b^12)*d^4)) + a^3 - 3*a*b^2)/((a^6 + 3*a^4*b^2 + 3*a^2*b 
^4 + b^6)*d^2))*log(-(3*a^2 - b^2)*sqrt(b*tan(d*x + c) + a) - (2*(a^7 + 3* 
a^5*b^2 + 3*a^3*b^4 + a*b^6)*d^3*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/((a^1 
2 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12) 
*d^4)) - (3*a^4 - 4*a^2*b^2 + b^4)*d)*sqrt(((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + 
 b^6)*d^2*sqrt(-(9*a^4*b^2 - 6*a^2*b^4 + b^6)/((a^12 + 6*a^10*b^2 + 15*a^8 
*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) + a^3 - 3*a*b...
 

Sympy [F]

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

integrate(tan(d*x+c)**5/(a+b*tan(d*x+c))**(3/2),x)
 

Output:

Integral(tan(c + d*x)**5/(a + b*tan(c + d*x))**(3/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

integrate(tan(d*x + c)^5/(b*tan(d*x + c) + a)^(3/2), x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 13.08 (sec) , antiderivative size = 2930, normalized size of antiderivative = 10.39 \[ \int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

int(tan(c + d*x)^5/(a + b*tan(c + d*x))^(3/2),x)
 

Output:

((8*a^2)/(b^4*d) - (2*(a^2 + b^2))/(b^4*d))*(a + b*tan(c + d*x))^(1/2) - a 
tan(((1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i))^(1/2)*(((1/(a 
^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i))^(1/2)*(32*a^6*b^6*d^4 - 
 48*a^2*b^10*d^4 - 32*a^4*b^8*d^4 - 16*b^12*d^4 + 48*a^8*b^4*d^4 + 16*a^10 
*b^2*d^4 + ((1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i))^(1/2)* 
(a + b*tan(c + d*x))^(1/2)*(64*a*b^12*d^5 + 320*a^3*b^10*d^5 + 640*a^5*b^8 
*d^5 + 640*a^7*b^6*d^5 + 320*a^9*b^4*d^5 + 64*a^11*b^2*d^5))/4))/2 + ((a + 
 b*tan(c + d*x))^(1/2)*(16*b^10*d^3 + 32*a^2*b^8*d^3 - 32*a^6*b^4*d^3 - 16 
*a^8*b^2*d^3))/2)*1i + (1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2* 
3i))^(1/2)*(((1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2*b*d^2*3i))^(1/2) 
*(16*b^12*d^4 + 48*a^2*b^10*d^4 + 32*a^4*b^8*d^4 - 32*a^6*b^6*d^4 - 48*a^8 
*b^4*d^4 - 16*a^10*b^2*d^4 + ((1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^2 
*b*d^2*3i))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^12*d^5 + 320*a^3*b^10 
*d^5 + 640*a^5*b^8*d^5 + 640*a^7*b^6*d^5 + 320*a^9*b^4*d^5 + 64*a^11*b^2*d 
^5))/4))/2 + ((a + b*tan(c + d*x))^(1/2)*(16*b^10*d^3 + 32*a^2*b^8*d^3 - 3 
2*a^6*b^4*d^3 - 16*a^8*b^2*d^3))/2)*1i)/((1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^ 
2*d^2 - a^2*b*d^2*3i))^(1/2)*(((1/(a^3*d^2 + b^3*d^2*1i - 3*a*b^2*d^2 - a^ 
2*b*d^2*3i))^(1/2)*(16*b^12*d^4 + 48*a^2*b^10*d^4 + 32*a^4*b^8*d^4 - 32*a^ 
6*b^6*d^4 - 48*a^8*b^4*d^4 - 16*a^10*b^2*d^4 + ((1/(a^3*d^2 + b^3*d^2*1i - 
 3*a*b^2*d^2 - a^2*b*d^2*3i))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^...
 

Reduce [F]

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

int(tan(d*x+c)^5/(a+b*tan(d*x+c))^(3/2),x)
 

Output:

int((sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**5)/(tan(c + d*x)**2*b**2 + 2*t 
an(c + d*x)*a*b + a**2),x)