\(\int (d \cot (e+f x))^{3/2} \tan ^5(e+f x) \, dx\) [199]

Optimal result

Integrand size = 21, antiderivative size = 234 \[ \int (d \cot (e+f x))^{3/2} \tan ^5(e+f x) \, dx=\frac {d^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}-\frac {d^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}+\frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}-\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}+\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f} \]

[Out]

2/5*d^4/f/(d*cot(f*x+e))^(5/2)+1/2*d^(3/2)*arctan(1-2^(1/2)*(d*cot(f*x+e))^(1/2)/d^(1/2))/f*2^(1/2)-1/2*d^(3/2
)*arctan(1+2^(1/2)*(d*cot(f*x+e))^(1/2)/d^(1/2))/f*2^(1/2)-1/4*d^(3/2)*ln(d^(1/2)+cot(f*x+e)*d^(1/2)-2^(1/2)*(
d*cot(f*x+e))^(1/2))/f*2^(1/2)+1/4*d^(3/2)*ln(d^(1/2)+cot(f*x+e)*d^(1/2)+2^(1/2)*(d*cot(f*x+e))^(1/2))/f*2^(1/
2)-2*d^2/f/(d*cot(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {16, 3555, 3557, 335, 303, 1176, 631, 210, 1179, 642} \[ \int (d \cot (e+f x))^{3/2} \tan ^5(e+f x) \, dx=\frac {d^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}-\frac {d^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} f}-\frac {d^{3/2} \log \left (\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f}+\frac {d^{3/2} \log \left (\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f}+\frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}} \]

[In]

Int[(d*Cot[e + f*x])^(3/2)*Tan[e + f*x]^5,x]

[Out]

(d^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Cot[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) - (d^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d
*Cot[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) + (2*d^4)/(5*f*(d*Cot[e + f*x])^(5/2)) - (2*d^2)/(f*Sqrt[d*Cot[e + f*x]]
) - (d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Cot[e + f*x] - Sqrt[2]*Sqrt[d*Cot[e + f*x]]])/(2*Sqrt[2]*f) + (d^(3/2)*Log[
Sqrt[d] + Sqrt[d]*Cot[e + f*x] + Sqrt[2]*Sqrt[d*Cot[e + f*x]]])/(2*Sqrt[2]*f)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 210

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

Rule 303

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 335

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 631

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 642

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 1176

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 1179

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 3555

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 3557

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]

Rubi steps \begin{align*} \text {integral}& = d^5 \int \frac {1}{(d \cot (e+f x))^{7/2}} \, dx \\ & = \frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-d^3 \int \frac {1}{(d \cot (e+f x))^{3/2}} \, dx \\ & = \frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}+d \int \sqrt {d \cot (e+f x)} \, dx \\ & = \frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}-\frac {d^2 \text {Subst}\left (\int \frac {\sqrt {x}}{d^2+x^2} \, dx,x,d \cot (e+f x)\right )}{f} \\ & = \frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}-\frac {\left (2 d^2\right ) \text {Subst}\left (\int \frac {x^2}{d^2+x^4} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{f} \\ & = \frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}+\frac {d^2 \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{f}-\frac {d^2 \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{f} \\ & = \frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}-\frac {d^{3/2} \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}-\frac {d^{3/2} \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}-\frac {d^2 \text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{2 f}-\frac {d^2 \text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{2 f} \\ & = \frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}-\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}+\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}-\frac {d^{3/2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}+\frac {d^{3/2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f} \\ & = \frac {d^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}-\frac {d^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}+\frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}-\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}+\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.47 \[ \int (d \cot (e+f x))^{3/2} \tan ^5(e+f x) \, dx=-\frac {(d \cot (e+f x))^{3/2} \left (-2+10 \cot ^2(e+f x)+5 \arctan \left (\sqrt [4]{-\cot ^2(e+f x)}\right ) \sqrt [4]{-\cot (e+f x)} \cot ^{\frac {9}{4}}(e+f x)+5 \text {arctanh}\left (\sqrt [4]{-\cot ^2(e+f x)}\right ) \left (-\cot ^2(e+f x)\right )^{5/4}\right ) \tan ^4(e+f x)}{5 f} \]

[In]

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

[Out]

-1/5*((d*Cot[e + f*x])^(3/2)*(-2 + 10*Cot[e + f*x]^2 + 5*ArcTan[(-Cot[e + f*x]^2)^(1/4)]*(-Cot[e + f*x])^(1/4)
*Cot[e + f*x]^(9/4) + 5*ArcTanh[(-Cot[e + f*x]^2)^(1/4)]*(-Cot[e + f*x]^2)^(5/4))*Tan[e + f*x]^4)/f

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(631\) vs. \(2(181)=362\).

Time = 3.41 (sec) , antiderivative size = 632, normalized size of antiderivative = 2.70

method	result	size
default	\(\frac {\left (\sec ^{3}\left (f x +e \right )\right ) \csc \left (f x +e \right ) \left (5 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \ln \left (-\frac {\cot \left (f x +e \right ) \cos \left (f x +e \right )-2 \cot \left (f x +e \right )+2 \sin \left (f x +e \right ) \sqrt {-\left (\cot ^{3}\left (f x +e \right )\right )+3 \left (\cot ^{2}\left (f x +e \right )\right ) \csc \left (f x +e \right )-3 \left (\csc ^{2}\left (f x +e \right )\right ) \cot \left (f x +e \right )+\csc ^{3}\left (f x +e \right )+\cot \left (f x +e \right )-\csc \left (f x +e \right )}-2 \cos \left (f x +e \right )-\sin \left (f x +e \right )+\csc \left (f x +e \right )+2}{\cos \left (f x +e \right )-1}\right ) \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-5 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \ln \left (-\frac {\cot \left (f x +e \right ) \cos \left (f x +e \right )-2 \cot \left (f x +e \right )-2 \sin \left (f x +e \right ) \sqrt {-\left (\cot ^{3}\left (f x +e \right )\right )+3 \left (\cot ^{2}\left (f x +e \right )\right ) \csc \left (f x +e \right )-3 \left (\csc ^{2}\left (f x +e \right )\right ) \cot \left (f x +e \right )+\csc ^{3}\left (f x +e \right )+\cot \left (f x +e \right )-\csc \left (f x +e \right )}-2 \cos \left (f x +e \right )-\sin \left (f x +e \right )+\csc \left (f x +e \right )+2}{\cos \left (f x +e \right )-1}\right ) \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+10 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )+\cos \left (f x +e \right )-1}{\cos \left (f x +e \right )-1}\right ) \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-10 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \arctan \left (\frac {-\sqrt {2}\, \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )+\cos \left (f x +e \right )-1}{\cos \left (f x +e \right )-1}\right ) \sqrt {-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+24 \sqrt {2}\, \left (\cos ^{3}\left (f x +e \right )\right )-24 \sqrt {2}\, \left (\cos ^{2}\left (f x +e \right )\right )-4 \sqrt {2}\, \cos \left (f x +e \right )+4 \sqrt {2}\right ) \left (\cos \left (f x +e \right )+1\right ) d \sqrt {\cot \left (f x +e \right ) d}\, \sqrt {2}}{20 f}\)	\(632\)

[In]

int((cot(f*x+e)*d)^(3/2)*tan(f*x+e)^5,x,method=_RETURNVERBOSE)

[Out]

1/20/f*sec(f*x+e)^3*csc(f*x+e)*(5*cos(f*x+e)^2*sin(f*x+e)*ln(-(cot(f*x+e)*cos(f*x+e)-2*cot(f*x+e)+2*sin(f*x+e)
*(-cot(f*x+e)^3+3*cot(f*x+e)^2*csc(f*x+e)-3*csc(f*x+e)^2*cot(f*x+e)+csc(f*x+e)^3+cot(f*x+e)-csc(f*x+e))^(1/2)-
2*cos(f*x+e)-sin(f*x+e)+csc(f*x+e)+2)/(cos(f*x+e)-1))*(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-5*cos(f*
x+e)^2*sin(f*x+e)*ln(-(cot(f*x+e)*cos(f*x+e)-2*cot(f*x+e)-2*sin(f*x+e)*(-cot(f*x+e)^3+3*cot(f*x+e)^2*csc(f*x+e
)-3*csc(f*x+e)^2*cot(f*x+e)+csc(f*x+e)^3+cot(f*x+e)-csc(f*x+e))^(1/2)-2*cos(f*x+e)-sin(f*x+e)+csc(f*x+e)+2)/(c
os(f*x+e)-1))*(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+10*cos(f*x+e)^2*sin(f*x+e)*arctan((2^(1/2)*(-sin
(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/(cos(f*x+e)-1))*(-sin(f*x+e)*cos(f*x+e)/(c
os(f*x+e)+1)^2)^(1/2)-10*cos(f*x+e)^2*sin(f*x+e)*arctan((-2^(1/2)*(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1
/2)*sin(f*x+e)+cos(f*x+e)-1)/(cos(f*x+e)-1))*(-sin(f*x+e)*cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+24*2^(1/2)*cos(f*
x+e)^3-24*2^(1/2)*cos(f*x+e)^2-4*2^(1/2)*cos(f*x+e)+4*2^(1/2))*(cos(f*x+e)+1)*d*(cot(f*x+e)*d)^(1/2)*2^(1/2)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.95 \[ \int (d \cot (e+f x))^{3/2} \tan ^5(e+f x) \, dx=-\frac {5 \, \left (-\frac {d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \log \left (d^{4} \sqrt {\frac {d}{\tan \left (f x + e\right )}} + \left (-\frac {d^{6}}{f^{4}}\right )^{\frac {3}{4}} f^{3}\right ) - 5 i \, \left (-\frac {d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \log \left (d^{4} \sqrt {\frac {d}{\tan \left (f x + e\right )}} + i \, \left (-\frac {d^{6}}{f^{4}}\right )^{\frac {3}{4}} f^{3}\right ) + 5 i \, \left (-\frac {d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \log \left (d^{4} \sqrt {\frac {d}{\tan \left (f x + e\right )}} - i \, \left (-\frac {d^{6}}{f^{4}}\right )^{\frac {3}{4}} f^{3}\right ) - 5 \, \left (-\frac {d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \log \left (d^{4} \sqrt {\frac {d}{\tan \left (f x + e\right )}} - \left (-\frac {d^{6}}{f^{4}}\right )^{\frac {3}{4}} f^{3}\right ) - 4 \, {\left (d \tan \left (f x + e\right )^{3} - 5 \, d \tan \left (f x + e\right )\right )} \sqrt {\frac {d}{\tan \left (f x + e\right )}}}{10 \, f} \]

[In]

integrate((d*cot(f*x+e))^(3/2)*tan(f*x+e)^5,x, algorithm="fricas")

[Out]

-1/10*(5*(-d^6/f^4)^(1/4)*f*log(d^4*sqrt(d/tan(f*x + e)) + (-d^6/f^4)^(3/4)*f^3) - 5*I*(-d^6/f^4)^(1/4)*f*log(
d^4*sqrt(d/tan(f*x + e)) + I*(-d^6/f^4)^(3/4)*f^3) + 5*I*(-d^6/f^4)^(1/4)*f*log(d^4*sqrt(d/tan(f*x + e)) - I*(
-d^6/f^4)^(3/4)*f^3) - 5*(-d^6/f^4)^(1/4)*f*log(d^4*sqrt(d/tan(f*x + e)) - (-d^6/f^4)^(3/4)*f^3) - 4*(d*tan(f*
x + e)^3 - 5*d*tan(f*x + e))*sqrt(d/tan(f*x + e)))/f

Sympy [F]

\[ \int (d \cot (e+f x))^{3/2} \tan ^5(e+f x) \, dx=\int \left (d \cot {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{5}{\left (e + f x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.88 \[ \int (d \cot (e+f x))^{3/2} \tan ^5(e+f x) \, dx=-\frac {d^{6} {\left (\frac {5 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {\frac {d}{\tan \left (f x + e\right )}}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {\frac {d}{\tan \left (f x + e\right )}}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {d} \sqrt {\frac {d}{\tan \left (f x + e\right )}} + d + \frac {d}{\tan \left (f x + e\right )}\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {d} \sqrt {\frac {d}{\tan \left (f x + e\right )}} + d + \frac {d}{\tan \left (f x + e\right )}\right )}{\sqrt {d}}\right )}}{d^{4}} - \frac {8 \, {\left (d^{2} - \frac {5 \, d^{2}}{\tan \left (f x + e\right )^{2}}\right )}}{d^{4} \left (\frac {d}{\tan \left (f x + e\right )}\right )^{\frac {5}{2}}}\right )}}{20 \, f} \]

[In]

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

[Out]

-1/20*d^6*(5*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d/tan(f*x + e)))/sqrt(d))/sqrt(d) + 2*sqr
t(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d/tan(f*x + e)))/sqrt(d))/sqrt(d) - sqrt(2)*log(sqrt(2)*sqr
t(d)*sqrt(d/tan(f*x + e)) + d + d/tan(f*x + e))/sqrt(d) + sqrt(2)*log(-sqrt(2)*sqrt(d)*sqrt(d/tan(f*x + e)) +
d + d/tan(f*x + e))/sqrt(d))/d^4 - 8*(d^2 - 5*d^2/tan(f*x + e)^2)/(d^4*(d/tan(f*x + e))^(5/2)))/f

Giac [F]

\[ \int (d \cot (e+f x))^{3/2} \tan ^5(e+f x) \, dx=\int { \left (d \cot \left (f x + e\right )\right )^{\frac {3}{2}} \tan \left (f x + e\right )^{5} \,d x } \]

[In]

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

[Out]

integrate((d*cot(f*x + e))^(3/2)*tan(f*x + e)^5, x)

Mupad [B] (verification not implemented)

Time = 2.78 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.41 \[ \int (d \cot (e+f x))^{3/2} \tan ^5(e+f x) \, dx=\frac {\frac {2\,d^4}{5}-\frac {2\,d^4}{{\mathrm {tan}\left (e+f\,x\right )}^2}}{f\,{\left (\frac {d}{\mathrm {tan}\left (e+f\,x\right )}\right )}^{5/2}}-\frac {{\left (-1\right )}^{1/4}\,d^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {\frac {d}{\mathrm {tan}\left (e+f\,x\right )}}}{\sqrt {d}}\right )}{f}+\frac {{\left (-1\right )}^{1/4}\,d^{3/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {\frac {d}{\mathrm {tan}\left (e+f\,x\right )}}}{\sqrt {d}}\right )}{f} \]

[In]

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

[Out]

((2*d^4)/5 - (2*d^4)/tan(e + f*x)^2)/(f*(d/tan(e + f*x))^(5/2)) - ((-1)^(1/4)*d^(3/2)*atan(((-1)^(1/4)*(d/tan(
e + f*x))^(1/2))/d^(1/2)))/f + ((-1)^(1/4)*d^(3/2)*atanh(((-1)^(1/4)*(d/tan(e + f*x))^(1/2))/d^(1/2)))/f