3.7 \(\int (b \tan ^3(e+f x))^{5/2} \, dx\)

Optimal. Leaf size=364 \[ \frac{2 b^2 \tan ^5(e+f x) \sqrt{b \tan ^3(e+f x)}}{13 f}-\frac{2 b^2 \tan ^3(e+f x) \sqrt{b \tan ^3(e+f x)}}{9 f}+\frac{2 b^2 \tan (e+f x) \sqrt{b \tan ^3(e+f x)}}{5 f}-\frac{b^2 \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (e+f x)}\right ) \sqrt{b \tan ^3(e+f x)}}{\sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}+\frac{b^2 \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (e+f x)}+1\right ) \sqrt{b \tan ^3(e+f x)}}{\sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}-\frac{b^2 \sqrt{b \tan ^3(e+f x)} \log \left (\tan (e+f x)-\sqrt{2} \sqrt{\tan (e+f x)}+1\right )}{2 \sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}+\frac{b^2 \sqrt{b \tan ^3(e+f x)} \log \left (\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}+1\right )}{2 \sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}-\frac{2 b^2 \cot (e+f x) \sqrt{b \tan ^3(e+f x)}}{f} \]

[Out]

(-2*b^2*Cot[e + f*x]*Sqrt[b*Tan[e + f*x]^3])/f - (b^2*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]]*Sqrt[b*Tan[e + f*
x]^3])/(Sqrt[2]*f*Tan[e + f*x]^(3/2)) + (b^2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]]*Sqrt[b*Tan[e + f*x]^3])/(S
qrt[2]*f*Tan[e + f*x]^(3/2)) - (b^2*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]]*Sqrt[b*Tan[e + f*x]^3])
/(2*Sqrt[2]*f*Tan[e + f*x]^(3/2)) + (b^2*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]]*Sqrt[b*Tan[e + f*x
]^3])/(2*Sqrt[2]*f*Tan[e + f*x]^(3/2)) + (2*b^2*Tan[e + f*x]*Sqrt[b*Tan[e + f*x]^3])/(5*f) - (2*b^2*Tan[e + f*
x]^3*Sqrt[b*Tan[e + f*x]^3])/(9*f) + (2*b^2*Tan[e + f*x]^5*Sqrt[b*Tan[e + f*x]^3])/(13*f)

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Rubi [A]  time = 0.145777, antiderivative size = 364, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 10, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.714, Rules used = {3658, 3473, 3476, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{2 b^2 \tan ^5(e+f x) \sqrt{b \tan ^3(e+f x)}}{13 f}-\frac{2 b^2 \tan ^3(e+f x) \sqrt{b \tan ^3(e+f x)}}{9 f}+\frac{2 b^2 \tan (e+f x) \sqrt{b \tan ^3(e+f x)}}{5 f}-\frac{b^2 \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (e+f x)}\right ) \sqrt{b \tan ^3(e+f x)}}{\sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}+\frac{b^2 \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (e+f x)}+1\right ) \sqrt{b \tan ^3(e+f x)}}{\sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}-\frac{b^2 \sqrt{b \tan ^3(e+f x)} \log \left (\tan (e+f x)-\sqrt{2} \sqrt{\tan (e+f x)}+1\right )}{2 \sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}+\frac{b^2 \sqrt{b \tan ^3(e+f x)} \log \left (\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}+1\right )}{2 \sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}-\frac{2 b^2 \cot (e+f x) \sqrt{b \tan ^3(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^3)^(5/2),x]

[Out]

(-2*b^2*Cot[e + f*x]*Sqrt[b*Tan[e + f*x]^3])/f - (b^2*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]]*Sqrt[b*Tan[e + f*
x]^3])/(Sqrt[2]*f*Tan[e + f*x]^(3/2)) + (b^2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]]*Sqrt[b*Tan[e + f*x]^3])/(S
qrt[2]*f*Tan[e + f*x]^(3/2)) - (b^2*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]]*Sqrt[b*Tan[e + f*x]^3])
/(2*Sqrt[2]*f*Tan[e + f*x]^(3/2)) + (b^2*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]]*Sqrt[b*Tan[e + f*x
]^3])/(2*Sqrt[2]*f*Tan[e + f*x]^(3/2)) + (2*b^2*Tan[e + f*x]*Sqrt[b*Tan[e + f*x]^3])/(5*f) - (2*b^2*Tan[e + f*
x]^3*Sqrt[b*Tan[e + f*x]^3])/(9*f) + (2*b^2*Tan[e + f*x]^5*Sqrt[b*Tan[e + f*x]^3])/(13*f)

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[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
]])

Rule 3473

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

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[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 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), 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 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

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 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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 617

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

Rule 204

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

Rubi steps

\begin{align*} \int \left (b \tan ^3(e+f x)\right )^{5/2} \, dx &=\frac{\left (b^2 \sqrt{b \tan ^3(e+f x)}\right ) \int \tan ^{\frac{15}{2}}(e+f x) \, dx}{\tan ^{\frac{3}{2}}(e+f x)}\\ &=\frac{2 b^2 \tan ^5(e+f x) \sqrt{b \tan ^3(e+f x)}}{13 f}-\frac{\left (b^2 \sqrt{b \tan ^3(e+f x)}\right ) \int \tan ^{\frac{11}{2}}(e+f x) \, dx}{\tan ^{\frac{3}{2}}(e+f x)}\\ &=-\frac{2 b^2 \tan ^3(e+f x) \sqrt{b \tan ^3(e+f x)}}{9 f}+\frac{2 b^2 \tan ^5(e+f x) \sqrt{b \tan ^3(e+f x)}}{13 f}+\frac{\left (b^2 \sqrt{b \tan ^3(e+f x)}\right ) \int \tan ^{\frac{7}{2}}(e+f x) \, dx}{\tan ^{\frac{3}{2}}(e+f x)}\\ &=\frac{2 b^2 \tan (e+f x) \sqrt{b \tan ^3(e+f x)}}{5 f}-\frac{2 b^2 \tan ^3(e+f x) \sqrt{b \tan ^3(e+f x)}}{9 f}+\frac{2 b^2 \tan ^5(e+f x) \sqrt{b \tan ^3(e+f x)}}{13 f}-\frac{\left (b^2 \sqrt{b \tan ^3(e+f x)}\right ) \int \tan ^{\frac{3}{2}}(e+f x) \, dx}{\tan ^{\frac{3}{2}}(e+f x)}\\ &=-\frac{2 b^2 \cot (e+f x) \sqrt{b \tan ^3(e+f x)}}{f}+\frac{2 b^2 \tan (e+f x) \sqrt{b \tan ^3(e+f x)}}{5 f}-\frac{2 b^2 \tan ^3(e+f x) \sqrt{b \tan ^3(e+f x)}}{9 f}+\frac{2 b^2 \tan ^5(e+f x) \sqrt{b \tan ^3(e+f x)}}{13 f}+\frac{\left (b^2 \sqrt{b \tan ^3(e+f x)}\right ) \int \frac{1}{\sqrt{\tan (e+f x)}} \, dx}{\tan ^{\frac{3}{2}}(e+f x)}\\ &=-\frac{2 b^2 \cot (e+f x) \sqrt{b \tan ^3(e+f x)}}{f}+\frac{2 b^2 \tan (e+f x) \sqrt{b \tan ^3(e+f x)}}{5 f}-\frac{2 b^2 \tan ^3(e+f x) \sqrt{b \tan ^3(e+f x)}}{9 f}+\frac{2 b^2 \tan ^5(e+f x) \sqrt{b \tan ^3(e+f x)}}{13 f}+\frac{\left (b^2 \sqrt{b \tan ^3(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f \tan ^{\frac{3}{2}}(e+f x)}\\ &=-\frac{2 b^2 \cot (e+f x) \sqrt{b \tan ^3(e+f x)}}{f}+\frac{2 b^2 \tan (e+f x) \sqrt{b \tan ^3(e+f x)}}{5 f}-\frac{2 b^2 \tan ^3(e+f x) \sqrt{b \tan ^3(e+f x)}}{9 f}+\frac{2 b^2 \tan ^5(e+f x) \sqrt{b \tan ^3(e+f x)}}{13 f}+\frac{\left (2 b^2 \sqrt{b \tan ^3(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^4} \, dx,x,\sqrt{\tan (e+f x)}\right )}{f \tan ^{\frac{3}{2}}(e+f x)}\\ &=-\frac{2 b^2 \cot (e+f x) \sqrt{b \tan ^3(e+f x)}}{f}+\frac{2 b^2 \tan (e+f x) \sqrt{b \tan ^3(e+f x)}}{5 f}-\frac{2 b^2 \tan ^3(e+f x) \sqrt{b \tan ^3(e+f x)}}{9 f}+\frac{2 b^2 \tan ^5(e+f x) \sqrt{b \tan ^3(e+f x)}}{13 f}+\frac{\left (b^2 \sqrt{b \tan ^3(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (e+f x)}\right )}{f \tan ^{\frac{3}{2}}(e+f x)}+\frac{\left (b^2 \sqrt{b \tan ^3(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (e+f x)}\right )}{f \tan ^{\frac{3}{2}}(e+f x)}\\ &=-\frac{2 b^2 \cot (e+f x) \sqrt{b \tan ^3(e+f x)}}{f}+\frac{2 b^2 \tan (e+f x) \sqrt{b \tan ^3(e+f x)}}{5 f}-\frac{2 b^2 \tan ^3(e+f x) \sqrt{b \tan ^3(e+f x)}}{9 f}+\frac{2 b^2 \tan ^5(e+f x) \sqrt{b \tan ^3(e+f x)}}{13 f}+\frac{\left (b^2 \sqrt{b \tan ^3(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (e+f x)}\right )}{2 f \tan ^{\frac{3}{2}}(e+f x)}+\frac{\left (b^2 \sqrt{b \tan ^3(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (e+f x)}\right )}{2 f \tan ^{\frac{3}{2}}(e+f x)}-\frac{\left (b^2 \sqrt{b \tan ^3(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (e+f x)}\right )}{2 \sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}-\frac{\left (b^2 \sqrt{b \tan ^3(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (e+f x)}\right )}{2 \sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}\\ &=-\frac{2 b^2 \cot (e+f x) \sqrt{b \tan ^3(e+f x)}}{f}-\frac{b^2 \log \left (1-\sqrt{2} \sqrt{\tan (e+f x)}+\tan (e+f x)\right ) \sqrt{b \tan ^3(e+f x)}}{2 \sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}+\frac{b^2 \log \left (1+\sqrt{2} \sqrt{\tan (e+f x)}+\tan (e+f x)\right ) \sqrt{b \tan ^3(e+f x)}}{2 \sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}+\frac{2 b^2 \tan (e+f x) \sqrt{b \tan ^3(e+f x)}}{5 f}-\frac{2 b^2 \tan ^3(e+f x) \sqrt{b \tan ^3(e+f x)}}{9 f}+\frac{2 b^2 \tan ^5(e+f x) \sqrt{b \tan ^3(e+f x)}}{13 f}+\frac{\left (b^2 \sqrt{b \tan ^3(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (e+f x)}\right )}{\sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}-\frac{\left (b^2 \sqrt{b \tan ^3(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (e+f x)}\right )}{\sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}\\ &=-\frac{2 b^2 \cot (e+f x) \sqrt{b \tan ^3(e+f x)}}{f}-\frac{b^2 \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (e+f x)}\right ) \sqrt{b \tan ^3(e+f x)}}{\sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}+\frac{b^2 \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (e+f x)}\right ) \sqrt{b \tan ^3(e+f x)}}{\sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}-\frac{b^2 \log \left (1-\sqrt{2} \sqrt{\tan (e+f x)}+\tan (e+f x)\right ) \sqrt{b \tan ^3(e+f x)}}{2 \sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}+\frac{b^2 \log \left (1+\sqrt{2} \sqrt{\tan (e+f x)}+\tan (e+f x)\right ) \sqrt{b \tan ^3(e+f x)}}{2 \sqrt{2} f \tan ^{\frac{3}{2}}(e+f x)}+\frac{2 b^2 \tan (e+f x) \sqrt{b \tan ^3(e+f x)}}{5 f}-\frac{2 b^2 \tan ^3(e+f x) \sqrt{b \tan ^3(e+f x)}}{9 f}+\frac{2 b^2 \tan ^5(e+f x) \sqrt{b \tan ^3(e+f x)}}{13 f}\\ \end{align*}

Mathematica [A]  time = 0.829392, size = 199, normalized size = 0.55 \[ \frac{b \left (b \tan ^3(e+f x)\right )^{3/2} \left (360 \tan ^{\frac{13}{2}}(e+f x)-520 \tan ^{\frac{9}{2}}(e+f x)+936 \tan ^{\frac{5}{2}}(e+f x)-1170 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (e+f x)}\right )+1170 \sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (e+f x)}+1\right )-4680 \sqrt{\tan (e+f x)}-585 \sqrt{2} \log \left (\tan (e+f x)-\sqrt{2} \sqrt{\tan (e+f x)}+1\right )+585 \sqrt{2} \log \left (\tan (e+f x)+\sqrt{2} \sqrt{\tan (e+f x)}+1\right )\right )}{2340 f \tan ^{\frac{9}{2}}(e+f x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x]^3)^(5/2),x]

[Out]

(b*(b*Tan[e + f*x]^3)^(3/2)*(-1170*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]] + 1170*Sqrt[2]*ArcTan[1 + Sq
rt[2]*Sqrt[Tan[e + f*x]]] - 585*Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] + 585*Sqrt[2]*Log[1
+ Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - 4680*Sqrt[Tan[e + f*x]] + 936*Tan[e + f*x]^(5/2) - 520*Tan[e +
f*x]^(9/2) + 360*Tan[e + f*x]^(13/2)))/(2340*f*Tan[e + f*x]^(9/2))

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Maple [A]  time = 0.044, size = 266, normalized size = 0.7 \begin{align*}{\frac{1}{2340\,f \left ( \tan \left ( fx+e \right ) \right ) ^{5}{b}^{4}} \left ( b \left ( \tan \left ( fx+e \right ) \right ) ^{3} \right ) ^{{\frac{5}{2}}} \left ( 360\, \left ( b\tan \left ( fx+e \right ) \right ) ^{13/2}-520\,{b}^{2} \left ( b\tan \left ( fx+e \right ) \right ) ^{9/2}+585\,{b}^{6}\sqrt [4]{{b}^{2}}\sqrt{2}\ln \left ( -{\frac{b\tan \left ( fx+e \right ) +\sqrt [4]{{b}^{2}}\sqrt{b\tan \left ( fx+e \right ) }\sqrt{2}+\sqrt{{b}^{2}}}{\sqrt [4]{{b}^{2}}\sqrt{b\tan \left ( fx+e \right ) }\sqrt{2}-b\tan \left ( fx+e \right ) -\sqrt{{b}^{2}}}} \right ) +1170\,{b}^{6}\sqrt [4]{{b}^{2}}\sqrt{2}\arctan \left ({\frac{\sqrt{2}\sqrt{b\tan \left ( fx+e \right ) }+\sqrt [4]{{b}^{2}}}{\sqrt [4]{{b}^{2}}}} \right ) +1170\,{b}^{6}\sqrt [4]{{b}^{2}}\sqrt{2}\arctan \left ({\frac{\sqrt{2}\sqrt{b\tan \left ( fx+e \right ) }-\sqrt [4]{{b}^{2}}}{\sqrt [4]{{b}^{2}}}} \right ) +936\,{b}^{4} \left ( b\tan \left ( fx+e \right ) \right ) ^{5/2}-4680\,{b}^{6}\sqrt{b\tan \left ( fx+e \right ) } \right ) \left ( b\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e)^3)^(5/2),x)

[Out]

1/2340/f*(b*tan(f*x+e)^3)^(5/2)*(360*(b*tan(f*x+e))^(13/2)-520*b^2*(b*tan(f*x+e))^(9/2)+585*b^6*(b^2)^(1/4)*2^
(1/2)*ln(-(b*tan(f*x+e)+(b^2)^(1/4)*(b*tan(f*x+e))^(1/2)*2^(1/2)+(b^2)^(1/2))/((b^2)^(1/4)*(b*tan(f*x+e))^(1/2
)*2^(1/2)-b*tan(f*x+e)-(b^2)^(1/2)))+1170*b^6*(b^2)^(1/4)*2^(1/2)*arctan((2^(1/2)*(b*tan(f*x+e))^(1/2)+(b^2)^(
1/4))/(b^2)^(1/4))+1170*b^6*(b^2)^(1/4)*2^(1/2)*arctan((2^(1/2)*(b*tan(f*x+e))^(1/2)-(b^2)^(1/4))/(b^2)^(1/4))
+936*b^4*(b*tan(f*x+e))^(5/2)-4680*b^6*(b*tan(f*x+e))^(1/2))/tan(f*x+e)^5/(b*tan(f*x+e))^(5/2)/b^4

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Maxima [A]  time = 1.67078, size = 240, normalized size = 0.66 \begin{align*} \frac{360 \, b^{\frac{5}{2}} \tan \left (f x + e\right )^{\frac{13}{2}} - 520 \, b^{\frac{5}{2}} \tan \left (f x + e\right )^{\frac{9}{2}} + 936 \, b^{\frac{5}{2}} \tan \left (f x + e\right )^{\frac{5}{2}} + 585 \,{\left (2 \, \sqrt{2} \sqrt{b} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (f x + e\right )}\right )}\right ) + 2 \, \sqrt{2} \sqrt{b} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (f x + e\right )}\right )}\right ) + \sqrt{2} \sqrt{b} \log \left (\sqrt{2} \sqrt{\tan \left (f x + e\right )} + \tan \left (f x + e\right ) + 1\right ) - \sqrt{2} \sqrt{b} \log \left (-\sqrt{2} \sqrt{\tan \left (f x + e\right )} + \tan \left (f x + e\right ) + 1\right )\right )} b^{2} - 4680 \, b^{\frac{5}{2}} \sqrt{\tan \left (f x + e\right )}}{2340 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^3)^(5/2),x, algorithm="maxima")

[Out]

1/2340*(360*b^(5/2)*tan(f*x + e)^(13/2) - 520*b^(5/2)*tan(f*x + e)^(9/2) + 936*b^(5/2)*tan(f*x + e)^(5/2) + 58
5*(2*sqrt(2)*sqrt(b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(f*x + e)))) + 2*sqrt(2)*sqrt(b)*arctan(-1/2*sqrt
(2)*(sqrt(2) - 2*sqrt(tan(f*x + e)))) + sqrt(2)*sqrt(b)*log(sqrt(2)*sqrt(tan(f*x + e)) + tan(f*x + e) + 1) - s
qrt(2)*sqrt(b)*log(-sqrt(2)*sqrt(tan(f*x + e)) + tan(f*x + e) + 1))*b^2 - 4680*b^(5/2)*sqrt(tan(f*x + e)))/f

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Fricas [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((b*tan(f*x+e)^3)^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan ^{3}{\left (e + f x \right )}\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)**3)**(5/2),x)

[Out]

Integral((b*tan(e + f*x)**3)**(5/2), x)

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Giac [A]  time = 1.73344, size = 412, normalized size = 1.13 \begin{align*} \frac{1}{2340} \,{\left (\frac{1170 \, \sqrt{2} b \sqrt{{\left | b \right |}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | b \right |}} + 2 \, \sqrt{b \tan \left (f x + e\right )}\right )}}{2 \, \sqrt{{\left | b \right |}}}\right )}{f} + \frac{1170 \, \sqrt{2} b \sqrt{{\left | b \right |}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | b \right |}} - 2 \, \sqrt{b \tan \left (f x + e\right )}\right )}}{2 \, \sqrt{{\left | b \right |}}}\right )}{f} + \frac{585 \, \sqrt{2} b \sqrt{{\left | b \right |}} \log \left (b \tan \left (f x + e\right ) + \sqrt{2} \sqrt{b \tan \left (f x + e\right )} \sqrt{{\left | b \right |}} +{\left | b \right |}\right )}{f} - \frac{585 \, \sqrt{2} b \sqrt{{\left | b \right |}} \log \left (b \tan \left (f x + e\right ) - \sqrt{2} \sqrt{b \tan \left (f x + e\right )} \sqrt{{\left | b \right |}} +{\left | b \right |}\right )}{f} + \frac{8 \,{\left (45 \, \sqrt{b \tan \left (f x + e\right )} b^{66} f^{12} \tan \left (f x + e\right )^{6} - 65 \, \sqrt{b \tan \left (f x + e\right )} b^{66} f^{12} \tan \left (f x + e\right )^{4} + 117 \, \sqrt{b \tan \left (f x + e\right )} b^{66} f^{12} \tan \left (f x + e\right )^{2} - 585 \, \sqrt{b \tan \left (f x + e\right )} b^{66} f^{12}\right )}}{b^{65} f^{13}}\right )} b \mathrm{sgn}\left (\tan \left (f x + e\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e)^3)^(5/2),x, algorithm="giac")

[Out]

1/2340*(1170*sqrt(2)*b*sqrt(abs(b))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) + 2*sqrt(b*tan(f*x + e)))/sqrt(ab
s(b)))/f + 1170*sqrt(2)*b*sqrt(abs(b))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) - 2*sqrt(b*tan(f*x + e)))/sqr
t(abs(b)))/f + 585*sqrt(2)*b*sqrt(abs(b))*log(b*tan(f*x + e) + sqrt(2)*sqrt(b*tan(f*x + e))*sqrt(abs(b)) + abs
(b))/f - 585*sqrt(2)*b*sqrt(abs(b))*log(b*tan(f*x + e) - sqrt(2)*sqrt(b*tan(f*x + e))*sqrt(abs(b)) + abs(b))/f
+ 8*(45*sqrt(b*tan(f*x + e))*b^66*f^12*tan(f*x + e)^6 - 65*sqrt(b*tan(f*x + e))*b^66*f^12*tan(f*x + e)^4 + 11
7*sqrt(b*tan(f*x + e))*b^66*f^12*tan(f*x + e)^2 - 585*sqrt(b*tan(f*x + e))*b^66*f^12)/(b^65*f^13))*b*sgn(tan(f
*x + e))