3.203 \(\int (d \cot (e+f x))^{3/2} \tan (e+f x) \, dx\)

Optimal. Leaf size=192 \[ -\frac{d^{3/2} \log \left (\sqrt{d} \cot (e+f x)-\sqrt{2} \sqrt{d \cot (e+f x)}+\sqrt{d}\right )}{2 \sqrt{2} f}+\frac{d^{3/2} \log \left (\sqrt{d} \cot (e+f x)+\sqrt{2} \sqrt{d \cot (e+f x)}+\sqrt{d}\right )}{2 \sqrt{2} f}+\frac{d^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \cot (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} f}-\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \cot (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} f} \]

[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) - (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)

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Rubi [A]  time = 0.137959, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.474, Rules used = {16, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{d^{3/2} \log \left (\sqrt{d} \cot (e+f x)-\sqrt{2} \sqrt{d \cot (e+f x)}+\sqrt{d}\right )}{2 \sqrt{2} f}+\frac{d^{3/2} \log \left (\sqrt{d} \cot (e+f x)+\sqrt{2} \sqrt{d \cot (e+f x)}+\sqrt{d}\right )}{2 \sqrt{2} f}+\frac{d^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \cot (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} f}-\frac{d^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \cot (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} f} \]

Antiderivative was successfully verified.

[In]

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

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]

Rubi steps

\begin{align*} \int (d \cot (e+f x))^{3/2} \tan (e+f x) \, dx &=d \int \sqrt{d \cot (e+f x)} \, dx\\ &=-\frac{d^2 \operatorname{Subst}\left (\int \frac{\sqrt{x}}{d^2+x^2} \, dx,x,d \cot (e+f x)\right )}{f}\\ &=-\frac{\left (2 d^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{d^2+x^4} \, dx,x,\sqrt{d \cot (e+f x)}\right )}{f}\\ &=\frac{d^2 \operatorname{Subst}\left (\int \frac{d-x^2}{d^2+x^4} \, dx,x,\sqrt{d \cot (e+f x)}\right )}{f}-\frac{d^2 \operatorname{Subst}\left (\int \frac{d+x^2}{d^2+x^4} \, dx,x,\sqrt{d \cot (e+f x)}\right )}{f}\\ &=-\frac{d^{3/2} \operatorname{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} \operatorname{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 \operatorname{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 \operatorname{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^{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} \operatorname{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} \operatorname{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} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \cot (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} f}-\frac{d^{3/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \cot (e+f x)}}{\sqrt{d}}\right )}{\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} \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 [C]  time = 0.0287894, size = 37, normalized size = 0.19 \[ -\frac{2 (d \cot (e+f x))^{3/2} \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\cot ^2(e+f x)\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d*Cot[e + f*x])^(3/2)*Hypergeometric2F1[3/4, 1, 7/4, -Cot[e + f*x]^2])/(3*f)

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Maple [C]  time = 0.136, size = 318, normalized size = 1.7 \begin{align*} -{\frac{\sqrt{2} \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2} \left ( \cos \left ( fx+e \right ) -1 \right ) }{2\,f\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( i{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -i{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -2\,{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},1/2\,\sqrt{2} \right ) +{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) +{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) \right ) \sqrt{{\frac{\cos \left ( fx+e \right ) -1}{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{\cos \left ( fx+e \right ) -1+\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}}\sqrt{{\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }}} \left ({\frac{d\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*2^(1/2)*(cos(f*x+e)+1)^2*(I*EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/
2))-I*EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))-2*EllipticF(((1-cos(f*x+e
)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))+EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I
,1/2*2^(1/2))+EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2)))*((cos(f*x+e)-1)/
sin(f*x+e))^(1/2)*((cos(f*x+e)-1+sin(f*x+e))/sin(f*x+e))^(1/2)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(c
os(f*x+e)-1)*(d*cos(f*x+e)/sin(f*x+e))^(3/2)/sin(f*x+e)/cos(f*x+e)^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.89237, size = 1295, normalized size = 6.74 \begin{align*} \sqrt{2} \left (\frac{d^{6}}{f^{4}}\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{2} \left (\frac{d^{6}}{f^{4}}\right )^{\frac{1}{4}} d^{4} f \sqrt{\frac{d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} + d^{6} - \sqrt{2} \left (\frac{d^{6}}{f^{4}}\right )^{\frac{1}{4}} f \sqrt{\frac{d^{9} \cos \left (f x + e\right ) + \sqrt{2} \left (\frac{d^{6}}{f^{4}}\right )^{\frac{3}{4}} d^{4} f^{3} \sqrt{\frac{d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \sin \left (f x + e\right ) + \sqrt{\frac{d^{6}}{f^{4}}} d^{6} f^{2} \sin \left (f x + e\right )}{\sin \left (f x + e\right )}}}{d^{6}}\right ) + \sqrt{2} \left (\frac{d^{6}}{f^{4}}\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{2} \left (\frac{d^{6}}{f^{4}}\right )^{\frac{1}{4}} d^{4} f \sqrt{\frac{d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} - d^{6} - \sqrt{2} \left (\frac{d^{6}}{f^{4}}\right )^{\frac{1}{4}} f \sqrt{\frac{d^{9} \cos \left (f x + e\right ) - \sqrt{2} \left (\frac{d^{6}}{f^{4}}\right )^{\frac{3}{4}} d^{4} f^{3} \sqrt{\frac{d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \sin \left (f x + e\right ) + \sqrt{\frac{d^{6}}{f^{4}}} d^{6} f^{2} \sin \left (f x + e\right )}{\sin \left (f x + e\right )}}}{d^{6}}\right ) + \frac{1}{4} \, \sqrt{2} \left (\frac{d^{6}}{f^{4}}\right )^{\frac{1}{4}} \log \left (\frac{d^{9} \cos \left (f x + e\right ) + \sqrt{2} \left (\frac{d^{6}}{f^{4}}\right )^{\frac{3}{4}} d^{4} f^{3} \sqrt{\frac{d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \sin \left (f x + e\right ) + \sqrt{\frac{d^{6}}{f^{4}}} d^{6} f^{2} \sin \left (f x + e\right )}{\sin \left (f x + e\right )}\right ) - \frac{1}{4} \, \sqrt{2} \left (\frac{d^{6}}{f^{4}}\right )^{\frac{1}{4}} \log \left (\frac{d^{9} \cos \left (f x + e\right ) - \sqrt{2} \left (\frac{d^{6}}{f^{4}}\right )^{\frac{3}{4}} d^{4} f^{3} \sqrt{\frac{d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \sin \left (f x + e\right ) + \sqrt{\frac{d^{6}}{f^{4}}} d^{6} f^{2} \sin \left (f x + e\right )}{\sin \left (f x + e\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*(d^6/f^4)^(1/4)*arctan(-(sqrt(2)*(d^6/f^4)^(1/4)*d^4*f*sqrt(d*cos(f*x + e)/sin(f*x + e)) + d^6 - sqrt(
2)*(d^6/f^4)^(1/4)*f*sqrt((d^9*cos(f*x + e) + sqrt(2)*(d^6/f^4)^(3/4)*d^4*f^3*sqrt(d*cos(f*x + e)/sin(f*x + e)
)*sin(f*x + e) + sqrt(d^6/f^4)*d^6*f^2*sin(f*x + e))/sin(f*x + e)))/d^6) + sqrt(2)*(d^6/f^4)^(1/4)*arctan(-(sq
rt(2)*(d^6/f^4)^(1/4)*d^4*f*sqrt(d*cos(f*x + e)/sin(f*x + e)) - d^6 - sqrt(2)*(d^6/f^4)^(1/4)*f*sqrt((d^9*cos(
f*x + e) - sqrt(2)*(d^6/f^4)^(3/4)*d^4*f^3*sqrt(d*cos(f*x + e)/sin(f*x + e))*sin(f*x + e) + sqrt(d^6/f^4)*d^6*
f^2*sin(f*x + e))/sin(f*x + e)))/d^6) + 1/4*sqrt(2)*(d^6/f^4)^(1/4)*log((d^9*cos(f*x + e) + sqrt(2)*(d^6/f^4)^
(3/4)*d^4*f^3*sqrt(d*cos(f*x + e)/sin(f*x + e))*sin(f*x + e) + sqrt(d^6/f^4)*d^6*f^2*sin(f*x + e))/sin(f*x + e
)) - 1/4*sqrt(2)*(d^6/f^4)^(1/4)*log((d^9*cos(f*x + e) - sqrt(2)*(d^6/f^4)^(3/4)*d^4*f^3*sqrt(d*cos(f*x + e)/s
in(f*x + e))*sin(f*x + e) + sqrt(d^6/f^4)*d^6*f^2*sin(f*x + e))/sin(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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