3.191 \(\int \sqrt{d \cot (e+f x)} \tan ^4(e+f x) \, dx\)

Optimal. Leaf size=232 \[ \frac{2 d^3}{5 f (d \cot (e+f x))^{5/2}}-\frac{2 d}{f \sqrt{d \cot (e+f x)}}-\frac{\sqrt{d} \log \left (\sqrt{d} \cot (e+f x)-\sqrt{2} \sqrt{d \cot (e+f x)}+\sqrt{d}\right )}{2 \sqrt{2} f}+\frac{\sqrt{d} \log \left (\sqrt{d} \cot (e+f x)+\sqrt{2} \sqrt{d \cot (e+f x)}+\sqrt{d}\right )}{2 \sqrt{2} f}+\frac{\sqrt{d} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \cot (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} f}-\frac{\sqrt{d} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \cot (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} f} \]

[Out]

(Sqrt[d]*ArcTan[1 - (Sqrt[2]*Sqrt[d*Cot[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) - (Sqrt[d]*ArcTan[1 + (Sqrt[2]*Sqrt[d
*Cot[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) + (2*d^3)/(5*f*(d*Cot[e + f*x])^(5/2)) - (2*d)/(f*Sqrt[d*Cot[e + f*x]])
- (Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Cot[e + f*x] - Sqrt[2]*Sqrt[d*Cot[e + f*x]]])/(2*Sqrt[2]*f) + (Sqrt[d]*Log[Sq
rt[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 = 0.227133, antiderivative size = 232, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {16, 3474, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{2 d^3}{5 f (d \cot (e+f x))^{5/2}}-\frac{2 d}{f \sqrt{d \cot (e+f x)}}-\frac{\sqrt{d} \log \left (\sqrt{d} \cot (e+f x)-\sqrt{2} \sqrt{d \cot (e+f x)}+\sqrt{d}\right )}{2 \sqrt{2} f}+\frac{\sqrt{d} \log \left (\sqrt{d} \cot (e+f x)+\sqrt{2} \sqrt{d \cot (e+f x)}+\sqrt{d}\right )}{2 \sqrt{2} f}+\frac{\sqrt{d} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \cot (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} f}-\frac{\sqrt{d} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \cot (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*Cot[e + f*x]]*Tan[e + f*x]^4,x]

[Out]

(Sqrt[d]*ArcTan[1 - (Sqrt[2]*Sqrt[d*Cot[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) - (Sqrt[d]*ArcTan[1 + (Sqrt[2]*Sqrt[d
*Cot[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) + (2*d^3)/(5*f*(d*Cot[e + f*x])^(5/2)) - (2*d)/(f*Sqrt[d*Cot[e + f*x]])
- (Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Cot[e + f*x] - Sqrt[2]*Sqrt[d*Cot[e + f*x]]])/(2*Sqrt[2]*f) + (Sqrt[d]*Log[Sq
rt[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 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 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 \sqrt{d \cot (e+f x)} \tan ^4(e+f x) \, dx &=d^4 \int \frac{1}{(d \cot (e+f x))^{7/2}} \, dx\\ &=\frac{2 d^3}{5 f (d \cot (e+f x))^{5/2}}-d^2 \int \frac{1}{(d \cot (e+f x))^{3/2}} \, dx\\ &=\frac{2 d^3}{5 f (d \cot (e+f x))^{5/2}}-\frac{2 d}{f \sqrt{d \cot (e+f x)}}+\int \sqrt{d \cot (e+f x)} \, dx\\ &=\frac{2 d^3}{5 f (d \cot (e+f x))^{5/2}}-\frac{2 d}{f \sqrt{d \cot (e+f x)}}-\frac{d \operatorname{Subst}\left (\int \frac{\sqrt{x}}{d^2+x^2} \, dx,x,d \cot (e+f x)\right )}{f}\\ &=\frac{2 d^3}{5 f (d \cot (e+f x))^{5/2}}-\frac{2 d}{f \sqrt{d \cot (e+f x)}}-\frac{(2 d) \operatorname{Subst}\left (\int \frac{x^2}{d^2+x^4} \, dx,x,\sqrt{d \cot (e+f x)}\right )}{f}\\ &=\frac{2 d^3}{5 f (d \cot (e+f x))^{5/2}}-\frac{2 d}{f \sqrt{d \cot (e+f x)}}+\frac{d \operatorname{Subst}\left (\int \frac{d-x^2}{d^2+x^4} \, dx,x,\sqrt{d \cot (e+f x)}\right )}{f}-\frac{d \operatorname{Subst}\left (\int \frac{d+x^2}{d^2+x^4} \, dx,x,\sqrt{d \cot (e+f x)}\right )}{f}\\ &=\frac{2 d^3}{5 f (d \cot (e+f x))^{5/2}}-\frac{2 d}{f \sqrt{d \cot (e+f x)}}-\frac{\sqrt{d} \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{\sqrt{d} \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 \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 \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{2 d^3}{5 f (d \cot (e+f x))^{5/2}}-\frac{2 d}{f \sqrt{d \cot (e+f x)}}-\frac{\sqrt{d} \log \left (\sqrt{d}+\sqrt{d} \cot (e+f x)-\sqrt{2} \sqrt{d \cot (e+f x)}\right )}{2 \sqrt{2} f}+\frac{\sqrt{d} \log \left (\sqrt{d}+\sqrt{d} \cot (e+f x)+\sqrt{2} \sqrt{d \cot (e+f x)}\right )}{2 \sqrt{2} f}-\frac{\sqrt{d} \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{\sqrt{d} \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{\sqrt{d} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \cot (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} f}-\frac{\sqrt{d} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \cot (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} f}+\frac{2 d^3}{5 f (d \cot (e+f x))^{5/2}}-\frac{2 d}{f \sqrt{d \cot (e+f x)}}-\frac{\sqrt{d} \log \left (\sqrt{d}+\sqrt{d} \cot (e+f x)-\sqrt{2} \sqrt{d \cot (e+f x)}\right )}{2 \sqrt{2} f}+\frac{\sqrt{d} \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 = 0.0789621, size = 45, normalized size = 0.19 \[ \frac{2 \tan ^3(e+f x) \sqrt{d \cot (e+f x)} \, _2F_1\left (-\frac{5}{4},1;-\frac{1}{4};-\cot ^2(e+f x)\right )}{5 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*Cot[e + f*x]]*Tan[e + f*x]^4,x]

[Out]

(2*Sqrt[d*Cot[e + f*x]]*Hypergeometric2F1[-5/4, 1, -1/4, -Cot[e + f*x]^2]*Tan[e + f*x]^3)/(5*f)

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Maple [C]  time = 0.258, size = 728, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cot(f*x+e))^(1/2)*tan(f*x+e)^4,x)

[Out]

1/10/f*2^(1/2)*(cos(f*x+e)-1)*(5*I*cos(f*x+e)^2*sin(f*x+e)*((cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((cos(f*x+e)-1+si
n(f*x+e))/sin(f*x+e))^(1/2)*(-(cos(f*x+e)-1-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(cos(f*x+e)-1-sin(f*x+e
))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))-5*I*cos(f*x+e)^2*sin(f*x+e)*((cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((co
s(f*x+e)-1+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(cos(f*x+e)-1-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(cos(f*x+e
)-1-sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))-5*cos(f*x+e)^2*sin(f*x+e)*((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)*(-(cos(f*x+e)-1-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((
-(cos(f*x+e)-1-sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))+10*cos(f*x+e)^2*sin(f*x+e)*((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)*(-(cos(f*x+e)-1-sin(f*x+e))/sin(f*x+e))^(1/2)*
EllipticF((-(cos(f*x+e)-1-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))-5*cos(f*x+e)^2*sin(f*x+e)*((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)*(-(cos(f*x+e)-1-sin(f*x+e))/sin(f*x+e))^(1/2)*
EllipticPi((-(cos(f*x+e)-1-sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))-12*cos(f*x+e)^3*2^(1/2)+12*cos
(f*x+e)^2*2^(1/2)+2*cos(f*x+e)*2^(1/2)-2*2^(1/2))*(cos(f*x+e)+1)^2*(d*cos(f*x+e)/sin(f*x+e))^(1/2)/cos(f*x+e)^
3/sin(f*x+e)^3

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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))^(1/2)*tan(f*x+e)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.99986, size = 1511, normalized size = 6.51 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cot(f*x+e))^(1/2)*tan(f*x+e)^4,x, algorithm="fricas")

[Out]

1/20*(20*sqrt(2)*f*(d^2/f^4)^(1/4)*arctan(-(sqrt(2)*d*f*sqrt(d*cos(f*x + e)/sin(f*x + e))*(d^2/f^4)^(1/4) - sq
rt(2)*f*sqrt((sqrt(2)*d*f^3*sqrt(d*cos(f*x + e)/sin(f*x + e))*(d^2/f^4)^(3/4)*sin(f*x + e) + d^2*f^2*sqrt(d^2/
f^4)*sin(f*x + e) + d^3*cos(f*x + e))/sin(f*x + e))*(d^2/f^4)^(1/4) + d^2)/d^2)*cos(f*x + e)^3 + 20*sqrt(2)*f*
(d^2/f^4)^(1/4)*arctan(-(sqrt(2)*d*f*sqrt(d*cos(f*x + e)/sin(f*x + e))*(d^2/f^4)^(1/4) - sqrt(2)*f*sqrt(-(sqrt
(2)*d*f^3*sqrt(d*cos(f*x + e)/sin(f*x + e))*(d^2/f^4)^(3/4)*sin(f*x + e) - d^2*f^2*sqrt(d^2/f^4)*sin(f*x + e)
- d^3*cos(f*x + e))/sin(f*x + e))*(d^2/f^4)^(1/4) - d^2)/d^2)*cos(f*x + e)^3 + 5*sqrt(2)*f*(d^2/f^4)^(1/4)*cos
(f*x + e)^3*log((sqrt(2)*d*f^3*sqrt(d*cos(f*x + e)/sin(f*x + e))*(d^2/f^4)^(3/4)*sin(f*x + e) + d^2*f^2*sqrt(d
^2/f^4)*sin(f*x + e) + d^3*cos(f*x + e))/sin(f*x + e)) - 5*sqrt(2)*f*(d^2/f^4)^(1/4)*cos(f*x + e)^3*log(-(sqrt
(2)*d*f^3*sqrt(d*cos(f*x + e)/sin(f*x + e))*(d^2/f^4)^(3/4)*sin(f*x + e) - d^2*f^2*sqrt(d^2/f^4)*sin(f*x + e)
- d^3*cos(f*x + e))/sin(f*x + e)) - 8*(6*cos(f*x + e)^2 - 1)*sqrt(d*cos(f*x + e)/sin(f*x + e))*sin(f*x + e))/(
f*cos(f*x + e)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d \cot{\left (e + f x \right )}} \tan ^{4}{\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))**(1/2)*tan(f*x+e)**4,x)

[Out]

Integral(sqrt(d*cot(e + f*x))*tan(e + f*x)**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d \cot \left (f x + e\right )} \tan \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cot(f*x+e))^(1/2)*tan(f*x+e)^4,x, algorithm="giac")

[Out]

integrate(sqrt(d*cot(f*x + e))*tan(f*x + e)^4, x)