\(\int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [383]

Optimal result

Integrand size = 31, antiderivative size = 177 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(b (A-B)+a (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(b (A-B)+a (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {2 a A}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (A b+a B)}{d \sqrt {\tan (c+d x)}} \] Output:

-1/2*(b*(A-B)+a*(A+B))*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/d-1/2*( 
b*(A-B)+a*(A+B))*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/d-1/2*(a*(A-B) 
-b*(A+B))*arctanh(2^(1/2)*tan(d*x+c)^(1/2)/(1+tan(d*x+c)))*2^(1/2)/d-2/3*a 
*A/d/tan(d*x+c)^(3/2)-2*(A*b+B*a)/d/tan(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {6 \sqrt {2} (b (A-B)+a (A+B)) \left (\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )\right )+3 \sqrt {2} (a (A-B)-b (A+B)) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )-\frac {8 a A}{\tan ^{\frac {3}{2}}(c+d x)}-\frac {24 (A b+a B)}{\sqrt {\tan (c+d x)}}}{12 d} \] Input:

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

Output:

(6*Sqrt[2]*(b*(A - B) + a*(A + B))*(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] 
 - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]) + 3*Sqrt[2]*(a*(A - B) - b*(A + 
 B))*(Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Log[1 + Sqrt[2] 
*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]) - (8*a*A)/Tan[c + d*x]^(3/2) - (24*(A 
*b + a*B))/Sqrt[Tan[c + d*x]])/(12*d)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.16, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 4074, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4074

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

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

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(-2*(((b*(A - B) + a*(A + B))*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sq 
rt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]))/2 + ((a*(A - B) 
- b*(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*A)/(3*d*Tan[c + d*x]^(3/2)) - (2*(A*b + 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 4074

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.25

method	result	size
derivativedivides	\(\frac {\frac {\left (-a A +B b \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {2 \left (A b +B a \right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {2 a A}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}}{d}\)	\(222\)
default	\(\frac {\frac {\left (-a A +B b \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {2 \left (A b +B a \right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {2 a A}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}}{d}\)	\(222\)
parts	\(\frac {\left (A b +B a \right ) \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {B b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4 d}+\frac {a A \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}\right )}{d}\)	\(301\)

Input:

int((a+b*tan(d*x+c))*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x,method=_RETURNVER 
BOSE)
 

Output:

1/d*(1/4*(-A*a+B*b)*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)))+1/4*(-A*b-B*a)*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) 
))-2*(A*b+B*a)/tan(d*x+c)^(1/2)-2/3*a*A/tan(d*x+c)^(3/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 944 vs. \(2 (153) = 306\).

Time = 0.10 (sec) , antiderivative size = 944, normalized size of antiderivative = 5.33 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/6*(6*sqrt(1/2)*d*sqrt(((A^2 + 2*A*B + B^2)*a^2 + 2*(A^2 - B^2)*a*b + (A^ 
2 - 2*A*B + B^2)*b^2)/d^2)*arctan((2*sqrt(1/2)*((A - B)*a - (A + B)*b)*d*s 
qrt(((A^2 + 2*A*B + B^2)*a^2 + 2*(A^2 - B^2)*a*b + (A^2 - 2*A*B + B^2)*b^2 
)/d^2)*sqrt(tan(d*x + c)) + d^2*sqrt(((A^2 + 2*A*B + B^2)*a^2 + 2*(A^2 - B 
^2)*a*b + (A^2 - 2*A*B + B^2)*b^2)/d^2)*sqrt(((A^2 - 2*A*B + B^2)*a^2 - 2* 
(A^2 - B^2)*a*b + (A^2 + 2*A*B + B^2)*b^2)/d^2))/(4*A*B*a*b - (A^2 - B^2)* 
a^2 + (A^2 - B^2)*b^2))*tan(d*x + c)^2 + 6*sqrt(1/2)*d*sqrt(((A^2 + 2*A*B 
+ B^2)*a^2 + 2*(A^2 - B^2)*a*b + (A^2 - 2*A*B + B^2)*b^2)/d^2)*arctan((2*s 
qrt(1/2)*((A - B)*a - (A + B)*b)*d*sqrt(((A^2 + 2*A*B + B^2)*a^2 + 2*(A^2 
- B^2)*a*b + (A^2 - 2*A*B + B^2)*b^2)/d^2)*sqrt(tan(d*x + c)) - d^2*sqrt(( 
(A^2 + 2*A*B + B^2)*a^2 + 2*(A^2 - B^2)*a*b + (A^2 - 2*A*B + B^2)*b^2)/d^2 
)*sqrt(((A^2 - 2*A*B + B^2)*a^2 - 2*(A^2 - B^2)*a*b + (A^2 + 2*A*B + B^2)* 
b^2)/d^2))/(4*A*B*a*b - (A^2 - B^2)*a^2 + (A^2 - B^2)*b^2))*tan(d*x + c)^2 
 + 3*sqrt(1/2)*d*sqrt(((A^2 - 2*A*B + B^2)*a^2 - 2*(A^2 - B^2)*a*b + (A^2 
+ 2*A*B + B^2)*b^2)/d^2)*log(2*sqrt(1/2)*d*sqrt(((A^2 - 2*A*B + B^2)*a^2 - 
 2*(A^2 - B^2)*a*b + (A^2 + 2*A*B + B^2)*b^2)/d^2)*sqrt(tan(d*x + c)) - (A 
 - B)*a + (A + B)*b - ((A - B)*a - (A + B)*b)*tan(d*x + c))*tan(d*x + c)^2 
 - 3*sqrt(1/2)*d*sqrt(((A^2 - 2*A*B + B^2)*a^2 - 2*(A^2 - B^2)*a*b + (A^2 
+ 2*A*B + B^2)*b^2)/d^2)*log(-2*sqrt(1/2)*d*sqrt(((A^2 - 2*A*B + B^2)*a^2 
- 2*(A^2 - B^2)*a*b + (A^2 + 2*A*B + B^2)*b^2)/d^2)*sqrt(tan(d*x + c)) ...
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {6 \, \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 6 \, \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 3 \, \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 3 \, \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \frac {8 \, {\left (A a + 3 \, {\left (B a + A b\right )} \tan \left (d x + c\right )\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \] Input:

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

Output:

-1/12*(6*sqrt(2)*((A + B)*a + (A - B)*b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*s 
qrt(tan(d*x + c)))) + 6*sqrt(2)*((A + B)*a + (A - B)*b)*arctan(-1/2*sqrt(2 
)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + 3*sqrt(2)*((A - B)*a - (A + B)*b)*lo 
g(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - 3*sqrt(2)*((A - B)*a - 
(A + B)*b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 8*(A*a + 
3*(B*a + A*b)*tan(d*x + c))/tan(d*x + c)^(3/2))/d
 

Giac [F]

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

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

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 7.17 (sec) , antiderivative size = 1448, normalized size of antiderivative = 8.18 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

2*atanh((32*A^2*a^2*d^3*tan(c + d*x)^(1/2)*(- (2*A^4*a^2*b^2*d^4 - A^4*b^4 
*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4) - (A^2*a*b)/(2*d^2))^(1/2))/(16*A*a*(2*A 
^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2) - 16*A^3*b^3*d^2 + 16*A^ 
3*a^2*b*d^2) - (32*A^2*b^2*d^3*tan(c + d*x)^(1/2)*(- (2*A^4*a^2*b^2*d^4 - 
A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4) - (A^2*a*b)/(2*d^2))^(1/2))/(16*A 
*a*(2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2) - 16*A^3*b^3*d^2 
+ 16*A^3*a^2*b*d^2))*(- (2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1 
/2)/(4*d^4) - (A^2*a*b)/(2*d^2))^(1/2) - 2*atanh((32*A^2*a^2*d^3*tan(c + d 
*x)^(1/2)*((2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4) - 
 (A^2*a*b)/(2*d^2))^(1/2))/(16*A*a*(2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4* 
a^4*d^4)^(1/2) + 16*A^3*b^3*d^2 - 16*A^3*a^2*b*d^2) - (32*A^2*b^2*d^3*tan( 
c + d*x)^(1/2)*((2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d 
^4) - (A^2*a*b)/(2*d^2))^(1/2))/(16*A*a*(2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - 
 A^4*a^4*d^4)^(1/2) + 16*A^3*b^3*d^2 - 16*A^3*a^2*b*d^2))*((2*A^4*a^2*b^2* 
d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4) - (A^2*a*b)/(2*d^2))^(1/2) 
+ 2*atanh((32*B^2*a^2*d^3*tan(c + d*x)^(1/2)*((B^2*a*b)/(2*d^2) - (2*B^4*a 
^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^4*d^4)^(1/2)/(4*d^4))^(1/2))/(16*B*b*(2*B 
^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^4*d^4)^(1/2) + 16*B^3*a^3*d^2 - 16*B^ 
3*a*b^2*d^2) - (32*B^2*b^2*d^3*tan(c + d*x)^(1/2)*((B^2*a*b)/(2*d^2) - (2* 
B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^4*d^4)^(1/2)/(4*d^4))^(1/2))/(16*...
 

Reduce [F]

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

int((a+b*tan(d*x+c))*(A+B*tan(d*x+c))/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