\(\int \frac {a B+b B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx\) [425]

Optimal result

Integrand size = 36, antiderivative size = 185 \[ \int \frac {a B+b B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx=-\frac {(a-b) B \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a-b) B \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 b^{3/2} B \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} \left (a^2+b^2\right ) d}+\frac {(a+b) B \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d} \] Output:

1/2*(a-b)*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/(a^2+b^2)/d+1/2*(a 
-b)*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/(a^2+b^2)/d+2*b^(3/2)*B*a 
rctan(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/a^(1/2)/(a^2+b^2)/d+1/2*(a+b)*B*ar 
ctanh(2^(1/2)*tan(d*x+c)^(1/2)/(1+tan(d*x+c)))*2^(1/2)/(a^2+b^2)/d
 

Mathematica [C] (verified)

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

Time = 0.08 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.24 \[ \int \frac {a B+b B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx=B \left (\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} \left (a^2+b^2\right ) d}-\frac {a \left (2 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-2 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )+\sqrt {2} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )}{4 \left (a^2+b^2\right ) d}-\frac {2 b \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},-\tan ^2(c+d x)\right ) \tan ^{\frac {3}{2}}(c+d x)}{3 \left (a^2+b^2\right ) d}\right ) \] Input:

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

Output:

B*((2*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt[a]*(a^2 
+ b^2)*d) - (a*(2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 2*Sqrt[ 
2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[T 
an[c + d*x]] + Tan[c + d*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] 
+ Tan[c + d*x]]))/(4*(a^2 + b^2)*d) - (2*b*Hypergeometric2F1[3/4, 1, 7/4, 
-Tan[c + d*x]^2]*Tan[c + d*x]^(3/2))/(3*(a^2 + b^2)*d))
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.10, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {2011, 3042, 4057, 3042, 4017, 1482, 1476, 1082, 217, 1479, 25, 27, 1103, 4117, 73, 218}

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.

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

\(\Big \downarrow \) 2011

\(\displaystyle B \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle B \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx\)

\(\Big \downarrow \) 4057

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

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

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

B*((2*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt[a]*(a^2 
+ b^2)*d) + (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*Lo 
g[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))/((a^2 + b^2)*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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, 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 218

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

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 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

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 4057

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

rule 4117

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
 Simp[A/f   Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; 
FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
 

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.22

method	result	size
derivativedivides	\(\frac {B \left (\frac {2 b^{2} \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right ) \sqrt {a b}}+\frac {\frac {a \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{a^{2}+b^{2}}\right )}{d}\)	\(226\)
default	\(\frac {B \left (\frac {2 b^{2} \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right ) \sqrt {a b}}+\frac {\frac {a \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{a^{2}+b^{2}}\right )}{d}\)	\(226\)

Input:

int((B*a+b*B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^2,x,method=_RET 
URNVERBOSE)
 

Output:

1/d*B*(2*b^2/(a^2+b^2)/(a*b)^(1/2)*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))+ 
2/(a^2+b^2)*(1/8*a*2^(1/2)*(ln((tan(d*x+c)+2^(1/2)*tan(d*x+c)^(1/2)+1)/(ta 
n(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)))-1/8*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)))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 875 vs. \(2 (158) = 316\).

Time = 0.14 (sec) , antiderivative size = 1780, normalized size of antiderivative = 9.62 \[ \int \frac {a B+b B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:

integrate((B*a+b*B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^2,x, algo 
rithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {a B+b B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx=B \int \frac {1}{a \sqrt {\tan {\left (c + d x \right )}} + b \tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \] Input:

integrate((B*a+b*B*tan(d*x+c))/tan(d*x+c)**(1/2)/(a+b*tan(d*x+c))**2,x)
 

Output:

B*Integral(1/(a*sqrt(tan(c + d*x)) + b*tan(c + d*x)**(3/2)), x)
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.93 \[ \int \frac {a B+b B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx=\frac {\frac {8 \, B b^{2} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{2} + b^{2}\right )} \sqrt {a b}} + \frac {{\left (2 \, \sqrt {2} {\left (a - b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a - b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (a + b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (a + b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} B}{a^{2} + b^{2}}}{4 \, d} \] Input:

integrate((B*a+b*B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^2,x, algo 
rithm="maxima")
 

Output:

1/4*(8*B*b^2*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^2 + b^2)*sqrt(a*b) 
) + (2*sqrt(2)*(a - b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c))) 
) + 2*sqrt(2)*(a - b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c))) 
) + sqrt(2)*(a + b)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - s 
qrt(2)*(a + b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))*B/(a^2 
 + b^2))/d
 

Giac [F(-2)]

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

integrate((B*a+b*B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^2,x, algo 
rithm="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 = 30.23 (sec) , antiderivative size = 16598, normalized size of antiderivative = 89.72 \[ \int \frac {a B+b B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:

int((B*a + B*b*tan(c + d*x))/(tan(c + d*x)^(1/2)*(a + b*tan(c + d*x))^2),x 
)
 

Output:

(log((((((((((128*B*b^2*(2*b^6 - a^6 + 9*a^2*b^4 + 6*a^4*b^2))/d + 128*b^3 
*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)^2*((4*(-B^4*a^4*d^4*(a^4 + b^4 
 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a^3*b^3*d^2 + 16*B^2*a^5*b*d^2)/(d^4*(a^2 
+ b^2)^4))^(1/2))*((4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16* 
B^2*a^3*b^3*d^2 + 16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (64*B^ 
2*a*b^2*tan(c + d*x)^(1/2)*(2*b^8 - a^8 + 5*a^2*b^6 + 67*a^4*b^4 - a^6*b^2 
))/(d^2*(a^2 + b^2)^2))*((4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) 
 - 16*B^2*a^3*b^3*d^2 + 16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 - 
(32*B^3*a*b^5*(25*a^6 + b^6 - 13*a^2*b^4 - 85*a^4*b^2))/(d^3*(a^2 + b^2)^3 
))*((4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a^3*b^3*d^2 
 + 16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (16*B^4*a^2*b^5*tan(c 
 + d*x)^(1/2)*(b^6 - 27*a^6 + 7*a^2*b^4 + 11*a^4*b^2))/(d^4*(a^2 + b^2)^4) 
)*((4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a^3*b^3*d^2 
+ 16*B^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (16*B^5*a^4*b^6*(5*a^2 
 + b^2))/(d^5*(a^2 + b^2)^4))*(((192*B^4*a^6*b^6*d^4 - 16*B^4*a^4*b^8*d^4 
- 16*B^4*a^12*d^4 - 608*B^4*a^8*b^4*d^4 + 192*B^4*a^10*b^2*d^4)^(1/2) - 16 
*B^2*a^3*b^3*d^2 + 16*B^2*a^5*b*d^2)/(a^8*d^4 + b^8*d^4 + 4*a^2*b^6*d^4 + 
6*a^4*b^4*d^4 + 4*a^6*b^2*d^4))^(1/2))/4 + (log((((((((((128*B*b^2*(2*b^6 
- a^6 + 9*a^2*b^4 + 6*a^4*b^2))/d + 128*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2) 
*(a^2 + b^2)^2*(-(4*(-B^4*a^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) + 16...
 

Reduce [F]

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

int((B*a+b*B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^2,x)
 

Output:

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