3.12 \(\int \frac {1}{(b \tan ^3(e+f x))^{5/2}} \, dx\)

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

[Out]

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

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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, 3474, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right ) \tan ^{\frac {3}{2}}(e+f x)}{\sqrt {2} b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {\tan (e+f x)}+1\right ) \tan ^{\frac {3}{2}}(e+f x)}{\sqrt {2} b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \tan (e+f x)}{b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\tan ^{\frac {3}{2}}(e+f x) \log \left (\tan (e+f x)-\sqrt {2} \sqrt {\tan (e+f x)}+1\right )}{2 \sqrt {2} b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {\tan ^{\frac {3}{2}}(e+f x) \log \left (\tan (e+f x)+\sqrt {2} \sqrt {\tan (e+f x)}+1\right )}{2 \sqrt {2} b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {2 \cot ^5(e+f x)}{13 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \cot ^3(e+f x)}{9 b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {2 \cot (e+f x)}{5 b^2 f \sqrt {b \tan ^3(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x]^3)^(-5/2),x]

[Out]

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

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 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{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 3474

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x])^(n + 1)/(b*d*(n + 1)), x] - Dist[
1/b^2, Int[(b*Tan[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 \frac {1}{\left (b \tan ^3(e+f x)\right )^{5/2}} \, dx &=\frac {\tan ^{\frac {3}{2}}(e+f x) \int \frac {1}{\tan ^{\frac {15}{2}}(e+f x)} \, dx}{b^2 \sqrt {b \tan ^3(e+f x)}}\\ &=-\frac {2 \cot ^5(e+f x)}{13 b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {\tan ^{\frac {3}{2}}(e+f x) \int \frac {1}{\tan ^{\frac {11}{2}}(e+f x)} \, dx}{b^2 \sqrt {b \tan ^3(e+f x)}}\\ &=\frac {2 \cot ^3(e+f x)}{9 b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {2 \cot ^5(e+f x)}{13 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\tan ^{\frac {3}{2}}(e+f x) \int \frac {1}{\tan ^{\frac {7}{2}}(e+f x)} \, dx}{b^2 \sqrt {b \tan ^3(e+f x)}}\\ &=-\frac {2 \cot (e+f x)}{5 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \cot ^3(e+f x)}{9 b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {2 \cot ^5(e+f x)}{13 b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {\tan ^{\frac {3}{2}}(e+f x) \int \frac {1}{\tan ^{\frac {3}{2}}(e+f x)} \, dx}{b^2 \sqrt {b \tan ^3(e+f x)}}\\ &=-\frac {2 \cot (e+f x)}{5 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \cot ^3(e+f x)}{9 b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {2 \cot ^5(e+f x)}{13 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \tan (e+f x)}{b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\tan ^{\frac {3}{2}}(e+f x) \int \sqrt {\tan (e+f x)} \, dx}{b^2 \sqrt {b \tan ^3(e+f x)}}\\ &=-\frac {2 \cot (e+f x)}{5 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \cot ^3(e+f x)}{9 b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {2 \cot ^5(e+f x)}{13 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \tan (e+f x)}{b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\tan ^{\frac {3}{2}}(e+f x) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{b^2 f \sqrt {b \tan ^3(e+f x)}}\\ &=-\frac {2 \cot (e+f x)}{5 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \cot ^3(e+f x)}{9 b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {2 \cot ^5(e+f x)}{13 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \tan (e+f x)}{b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\left (2 \tan ^{\frac {3}{2}}(e+f x)\right ) \operatorname {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\sqrt {\tan (e+f x)}\right )}{b^2 f \sqrt {b \tan ^3(e+f x)}}\\ &=-\frac {2 \cot (e+f x)}{5 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \cot ^3(e+f x)}{9 b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {2 \cot ^5(e+f x)}{13 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \tan (e+f x)}{b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {\tan ^{\frac {3}{2}}(e+f x) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (e+f x)}\right )}{b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\tan ^{\frac {3}{2}}(e+f x) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (e+f x)}\right )}{b^2 f \sqrt {b \tan ^3(e+f x)}}\\ &=-\frac {2 \cot (e+f x)}{5 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \cot ^3(e+f x)}{9 b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {2 \cot ^5(e+f x)}{13 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \tan (e+f x)}{b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\tan ^{\frac {3}{2}}(e+f x) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (e+f x)}\right )}{2 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\tan ^{\frac {3}{2}}(e+f x) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (e+f x)}\right )}{2 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\tan ^{\frac {3}{2}}(e+f x) \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} b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\tan ^{\frac {3}{2}}(e+f x) \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} b^2 f \sqrt {b \tan ^3(e+f x)}}\\ &=-\frac {2 \cot (e+f x)}{5 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \cot ^3(e+f x)}{9 b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {2 \cot ^5(e+f x)}{13 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \tan (e+f x)}{b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\log \left (1-\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right ) \tan ^{\frac {3}{2}}(e+f x)}{2 \sqrt {2} b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {\log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right ) \tan ^{\frac {3}{2}}(e+f x)}{2 \sqrt {2} b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\tan ^{\frac {3}{2}}(e+f x) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (e+f x)}\right )}{\sqrt {2} b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {\tan ^{\frac {3}{2}}(e+f x) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (e+f x)}\right )}{\sqrt {2} b^2 f \sqrt {b \tan ^3(e+f x)}}\\ &=-\frac {2 \cot (e+f x)}{5 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \cot ^3(e+f x)}{9 b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {2 \cot ^5(e+f x)}{13 b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {2 \tan (e+f x)}{b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right ) \tan ^{\frac {3}{2}}(e+f x)}{\sqrt {2} b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right ) \tan ^{\frac {3}{2}}(e+f x)}{\sqrt {2} b^2 f \sqrt {b \tan ^3(e+f x)}}+\frac {\log \left (1-\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right ) \tan ^{\frac {3}{2}}(e+f x)}{2 \sqrt {2} b^2 f \sqrt {b \tan ^3(e+f x)}}-\frac {\log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right ) \tan ^{\frac {3}{2}}(e+f x)}{2 \sqrt {2} b^2 f \sqrt {b \tan ^3(e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 45, normalized size = 0.12 \[ -\frac {2 \tan (e+f x) \, _2F_1\left (-\frac {13}{4},1;-\frac {9}{4};-\tan ^2(e+f x)\right )}{13 f \left (b \tan ^3(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x]^3)^(-5/2),x]

[Out]

(-2*Hypergeometric2F1[-13/4, 1, -9/4, -Tan[e + f*x]^2]*Tan[e + f*x])/(13*f*(b*Tan[e + f*x]^3)^(5/2))

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tan \left (f x + e\right )^{3}\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^3)^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.23, size = 272, normalized size = 0.75 \[ \frac {\tan \left (f x +e \right ) \left (585 \sqrt {2}\, \left (b \tan \left (f x +e \right )\right )^{\frac {13}{2}} \ln \left (-\frac {\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (f x +e \right )}\, \sqrt {2}-b \tan \left (f x +e \right )-\sqrt {b^{2}}}{b \tan \left (f x +e \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+1170 \sqrt {2}\, \left (b \tan \left (f x +e \right )\right )^{\frac {13}{2}} \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (f x +e \right )}+\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+1170 \sqrt {2}\, \left (b \tan \left (f x +e \right )\right )^{\frac {13}{2}} \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (f x +e \right )}-\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+4680 \left (b^{2}\right )^{\frac {1}{4}} \left (\tan ^{6}\left (f x +e \right )\right ) b^{6}-936 b^{6} \left (b^{2}\right )^{\frac {1}{4}} \left (\tan ^{4}\left (f x +e \right )\right )+520 b^{6} \left (b^{2}\right )^{\frac {1}{4}} \left (\tan ^{2}\left (f x +e \right )\right )-360 b^{6} \left (b^{2}\right )^{\frac {1}{4}}\right )}{2340 f \,b^{6} \left (b \left (\tan ^{3}\left (f x +e \right )\right )\right )^{\frac {5}{2}} \left (b^{2}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(f*x+e)^3)^(5/2),x)

[Out]

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

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maxima [A]  time = 1.92, size = 172, normalized size = 0.47 \[ \frac {\frac {585 \, {\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (f x + e\right )}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (f x + e\right )}\right )}\right ) - \sqrt {2} \log \left (\sqrt {2} \sqrt {\tan \left (f x + e\right )} + \tan \left (f x + e\right ) + 1\right ) + \sqrt {2} \log \left (-\sqrt {2} \sqrt {\tan \left (f x + e\right )} + \tan \left (f x + e\right ) + 1\right )\right )}}{b^{\frac {5}{2}}} + \frac {8 \, {\left (\frac {585 \, \sqrt {b}}{\sqrt {\tan \left (f x + e\right )}} - \frac {117 \, \sqrt {b}}{\tan \left (f x + e\right )^{\frac {5}{2}}} + \frac {65 \, \sqrt {b}}{\tan \left (f x + e\right )^{\frac {9}{2}}} - \frac {45 \, \sqrt {b}}{\tan \left (f x + e\right )^{\frac {13}{2}}}\right )}}{b^{3}}}{2340 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)^3)^(5/2),x, algorithm="maxima")

[Out]

1/2340*(585*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(f*x + e)))) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(s
qrt(2) - 2*sqrt(tan(f*x + e)))) - sqrt(2)*log(sqrt(2)*sqrt(tan(f*x + e)) + tan(f*x + e) + 1) + sqrt(2)*log(-sq
rt(2)*sqrt(tan(f*x + e)) + tan(f*x + e) + 1))/b^(5/2) + 8*(585*sqrt(b)/sqrt(tan(f*x + e)) - 117*sqrt(b)/tan(f*
x + e)^(5/2) + 65*sqrt(b)/tan(f*x + e)^(9/2) - 45*sqrt(b)/tan(f*x + e)^(13/2))/b^3)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(e + f*x)^3)^(5/2),x)

[Out]

int(1/(b*tan(e + f*x)^3)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tan ^{3}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(f*x+e)**3)**(5/2),x)

[Out]

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

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