3.395 \(\int \tan ^4(e+f x) (1+\tan (e+f x))^{3/2} \, dx\)
Optimal. Leaf size=227 \[ \frac{2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}-\frac{\sqrt{\sqrt{2}-1} \tan ^{-1}\left (\frac{\left (1-\sqrt{2}\right ) \tan (e+f x)-2 \sqrt{2}+3}{\sqrt{2 \left (5 \sqrt{2}-7\right )} \sqrt{\tan (e+f x)+1}}\right )}{f}-\frac{8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{63 f}-\frac{22 (\tan (e+f x)+1)^{5/2}}{63 f}+\frac{2 \sqrt{\tan (e+f x)+1}}{f}-\frac{\sqrt{1+\sqrt{2}} \tanh ^{-1}\left (\frac{\left (1+\sqrt{2}\right ) \tan (e+f x)+2 \sqrt{2}+3}{\sqrt{2 \left (7+5 \sqrt{2}\right )} \sqrt{\tan (e+f x)+1}}\right )}{f} \]
[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)
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Rubi [A] time = 0.382771, antiderivative size = 227, normalized size of antiderivative = 1.,
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
= {3566, 3647, 3631, 3482, 12, 3536, 3535, 203, 207} \[ \frac{2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}-\frac{\sqrt{\sqrt{2}-1} \tan ^{-1}\left (\frac{\left (1-\sqrt{2}\right ) \tan (e+f x)-2 \sqrt{2}+3}{\sqrt{2 \left (5 \sqrt{2}-7\right )} \sqrt{\tan (e+f x)+1}}\right )}{f}-\frac{8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{63 f}-\frac{22 (\tan (e+f x)+1)^{5/2}}{63 f}+\frac{2 \sqrt{\tan (e+f x)+1}}{f}-\frac{\sqrt{1+\sqrt{2}} \tanh ^{-1}\left (\frac{\left (1+\sqrt{2}\right ) \tan (e+f x)+2 \sqrt{2}+3}{\sqrt{2 \left (7+5 \sqrt{2}\right )} \sqrt{\tan (e+f x)+1}}\right )}{f} \]
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 3566
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 3647
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])))
Rule 3631
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 3482
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 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 3536
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 3535
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 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 207
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 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 ) \operatorname{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 ) \operatorname{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] time = 1.55359, size = 122, normalized size = 0.54 \[ \frac{2 \sqrt{\tan (e+f x)+1} \left (7 \tan ^4(e+f x)+10 \tan ^3(e+f x)-12 \tan ^2(e+f x)-26 \tan (e+f x)+52\right )-\frac{126 \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1-i}}\right )}{\sqrt{1-i}}-\frac{126 \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1+i}}\right )}{\sqrt{1+i}}}{63 f} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[e + f*x]^4*(1 + Tan[e + f*x])^(3/2),x]
[Out]
((-126*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]])/Sqrt[1 - I] - (126*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 +
I]])/Sqrt[1 + I] + 2*Sqrt[1 + Tan[e + f*x]]*(52 - 26*Tan[e + f*x] - 12*Tan[e + f*x]^2 + 10*Tan[e + f*x]^3 + 7
*Tan[e + f*x]^4))/(63*f)
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Maple [A] time = 0.03, size = 344, normalized size = 1.5 \begin{align*}{\frac{2}{9\,f} \left ( 1+\tan \left ( fx+e \right ) \right ) ^{{\frac{9}{2}}}}-{\frac{4}{7\,f} \left ( 1+\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}}}+2\,{\frac{\sqrt{1+\tan \left ( fx+e \right ) }}{f}}+{\frac{\sqrt{2\,\sqrt{2}+2}\sqrt{2}}{4\,f}\ln \left ( 1+\sqrt{2}-\sqrt{2\,\sqrt{2}+2}\sqrt{1+\tan \left ( fx+e \right ) }+\tan \left ( fx+e \right ) \right ) }-{\frac{\sqrt{2}}{f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+\tan \left ( fx+e \right ) }-\sqrt{2\,\sqrt{2}+2} \right ) } \right ) }+2\,{\frac{1}{f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{2\,\sqrt{1+\tan \left ( fx+e \right ) }-\sqrt{2\,\sqrt{2}+2}}{\sqrt{-2+2\,\sqrt{2}}}} \right ) }-{\frac{\sqrt{2\,\sqrt{2}+2}\sqrt{2}}{4\,f}\ln \left ( 1+\sqrt{2}+\sqrt{2\,\sqrt{2}+2}\sqrt{1+\tan \left ( fx+e \right ) }+\tan \left ( fx+e \right ) \right ) }-{\frac{\sqrt{2}}{f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( \sqrt{2\,\sqrt{2}+2}+2\,\sqrt{1+\tan \left ( fx+e \right ) } \right ) } \right ) }+2\,{\frac{1}{f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{\sqrt{2\,\sqrt{2}+2}+2\,\sqrt{1+\tan \left ( fx+e \right ) }}{\sqrt{-2+2\,\sqrt{2}}}} \right ) } \end{align*}
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)
[Out]
2/9/f*(1+tan(f*x+e))^(9/2)-4/7/f*(1+tan(f*x+e))^(7/2)+2*(1+tan(f*x+e))^(1/2)/f+1/4/f*(2*2^(1/2)+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))-1/f/(-2+2*2^(1/2))^(1/2)*arctan((2*(1+tan
(f*x+e))^(1/2)-(2*2^(1/2)+2)^(1/2))/(-2+2*2^(1/2))^(1/2))*2^(1/2)+2/f/(-2+2*2^(1/2))^(1/2)*arctan((2*(1+tan(f*
x+e))^(1/2)-(2*2^(1/2)+2)^(1/2))/(-2+2*2^(1/2))^(1/2))-1/4/f*(2*2^(1/2)+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))-1/f/(-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))*2^(1/2)+2/f/(-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))
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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]
Timed out
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Fricas [B] time = 2.14938, size = 3069, normalized size = 13.52 \begin{align*} \text{result too large to display} \end{align*}
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)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\tan{\left (e + f x \right )} + 1\right )^{\frac{3}{2}} \tan ^{4}{\left (e + f x \right )}\, dx \end{align*}
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)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\tan \left (f x + e\right ) + 1\right )}^{\frac{3}{2}} \tan \left (f x + e\right )^{4}\,{d x} \end{align*}
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]
integrate((tan(f*x + e) + 1)^(3/2)*tan(f*x + e)^4, x)