\(\int \sqrt {b \tan ^3(c+d x)} \, dx\) [32]

Optimal result

Integrand size = 14, antiderivative size = 193 \[ \int \sqrt {b \tan ^3(c+d x)} \, dx=\frac {2 \cot (c+d x) \sqrt {b \tan ^3(c+d x)}}{d}+\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right ) \sqrt {b \tan ^3(c+d x)}}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right ) \sqrt {b \tan ^3(c+d x)}}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right ) \sqrt {b \tan ^3(c+d x)}}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.84 \[ \int \sqrt {b \tan ^3(c+d x)} \, dx=\frac {\left (\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}+\frac {\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2}}+2 \sqrt {\tan (c+d x)}\right ) \sqrt {b \tan ^3(c+d x)}}{d \tan ^{\frac {3}{2}}(c+d x)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.92, 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, 3954, 3042, 3957, 266, 755, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4141

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3954

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[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 + (-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*Sqrt[Tan[c + d*x]])/d)*Sqrt[b*Tan[c + d*x] 
^3])/Tan[c + d*x]^(3/2)
 

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 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   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 3954

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d 
*x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2   Int[(b*Tan[c + d*x])^(n - 2), x] 
, x] /; FreeQ[{b, c, d}, x] && GtQ[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 = 0.05 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(-\frac {\sqrt {b \tan \left (d x +c \right )^{3}}\, \left (\left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (d x +c \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}+\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+2 \left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}-\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )-8 \sqrt {b \tan \left (d x +c \right )}\right )}{4 d \tan \left (d x +c \right ) \sqrt {b \tan \left (d x +c \right )}}\)	\(205\)
default	\(-\frac {\sqrt {b \tan \left (d x +c \right )^{3}}\, \left (\left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (d x +c \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}+\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+2 \left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}-\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )-8 \sqrt {b \tan \left (d x +c \right )}\right )}{4 d \tan \left (d x +c \right ) \sqrt {b \tan \left (d x +c \right )}}\)	\(205\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.37 \[ \int \sqrt {b \tan ^3(c+d x)} \, dx=-\frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )^{3}} \sqrt {b}}{b \tan \left (d x + c\right )}\right ) \tan \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {b} \arctan \left (-\frac {b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )^{3}} \sqrt {b}}{b \tan \left (d x + c\right )}\right ) \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b} \log \left (\frac {b \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )^{3}} \sqrt {b}}{\tan \left (d x + c\right )}\right ) \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b} \log \left (\frac {b \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )^{3}} \sqrt {b}}{\tan \left (d x + c\right )}\right ) \tan \left (d x + c\right ) - 8 \, \sqrt {b \tan \left (d x + c\right )^{3}}}{4 \, d \tan \left (d x + c\right )} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.69 \[ \int \sqrt {b \tan ^3(c+d x)} \, dx=-\frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} \sqrt {b} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} \sqrt {b} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} \sqrt {b} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 8 \, \sqrt {b} \sqrt {\tan \left (d x + c\right )}}{4 \, d} \] Input:

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

Output:

-1/4*(2*sqrt(2)*sqrt(b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)) 
)) + 2*sqrt(2)*sqrt(b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)) 
)) + sqrt(2)*sqrt(b)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - 
sqrt(2)*sqrt(b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - 8*sq 
rt(b)*sqrt(tan(d*x + c)))/d
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.01 \[ \int \sqrt {b \tan ^3(c+d x)} \, dx=-\frac {1}{4} \, {\left (\frac {2 \, \sqrt {2} \sqrt {{\left | b \right |}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} + 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{d} + \frac {2 \, \sqrt {2} \sqrt {{\left | b \right |}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} - 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{d} + \frac {\sqrt {2} \sqrt {{\left | b \right |}} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{d} - \frac {\sqrt {2} \sqrt {{\left | b \right |}} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{d} - \frac {8 \, \sqrt {b \tan \left (d x + c\right )}}{d}\right )} \mathrm {sgn}\left (\tan \left (d x + c\right )\right ) \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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