\(\int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\) [418]

Optimal result

Integrand size = 36, antiderivative size = 99 \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {B \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {B \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {B \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} d} \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.61 \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {B \left (\arctan \left (\sqrt [4]{-\tan ^2(c+d x)}\right )-\text {arctanh}\left (\sqrt [4]{-\tan ^2(c+d x)}\right )\right ) \sqrt [4]{-\tan (c+d x)}}{d \sqrt [4]{\tan (c+d x)}} \] Input:

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

Output:

(B*(ArcTan[(-Tan[c + d*x]^2)^(1/4)] - ArcTanh[(-Tan[c + d*x]^2)^(1/4)])*(- 
Tan[c + d*x])^(1/4))/(d*Tan[c + d*x]^(1/4))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.39, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2011, 3042, 3957, 266, 826, 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.

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

\(\Big \downarrow \) 2011

\(\displaystyle B \int \sqrt {\tan (c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle B \int \sqrt {\tan (c+d x)}dx\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {B \int \frac {\sqrt {\tan (c+d x)}}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 B \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}\)

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(2*B*((-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt 
[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2])/2 + (Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + 
 Tan[c + d*x]]/(2*Sqrt[2]) - Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + 
d*x]]/(2*Sqrt[2]))/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 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 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, 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 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 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {B \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4 d}\)	\(90\)
default	\(\frac {B \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4 d}\)	\(90\)

Input:

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

Output:

1/4/d*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 [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {2 \, \sqrt {2} B \arctan \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + 1\right ) + 2 \, \sqrt {2} B \arctan \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} - 1\right ) - \sqrt {2} B \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} B \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \, d} \] Input:

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

Output:

1/4*(2*sqrt(2)*B*arctan(sqrt(2)*sqrt(tan(d*x + c)) + 1) + 2*sqrt(2)*B*arct 
an(sqrt(2)*sqrt(tan(d*x + c)) - 1) - sqrt(2)*B*log(sqrt(2)*sqrt(tan(d*x + 
c)) + tan(d*x + c) + 1) + sqrt(2)*B*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan( 
d*x + c) + 1))/d
 

Sympy [F]

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

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

Output:

B*Integral(sqrt(tan(c + d*x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {{\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} B}{4 \, d} \] Input:

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

Output:

1/4*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sq 
rt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*log( 
sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*log(-sqrt(2)*sqrt 
(tan(d*x + c)) + tan(d*x + c) + 1))*B/d
 

Giac [F(-2)]

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

integrate(tan(d*x+c)^(1/2)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algori 
thm="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 = 8.46 (sec) , antiderivative size = 15753, normalized size of antiderivative = 159.12 \[ \int \frac {\sqrt {\tan (c+d x)} (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

atan(((((32*(13*B^3*a^5*b^4*d^2 + B^3*a^7*b^2*d^2))/d^5 + (((32*(12*B*a^2* 
b^7*d^4 + 24*B*a^4*b^5*d^4 + 12*B*a^6*b^3*d^4))/d^5 - (32*tan(c + d*x)^(1/ 
2)*(((64*B^4*a^6*b^2*d^4 - B^4*a^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d 
^4))^(1/2) - 8*B^2*a^3*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1 
/2)*(16*b^9*d^4 + 16*a^2*b^7*d^4 - 16*a^4*b^5*d^4 - 16*a^6*b^3*d^4))/d^4)* 
(((64*B^4*a^6*b^2*d^4 - B^4*a^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^2*b^2*d^4) 
)^(1/2) - 8*B^2*a^3*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^4)))^(1/2) 
 + (32*tan(c + d*x)^(1/2)*(20*B^2*a^5*b^4*d^2 - 14*B^2*a^3*b^6*d^2 + 2*B^2 
*a^7*b^2*d^2))/d^4)*(((64*B^4*a^6*b^2*d^4 - B^4*a^4*(16*a^4*d^4 + 16*b^4*d 
^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a^3*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2* 
a^2*b^2*d^4)))^(1/2))*(((64*B^4*a^6*b^2*d^4 - B^4*a^4*(16*a^4*d^4 + 16*b^4 
*d^4 + 32*a^2*b^2*d^4))^(1/2) - 8*B^2*a^3*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 
2*a^2*b^2*d^4)))^(1/2) - (32*tan(c + d*x)^(1/2)*(B^4*a^4*b^5 - 2*B^4*a^6*b 
^3))/d^4)*(((64*B^4*a^6*b^2*d^4 - B^4*a^4*(16*a^4*d^4 + 16*b^4*d^4 + 32*a^ 
2*b^2*d^4))^(1/2) - 8*B^2*a^3*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2*d^ 
4)))^(1/2)*1i - (((32*(13*B^3*a^5*b^4*d^2 + B^3*a^7*b^2*d^2))/d^5 + (((32* 
(12*B*a^2*b^7*d^4 + 24*B*a^4*b^5*d^4 + 12*B*a^6*b^3*d^4))/d^5 + (32*tan(c 
+ d*x)^(1/2)*(((64*B^4*a^6*b^2*d^4 - B^4*a^4*(16*a^4*d^4 + 16*b^4*d^4 + 32 
*a^2*b^2*d^4))^(1/2) - 8*B^2*a^3*b*d^2)/(16*(a^4*d^4 + b^4*d^4 + 2*a^2*b^2 
*d^4)))^(1/2)*(16*b^9*d^4 + 16*a^2*b^7*d^4 - 16*a^4*b^5*d^4 - 16*a^6*b^...
 

Reduce [F]

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

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

Output:

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