\(\int \frac {\tan ^5(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\) [528]

Optimal result

Integrand size = 23, antiderivative size = 229 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b} d}-\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{105 b^4 d}+\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{105 b^3 d}-\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{35 b^2 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d} \] Output:

-arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(1/2)/d-arctanh((a+ 
b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(1/2)/d-4/105*a*(24*a^2-35*b^2) 
*(a+b*tan(d*x+c))^(1/2)/b^4/d+2/105*(24*a^2-35*b^2)*tan(d*x+c)*(a+b*tan(d* 
x+c))^(1/2)/b^3/d-12/35*a*tan(d*x+c)^2*(a+b*tan(d*x+c))^(1/2)/b^2/d+2/7*ta 
n(d*x+c)^3*(a+b*tan(d*x+c))^(1/2)/b/d
 

Mathematica [A] (verified)

Time = 3.37 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.71 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=-\frac {\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b}}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b}}+\frac {2 \sqrt {a+b \tan (c+d x)} \left (48 a^3-88 a b^2+\left (-24 a^2 b+50 b^3\right ) \tan (c+d x)+3 b^2 \sec ^2(c+d x) (6 a-5 b \tan (c+d x))\right )}{105 b^4}}{d} \] Input:

Integrate[Tan[c + d*x]^5/Sqrt[a + b*Tan[c + d*x]],x]
 

Output:

-((ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]]/Sqrt[a - I*b] + ArcTanh 
[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]]/Sqrt[a + I*b] + (2*Sqrt[a + b*Tan 
[c + d*x]]*(48*a^3 - 88*a*b^2 + (-24*a^2*b + 50*b^3)*Tan[c + d*x] + 3*b^2* 
Sec[c + d*x]^2*(6*a - 5*b*Tan[c + d*x])))/(105*b^4))/d)
 

Rubi [A] (warning: unable to verify)

Time = 1.42 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.07, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.826, Rules used = {3042, 4049, 27, 3042, 4130, 27, 3042, 4131, 27, 3042, 4113, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

\(\displaystyle \frac {2 \int -\frac {\tan ^2(c+d x) \left (6 a \tan ^2(c+d x)+7 b \tan (c+d x)+6 a\right )}{2 \sqrt {a+b \tan (c+d x)}}dx}{7 b}+\frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\int \frac {\tan ^2(c+d x) \left (6 a \tan ^2(c+d x)+7 b \tan (c+d x)+6 a\right )}{\sqrt {a+b \tan (c+d x)}}dx}{7 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\int \frac {\tan (c+d x)^2 \left (6 a \tan (c+d x)^2+7 b \tan (c+d x)+6 a\right )}{\sqrt {a+b \tan (c+d x)}}dx}{7 b}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\frac {2 \int -\frac {\tan (c+d x) \left (24 a^2+\left (24 a^2-35 b^2\right ) \tan ^2(c+d x)\right )}{2 \sqrt {a+b \tan (c+d x)}}dx}{5 b}+\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}}{7 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\int \frac {\tan (c+d x) \left (24 a^2+\left (24 a^2-35 b^2\right ) \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}}dx}{5 b}}{7 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\int \frac {\tan (c+d x) \left (24 a^2+\left (24 a^2-35 b^2\right ) \tan (c+d x)^2\right )}{\sqrt {a+b \tan (c+d x)}}dx}{5 b}}{7 b}\)

\(\Big \downarrow \) 4131

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 \int -\frac {-105 \tan (c+d x) b^3+2 a \left (24 a^2-35 b^2\right ) \tan ^2(c+d x)+2 a \left (24 a^2-35 b^2\right )}{2 \sqrt {a+b \tan (c+d x)}}dx}{3 b}+\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}}{5 b}}{7 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\int \frac {-105 \tan (c+d x) b^3+2 a \left (24 a^2-35 b^2\right ) \tan ^2(c+d x)+2 a \left (24 a^2-35 b^2\right )}{\sqrt {a+b \tan (c+d x)}}dx}{3 b}}{5 b}}{7 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\int \frac {-105 \tan (c+d x) b^3+2 a \left (24 a^2-35 b^2\right ) \tan (c+d x)^2+2 a \left (24 a^2-35 b^2\right )}{\sqrt {a+b \tan (c+d x)}}dx}{3 b}}{5 b}}{7 b}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\int -\frac {105 b^3 \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}}{3 b}}{5 b}}{7 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}-105 b^3 \int \frac {\tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx}{3 b}}{5 b}}{7 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}-105 b^3 \int \frac {\tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx}{3 b}}{5 b}}{7 b}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}-105 b^3 \left (\frac {1}{2} i \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} i \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx\right )}{3 b}}{5 b}}{7 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}-105 b^3 \left (\frac {1}{2} i \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} i \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx\right )}{3 b}}{5 b}}{7 b}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}-105 b^3 \left (\frac {\int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {\int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}\right )}{3 b}}{5 b}}{7 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}-105 b^3 \left (-\frac {\int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {\int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}\right )}{3 b}}{5 b}}{7 b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}-105 b^3 \left (\frac {i \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}-\frac {i \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}\right )}{3 b}}{5 b}}{7 b}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)}}{7 b d}-\frac {\frac {12 a \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {\frac {2 \left (24 a^2-35 b^2\right ) \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d}-\frac {\frac {4 a \left (24 a^2-35 b^2\right ) \sqrt {a+b \tan (c+d x)}}{b d}-105 b^3 \left (\frac {i \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}-\frac {i \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}\right )}{3 b}}{5 b}}{7 b}\)

Input:

Int[Tan[c + d*x]^5/Sqrt[a + b*Tan[c + d*x]],x]
 

Output:

(2*Tan[c + d*x]^3*Sqrt[a + b*Tan[c + d*x]])/(7*b*d) - ((12*a*Tan[c + d*x]^ 
2*Sqrt[a + b*Tan[c + d*x]])/(5*b*d) - ((2*(24*a^2 - 35*b^2)*Tan[c + d*x]*S 
qrt[a + b*Tan[c + d*x]])/(3*b*d) - (-105*b^3*(((-I)*ArcTan[Tan[c + d*x]/Sq 
rt[a - I*b]])/(Sqrt[a - I*b]*d) + (I*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/( 
Sqrt[a + I*b]*d)) + (4*a*(24*a^2 - 35*b^2)*Sqrt[a + b*Tan[c + d*x]])/(b*d) 
)/(3*b))/(5*b))/(7*b)
 

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 4049

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])) 
)
 

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])))
 

rule 4131

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_)*((A_.) + (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 
- b*C)*(m + n + 1)*Tan[e + f*x] - C*m*(b*c - a*d)*Tan[e + f*x]^2, x], x], x 
] /; FreeQ[{a, b, c, d, e, f, A, 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. \(444\) vs. \(2(197)=394\).

Time = 0.23 (sec) , antiderivative size = 445, normalized size of antiderivative = 1.94

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {6 a \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+2 a^{2} \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}-\frac {2 b^{2} \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a^{3} \sqrt {a +b \tan \left (d x +c \right )}+2 a \,b^{2} \sqrt {a +b \tan \left (d x +c \right )}-2 b^{4} \left (\frac {-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (\sqrt {a^{2}+b^{2}}-a \right ) \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{4 \sqrt {a^{2}+b^{2}}}+\frac {\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{4 \sqrt {a^{2}+b^{2}}}\right )}{d \,b^{4}}\)	\(445\)
default	\(\frac {\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {6 a \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+2 a^{2} \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}-\frac {2 b^{2} \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a^{3} \sqrt {a +b \tan \left (d x +c \right )}+2 a \,b^{2} \sqrt {a +b \tan \left (d x +c \right )}-2 b^{4} \left (\frac {-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (\sqrt {a^{2}+b^{2}}-a \right ) \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{4 \sqrt {a^{2}+b^{2}}}+\frac {\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{4 \sqrt {a^{2}+b^{2}}}\right )}{d \,b^{4}}\)	\(445\)

Input:

int(tan(d*x+c)^5/(a+b*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 789 vs. \(2 (193) = 386\).

Time = 0.11 (sec) , antiderivative size = 789, normalized size of antiderivative = 3.45 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

1/210*(105*b^4*d*sqrt(((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)* 
d^4)) + a)/((a^2 + b^2)*d^2))*log(((a^2 + b^2)*d^3*sqrt(-b^2/((a^4 + 2*a^2 
*b^2 + b^4)*d^4)) - a*d)*sqrt(((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 
 + b^4)*d^4)) + a)/((a^2 + b^2)*d^2)) + sqrt(b*tan(d*x + c) + a)) - 105*b^ 
4*d*sqrt(((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + a)/(( 
a^2 + b^2)*d^2))*log(-((a^2 + b^2)*d^3*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)* 
d^4)) - a*d)*sqrt(((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4) 
) + a)/((a^2 + b^2)*d^2)) + sqrt(b*tan(d*x + c) + a)) - 105*b^4*d*sqrt(-(( 
a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a)/((a^2 + b^2)* 
d^2))*log(((a^2 + b^2)*d^3*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + a*d) 
*sqrt(-((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a)/((a^ 
2 + b^2)*d^2)) + sqrt(b*tan(d*x + c) + a)) + 105*b^4*d*sqrt(-((a^2 + b^2)* 
d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a)/((a^2 + b^2)*d^2))*log(- 
((a^2 + b^2)*d^3*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + a*d)*sqrt(-((a 
^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a)/((a^2 + b^2)*d 
^2)) + sqrt(b*tan(d*x + c) + a)) + 4*(15*b^3*tan(d*x + c)^3 - 18*a*b^2*tan 
(d*x + c)^2 - 48*a^3 + 70*a*b^2 + (24*a^2*b - 35*b^3)*tan(d*x + c))*sqrt(b 
*tan(d*x + c) + a))/(b^4*d)
 

Sympy [F]

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

integrate(tan(d*x+c)**5/(a+b*tan(d*x+c))**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

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

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[0,24,9]%%%}+%%%{10,[0,22,9]%%%}+%%%{45,[0,20,9]%%%} 
+%%%{120,
 

Mupad [B] (verification not implemented)

Time = 9.49 (sec) , antiderivative size = 938, normalized size of antiderivative = 4.10 \[ \int \frac {\tan ^5(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx =\text {Too large to display} \] Input:

int(tan(c + d*x)^5/(a + b*tan(c + d*x))^(1/2),x)
 

Output:

(a + b*tan(c + d*x))^(1/2)*(2*a*((8*a^2)/(b^4*d) - (2*(a^2 + b^2))/(b^4*d) 
) - (20*a^3)/(b^4*d) + (6*a*(a^2 + b^2))/(b^4*d)) - atan((a^2*b^2*(a/(4*a^ 
2*d^2 + 4*b^2*d^2) - (b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*(a + b*tan(c + 
d*x))^(1/2)*128i)/((64*b^4)/d + (64*a^2*b^2)/d - (256*a^2*b^4*d^2)/(4*a^2* 
d^3 + 4*b^2*d^3) + (a^3*b^3*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3) - (256*a^4*b 
^2*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a*b^5*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3) 
) - (b^2*(a/(4*a^2*d^2 + 4*b^2*d^2) - (b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2 
)*(a + b*tan(c + d*x))^(1/2)*32i)/((16*b^2)/d - (64*a^2*b^2*d^2)/(4*a^2*d^ 
3 + 4*b^2*d^3) + (a*b^3*d^2*64i)/(4*a^2*d^3 + 4*b^2*d^3)) + (128*a*b^3*(a/ 
(4*a^2*d^2 + 4*b^2*d^2) - (b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*(a + b*tan 
(c + d*x))^(1/2))/((64*b^4)/d + (64*a^2*b^2)/d - (256*a^2*b^4*d^2)/(4*a^2* 
d^3 + 4*b^2*d^3) + (a^3*b^3*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3) - (256*a^4*b 
^2*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a*b^5*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3) 
))*((a - b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*2i + ((8*a^2)/(3*b^4*d) - (2 
*(a^2 + b^2))/(3*b^4*d))*(a + b*tan(c + d*x))^(3/2) + atan((b^2*(1/(a*d^2 
- b*d^2*1i))^(1/2)*(a + b*tan(c + d*x))^(1/2)*16i)/((16*b^2)/d - (16*a*b^2 
*d^2)/(a*d^3 - b*d^3*1i)) + (a*b^2*(1/(a*d^2 - b*d^2*1i))^(1/2)*(a + b*tan 
(c + d*x))^(1/2)*16i)/((b^3*16i)/d - (16*a*b^2)/d - (a*b^3*d^2*16i)/(a*d^3 
 - b*d^3*1i) + (16*a^2*b^2*d^2)/(a*d^3 - b*d^3*1i)))*(1/(a*d^2 - b*d^2*1i) 
)^(1/2)*1i + (2*(a + b*tan(c + d*x))^(7/2))/(7*b^4*d) - (6*a*(a + b*tan...
 

Reduce [F]

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

int(tan(d*x+c)^5/(a+b*tan(d*x+c))^(1/2),x)
 

Output:

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