\(\int \frac {(a+b \tan (c+d x))^2}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [567]

Optimal result

Integrand size = 23, antiderivative size = 172 \[ \int \frac {(a+b \tan (c+d x))^2}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\left (a^2+2 a b-b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2-2 a b-b^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {2 a^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {4 a b}{d \sqrt {\tan (c+d x)}} \] 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-2/3*a^2/d/t 
an(d*x+c)^(3/2)-4*a*b/d/tan(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 0.14 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.45 \[ \int \frac {(a+b \tan (c+d x))^2}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \left (\left (a^2-b^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},-\tan ^2(c+d x)\right )+b \left (b+6 a \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},-\tan ^2(c+d x)\right ) \tan (c+d x)\right )\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)} \] Input:

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

Output:

(-2*((a^2 - b^2)*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[c + d*x]^2] + b*(b + 
 6*a*Hypergeometric2F1[-1/4, 1, 3/4, -Tan[c + d*x]^2]*Tan[c + d*x])))/(3*d 
*Tan[c + d*x]^(3/2))
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.17, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 4025, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4025

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

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

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(-2*(((a^2 + 2*a*b - b^2)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2 
]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]))/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] + Lo 
g[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2])))/2))/d - (2* 
a^2)/(3*d*Tan[c + d*x]^(3/2)) - (4*a*b)/(d*Sqrt[Tan[c + d*x]])
 

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]
 

rule 4025

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(\frac {\frac {\left (-a^{2}+b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{2}-\frac {2 a^{2}}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {4 a b}{\sqrt {\tan \left (d x +c \right )}}}{d}\)	\(212\)
default	\(\frac {\frac {\left (-a^{2}+b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {a b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{2}-\frac {2 a^{2}}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {4 a b}{\sqrt {\tan \left (d x +c \right )}}}{d}\)	\(212\)
parts	\(\frac {a^{2} \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}+\frac {b^{2} \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4 d}+\frac {2 a b \left (-\frac {2}{\sqrt {\tan \left (d x +c \right )}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(299\)

Input:

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

Output:

1/d*(1/4*(-a^2+b^2)*2^(1/2)*(ln((tan(d*x+c)+2^(1/2)*tan(d*x+c)^(1/2)+1)/(t 
an(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)))-1/2*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*arct 
an(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^2/tan(d*x+c)^(3/2)-4*a*b/tan(d*x+c)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 693 vs. \(2 (148) = 296\).

Time = 0.09 (sec) , antiderivative size = 693, normalized size of antiderivative = 4.03 \[ \int \frac {(a+b \tan (c+d x))^2}{\tan ^{\frac {5}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

1/6*(6*sqrt(1/2)*d*sqrt((a^4 + 4*a^3*b + 2*a^2*b^2 - 4*a*b^3 + b^4)/d^2)*a 
rctan(-(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))*tan(d*x + c)^2 + 6*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(ta 
n(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)*sq 
rt((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))*tan(d*x + c)^2 + 3*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*sqrt(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))*tan(d*x + c)^2 - 3*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*sqrt(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))*tan(d*x + c)^2 - 4*(6*a*b*tan(d*x + c) + a^2)*sqr 
t(tan(d*x + c)))/(d*tan(d*x + c)^2)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b \tan (c+d x))^2}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {6 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 6 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 3 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 3 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \frac {8 \, {\left (6 \, a b \tan \left (d x + c\right ) + a^{2}\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \] Input:

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

Output:

-1/12*(6*sqrt(2)*(a^2 + 2*a*b - b^2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt( 
tan(d*x + c)))) + 6*sqrt(2)*(a^2 + 2*a*b - b^2)*arctan(-1/2*sqrt(2)*(sqrt( 
2) - 2*sqrt(tan(d*x + c)))) + 3*sqrt(2)*(a^2 - 2*a*b - b^2)*log(sqrt(2)*sq 
rt(tan(d*x + c)) + tan(d*x + c) + 1) - 3*sqrt(2)*(a^2 - 2*a*b - b^2)*log(- 
sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 8*(6*a*b*tan(d*x + c) + a 
^2)/tan(d*x + c)^(3/2))/d
 

Giac [F]

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

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

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 2.19 (sec) , antiderivative size = 968, normalized size of antiderivative = 5.63 \[ \int \frac {(a+b \tan (c+d x))^2}{\tan ^{\frac {5}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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