\(\int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx\) [390]

Optimal result

Integrand size = 21, antiderivative size = 315 \[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=-\frac {\sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{f}+\frac {\sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{f}-\frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {1+\sqrt {2}} f}+\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {1+\sqrt {2}} f}-\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {2 (1+\tan (e+f x))^{3/2}}{3 f}-\frac {4 (1+\tan (e+f x))^{5/2}}{35 f}+\frac {2 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{7 f} \]

[Out]

-1/2*ln(1+2^(1/2)-(2+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))/f/(1+2^(1/2))^(1/2)+1/2*ln(1+2^(1/2)+(2
+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))/f/(1+2^(1/2))^(1/2)-arctan(((2+2*2^(1/2))^(1/2)-2*(1+tan(f*
x+e))^(1/2))/(-2+2*2^(1/2))^(1/2))*(1+2^(1/2))^(1/2)/f+arctan(((2+2*2^(1/2))^(1/2)+2*(1+tan(f*x+e))^(1/2))/(-2
+2*2^(1/2))^(1/2))*(1+2^(1/2))^(1/2)/f-2*(1+tan(f*x+e))^(1/2)/f-2/3*(1+tan(f*x+e))^(3/2)/f-4/35*(1+tan(f*x+e))
^(5/2)/f+2/7*tan(f*x+e)*(1+tan(f*x+e))^(5/2)/f

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3647, 3711, 12, 3609, 3566, 722, 1108, 648, 632, 210, 642} \[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=-\frac {\sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}+\frac {\sqrt {1+\sqrt {2}} \arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}+\frac {2 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}-\frac {4 (\tan (e+f x)+1)^{5/2}}{35 f}-\frac {2 (\tan (e+f x)+1)^{3/2}}{3 f}-\frac {2 \sqrt {\tan (e+f x)+1}}{f}-\frac {\log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {1+\sqrt {2}} f}+\frac {\log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {1+\sqrt {2}} f} \]

[In]

Int[Tan[e + f*x]^3*(1 + Tan[e + f*x])^(3/2),x]

[Out]

-((Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/f) + (
Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/f - Log[1
+ Sqrt[2] + Tan[e + f*x] - Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*x]]]/(2*Sqrt[1 + Sqrt[2]]*f) + Log[1 + Sq
rt[2] + Tan[e + f*x] + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*x]]]/(2*Sqrt[1 + Sqrt[2]]*f) - (2*Sqrt[1 + Tan
[e + f*x]])/f - (2*(1 + Tan[e + f*x])^(3/2))/(3*f) - (4*(1 + Tan[e + f*x])^(5/2))/(35*f) + (2*Tan[e + f*x]*(1
+ Tan[e + f*x])^(5/2))/(7*f)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

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 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 722

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2*e, Subst[Int[1/(c*d^2 + a*e^2 - 2*c
*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0]

Rule 1108

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/c, 2]}
, Dist[1/(2*c*q*r), Int[(r - x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(r + x)/(q + r*x + x^2), x], x
]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[b^2 - 4*a*c]

Rule 3566

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[(a + x)^n/(b^2 + x^2), x], x
, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 + b^2, 0]

Rule 3609

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

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3711

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{7 f}+\frac {2}{7} \int (1+\tan (e+f x))^{3/2} \left (-1-\frac {7}{2} \tan (e+f x)-\tan ^2(e+f x)\right ) \, dx \\ & = -\frac {4 (1+\tan (e+f x))^{5/2}}{35 f}+\frac {2 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{7 f}+\frac {2}{7} \int -\frac {7}{2} \tan (e+f x) (1+\tan (e+f x))^{3/2} \, dx \\ & = -\frac {4 (1+\tan (e+f x))^{5/2}}{35 f}+\frac {2 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{7 f}-\int \tan (e+f x) (1+\tan (e+f x))^{3/2} \, dx \\ & = -\frac {2 (1+\tan (e+f x))^{3/2}}{3 f}-\frac {4 (1+\tan (e+f x))^{5/2}}{35 f}+\frac {2 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{7 f}-\int (-1+\tan (e+f x)) \sqrt {1+\tan (e+f x)} \, dx \\ & = -\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {2 (1+\tan (e+f x))^{3/2}}{3 f}-\frac {4 (1+\tan (e+f x))^{5/2}}{35 f}+\frac {2 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{7 f}-\int -\frac {2}{\sqrt {1+\tan (e+f x)}} \, dx \\ & = -\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {2 (1+\tan (e+f x))^{3/2}}{3 f}-\frac {4 (1+\tan (e+f x))^{5/2}}{35 f}+\frac {2 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{7 f}+2 \int \frac {1}{\sqrt {1+\tan (e+f x)}} \, dx \\ & = -\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {2 (1+\tan (e+f x))^{3/2}}{3 f}-\frac {4 (1+\tan (e+f x))^{5/2}}{35 f}+\frac {2 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{7 f}+\frac {2 \text {Subst}\left (\int \frac {1}{\sqrt {1+x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {2 (1+\tan (e+f x))^{3/2}}{3 f}-\frac {4 (1+\tan (e+f x))^{5/2}}{35 f}+\frac {2 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{7 f}+\frac {4 \text {Subst}\left (\int \frac {1}{2-2 x^2+x^4} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{f} \\ & = -\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {2 (1+\tan (e+f x))^{3/2}}{3 f}-\frac {4 (1+\tan (e+f x))^{5/2}}{35 f}+\frac {2 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{7 f}+\frac {\text {Subst}\left (\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-x}{\sqrt {2}-\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{\sqrt {1+\sqrt {2}} f}+\frac {\text {Subst}\left (\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}+x}{\sqrt {2}+\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{\sqrt {1+\sqrt {2}} f} \\ & = -\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {2 (1+\tan (e+f x))^{3/2}}{3 f}-\frac {4 (1+\tan (e+f x))^{5/2}}{35 f}+\frac {2 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{7 f}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {2}-\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{\sqrt {2} f}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {2}+\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{\sqrt {2} f}-\frac {\text {Subst}\left (\int \frac {-\sqrt {2 \left (1+\sqrt {2}\right )}+2 x}{\sqrt {2}-\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {1+\sqrt {2}} f}+\frac {\text {Subst}\left (\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 x}{\sqrt {2}+\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {1+\sqrt {2}} f} \\ & = -\frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {1+\sqrt {2}} f}+\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {1+\sqrt {2}} f}-\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {2 (1+\tan (e+f x))^{3/2}}{3 f}-\frac {4 (1+\tan (e+f x))^{5/2}}{35 f}+\frac {2 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{7 f}-\frac {\sqrt {2} \text {Subst}\left (\int \frac {1}{2 \left (1-\sqrt {2}\right )-x^2} \, dx,x,-\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}\right )}{f}-\frac {\sqrt {2} \text {Subst}\left (\int \frac {1}{2 \left (1-\sqrt {2}\right )-x^2} \, dx,x,\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}\right )}{f} \\ & = -\frac {\arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{\sqrt {-1+\sqrt {2}} f}+\frac {\arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{\sqrt {-1+\sqrt {2}} f}-\frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {1+\sqrt {2}} f}+\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {1+\sqrt {2}} f}-\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {2 (1+\tan (e+f x))^{3/2}}{3 f}-\frac {4 (1+\tan (e+f x))^{5/2}}{35 f}+\frac {2 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{7 f} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.66 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.36 \[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\frac {105 (1-i)^{3/2} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )+105 (1+i)^{3/2} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )+2 \sqrt {1+\tan (e+f x)} \left (-146-32 \tan (e+f x)+24 \tan ^2(e+f x)+15 \tan ^3(e+f x)\right )}{105 f} \]

[In]

Integrate[Tan[e + f*x]^3*(1 + Tan[e + f*x])^(3/2),x]

[Out]

(105*(1 - I)^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]] + 105*(1 + I)^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*
x]]/Sqrt[1 + I]] + 2*Sqrt[1 + Tan[e + f*x]]*(-146 - 32*Tan[e + f*x] + 24*Tan[e + f*x]^2 + 15*Tan[e + f*x]^3))/
(105*f)

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(\frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7 f}-\frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}-\frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f}-\frac {2 \sqrt {1+\tan \left (f x +e \right )}}{f}-\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2 f}+\frac {\arctan \left (\frac {\sqrt {2+2 \sqrt {2}}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2 f}+\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{f \sqrt {-2+2 \sqrt {2}}}\)	\(354\)
default	\(\frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7 f}-\frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}-\frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f}-\frac {2 \sqrt {1+\tan \left (f x +e \right )}}{f}-\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2 f}+\frac {\arctan \left (\frac {\sqrt {2+2 \sqrt {2}}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2 f}+\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{f \sqrt {-2+2 \sqrt {2}}}\)	\(354\)

[In]

int(tan(f*x+e)^3*(1+tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/7/f*(1+tan(f*x+e))^(7/2)-2/5*(1+tan(f*x+e))^(5/2)/f-2/3*(1+tan(f*x+e))^(3/2)/f-2*(1+tan(f*x+e))^(1/2)/f-1/4/
f*(2+2*2^(1/2))^(1/2)*2^(1/2)*ln(1+2^(1/2)+(2+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))+1/2/f*(2+2*2^(
1/2))^(1/2)*ln(1+2^(1/2)+(2+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))+1/f/(-2+2*2^(1/2))^(1/2)*arctan(
((2+2*2^(1/2))^(1/2)+2*(1+tan(f*x+e))^(1/2))/(-2+2*2^(1/2))^(1/2))*2^(1/2)+1/4/f*(2+2*2^(1/2))^(1/2)*2^(1/2)*l
n(1+2^(1/2)-(2+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))-1/2/f*(2+2*2^(1/2))^(1/2)*ln(1+2^(1/2)-(2+2*2
^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))+1/f/(-2+2*2^(1/2))^(1/2)*arctan((2*(1+tan(f*x+e))^(1/2)-(2+2*2^
(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2))*2^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.14 \[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\frac {105 \, \sqrt {2} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 105 \, \sqrt {2} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 105 \, \sqrt {2} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 105 \, \sqrt {2} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 4 \, {\left (15 \, \tan \left (f x + e\right )^{3} + 24 \, \tan \left (f x + e\right )^{2} - 32 \, \tan \left (f x + e\right ) - 146\right )} \sqrt {\tan \left (f x + e\right ) + 1}}{210 \, f} \]

[In]

integrate(tan(f*x+e)^3*(1+tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/210*(105*sqrt(2)*f*sqrt(-(f^2*sqrt(-1/f^4) + 1)/f^2)*log(sqrt(2)*(f^3*sqrt(-1/f^4) + f)*sqrt(-(f^2*sqrt(-1/f
^4) + 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) - 105*sqrt(2)*f*sqrt(-(f^2*sqrt(-1/f^4) + 1)/f^2)*log(-sqrt(2)*(f^3*
sqrt(-1/f^4) + f)*sqrt(-(f^2*sqrt(-1/f^4) + 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) - 105*sqrt(2)*f*sqrt((f^2*sqrt
(-1/f^4) - 1)/f^2)*log(sqrt(2)*(f^3*sqrt(-1/f^4) - f)*sqrt((f^2*sqrt(-1/f^4) - 1)/f^2) + 2*sqrt(tan(f*x + e) +
1)) + 105*sqrt(2)*f*sqrt((f^2*sqrt(-1/f^4) - 1)/f^2)*log(-sqrt(2)*(f^3*sqrt(-1/f^4) - f)*sqrt((f^2*sqrt(-1/f^
4) - 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) + 4*(15*tan(f*x + e)^3 + 24*tan(f*x + e)^2 - 32*tan(f*x + e) - 146)*s
qrt(tan(f*x + e) + 1))/f

Sympy [F]

\[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\int \left (\tan {\left (e + f x \right )} + 1\right )^{\frac {3}{2}} \tan ^{3}{\left (e + f x \right )}\, dx \]

[In]

integrate(tan(f*x+e)**3*(1+tan(f*x+e))**(3/2),x)

[Out]

Integral((tan(e + f*x) + 1)**(3/2)*tan(e + f*x)**3, x)

Maxima [F]

\[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\int { {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{3} \,d x } \]

[In]

integrate(tan(f*x+e)^3*(1+tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((tan(f*x + e) + 1)^(3/2)*tan(f*x + e)^3, x)

Giac [F(-1)]

Timed out. \[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\text {Timed out} \]

[In]

integrate(tan(f*x+e)^3*(1+tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 5.90 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.41 \[ \int \tan ^3(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{7/2}}{7\,f}-\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{3/2}}{3\,f}-\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{5/2}}{5\,f}-\frac {2\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f}+\mathrm {atan}\left (f\,\sqrt {\frac {-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (f\,\sqrt {\frac {-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \]

[In]

int(tan(e + f*x)^3*(tan(e + f*x) + 1)^(3/2),x)

[Out]

(2*(tan(e + f*x) + 1)^(7/2))/(7*f) - (2*(tan(e + f*x) + 1)^(3/2))/(3*f) - (2*(tan(e + f*x) + 1)^(5/2))/(5*f) -
(2*(tan(e + f*x) + 1)^(1/2))/f + atan(f*((- 1/2 - 1i/2)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2))*((- 1/2 - 1i/2)/
f^2)^(1/2)*2i - atan(f*((- 1/2 + 1i/2)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2))*((- 1/2 + 1i/2)/f^2)^(1/2)*2i