3.4.90 \(\int \frac {(a+b \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [390]

3.4.90.1 Optimal result

Integrand size = 33, antiderivative size = 283 \[ \int \frac {(a+b \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (2 a b (A-B)+a^2 (A+B)-b^2 (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (a^2 (A-B)-b^2 (A-B)-2 a b (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a^2 A}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a (2 A b+a B)}{d \sqrt {\tan (c+d x)}} \]

output

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

3.4.90.2 Mathematica [C] (verified)

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

Time = 0.74 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.42 \[ \int \frac {(a+b \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \left (-a^2 A+A b^2+2 a b B\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},-\tan ^2(c+d x)\right )-6 \left (2 a A b+a^2 B-b^2 B\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},-\tan ^2(c+d x)\right ) \tan (c+d x)-2 b (A b+2 a B+3 b B \tan (c+d x))}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]

input

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

output

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

3.4.90.3 Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.83, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4087, 3042, 4111, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4087

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4111

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

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

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

input

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

output

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

3.4.90.3.1 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 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 4087

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

rule 4111

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

3.4.90.4 Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {-\frac {2 A \,a^{2}}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 a \left (2 A b +B a \right )}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (-A \,a^{2}+A \,b^{2}+2 B a b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-2 A a b -B \,a^{2}+B \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(243\)
default	\(\frac {-\frac {2 A \,a^{2}}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 a \left (2 A b +B a \right )}{\sqrt {\tan \left (d x +c \right )}}+\frac {\left (-A \,a^{2}+A \,b^{2}+2 B a b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-2 A a b -B \,a^{2}+B \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(243\)
parts	\(\frac {\left (A \,b^{2}+2 B a b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {\left (2 A a b +B \,a^{2}\right ) \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {A \,a^{2} \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {B \,b^{2} \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}\)	\(408\)

input

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

output

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

3.4.90.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4313 vs. \(2 (249) = 498\).

Time = 0.72 (sec) , antiderivative size = 4313, normalized size of antiderivative = 15.24 \[ \int \frac {(a+b \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \]

input

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

output

1/6*(3*d*sqrt(-(2*A*B*a^4 - 12*A*B*a^2*b^2 + 2*A*B*b^4 + 4*(A^2 - B^2)*a^3 
*b - 4*(A^2 - B^2)*a*b^3 + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*(A^ 
3*B - A*B^3)*a^7*b - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*b^2 + 112*(A^3*B - 
 A*B^3)*a^5*b^3 + 2*(19*A^4 - 102*A^2*B^2 + 19*B^4)*a^4*b^4 - 112*(A^3*B - 
 A*B^3)*a^3*b^5 - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A*B 
^3)*a*b^7 + (A^4 - 2*A^2*B^2 + B^4)*b^8)/d^4))/d^2)*log(((B*a^2 + 2*A*a*b 
- B*b^2)*d^3*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*(A^3*B - A*B^3)*a^7*b 
 - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*b^2 + 112*(A^3*B - A*B^3)*a^5*b^3 + 
2*(19*A^4 - 102*A^2*B^2 + 19*B^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5 - 
4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A*B^3)*a*b^7 + (A^4 - 
 2*A^2*B^2 + B^4)*b^8)/d^4) + ((A^3 - A*B^2)*a^6 - 2*(5*A^2*B - B^3)*a^5*b 
 - (7*A^3 - 23*A*B^2)*a^4*b^2 + 4*(7*A^2*B - 3*B^3)*a^3*b^3 + (7*A^3 - 23* 
A*B^2)*a^2*b^4 - 2*(5*A^2*B - B^3)*a*b^5 - (A^3 - A*B^2)*b^6)*d)*sqrt(-(2* 
A*B*a^4 - 12*A*B*a^2*b^2 + 2*A*B*b^4 + 4*(A^2 - B^2)*a^3*b - 4*(A^2 - B^2) 
*a*b^3 + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^8 - 16*(A^3*B - A*B^3)*a^7*b 
 - 4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^6*b^2 + 112*(A^3*B - A*B^3)*a^5*b^3 + 
2*(19*A^4 - 102*A^2*B^2 + 19*B^4)*a^4*b^4 - 112*(A^3*B - A*B^3)*a^3*b^5 - 
4*(3*A^4 - 22*A^2*B^2 + 3*B^4)*a^2*b^6 + 16*(A^3*B - A*B^3)*a*b^7 + (A^4 - 
 2*A^2*B^2 + B^4)*b^8)/d^4))/d^2) - ((A^4 - B^4)*a^8 - 8*(A^3*B + A*B^3)*a 
^7*b - 4*(A^4 - B^4)*a^6*b^2 - 8*(A^3*B + A*B^3)*a^5*b^3 - 10*(A^4 - B^...
 

3.4.90.6 Sympy [F]

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

input

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

3.4.90.7 Maxima [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {6 \, \sqrt {2} {\left ({\left (A + B\right )} a^{2} + 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 6 \, \sqrt {2} {\left ({\left (A + B\right )} a^{2} + 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 3 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} - 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 3 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} - 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \frac {8 \, {\left (A a^{2} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \]

input

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

output

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

3.4.90.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.4.90.9 Mupad [B] (verification not implemented)

Time = 11.81 (sec) , antiderivative size = 3745, normalized size of antiderivative = 13.23 \[ \int \frac {(a+b \tan (c+d x))^2 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \]

input

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

output

2*atanh((32*B^2*a^4*d^3*tan(c + d*x)^(1/2)*((B^2*a^3*b)/d^2 - (B^2*a*b^3)/ 
d^2 - (12*B^4*a^2*b^6*d^4 - B^4*b^8*d^4 - B^4*a^8*d^4 - 38*B^4*a^4*b^4*d^4 
 + 12*B^4*a^6*b^2*d^4)^(1/2)/(4*d^4))^(1/2))/(16*B^3*a^6*d^2 - 16*B^3*b^6* 
d^2 + 32*B*a*b*(12*B^4*a^2*b^6*d^4 - B^4*b^8*d^4 - B^4*a^8*d^4 - 38*B^4*a^ 
4*b^4*d^4 + 12*B^4*a^6*b^2*d^4)^(1/2) + 112*B^3*a^2*b^4*d^2 - 112*B^3*a^4* 
b^2*d^2) + (32*B^2*b^4*d^3*tan(c + d*x)^(1/2)*((B^2*a^3*b)/d^2 - (B^2*a*b^ 
3)/d^2 - (12*B^4*a^2*b^6*d^4 - B^4*b^8*d^4 - B^4*a^8*d^4 - 38*B^4*a^4*b^4* 
d^4 + 12*B^4*a^6*b^2*d^4)^(1/2)/(4*d^4))^(1/2))/(16*B^3*a^6*d^2 - 16*B^3*b 
^6*d^2 + 32*B*a*b*(12*B^4*a^2*b^6*d^4 - B^4*b^8*d^4 - B^4*a^8*d^4 - 38*B^4 
*a^4*b^4*d^4 + 12*B^4*a^6*b^2*d^4)^(1/2) + 112*B^3*a^2*b^4*d^2 - 112*B^3*a 
^4*b^2*d^2) - (192*B^2*a^2*b^2*d^3*tan(c + d*x)^(1/2)*((B^2*a^3*b)/d^2 - ( 
B^2*a*b^3)/d^2 - (12*B^4*a^2*b^6*d^4 - B^4*b^8*d^4 - B^4*a^8*d^4 - 38*B^4* 
a^4*b^4*d^4 + 12*B^4*a^6*b^2*d^4)^(1/2)/(4*d^4))^(1/2))/(16*B^3*a^6*d^2 - 
16*B^3*b^6*d^2 + 32*B*a*b*(12*B^4*a^2*b^6*d^4 - B^4*b^8*d^4 - B^4*a^8*d^4 
- 38*B^4*a^4*b^4*d^4 + 12*B^4*a^6*b^2*d^4)^(1/2) + 112*B^3*a^2*b^4*d^2 - 1 
12*B^3*a^4*b^2*d^2))*((B^2*a^3*b)/d^2 - (B^2*a*b^3)/d^2 - (12*B^4*a^2*b^6* 
d^4 - B^4*b^8*d^4 - B^4*a^8*d^4 - 38*B^4*a^4*b^4*d^4 + 12*B^4*a^6*b^2*d^4) 
^(1/2)/(4*d^4))^(1/2) - 2*atanh((32*B^2*a^4*d^3*tan(c + d*x)^(1/2)*((12*B^ 
4*a^2*b^6*d^4 - B^4*b^8*d^4 - B^4*a^8*d^4 - 38*B^4*a^4*b^4*d^4 + 12*B^4*a^ 
6*b^2*d^4)^(1/2)/(4*d^4) - (B^2*a*b^3)/d^2 + (B^2*a^3*b)/d^2)^(1/2))/(1...