3.6.60 \(\int \frac {a+b \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [560]

3.6.60.1 Optimal result
3.6.60.2 Mathematica [C] (verified)
3.6.60.3 Rubi [A] (verified)
3.6.60.4 Maple [A] (verified)
3.6.60.5 Fricas [B] (verification not implemented)
3.6.60.6 Sympy [F]
3.6.60.7 Maxima [A] (verification not implemented)
3.6.60.8 Giac [F(-1)]
3.6.60.9 Mupad [B] (verification not implemented)

3.6.60.1 Optimal result

Integrand size = 21, antiderivative size = 184 \[ \int \frac {a+b \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(a+b) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a+b) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a-b) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b}{d \sqrt {\tan (c+d x)}} \]

output
-1/2*(a+b)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)-1/2*(a+b)*arctan( 
1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)+1/4*(a-b)*ln(1-2^(1/2)*tan(d*x+c)^(1 
/2)+tan(d*x+c))/d*2^(1/2)-1/4*(a-b)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+ 
c))/d*2^(1/2)-2*b/d/tan(d*x+c)^(1/2)-2/3*a/d/tan(d*x+c)^(3/2)
 
3.6.60.2 Mathematica [C] (verified)

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

Time = 0.11 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.39 \[ \int \frac {a+b \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {-\left ((a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-i \tan (c+d x)\right )\right )-(a-i b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},i \tan (c+d x)\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]

input
Integrate[(a + b*Tan[c + d*x])/Tan[c + d*x]^(5/2),x]
 
output
(-((a + I*b)*Hypergeometric2F1[-3/2, 1, -1/2, (-I)*Tan[c + d*x]]) - (a - I 
*b)*Hypergeometric2F1[-3/2, 1, -1/2, I*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3 
/2))
 
3.6.60.3 Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.98, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4012, 3042, 4012, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

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

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

input
Int[(a + b*Tan[c + d*x])/Tan[c + d*x]^(5/2),x]
 
output
(-2*(((a + b)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[ 
1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]))/2 + ((a - b)*(-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 - (2*a)/(3*d*Tan[c + d*x]^(3/ 
2)) - (2*b)/(d*Sqrt[Tan[c + d*x]])
 

3.6.60.3.1 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 4012
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ 
(f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2)   Int[(a + b*Tan[e + f*x] 
)^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a 
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 
]
 

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]
 
3.6.60.4 Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.09

method result size
derivativedivides \(\frac {-\frac {a \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2 a}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 b}{\sqrt {\tan \left (d x +c \right )}}}{d}\) \(200\)
default \(\frac {-\frac {a \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2 a}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 b}{\sqrt {\tan \left (d x +c \right )}}}{d}\) \(200\)
parts \(\frac {a \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {b \left (-\frac {2}{\sqrt {\tan \left (d x +c \right )}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\) \(204\)

input
int((a+b*tan(d*x+c))/tan(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 
output
1/d*(-1/4*a*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2) 
*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arct 
an(-1+2^(1/2)*tan(d*x+c)^(1/2)))-1/4*b*2^(1/2)*(ln((1-2^(1/2)*tan(d*x+c)^( 
1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/ 
2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))-2/3*a/tan(d*x+ 
c)^(3/2)-2*b/tan(d*x+c)^(1/2))
 
3.6.60.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 631 vs. \(2 (150) = 300\).

Time = 0.24 (sec) , antiderivative size = 631, normalized size of antiderivative = 3.43 \[ \int \frac {a+b \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {3 \, d \sqrt {-\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} + 2 \, a b}{d^{2}}} \log \left ({\left (b d^{3} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} + {\left (a^{3} - a b^{2}\right )} d\right )} \sqrt {-\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} + 2 \, a b}{d^{2}}} - {\left (a^{4} - b^{4}\right )} \sqrt {\tan \left (d x + c\right )}\right ) \tan \left (d x + c\right )^{2} - 3 \, d \sqrt {-\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} + 2 \, a b}{d^{2}}} \log \left (-{\left (b d^{3} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} + {\left (a^{3} - a b^{2}\right )} d\right )} \sqrt {-\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} + 2 \, a b}{d^{2}}} - {\left (a^{4} - b^{4}\right )} \sqrt {\tan \left (d x + c\right )}\right ) \tan \left (d x + c\right )^{2} - 3 \, d \sqrt {\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} - 2 \, a b}{d^{2}}} \log \left ({\left (b d^{3} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} - {\left (a^{3} - a b^{2}\right )} d\right )} \sqrt {\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} - 2 \, a b}{d^{2}}} - {\left (a^{4} - b^{4}\right )} \sqrt {\tan \left (d x + c\right )}\right ) \tan \left (d x + c\right )^{2} + 3 \, d \sqrt {\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} - 2 \, a b}{d^{2}}} \log \left (-{\left (b d^{3} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} - {\left (a^{3} - a b^{2}\right )} d\right )} \sqrt {\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} - 2 \, a b}{d^{2}}} - {\left (a^{4} - b^{4}\right )} \sqrt {\tan \left (d x + c\right )}\right ) \tan \left (d x + c\right )^{2} - 4 \, {\left (3 \, b \tan \left (d x + c\right ) + a\right )} \sqrt {\tan \left (d x + c\right )}}{6 \, d \tan \left (d x + c\right )^{2}} \]

input
integrate((a+b*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="fricas")
 
output
1/6*(3*d*sqrt(-(d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/d^4) + 2*a*b)/d^2)*log(( 
b*d^3*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/d^4) + (a^3 - a*b^2)*d)*sqrt(-(d^2*sqr 
t(-(a^4 - 2*a^2*b^2 + b^4)/d^4) + 2*a*b)/d^2) - (a^4 - b^4)*sqrt(tan(d*x + 
 c)))*tan(d*x + c)^2 - 3*d*sqrt(-(d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/d^4) + 
 2*a*b)/d^2)*log(-(b*d^3*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/d^4) + (a^3 - a*b^2 
)*d)*sqrt(-(d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/d^4) + 2*a*b)/d^2) - (a^4 - 
b^4)*sqrt(tan(d*x + c)))*tan(d*x + c)^2 - 3*d*sqrt((d^2*sqrt(-(a^4 - 2*a^2 
*b^2 + b^4)/d^4) - 2*a*b)/d^2)*log((b*d^3*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/d^ 
4) - (a^3 - a*b^2)*d)*sqrt((d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/d^4) - 2*a*b 
)/d^2) - (a^4 - b^4)*sqrt(tan(d*x + c)))*tan(d*x + c)^2 + 3*d*sqrt((d^2*sq 
rt(-(a^4 - 2*a^2*b^2 + b^4)/d^4) - 2*a*b)/d^2)*log(-(b*d^3*sqrt(-(a^4 - 2* 
a^2*b^2 + b^4)/d^4) - (a^3 - a*b^2)*d)*sqrt((d^2*sqrt(-(a^4 - 2*a^2*b^2 + 
b^4)/d^4) - 2*a*b)/d^2) - (a^4 - b^4)*sqrt(tan(d*x + c)))*tan(d*x + c)^2 - 
 4*(3*b*tan(d*x + c) + a)*sqrt(tan(d*x + c)))/(d*tan(d*x + c)^2)
 
3.6.60.6 Sympy [F]

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

input
integrate((a+b*tan(d*x+c))/tan(d*x+c)**(5/2),x)
 
output
Integral((a + b*tan(c + d*x))/tan(c + d*x)**(5/2), x)
 
3.6.60.7 Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.79 \[ \int \frac {a+b \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {6 \, \sqrt {2} {\left (a + b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 6 \, \sqrt {2} {\left (a + b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 3 \, \sqrt {2} {\left (a - b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 3 \, \sqrt {2} {\left (a - b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \frac {8 \, {\left (3 \, b \tan \left (d x + c\right ) + a\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \]

input
integrate((a+b*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="maxima")
 
output
-1/12*(6*sqrt(2)*(a + b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c) 
))) + 6*sqrt(2)*(a + b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c) 
))) + 3*sqrt(2)*(a - b)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) 
 - 3*sqrt(2)*(a - b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 
 8*(3*b*tan(d*x + c) + a)/tan(d*x + c)^(3/2))/d
 
3.6.60.8 Giac [F(-1)]

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

input
integrate((a+b*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="giac")
 
output
Timed out
 
3.6.60.9 Mupad [B] (verification not implemented)

Time = 5.99 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.62 \[ \int \frac {a+b \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}-\frac {2\,b}{d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}-\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}-\frac {2\,a}{3\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )\,1{}\mathrm {i}}{d}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )\,1{}\mathrm {i}}{d} \]

input
int((a + b*tan(c + d*x))/tan(c + d*x)^(5/2),x)
 
output
((-1)^(1/4)*a*atan((-1)^(1/4)*tan(c + d*x)^(1/2))*1i)/d - (2*b)/(d*tan(c + 
 d*x)^(1/2)) - (2*a)/(3*d*tan(c + d*x)^(3/2)) + ((-1)^(1/4)*a*atanh((-1)^( 
1/4)*tan(c + d*x)^(1/2))*1i)/d - ((-1)^(1/4)*b*atan((-1)^(1/4)*tan(c + d*x 
)^(1/2)))/d + ((-1)^(1/4)*b*atanh((-1)^(1/4)*tan(c + d*x)^(1/2)))/d