\(\int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))} \, dx\) [420]

Optimal result

Integrand size = 36, antiderivative size = 114 \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\frac {B \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {B \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {B \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {2 B}{d \sqrt {\tan (c+d x)}} \] 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+ta 
n(d*x+c)))*2^(1/2)/d-2*B/d/tan(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.68 \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\frac {B \left (-2-\arctan \left (\sqrt [4]{-\tan ^2(c+d x)}\right ) \sqrt [4]{-\tan ^2(c+d x)}+\text {arctanh}\left (\sqrt [4]{-\tan ^2(c+d x)}\right ) \sqrt [4]{-\tan ^2(c+d x)}\right )}{d \sqrt {\tan (c+d x)}} \] Input:

Integrate[(a*B + b*B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x] 
)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.36, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {2011, 3042, 3955, 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.

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

\(\Big \downarrow \) 2011

\(\displaystyle B \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle B \int \frac {1}{\tan (c+d x)^{3/2}}dx\)

\(\Big \downarrow \) 3955

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[(a*B + b*B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])),x]
 

Output:

B*((-2*((-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sq 
rt[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 - 2/(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 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 3955

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x] 
)^(n + 1)/(b*d*(n + 1)), x] - Simp[1/b^2   Int[(b*Tan[c + d*x])^(n + 2), x] 
, x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1]
 

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 = 0.04 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {B \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}\)	\(102\)
default	\(\frac {B \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}-\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}\)	\(102\)

Input:

int((B*a+b*B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c)),x,method=_RETUR 
NVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.27 \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=-\frac {2 \, \sqrt {2} B \arctan \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + 1\right ) \tan \left (d x + c\right ) + 2 \, \sqrt {2} B \arctan \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} - 1\right ) \tan \left (d x + c\right ) - \sqrt {2} B \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) \tan \left (d x + c\right ) + \sqrt {2} B \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) \tan \left (d x + c\right ) + 8 \, B \sqrt {\tan \left (d x + c\right )}}{4 \, d \tan \left (d x + c\right )} \] Input:

integrate((B*a+b*B*tan(d*x+c))/tan(d*x+c)^(3/2)/(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)*tan(d*x + c) + 2* 
sqrt(2)*B*arctan(sqrt(2)*sqrt(tan(d*x + c)) - 1)*tan(d*x + c) - sqrt(2)*B* 
log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)*tan(d*x + c) + sqrt(2)* 
B*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)*tan(d*x + c) + 8*B*s 
qrt(tan(d*x + c)))/(d*tan(d*x + c))
 

Sympy [F]

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

integrate((B*a+b*B*tan(d*x+c))/tan(d*x+c)**(3/2)/(a+b*tan(d*x+c)),x)
 

Output:

B*Integral(tan(c + d*x)**(-3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.07 \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=-\frac {{\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} B + \frac {8 \, B}{\sqrt {\tan \left (d x + c\right )}}}{4 \, d} \] Input:

integrate((B*a+b*B*tan(d*x+c))/tan(d*x+c)^(3/2)/(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* 
sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*lo 
g(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*log(-sqrt(2)*sq 
rt(tan(d*x + c)) + tan(d*x + c) + 1))*B + 8*B/sqrt(tan(d*x + c)))/d
 

Giac [F]

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

integrate((B*a+b*B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c)),x, algori 
thm="giac")
 

Output:

undef
 

Mupad [B] (verification not implemented)

Time = 8.07 (sec) , antiderivative size = 15569, normalized size of antiderivative = 136.57 \[ \int \frac {a B+b B \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\text {Too large to display} \] Input:

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

Output:

atan(((tan(c + d*x)^(1/2)*(64*B^4*a^2*b^7*d^5 - 32*B^4*a^4*b^5*d^5) + (((6 
4*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)*((t 
an(c + d*x)^(1/2)*(128*B^2*a^5*b^4*d^7 - 448*B^2*a^3*b^6*d^7 + 64*B^2*a^7* 
b^2*d^7 + 512*B^2*a*b^8*d^7) - (((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)*(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)*(512*b^9*d^9 + 512*a^ 
2*b^7*d^9 - 512*a^4*b^5*d^9 - 512*a^6*b^3*d^9) - 512*B*b^9*d^8 - 640*B*a^2 
*b^7*d^8 + 256*B*a^4*b^5*d^8 + 384*B*a^6*b^3*d^8))*(((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*B^3*a^5*b^4*d^6 - 
 32*B^3*a^7*b^2*d^6 + 128*B^3*a*b^8*d^6))*(((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 + (tan(c + d*x)^(1/2)*(64*B^ 
4*a^2*b^7*d^5 - 32*B^4*a^4*b^5*d^5) + (((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)*((tan(c + d*x)^(1/2)*(128*B^2*a^5*b 
^4*d^7 - 448*B^2*a^3*b^6*d^7 + 64*B^2*a^7*b^2*d^7 + 512*B^2*a*b^8*d^7) ...
 

Reduce [F]

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

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

Output:

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