\(\int \frac {1}{\sqrt {b \tan ^3(c+d x)}} \, dx\) [33]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 14, antiderivative size = 192 \[ \int \frac {1}{\sqrt {b \tan ^3(c+d x)}} \, dx=-\frac {2 \tan (c+d x)}{d \sqrt {b \tan ^3(c+d x)}}+\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right ) \tan ^{\frac {3}{2}}(c+d x)}{\sqrt {2} d \sqrt {b \tan ^3(c+d x)}}-\frac {\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right ) \tan ^{\frac {3}{2}}(c+d x)}{\sqrt {2} d \sqrt {b \tan ^3(c+d x)}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right ) \tan ^{\frac {3}{2}}(c+d x)}{\sqrt {2} d \sqrt {b \tan ^3(c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.45 \[ \int \frac {1}{\sqrt {b \tan ^3(c+d x)}} \, dx=\frac {\tan (c+d x) \left (-2-\arctan \left (\sqrt [4]{-\tan ^2(c+d x)}\right ) \sqrt [4]{-\tan ^2(c+d x)}+\text {arctanh}\left (\sqrt [4]{-\tan ^2(c+d x)}\right ) \sqrt [4]{-\tan ^2(c+d x)}\right )}{d \sqrt {b \tan ^3(c+d x)}} \] Input:

Integrate[1/Sqrt[b*Tan[c + d*x]^3],x]
 

Output:

(Tan[c + d*x]*(-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[b*T 
an[c + d*x]^3])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.93, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.071, Rules used = {3042, 4141, 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.

\(\displaystyle \int \frac {1}{\sqrt {b \tan ^3(c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4141

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3955

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[1/Sqrt[b*Tan[c + d*x]^3],x]
 

Output:

(((-2*((-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqr 
t[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]]))*Tan[c + d*x]^(3/2))/ 
Sqrt[b*Tan[c + d*x]^3]
 

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 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]
 

rule 4141
Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff 
= FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ 
n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p]))   Int[ActivateTrig[u]*(Ta 
n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] 
 && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / 
; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
 
Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.10

method result size
derivativedivides \(-\frac {\tan \left (d x +c \right ) \left (\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}\, \ln \left (-\frac {\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}-b \tan \left (d x +c \right )-\sqrt {b^{2}}}{b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}+\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+2 \sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}-\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+8 \left (b^{2}\right )^{\frac {1}{4}}\right )}{4 d \sqrt {b \tan \left (d x +c \right )^{3}}\, \left (b^{2}\right )^{\frac {1}{4}}}\) \(211\)
default \(-\frac {\tan \left (d x +c \right ) \left (\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}\, \ln \left (-\frac {\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}-b \tan \left (d x +c \right )-\sqrt {b^{2}}}{b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}+\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+2 \sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}-\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+8 \left (b^{2}\right )^{\frac {1}{4}}\right )}{4 d \sqrt {b \tan \left (d x +c \right )^{3}}\, \left (b^{2}\right )^{\frac {1}{4}}}\) \(211\)

Input:

int(1/(b*tan(d*x+c)^3)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4/d*tan(d*x+c)*(2^(1/2)*(b*tan(d*x+c))^(1/2)*ln(-((b^2)^(1/4)*(b*tan(d* 
x+c))^(1/2)*2^(1/2)-b*tan(d*x+c)-(b^2)^(1/2))/(b*tan(d*x+c)+(b^2)^(1/4)*(b 
*tan(d*x+c))^(1/2)*2^(1/2)+(b^2)^(1/2)))+2*2^(1/2)*(b*tan(d*x+c))^(1/2)*ar 
ctan((2^(1/2)*(b*tan(d*x+c))^(1/2)+(b^2)^(1/4))/(b^2)^(1/4))+2*2^(1/2)*(b* 
tan(d*x+c))^(1/2)*arctan((2^(1/2)*(b*tan(d*x+c))^(1/2)-(b^2)^(1/4))/(b^2)^ 
(1/4))+8*(b^2)^(1/4))/(b*tan(d*x+c)^3)^(1/2)/(b^2)^(1/4)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.34 \[ \int \frac {1}{\sqrt {b \tan ^3(c+d x)}} \, dx=-\frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\frac {\sqrt {2} \sqrt {b \tan \left (d x + c\right )^{3}}}{\sqrt {b}} + \tan \left (d x + c\right )}{\tan \left (d x + c\right )}\right ) \tan \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\frac {\sqrt {2} \sqrt {b \tan \left (d x + c\right )^{3}}}{\sqrt {b}} - \tan \left (d x + c\right )}{\tan \left (d x + c\right )}\right ) \tan \left (d x + c\right )^{2} - \sqrt {2} \sqrt {b} \log \left (\frac {\tan \left (d x + c\right )^{2} + \frac {\sqrt {2} \sqrt {b \tan \left (d x + c\right )^{3}}}{\sqrt {b}} + \tan \left (d x + c\right )}{\tan \left (d x + c\right )}\right ) \tan \left (d x + c\right )^{2} + \sqrt {2} \sqrt {b} \log \left (\frac {\tan \left (d x + c\right )^{2} - \frac {\sqrt {2} \sqrt {b \tan \left (d x + c\right )^{3}}}{\sqrt {b}} + \tan \left (d x + c\right )}{\tan \left (d x + c\right )}\right ) \tan \left (d x + c\right )^{2} + 8 \, \sqrt {b \tan \left (d x + c\right )^{3}}}{4 \, b d \tan \left (d x + c\right )^{2}} \] Input:

integrate(1/(b*tan(d*x+c)^3)^(1/2),x, algorithm="fricas")
 

Output:

-1/4*(2*sqrt(2)*sqrt(b)*arctan((sqrt(2)*sqrt(b*tan(d*x + c)^3)/sqrt(b) + t 
an(d*x + c))/tan(d*x + c))*tan(d*x + c)^2 + 2*sqrt(2)*sqrt(b)*arctan((sqrt 
(2)*sqrt(b*tan(d*x + c)^3)/sqrt(b) - tan(d*x + c))/tan(d*x + c))*tan(d*x + 
 c)^2 - sqrt(2)*sqrt(b)*log((tan(d*x + c)^2 + sqrt(2)*sqrt(b*tan(d*x + c)^ 
3)/sqrt(b) + tan(d*x + c))/tan(d*x + c))*tan(d*x + c)^2 + sqrt(2)*sqrt(b)* 
log((tan(d*x + c)^2 - sqrt(2)*sqrt(b*tan(d*x + c)^3)/sqrt(b) + tan(d*x + c 
))/tan(d*x + c))*tan(d*x + c)^2 + 8*sqrt(b*tan(d*x + c)^3))/(b*d*tan(d*x + 
 c)^2)
 

Sympy [F]

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

integrate(1/(b*tan(d*x+c)**3)**(1/2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\sqrt {b \tan ^3(c+d x)}} \, dx=-\frac {\frac {2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{\sqrt {b}} + \frac {8}{\sqrt {b} \sqrt {\tan \left (d x + c\right )}}}{4 \, d} \] Input:

integrate(1/(b*tan(d*x+c)^3)^(1/2),x, algorithm="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))/sqrt(b) + 8/(sqrt(b)*sqrt(tan(d*x + 
c))))/d
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.31 \[ \int \frac {1}{\sqrt {b \tan ^3(c+d x)}} \, dx=-\frac {1}{4} \, b^{2} {\left (\frac {2 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} + 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{b^{4} d \mathrm {sgn}\left (\tan \left (d x + c\right )\right )} + \frac {2 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} - 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{b^{4} d \mathrm {sgn}\left (\tan \left (d x + c\right )\right )} - \frac {\sqrt {2} {\left | b \right |}^{\frac {3}{2}} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{b^{4} d \mathrm {sgn}\left (\tan \left (d x + c\right )\right )} + \frac {\sqrt {2} {\left | b \right |}^{\frac {3}{2}} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{b^{4} d \mathrm {sgn}\left (\tan \left (d x + c\right )\right )} + \frac {8}{\sqrt {b \tan \left (d x + c\right )} b^{2} d \mathrm {sgn}\left (\tan \left (d x + c\right )\right )}\right )} \] Input:

integrate(1/(b*tan(d*x+c)^3)^(1/2),x, algorithm="giac")
 

Output:

-1/4*b^2*(2*sqrt(2)*abs(b)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) 
+ 2*sqrt(b*tan(d*x + c)))/sqrt(abs(b)))/(b^4*d*sgn(tan(d*x + c))) + 2*sqrt 
(2)*abs(b)^(3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) - 2*sqrt(b*tan( 
d*x + c)))/sqrt(abs(b)))/(b^4*d*sgn(tan(d*x + c))) - sqrt(2)*abs(b)^(3/2)* 
log(b*tan(d*x + c) + sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(abs(b)) + abs(b))/( 
b^4*d*sgn(tan(d*x + c))) + sqrt(2)*abs(b)^(3/2)*log(b*tan(d*x + c) - sqrt( 
2)*sqrt(b*tan(d*x + c))*sqrt(abs(b)) + abs(b))/(b^4*d*sgn(tan(d*x + c))) + 
 8/(sqrt(b*tan(d*x + c))*b^2*d*sgn(tan(d*x + c))))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {b \tan ^3(c+d x)}} \, dx=\int \frac {1}{\sqrt {b\,{\mathrm {tan}\left (c+d\,x\right )}^3}} \,d x \] Input:

int(1/(b*tan(c + d*x)^3)^(1/2),x)
 

Output:

int(1/(b*tan(c + d*x)^3)^(1/2), x)
 

Reduce [F]

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

int(1/(b*tan(d*x+c)^3)^(1/2),x)
 

Output:

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