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

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

[Out]

-2*b^2*cot(f*x+e)*(b*tan(f*x+e)^3)^(1/2)/f+1/2*b^2*arctan(-1+2^(1/2)*tan(f*x+e)^(1/2))*(b*tan(f*x+e)^3)^(1/2)/
f*2^(1/2)/tan(f*x+e)^(3/2)+1/2*b^2*arctan(1+2^(1/2)*tan(f*x+e)^(1/2))*(b*tan(f*x+e)^3)^(1/2)/f*2^(1/2)/tan(f*x
+e)^(3/2)-1/4*b^2*ln(1-2^(1/2)*tan(f*x+e)^(1/2)+tan(f*x+e))*(b*tan(f*x+e)^3)^(1/2)/f*2^(1/2)/tan(f*x+e)^(3/2)+
1/4*b^2*ln(1+2^(1/2)*tan(f*x+e)^(1/2)+tan(f*x+e))*(b*tan(f*x+e)^3)^(1/2)/f*2^(1/2)/tan(f*x+e)^(3/2)+2/5*b^2*(b
*tan(f*x+e)^3)^(1/2)*tan(f*x+e)/f-2/9*b^2*(b*tan(f*x+e)^3)^(1/2)*tan(f*x+e)^3/f+2/13*b^2*(b*tan(f*x+e)^3)^(1/2
)*tan(f*x+e)^5/f

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Rubi [A]  time = 0.15, antiderivative size = 364, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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 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 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 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
]])

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*}

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Mathematica [A]  time = 0.83, size = 199, normalized size = 0.55 \[ \frac {b \left (b \tan ^3(e+f x)\right )^{3/2} \left (-1170 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )+1170 \sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right )+360 \tan ^{\frac {13}{2}}(e+f x)-520 \tan ^{\frac {9}{2}}(e+f x)+936 \tan ^{\frac {5}{2}}(e+f x)-4680 \sqrt {\tan (e+f x)}-585 \sqrt {2} \log \left (\tan (e+f x)-\sqrt {2} \sqrt {\tan (e+f x)}+1\right )+585 \sqrt {2} \log \left (\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1\right )\right )}{2340 f \tan ^{\frac {9}{2}}(e+f x)} \]

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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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)2*b*((1/13*b^60*f^12*sqrt(b*tan(f*x+exp(1)))*(b*tan(f*x+exp(1)))^6-1/9*b^62*f^12*sqrt(b*tan(f*x+exp(1)))*(
b*tan(f*x+exp(1)))^4+1/5*b^64*f^12*sqrt(b*tan(f*x+exp(1)))*(b*tan(f*x+exp(1)))^2-b^66*f^12*sqrt(b*tan(f*x+exp(
1))))/b^65/f^13+1/2*b*sqrt(abs(b))*atan(sqrt(2)*(1/2*sqrt(2)*sqrt(abs(b))+sqrt(b*tan(f*x+exp(1))))/sqrt(abs(b)
))/sqrt(2)/f+1/2*b*sqrt(abs(b))*atan(sqrt(2)*(-1/2*sqrt(2)*sqrt(abs(b))+sqrt(b*tan(f*x+exp(1))))/sqrt(abs(b)))
/sqrt(2)/f+1/4*b*sqrt(abs(b))*ln(b*tan(f*x+exp(1))+sqrt(2)*sqrt(b*tan(f*x+exp(1)))*sqrt(abs(b))+abs(b))/sqrt(2
)/f-1/4*b*sqrt(abs(b))*ln(b*tan(f*x+exp(1))-sqrt(2)*sqrt(b*tan(f*x+exp(1)))*sqrt(abs(b))+abs(b))/sqrt(2)/f)*si
gn(tan(f*x+exp(1)))

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maple [A]  time = 0.25, size = 263, normalized size = 0.72 \[ \frac {\left (b \left (\tan ^{3}\left (f x +e \right )\right )\right )^{\frac {5}{2}} \left (360 \left (b \tan \left (f x +e \right )\right )^{\frac {13}{2}}-520 b^{2} \left (b \tan \left (f x +e \right )\right )^{\frac {9}{2}}+585 b^{6} \left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {b \tan \left (f x +e \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (f x +e \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+1170 b^{6} \left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (f x +e \right )}+\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+1170 b^{6} \left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (f x +e \right )}-\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+936 \left (b \tan \left (f x +e \right )\right )^{\frac {5}{2}} b^{4}-4680 b^{6} \sqrt {b \tan \left (f x +e \right )}\right )}{2340 f \tan \left (f x +e \right )^{5} \left (b \tan \left (f x +e \right )\right )^{\frac {5}{2}} b^{4}} \]

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*tan(f*x+e)-(b^2)^(1/4)*(b*tan(
f*x+e))^(1/2)*2^(1/2)+(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*tan(f*x+e))^(5/2)*b^4-4680*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 = 1.27, size = 178, normalized size = 0.49 \[ \frac {360 \, b^{\frac {5}{2}} \tan \left (f x + e\right )^{\frac {13}{2}} - 520 \, b^{\frac {5}{2}} \tan \left (f x + e\right )^{\frac {9}{2}} + 936 \, b^{\frac {5}{2}} \tan \left (f x + e\right )^{\frac {5}{2}} + 585 \, {\left (2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (f x + e\right )}\right )}\right ) + 2 \, \sqrt {2} \sqrt {b} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (f x + e\right )}\right )}\right ) + \sqrt {2} \sqrt {b} \log \left (\sqrt {2} \sqrt {\tan \left (f x + e\right )} + \tan \left (f x + e\right ) + 1\right ) - \sqrt {2} \sqrt {b} \log \left (-\sqrt {2} \sqrt {\tan \left (f x + e\right )} + \tan \left (f x + e\right ) + 1\right )\right )} b^{2} - 4680 \, b^{\frac {5}{2}} \sqrt {\tan \left (f x + e\right )}}{2340 \, f} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^3\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(e + f*x)^3)^(5/2),x)

[Out]

int((b*tan(e + f*x)^3)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \tan ^{3}{\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx \]

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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