\(\int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2 \, dx\) [344]

Optimal result

Integrand size = 25, antiderivative size = 192 \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2 \, dx=\frac {\sqrt {2} a^2 d^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {\sqrt {2} a^2 d^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {\sqrt {2} a^2 d^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{f}+\frac {4 a^2 (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f} \] Output:

2^(1/2)*a^2*d^(3/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/f-2^(1/ 
2)*a^2*d^(3/2)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/f+2^(1/2)*a^ 
2*d^(3/2)*arctanh(2^(1/2)*(d*tan(f*x+e))^(1/2)/(d^(1/2)+d^(1/2)*tan(f*x+e) 
))/f+4/3*a^2*(d*tan(f*x+e))^(3/2)/f+2/5*a^2*(d*tan(f*x+e))^(5/2)/d/f
 

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.58 \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2 \, dx=\frac {2 a^2 (d \tan (e+f x))^{3/2} \left (-15 \arctan \left (\sqrt [4]{-\tan ^2(e+f x)}\right ) \sqrt [4]{-\tan (e+f x)}+15 \text {arctanh}\left (\sqrt [4]{-\tan ^2(e+f x)}\right ) \sqrt [4]{-\tan (e+f x)}+\tan ^{\frac {7}{4}}(e+f x) (10+3 \tan (e+f x))\right )}{15 f \tan ^{\frac {7}{4}}(e+f x)} \] Input:

Integrate[(d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])^2,x]
 

Output:

(2*a^2*(d*Tan[e + f*x])^(3/2)*(-15*ArcTan[(-Tan[e + f*x]^2)^(1/4)]*(-Tan[e 
 + f*x])^(1/4) + 15*ArcTanh[(-Tan[e + f*x]^2)^(1/4)]*(-Tan[e + f*x])^(1/4) 
 + Tan[e + f*x]^(7/4)*(10 + 3*Tan[e + f*x])))/(15*f*Tan[e + f*x]^(7/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.64 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.20, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {3042, 4026, 27, 2030, 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.

\(\displaystyle \int (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \tan (e+f x)+a)^2 (d \tan (e+f x))^{3/2}dx\)

\(\Big \downarrow \) 4026

\(\displaystyle \int 2 a^2 \tan (e+f x) (d \tan (e+f x))^{3/2}dx+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle 2 a^2 \int \tan (e+f x) (d \tan (e+f x))^{3/2}dx+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 2030

\(\displaystyle \frac {2 a^2 \int (d \tan (e+f x))^{5/2}dx}{d}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a^2 \int (d \tan (e+f x))^{5/2}dx}{d}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 3954

\(\displaystyle \frac {2 a^2 \left (\frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-d^2 \int \sqrt {d \tan (e+f x)}dx\right )}{d}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a^2 \left (\frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-d^2 \int \sqrt {d \tan (e+f x)}dx\right )}{d}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {2 a^2 \left (\frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {d^3 \int \frac {\sqrt {d \tan (e+f x)}}{\tan ^2(e+f x) d^2+d^2}d(d \tan (e+f x))}{f}\right )}{d}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 a^2 \left (\frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \int \frac {d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{f}\right )}{d}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {2 a^2 \left (\frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \int \frac {d^2 \tan ^2(e+f x)+d}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}-\frac {1}{2} \int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}\right )}{d}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {2 a^2 \left (\frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}\right )-\frac {1}{2} \int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}\right )}{d}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {2 a^2 \left (\frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-d^2 \tan ^2(e+f x)-1}d\left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d^2 \tan ^2(e+f x)-1}d\left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}\right )-\frac {1}{2} \int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}\right )}{d}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {2 a^2 \left (\frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )-\frac {1}{2} \int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}\right )}{d}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {2 a^2 \left (\frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}\right )}{d}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 a^2 \left (\frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}\right )}{d}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 a^2 \left (\frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}\right )}{d}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {2 a^2 \left (\frac {2 d (d \tan (e+f x))^{3/2}}{3 f}-\frac {2 d^3 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\log \left (-\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}\right )}{d}+\frac {2 a^2 (d \tan (e+f x))^{5/2}}{5 d f}\)

Input:

Int[(d*Tan[e + f*x])^(3/2)*(a + a*Tan[e + f*x])^2,x]
 

Output:

(2*a^2*(d*Tan[e + f*x])^(5/2))/(5*d*f) + (2*a^2*((-2*d^3*((-(ArcTan[1 - Sq 
rt[2]*Sqrt[d]*Tan[e + f*x]]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + Sqrt[2]*Sqrt[d 
]*Tan[e + f*x]]/(Sqrt[2]*Sqrt[d]))/2 + (Log[d - Sqrt[2]*d^(3/2)*Tan[e + f* 
x] + d^2*Tan[e + f*x]^2]/(2*Sqrt[2]*Sqrt[d]) - Log[d + Sqrt[2]*d^(3/2)*Tan 
[e + f*x] + d^2*Tan[e + f*x]^2]/(2*Sqrt[2]*Sqrt[d]))/2))/f + (2*d*(d*Tan[e 
 + f*x])^(3/2))/(3*f)))/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 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

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 4026

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^2, x_Symbol] :> Simp[d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*( 
m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e + f* 
x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ 
[m, -1] &&  !(EqQ[m, 2] && EqQ[a, 0])
 

Maple [A] (verified)

Time = 1.55 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{3} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f d}\)	\(172\)
default	\(\frac {2 a^{2} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{3} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f d}\)	\(172\)
parts	\(\frac {2 a^{2} d \left (\sqrt {d \tan \left (f x +e \right )}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{f}+\frac {2 a^{2} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-d^{2} \sqrt {d \tan \left (f x +e \right )}+\frac {d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{f d}+\frac {2 a^{2} \left (\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{2} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}\)	\(481\)

Input:

int((d*tan(f*x+e))^(3/2)*(a+a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
 

Output:

2/f*a^2/d*(1/5*(d*tan(f*x+e))^(5/2)+2/3*d*(d*tan(f*x+e))^(3/2)-1/4*d^3/(d^ 
2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2 
)+(d^2)^(1/2))/(d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2 
)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2 
^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.02 \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2 \, dx=-\frac {30 \, \sqrt {2} a^{2} d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d}{d}\right ) + 30 \, \sqrt {2} a^{2} d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} - d}{d}\right ) - 15 \, \sqrt {2} a^{2} d^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right ) + 15 \, \sqrt {2} a^{2} d^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right ) - 4 \, {\left (3 \, a^{2} d \tan \left (f x + e\right )^{2} + 10 \, a^{2} d \tan \left (f x + e\right )\right )} \sqrt {d \tan \left (f x + e\right )}}{30 \, f} \] Input:

integrate((d*tan(f*x+e))^(3/2)*(a+a*tan(f*x+e))^2,x, algorithm="fricas")
 

Output:

-1/30*(30*sqrt(2)*a^2*d^(3/2)*arctan((sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) 
 + d)/d) + 30*sqrt(2)*a^2*d^(3/2)*arctan((sqrt(2)*sqrt(d*tan(f*x + e))*sqr 
t(d) - d)/d) - 15*sqrt(2)*a^2*d^(3/2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d* 
tan(f*x + e))*sqrt(d) + d) + 15*sqrt(2)*a^2*d^(3/2)*log(d*tan(f*x + e) - s 
qrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d) - 4*(3*a^2*d*tan(f*x + e)^2 + 10* 
a^2*d*tan(f*x + e))*sqrt(d*tan(f*x + e)))/f
 

Sympy [F]

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

integrate((d*tan(f*x+e))**(3/2)*(a+a*tan(f*x+e))**2,x)
 

Output:

a**2*(Integral((d*tan(e + f*x))**(3/2), x) + Integral(2*(d*tan(e + f*x))** 
(3/2)*tan(e + f*x), x) + Integral((d*tan(e + f*x))**(3/2)*tan(e + f*x)**2, 
 x))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.02 \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2 \, dx=-\frac {15 \, a^{2} d^{3} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - 12 \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a^{2} - 40 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{2} d}{30 \, d f} \] Input:

integrate((d*tan(f*x+e))^(3/2)*(a+a*tan(f*x+e))^2,x, algorithm="maxima")
 

Output:

-1/30*(15*a^2*d^3*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt( 
d*tan(f*x + e)))/sqrt(d))/sqrt(d) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) 
*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - sqrt(2)*log(d*tan(f* 
x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) + sqrt(2)*log(d 
*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d)) - 12*(d 
*tan(f*x + e))^(5/2)*a^2 - 40*(d*tan(f*x + e))^(3/2)*a^2*d)/(d*f)
 

Giac [F(-2)]

Exception generated. \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2 \, dx=\text {Exception raised: TypeError} \] Input:

integrate((d*tan(f*x+e))^(3/2)*(a+a*tan(f*x+e))^2,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[5,19]%%%}+%%%{8,[5,17]%%%}+%%%{28,[5,15]%%%}+%%%{56 
,[5,13]%%
 

Mupad [B] (verification not implemented)

Time = 1.77 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.54 \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2 \, dx=\frac {4\,a^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,f}+\frac {2\,a^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{5\,d\,f}-\frac {2\,{\left (-1\right )}^{1/4}\,a^2\,d^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{f}-\frac {{\left (-1\right )}^{1/4}\,a^2\,d^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{\sqrt {d}}\right )\,2{}\mathrm {i}}{f} \] Input:

int((d*tan(e + f*x))^(3/2)*(a + a*tan(e + f*x))^2,x)
 

Output:

(4*a^2*(d*tan(e + f*x))^(3/2))/(3*f) + (2*a^2*(d*tan(e + f*x))^(5/2))/(5*d 
*f) - (2*(-1)^(1/4)*a^2*d^(3/2)*atan(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d 
^(1/2)))/f - ((-1)^(1/4)*a^2*d^(3/2)*atan(((-1)^(1/4)*(d*tan(e + f*x))^(1/ 
2)*1i)/d^(1/2))*2i)/f
 

Reduce [F]

\[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x))^2 \, dx=\frac {2 \sqrt {d}\, a^{2} d \left (\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{2}+5 \left (\int \sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{2}d x \right ) f \right )}{5 f} \] Input:

int((d*tan(f*x+e))^(3/2)*(a+a*tan(f*x+e))^2,x)
 

Output:

(2*sqrt(d)*a**2*d*(sqrt(tan(e + f*x))*tan(e + f*x)**2 + 5*int(sqrt(tan(e + 
 f*x))*tan(e + f*x)**2,x)*f))/(5*f)