3.4.95 \(\int \tan ^4(e+f x) (1+\tan (e+f x))^{3/2} \, dx\) [395]

Optimal. Leaf size=227 \[ -\frac {\sqrt {-1+\sqrt {2}} \text {ArcTan}\left (\frac {3-2 \sqrt {2}+\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (-7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{f}-\frac {\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {3+2 \sqrt {2}+\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{f}+\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {22 (1+\tan (e+f x))^{5/2}}{63 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{63 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{9 f} \]

[Out]

-arctan((3-2*2^(1/2)+(1-2^(1/2))*tan(f*x+e))/(-14+10*2^(1/2))^(1/2)/(1+tan(f*x+e))^(1/2))*(2^(1/2)-1)^(1/2)/f-
arctanh((3+2*2^(1/2)+(1+2^(1/2))*tan(f*x+e))/(14+10*2^(1/2))^(1/2)/(1+tan(f*x+e))^(1/2))*(1+2^(1/2))^(1/2)/f+2
*(1+tan(f*x+e))^(1/2)/f-22/63*(1+tan(f*x+e))^(5/2)/f-8/63*tan(f*x+e)*(1+tan(f*x+e))^(5/2)/f+2/9*tan(f*x+e)^2*(
1+tan(f*x+e))^(5/2)/f

________________________________________________________________________________________

Rubi [A]
time = 0.29, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3647, 3728, 3712, 3563, 12, 3617, 3616, 209, 213} \begin {gather*} -\frac {\sqrt {\sqrt {2}-1} \text {ArcTan}\left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{f}+\frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{63 f}-\frac {22 (\tan (e+f x)+1)^{5/2}}{63 f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}-\frac {\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^4*(1 + Tan[e + f*x])^(3/2),x]

[Out]

-((Sqrt[-1 + Sqrt[2]]*ArcTan[(3 - 2*Sqrt[2] + (1 - Sqrt[2])*Tan[e + f*x])/(Sqrt[2*(-7 + 5*Sqrt[2])]*Sqrt[1 + T
an[e + f*x]])])/f) - (Sqrt[1 + Sqrt[2]]*ArcTanh[(3 + 2*Sqrt[2] + (1 + Sqrt[2])*Tan[e + f*x])/(Sqrt[2*(7 + 5*Sq
rt[2])]*Sqrt[1 + Tan[e + f*x]])])/f + (2*Sqrt[1 + Tan[e + f*x]])/f - (22*(1 + Tan[e + f*x])^(5/2))/(63*f) - (8
*Tan[e + f*x]*(1 + Tan[e + f*x])^(5/2))/(63*f) + (2*Tan[e + f*x]^2*(1 + Tan[e + f*x])^(5/2))/(9*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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 3563

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

Rule 3616

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

Rule 3617

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> With[{q =
Rt[a^2 + b^2, 2]}, Dist[1/(2*q), Int[(a*c + b*d + c*q + (b*c - a*d + d*q)*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*
x]], x], x] - Dist[1/(2*q), Int[(a*c + b*d - c*q + (b*c - a*d - d*q)*Tan[e + f*x])/Sqrt[a + b*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] && NeQ[c^2 + d^2, 0] && NeQ[2
*a*c*d - b*(c^2 - d^2), 0] && (PerfectSquareQ[a^2 + b^2] || RationalQ[a, b, c, d])

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 3712

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

Rule 3728

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((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 - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

\begin {align*} \int \tan ^4(e+f x) (1+\tan (e+f x))^{3/2} \, dx &=\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{9 f}+\frac {2}{9} \int \tan (e+f x) (1+\tan (e+f x))^{3/2} \left (-2-\frac {9}{2} \tan (e+f x)-2 \tan ^2(e+f x)\right ) \, dx\\ &=-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{63 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{9 f}+\frac {4}{63} \int (1+\tan (e+f x))^{3/2} \left (2-\frac {55}{4} \tan ^2(e+f x)\right ) \, dx\\ &=-\frac {22 (1+\tan (e+f x))^{5/2}}{63 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{63 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{9 f}+\int (1+\tan (e+f x))^{3/2} \, dx\\ &=\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {22 (1+\tan (e+f x))^{5/2}}{63 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{63 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{9 f}+\int \frac {2 \tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {22 (1+\tan (e+f x))^{5/2}}{63 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{63 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{9 f}+2 \int \frac {\tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {22 (1+\tan (e+f x))^{5/2}}{63 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{63 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{9 f}-\frac {\int \frac {1+\left (-1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx}{\sqrt {2}}+\frac {\int \frac {1+\left (-1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx}{\sqrt {2}}\\ &=\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {22 (1+\tan (e+f x))^{5/2}}{63 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{63 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{9 f}+\frac {\left (4-3 \sqrt {2}\right ) \text {Subst}\left (\int \frac {1}{2 \left (-1+\sqrt {2}\right )-4 \left (-1+\sqrt {2}\right )^2+x^2} \, dx,x,\frac {1-2 \left (-1+\sqrt {2}\right )-\left (-1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}}\right )}{f}+\frac {\left (4+3 \sqrt {2}\right ) \text {Subst}\left (\int \frac {1}{2 \left (-1-\sqrt {2}\right )-4 \left (-1-\sqrt {2}\right )^2+x^2} \, dx,x,\frac {1-2 \left (-1-\sqrt {2}\right )-\left (-1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}}\right )}{f}\\ &=-\frac {\sqrt {-1+\sqrt {2}} \tan ^{-1}\left (\frac {3-2 \sqrt {2}+\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (-7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{f}-\frac {\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {3+2 \sqrt {2}+\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{f}+\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {22 (1+\tan (e+f x))^{5/2}}{63 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{63 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{9 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 1.51, size = 155, normalized size = 0.68 \begin {gather*} \frac {2 \cos ^2(e+f x) \left (-63 \left (\frac {\tanh ^{-1}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )}{\sqrt {1-i}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )}{\sqrt {1+i}}\right ) (1+\tan (e+f x))^2+(1+\tan (e+f x))^{5/2} \left (71+7 \sec ^4(e+f x)-36 \tan (e+f x)+2 \sec ^2(e+f x) (-13+5 \tan (e+f x))\right )\right )}{63 f (\cos (e+f x)+\sin (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^4*(1 + Tan[e + f*x])^(3/2),x]

[Out]

(2*Cos[e + f*x]^2*(-63*(ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]]/Sqrt[1 - I] + ArcTanh[Sqrt[1 + Tan[e + f*x
]]/Sqrt[1 + I]]/Sqrt[1 + I])*(1 + Tan[e + f*x])^2 + (1 + Tan[e + f*x])^(5/2)*(71 + 7*Sec[e + f*x]^4 - 36*Tan[e
+ f*x] + 2*Sec[e + f*x]^2*(-13 + 5*Tan[e + f*x]))))/(63*f*(Cos[e + f*x] + Sin[e + f*x])^2)

________________________________________________________________________________________

Maple [A]
time = 0.16, size = 245, normalized size = 1.08

method	result	size
derivativedivides	\(\frac {\frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}-\frac {4 \left (1+\tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+2 \sqrt {1+\tan \left (f x +e \right )}-\frac {\sqrt {2}\, \left (-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}-\frac {\sqrt {2}\, \left (\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}}{f}\)	\(245\)
default	\(\frac {\frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}-\frac {4 \left (1+\tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+2 \sqrt {1+\tan \left (f x +e \right )}-\frac {\sqrt {2}\, \left (-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}-\frac {\sqrt {2}\, \left (\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}}{f}\)	\(245\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^4*(1+tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/f*(2/9*(1+tan(f*x+e))^(9/2)-4/7*(1+tan(f*x+e))^(7/2)+2*(1+tan(f*x+e))^(1/2)-1/2*2^(1/2)*(-1/2*(2*2^(1/2)+2)^
(1/2)*ln(1+2^(1/2)-(2*2^(1/2)+2)^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))+2*(1-2^(1/2))/(-2+2*2^(1/2))^(1/2)*arc
tan((2*(1+tan(f*x+e))^(1/2)-(2*2^(1/2)+2)^(1/2))/(-2+2*2^(1/2))^(1/2)))-1/2*2^(1/2)*(1/2*(2*2^(1/2)+2)^(1/2)*l
n(1+2^(1/2)+(2*2^(1/2)+2)^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))+2*(1-2^(1/2))/(-2+2*2^(1/2))^(1/2)*arctan(((2
*2^(1/2)+2)^(1/2)+2*(1+tan(f*x+e))^(1/2))/(-2+2*2^(1/2))^(1/2))))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4*(1+tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((tan(f*x + e) + 1)^(3/2)*tan(f*x + e)^4, x)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1105 vs. \(2 (190) = 380\).
time = 1.04, size = 1105, normalized size = 4.87 \begin {gather*} -\frac {252 \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} f \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (\frac {1}{16} \cdot 8^{\frac {3}{4}} \sqrt {2} {\left (2 \, f^{5} \sqrt {\frac {1}{f^{4}}} + \sqrt {2} f^{3}\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 8^{\frac {1}{4}} {\left (\sqrt {2} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + f \cos \left (f x + e\right )\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} + 2 \, \cos \left (f x + e\right ) + 2 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} - \frac {1}{8} \cdot 8^{\frac {3}{4}} {\left (2 \, f^{5} \sqrt {\frac {1}{f^{4}}} + \sqrt {2} f^{3}\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} - f^{2} \sqrt {\frac {1}{f^{4}}} - \sqrt {2}\right ) \cos \left (f x + e\right )^{4} + 252 \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} f \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (\frac {1}{16} \cdot 8^{\frac {3}{4}} \sqrt {2} {\left (2 \, f^{5} \sqrt {\frac {1}{f^{4}}} + \sqrt {2} f^{3}\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) - 8^{\frac {1}{4}} {\left (\sqrt {2} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + f \cos \left (f x + e\right )\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} + 2 \, \cos \left (f x + e\right ) + 2 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} - \frac {1}{8} \cdot 8^{\frac {3}{4}} {\left (2 \, f^{5} \sqrt {\frac {1}{f^{4}}} + \sqrt {2} f^{3}\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} + f^{2} \sqrt {\frac {1}{f^{4}}} + \sqrt {2}\right ) \cos \left (f x + e\right )^{4} + 63 \cdot 8^{\frac {1}{4}} {\left (\sqrt {2} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right )^{4} + 2 \, f \cos \left (f x + e\right )^{4}\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \frac {1}{f^{4}}^{\frac {1}{4}} \log \left (\frac {2 \, {\left (2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 8^{\frac {1}{4}} {\left (\sqrt {2} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + f \cos \left (f x + e\right )\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} + 2 \, \cos \left (f x + e\right ) + 2 \, \sin \left (f x + e\right )\right )}}{\cos \left (f x + e\right )}\right ) - 63 \cdot 8^{\frac {1}{4}} {\left (\sqrt {2} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right )^{4} + 2 \, f \cos \left (f x + e\right )^{4}\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \frac {1}{f^{4}}^{\frac {1}{4}} \log \left (\frac {2 \, {\left (2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) - 8^{\frac {1}{4}} {\left (\sqrt {2} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + f \cos \left (f x + e\right )\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} + 2 \, \cos \left (f x + e\right ) + 2 \, \sin \left (f x + e\right )\right )}}{\cos \left (f x + e\right )}\right ) - 16 \, {\left (71 \, \cos \left (f x + e\right )^{4} - 26 \, \cos \left (f x + e\right )^{2} - 2 \, {\left (18 \, \cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) + 7\right )} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{504 \, f \cos \left (f x + e\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4*(1+tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/504*(252*8^(1/4)*sqrt(2)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(1/16*8^(3/4)*sqrt(2)
*(2*f^5*sqrt(f^(-4)) + sqrt(2)*f^3)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((2*sqrt(2)*f^2*sqrt(f^(-4))*cos
(f*x + e) + 8^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4))
+ 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*
x + e))*(f^(-4))^(3/4) - 1/8*8^(3/4)*(2*f^5*sqrt(f^(-4)) + sqrt(2)*f^3)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*
sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) - f^2*sqrt(f^(-4)) - sqrt(2))*cos(f*x + e)^4 +
252*8^(1/4)*sqrt(2)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(1/16*8^(3/4)*sqrt(2)*(2*f^5
*sqrt(f^(-4)) + sqrt(2)*f^3)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x +
e) - 8^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sq
rt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x + e))
*(f^(-4))^(3/4) - 1/8*8^(3/4)*(2*f^5*sqrt(f^(-4)) + sqrt(2)*f^3)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((c
os(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) + f^2*sqrt(f^(-4)) + sqrt(2))*cos(f*x + e)^4 + 63*8^(
1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e)^4 + 2*f*cos(f*x + e)^4)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(f^(
-4))^(1/4)*log(2*(2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 8^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + f
*cos(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))
^(1/4) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x + e)) - 63*8^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e)^4
+ 2*f*cos(f*x + e)^4)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*log(2*(2*sqrt(2)*f^2*sqrt(f^(-4))*c
os(f*x + e) - 8^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)
) + 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(
f*x + e)) - 16*(71*cos(f*x + e)^4 - 26*cos(f*x + e)^2 - 2*(18*cos(f*x + e)^3 - 5*cos(f*x + e))*sin(f*x + e) +
7)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e)))/(f*cos(f*x + e)^4)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (\tan {\left (e + f x \right )} + 1\right )^{\frac {3}{2}} \tan ^{4}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**4*(1+tan(f*x+e))**(3/2),x)

[Out]

Integral((tan(e + f*x) + 1)**(3/2)*tan(e + f*x)**4, x)

________________________________________________________________________________________

Giac [A]
time = 0.84, size = 252, normalized size = 1.11 \begin {gather*} \frac {\sqrt {\sqrt {2} - 1} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{f} + \frac {\sqrt {\sqrt {2} - 1} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{f} - \frac {\sqrt {\sqrt {2} + 1} \log \left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{2 \, f} + \frac {\sqrt {\sqrt {2} + 1} \log \left (-2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{2 \, f} + \frac {2 \, {\left (7 \, f^{8} {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {9}{2}} - 18 \, f^{8} {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {7}{2}} + 63 \, f^{8} \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{63 \, f^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4*(1+tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

sqrt(sqrt(2) - 1)*arctan(1/2*2^(3/4)*(2^(1/4)*sqrt(sqrt(2) + 2) + 2*sqrt(tan(f*x + e) + 1))/sqrt(-sqrt(2) + 2)
)/f + sqrt(sqrt(2) - 1)*arctan(-1/2*2^(3/4)*(2^(1/4)*sqrt(sqrt(2) + 2) - 2*sqrt(tan(f*x + e) + 1))/sqrt(-sqrt(
2) + 2))/f - 1/2*sqrt(sqrt(2) + 1)*log(2^(1/4)*sqrt(sqrt(2) + 2)*sqrt(tan(f*x + e) + 1) + sqrt(2) + tan(f*x +
e) + 1)/f + 1/2*sqrt(sqrt(2) + 1)*log(-2^(1/4)*sqrt(sqrt(2) + 2)*sqrt(tan(f*x + e) + 1) + sqrt(2) + tan(f*x +
e) + 1)/f + 2/63*(7*f^8*(tan(f*x + e) + 1)^(9/2) - 18*f^8*(tan(f*x + e) + 1)^(7/2) + 63*f^8*sqrt(tan(f*x + e)
+ 1))/f^9

________________________________________________________________________________________

Mupad [B]
time = 6.00, size = 118, normalized size = 0.52 \begin {gather*} \frac {2\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f}-\frac {4\,{\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 )}^{9/2}}{9\,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}\,1{}\mathrm {i}\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}\,1{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^4*(tan(e + f*x) + 1)^(3/2),x)

[Out]

(2*(tan(e + f*x) + 1)^(1/2))/f - (4*(tan(e + f*x) + 1)^(7/2))/(7*f) + (2*(tan(e + f*x) + 1)^(9/2))/(9*f) + ata
n(f*((1/2 - 1i/2)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2)*1i)*((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)*1i)*((1/2 + 1i/2)/f^2)^(1/2)*2i

________________________________________________________________________________________