\(\int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx\) [562]

Optimal result

Integrand size = 23, antiderivative size = 216 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\left (a^2-2 a b-b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2-2 a b-b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2+2 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {4 a b \sqrt {\tan (c+d x)}}{d}+\frac {2 \left (a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a b \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 b^2 \tan ^{\frac {7}{2}}(c+d x)}{7 d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.77 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.62 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {-105 (-1)^{3/4} (a-i b)^2 \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+105 (-1)^{3/4} (a+i b)^2 \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 \sqrt {\tan (c+d x)} \left (-210 a b+35 \left (a^2-b^2\right ) \tan (c+d x)+42 a b \tan ^2(c+d x)+15 b^2 \tan ^3(c+d x)\right )}{105 d} \] Input:

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

Output:

(-105*(-1)^(3/4)*(a - I*b)^2*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 105*( 
-1)^(3/4)*(a + I*b)^2*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 2*Sqrt[Tan[ 
c + d*x]]*(-210*a*b + 35*(a^2 - b^2)*Tan[c + d*x] + 42*a*b*Tan[c + d*x]^2 
+ 15*b^2*Tan[c + d*x]^3))/(105*d)
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.14, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.783, Rules used = {3042, 4026, 3042, 4011, 3042, 4011, 3042, 4011, 3042, 4017, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4026

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^{5/2} \left (a^2+2 b \tan (c+d x) a-b^2\right )dx+\frac {2 b^2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

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

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

\(\displaystyle \frac {2 \left (\frac {1}{2} \left (a^2+2 a b-b^2\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )-\frac {1}{2} \left (a^2-2 a b-b^2\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 \left (a^2-b^2\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a b \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {4 a b \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}\)

Input:

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

Output:

(2*(-1/2*((a^2 - 2*a*b - b^2)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sq 
rt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2])) + ((a^2 + 2*a*b 
- b^2)*(-1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/Sqrt[2] + 
Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2])))/2))/d - ( 
4*a*b*Sqrt[Tan[c + d*x]])/d + (2*(a^2 - b^2)*Tan[c + d*x]^(3/2))/(3*d) + ( 
4*a*b*Tan[c + d*x]^(5/2))/(5*d) + (2*b^2*Tan[c + d*x]^(7/2))/(7*d)
 

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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 1482

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
a*c, 2]}, Simp[(d*q + a*e)/(2*a*c)   Int[(q + c*x^2)/(a + c*x^4), x], x] + 
Simp[(d*q - a*e)/(2*a*c)   Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a 
, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- 
a)*c]
 

rule 3042

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

rule 4011

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

rule 4017

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[2/f   Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq 
rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & 
& NeQ[c^2 + d^2, 0]
 

rule 4026

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.16

method	result	size
derivativedivides	\(\frac {\frac {2 b^{2} \tan \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {4 a b \tan \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 a^{2} \tan \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {2 b^{2} \tan \left (d x +c \right )^{\frac {3}{2}}}{3}-4 a b \sqrt {\tan \left (d x +c \right )}+\frac {a b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{2}+\frac {\left (-a^{2}+b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{d}\)	\(250\)
default	\(\frac {\frac {2 b^{2} \tan \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {4 a b \tan \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 a^{2} \tan \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {2 b^{2} \tan \left (d x +c \right )^{\frac {3}{2}}}{3}-4 a b \sqrt {\tan \left (d x +c \right )}+\frac {a b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{2}+\frac {\left (-a^{2}+b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{d}\)	\(250\)
parts	\(\frac {a^{2} \left (\frac {2 \tan \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}+\frac {b^{2} \left (\frac {2 \tan \left (d x +c \right )^{\frac {7}{2}}}{7}-\frac {2 \tan \left (d x +c \right )^{\frac {3}{2}}}{3}+\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}+\frac {2 a b \left (\frac {2 \tan \left (d x +c \right )^{\frac {5}{2}}}{5}-2 \sqrt {\tan \left (d x +c \right )}+\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(331\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 686 vs. \(2 (185) = 370\).

Time = 0.09 (sec) , antiderivative size = 686, normalized size of antiderivative = 3.18 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx =\text {Too large to display} \] Input:

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

Output:

1/210*(210*sqrt(1/2)*d*sqrt((a^4 - 4*a^3*b + 2*a^2*b^2 + 4*a*b^3 + b^4)/d^ 
2)*arctan(-(2*sqrt(1/2)*(a^2 + 2*a*b - b^2)*d*sqrt((a^4 - 4*a^3*b + 2*a^2* 
b^2 + 4*a*b^3 + b^4)/d^2)*sqrt(tan(d*x + c)) + d^2*sqrt((a^4 + 4*a^3*b + 2 
*a^2*b^2 - 4*a*b^3 + b^4)/d^2)*sqrt((a^4 - 4*a^3*b + 2*a^2*b^2 + 4*a*b^3 + 
 b^4)/d^2))/(a^4 - 6*a^2*b^2 + b^4)) + 210*sqrt(1/2)*d*sqrt((a^4 - 4*a^3*b 
 + 2*a^2*b^2 + 4*a*b^3 + b^4)/d^2)*arctan(-(2*sqrt(1/2)*(a^2 + 2*a*b - b^2 
)*d*sqrt((a^4 - 4*a^3*b + 2*a^2*b^2 + 4*a*b^3 + b^4)/d^2)*sqrt(tan(d*x + c 
)) - d^2*sqrt((a^4 + 4*a^3*b + 2*a^2*b^2 - 4*a*b^3 + b^4)/d^2)*sqrt((a^4 - 
 4*a^3*b + 2*a^2*b^2 + 4*a*b^3 + b^4)/d^2))/(a^4 - 6*a^2*b^2 + b^4)) - 105 
*sqrt(1/2)*d*sqrt((a^4 + 4*a^3*b + 2*a^2*b^2 - 4*a*b^3 + b^4)/d^2)*log(2*s 
qrt(1/2)*d*sqrt((a^4 + 4*a^3*b + 2*a^2*b^2 - 4*a*b^3 + b^4)/d^2)*sqrt(tan( 
d*x + c)) - a^2 - 2*a*b + b^2 - (a^2 + 2*a*b - b^2)*tan(d*x + c)) + 105*sq 
rt(1/2)*d*sqrt((a^4 + 4*a^3*b + 2*a^2*b^2 - 4*a*b^3 + b^4)/d^2)*log(-2*sqr 
t(1/2)*d*sqrt((a^4 + 4*a^3*b + 2*a^2*b^2 - 4*a*b^3 + b^4)/d^2)*sqrt(tan(d* 
x + c)) - a^2 - 2*a*b + b^2 - (a^2 + 2*a*b - b^2)*tan(d*x + c)) + 4*(15*b^ 
2*tan(d*x + c)^3 + 42*a*b*tan(d*x + c)^2 - 210*a*b + 35*(a^2 - b^2)*tan(d* 
x + c))*sqrt(tan(d*x + c)))/d
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {120 \, b^{2} \tan \left (d x + c\right )^{\frac {7}{2}} + 336 \, a b \tan \left (d x + c\right )^{\frac {5}{2}} - 210 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 210 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 105 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 105 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 1680 \, a b \sqrt {\tan \left (d x + c\right )} + 280 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{\frac {3}{2}}}{420 \, d} \] Input:

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

Output:

1/420*(120*b^2*tan(d*x + c)^(7/2) + 336*a*b*tan(d*x + c)^(5/2) - 210*sqrt( 
2)*(a^2 - 2*a*b - b^2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c))) 
) - 210*sqrt(2)*(a^2 - 2*a*b - b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt( 
tan(d*x + c)))) + 105*sqrt(2)*(a^2 + 2*a*b - b^2)*log(sqrt(2)*sqrt(tan(d*x 
 + c)) + tan(d*x + c) + 1) - 105*sqrt(2)*(a^2 + 2*a*b - b^2)*log(-sqrt(2)* 
sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - 1680*a*b*sqrt(tan(d*x + c)) + 280 
*(a^2 - b^2)*tan(d*x + c)^(3/2))/d
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 4.36 (sec) , antiderivative size = 995, normalized size of antiderivative = 4.61 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx=\text {Too large to display} \] Input:

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

Output:

tan(c + d*x)^(3/2)*((2*a^2)/(3*d) - (2*b^2)/(3*d)) - atan((a^4*tan(c + d*x 
)^(1/2)*((a^3*b)/d^2 - (b^4*1i)/(4*d^2) - (a*b^3)/d^2 - (a^4*1i)/(4*d^2) + 
 (a^2*b^2*3i)/(2*d^2))^(1/2)*32i)/((16*a^6)/d - (16*b^6)/d + (a*b^5*32i)/d 
 + (a^5*b*32i)/d + (112*a^2*b^4)/d - (a^3*b^3*192i)/d - (112*a^4*b^2)/d) + 
 (b^4*tan(c + d*x)^(1/2)*((a^3*b)/d^2 - (b^4*1i)/(4*d^2) - (a*b^3)/d^2 - ( 
a^4*1i)/(4*d^2) + (a^2*b^2*3i)/(2*d^2))^(1/2)*32i)/((16*a^6)/d - (16*b^6)/ 
d + (a*b^5*32i)/d + (a^5*b*32i)/d + (112*a^2*b^4)/d - (a^3*b^3*192i)/d - ( 
112*a^4*b^2)/d) - (a^2*b^2*tan(c + d*x)^(1/2)*((a^3*b)/d^2 - (b^4*1i)/(4*d 
^2) - (a*b^3)/d^2 - (a^4*1i)/(4*d^2) + (a^2*b^2*3i)/(2*d^2))^(1/2)*192i)/( 
(16*a^6)/d - (16*b^6)/d + (a*b^5*32i)/d + (a^5*b*32i)/d + (112*a^2*b^4)/d 
- (a^3*b^3*192i)/d - (112*a^4*b^2)/d))*(-(4*a*b^3 - 4*a^3*b + a^4*1i + b^4 
*1i - a^2*b^2*6i)/(4*d^2))^(1/2)*2i + atan((a^4*tan(c + d*x)^(1/2)*((a^4*1 
i)/(4*d^2) + (b^4*1i)/(4*d^2) - (a*b^3)/d^2 + (a^3*b)/d^2 - (a^2*b^2*3i)/( 
2*d^2))^(1/2)*32i)/((16*b^6)/d - (16*a^6)/d + (a*b^5*32i)/d + (a^5*b*32i)/ 
d - (112*a^2*b^4)/d - (a^3*b^3*192i)/d + (112*a^4*b^2)/d) + (b^4*tan(c + d 
*x)^(1/2)*((a^4*1i)/(4*d^2) + (b^4*1i)/(4*d^2) - (a*b^3)/d^2 + (a^3*b)/d^2 
 - (a^2*b^2*3i)/(2*d^2))^(1/2)*32i)/((16*b^6)/d - (16*a^6)/d + (a*b^5*32i) 
/d + (a^5*b*32i)/d - (112*a^2*b^4)/d - (a^3*b^3*192i)/d + (112*a^4*b^2)/d) 
 - (a^2*b^2*tan(c + d*x)^(1/2)*((a^4*1i)/(4*d^2) + (b^4*1i)/(4*d^2) - (a*b 
^3)/d^2 + (a^3*b)/d^2 - (a^2*b^2*3i)/(2*d^2))^(1/2)*192i)/((16*b^6)/d -...
 

Reduce [F]

\[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {4 \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2} a b -20 \sqrt {\tan \left (d x +c \right )}\, a b +10 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )}d x \right ) a b d +5 \left (\int \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{4}d x \right ) b^{2} d +5 \left (\int \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}d x \right ) a^{2} d}{5 d} \] Input:

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

Output:

(4*sqrt(tan(c + d*x))*tan(c + d*x)**2*a*b - 20*sqrt(tan(c + d*x))*a*b + 10 
*int(sqrt(tan(c + d*x))/tan(c + d*x),x)*a*b*d + 5*int(sqrt(tan(c + d*x))*t 
an(c + d*x)**4,x)*b**2*d + 5*int(sqrt(tan(c + d*x))*tan(c + d*x)**2,x)*a** 
2*d)/(5*d)