\(\int (b \tan ^3(c+d x))^{3/2} \, dx\) [31]

Optimal result

Integrand size = 14, antiderivative size = 223 \[ \int \left (b \tan ^3(c+d x)\right )^{3/2} \, dx=-\frac {2 b \sqrt {b \tan ^3(c+d x)}}{3 d}-\frac {b \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right ) \sqrt {b \tan ^3(c+d x)}}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}+\frac {b \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right ) \sqrt {b \tan ^3(c+d x)}}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}-\frac {b \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right ) \sqrt {b \tan ^3(c+d x)}}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 b \tan ^2(c+d x) \sqrt {b \tan ^3(c+d x)}}{7 d} \] Output:

-2/3*b*(b*tan(d*x+c)^3)^(1/2)/d+1/2*b*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*b*arctan(1+2^(1/2)*t 
an(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*b*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)+2/7*b*tan(d*x+c)^2*(b*tan(d*x+c)^3)^(1/2)/d
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.52 \[ \int \left (b \tan ^3(c+d x)\right )^{3/2} \, dx=\frac {b \sqrt {b \tan ^3(c+d x)} \left (21 \arctan \left (\sqrt [4]{-\tan ^2(c+d x)}\right ) \sqrt [4]{-\tan (c+d x)}-21 \text {arctanh}\left (\sqrt [4]{-\tan ^2(c+d x)}\right ) \sqrt [4]{-\tan (c+d x)}+2 \tan ^{\frac {7}{4}}(c+d x) \left (-7+3 \tan ^2(c+d x)\right )\right )}{21 d \tan ^{\frac {7}{4}}(c+d x)} \] Input:

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

Output:

(b*Sqrt[b*Tan[c + d*x]^3]*(21*ArcTan[(-Tan[c + d*x]^2)^(1/4)]*(-Tan[c + d* 
x])^(1/4) - 21*ArcTanh[(-Tan[c + d*x]^2)^(1/4)]*(-Tan[c + d*x])^(1/4) + 2* 
Tan[c + d*x]^(7/4)*(-7 + 3*Tan[c + d*x]^2)))/(21*d*Tan[c + d*x]^(7/4))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.89, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.214, Rules used = {3042, 4141, 3042, 3954, 3042, 3954, 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.

\(\displaystyle \int \left (b \tan ^3(c+d x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4141

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \int \tan ^{\frac {9}{2}}(c+d x)dx}{\tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \int \tan (c+d x)^{9/2}dx}{\tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3954

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \left (\frac {2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\int \tan ^{\frac {5}{2}}(c+d x)dx\right )}{\tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \left (\frac {2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\int \tan (c+d x)^{5/2}dx\right )}{\tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3954

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \left (\int \sqrt {\tan (c+d x)}dx+\frac {2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )}{\tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \left (\int \sqrt {\tan (c+d x)}dx+\frac {2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )}{\tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \left (\frac {\int \frac {\sqrt {\tan (c+d x)}}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}+\frac {2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )}{\tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \left (\frac {2 \int \frac {\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )}{\tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \left (\frac {2 \left (\frac {1}{2} \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}-\frac {1}{2} \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}+\frac {2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )}{\tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \left (\frac {2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )-\frac {1}{2} \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}+\frac {2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )}{\tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \left (\frac {2 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}+\frac {2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )}{\tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \left (\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}+\frac {2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )}{\tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \left (\frac {2 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )}{\tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \left (\frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )}{\tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \left (\frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )}{\tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {b \sqrt {b \tan ^3(c+d x)} \left (\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d}+\frac {2 \tan ^{\frac {7}{2}}(c+d x)}{7 d}-\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )}{\tan ^{\frac {3}{2}}(c+d x)}\)

Input:

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

Output:

(b*Sqrt[b*Tan[c + d*x]^3]*((2*((-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/S 
qrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2])/2 + (Log[1 - Sqr 
t[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*Tan[c + d*x]^(3/2)) 
/(3*d) + (2*Tan[c + d*x]^(7/2))/(7*d)))/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 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 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 = 236, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(\frac {\left (b \tan \left (d x +c \right )^{3}\right )^{\frac {3}{2}} \left (24 \left (b \tan \left (d x +c \right )\right )^{\frac {7}{2}} \left (b^{2}\right )^{\frac {1}{4}}+21 b^{4} \sqrt {2}\, \ln \left (-\frac {\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}-b \tan \left (d x +c \right )-\sqrt {b^{2}}}{b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+42 b^{4} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}+\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+42 b^{4} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}-\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )-56 b^{2} \left (b \tan \left (d x +c \right )\right )^{\frac {3}{2}} \left (b^{2}\right )^{\frac {1}{4}}\right )}{84 d \tan \left (d x +c \right )^{3} \left (b \tan \left (d x +c \right )\right )^{\frac {3}{2}} b^{2} \left (b^{2}\right )^{\frac {1}{4}}}\)	\(236\)
default	\(\frac {\left (b \tan \left (d x +c \right )^{3}\right )^{\frac {3}{2}} \left (24 \left (b \tan \left (d x +c \right )\right )^{\frac {7}{2}} \left (b^{2}\right )^{\frac {1}{4}}+21 b^{4} \sqrt {2}\, \ln \left (-\frac {\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}-b \tan \left (d x +c \right )-\sqrt {b^{2}}}{b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+42 b^{4} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}+\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+42 b^{4} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}-\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )-56 b^{2} \left (b \tan \left (d x +c \right )\right )^{\frac {3}{2}} \left (b^{2}\right )^{\frac {1}{4}}\right )}{84 d \tan \left (d x +c \right )^{3} \left (b \tan \left (d x +c \right )\right )^{\frac {3}{2}} b^{2} \left (b^{2}\right )^{\frac {1}{4}}}\)	\(236\)

Input:

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

Output:

1/84/d*(b*tan(d*x+c)^3)^(3/2)*(24*(b*tan(d*x+c))^(7/2)*(b^2)^(1/4)+21*b^4* 
2^(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))) 
+42*b^4*2^(1/2)*arctan((2^(1/2)*(b*tan(d*x+c))^(1/2)+(b^2)^(1/4))/(b^2)^(1 
/4))+42*b^4*2^(1/2)*arctan((2^(1/2)*(b*tan(d*x+c))^(1/2)-(b^2)^(1/4))/(b^2 
)^(1/4))-56*b^2*(b*tan(d*x+c))^(3/2)*(b^2)^(1/4))/tan(d*x+c)^3/(b*tan(d*x+ 
c))^(3/2)/b^2/(b^2)^(1/4)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.11 \[ \int \left (b \tan ^3(c+d x)\right )^{3/2} \, dx=\frac {42 \, \sqrt {2} b^{\frac {3}{2}} \arctan \left (\frac {b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )^{3}} \sqrt {b}}{b \tan \left (d x + c\right )}\right ) + 42 \, \sqrt {2} b^{\frac {3}{2}} \arctan \left (-\frac {b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )^{3}} \sqrt {b}}{b \tan \left (d x + c\right )}\right ) - 21 \, \sqrt {2} b^{\frac {3}{2}} \log \left (\frac {b \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )^{3}} \sqrt {b}}{\tan \left (d x + c\right )}\right ) + 21 \, \sqrt {2} b^{\frac {3}{2}} \log \left (\frac {b \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )^{3}} \sqrt {b}}{\tan \left (d x + c\right )}\right ) + 8 \, \sqrt {b \tan \left (d x + c\right )^{3}} {\left (3 \, b \tan \left (d x + c\right )^{2} - 7 \, b\right )}}{84 \, d} \] Input:

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

Output:

1/84*(42*sqrt(2)*b^(3/2)*arctan((b*tan(d*x + c) + sqrt(2)*sqrt(b*tan(d*x + 
 c)^3)*sqrt(b))/(b*tan(d*x + c))) + 42*sqrt(2)*b^(3/2)*arctan(-(b*tan(d*x 
+ c) - sqrt(2)*sqrt(b*tan(d*x + c)^3)*sqrt(b))/(b*tan(d*x + c))) - 21*sqrt 
(2)*b^(3/2)*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)) + 21*sqrt(2)*b^(3/2)*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)) 
 + 8*sqrt(b*tan(d*x + c)^3)*(3*b*tan(d*x + c)^2 - 7*b))/d
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.63 \[ \int \left (b \tan ^3(c+d x)\right )^{3/2} \, dx=\frac {24 \, b^{\frac {3}{2}} \tan \left (d x + c\right )^{\frac {7}{2}} - 56 \, b^{\frac {3}{2}} \tan \left (d x + c\right )^{\frac {3}{2}} + 21 \, {\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} b^{\frac {3}{2}}}{84 \, d} \] Input:

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

Output:

1/84*(24*b^(3/2)*tan(d*x + c)^(7/2) - 56*b^(3/2)*tan(d*x + c)^(3/2) + 21*( 
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)*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^(3/2))/d
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.13 \[ \int \left (b \tan ^3(c+d x)\right )^{3/2} \, dx=\frac {1}{84} \, b {\left (\frac {42 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} + 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{b d} + \frac {42 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} - 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{b d} - \frac {21 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{b d} + \frac {21 \, \sqrt {2} {\left | b \right |}^{\frac {3}{2}} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{b d} + \frac {8 \, {\left (3 \, \sqrt {b \tan \left (d x + c\right )} b^{21} d^{6} \tan \left (d x + c\right )^{3} - 7 \, \sqrt {b \tan \left (d x + c\right )} b^{21} d^{6} \tan \left (d x + c\right )\right )}}{b^{21} d^{7}}\right )} \mathrm {sgn}\left (\tan \left (d x + c\right )\right ) \] Input:

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

Output:

1/84*b*(42*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*d) + 42*sqrt(2)*abs(b)^(3/2)*arc 
tan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) - 2*sqrt(b*tan(d*x + c)))/sqrt(abs( 
b)))/(b*d) - 21*sqrt(2)*abs(b)^(3/2)*log(b*tan(d*x + c) + sqrt(2)*sqrt(b*t 
an(d*x + c))*sqrt(abs(b)) + abs(b))/(b*d) + 21*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*d) + 
 8*(3*sqrt(b*tan(d*x + c))*b^21*d^6*tan(d*x + c)^3 - 7*sqrt(b*tan(d*x + c) 
)*b^21*d^6*tan(d*x + c))/(b^21*d^7))*sgn(tan(d*x + c))
 

Mupad [F(-1)]

Timed out. \[ \int \left (b \tan ^3(c+d x)\right )^{3/2} \, dx=\int {\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \left (b \tan ^3(c+d x)\right )^{3/2} \, dx=\sqrt {b}\, \left (\int \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{4}d x \right ) b \] Input:

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

Output:

sqrt(b)*int(sqrt(tan(c + d*x))*tan(c + d*x)**4,x)*b