3.2.94 \(\int \sqrt {d \cot (e+f x)} \tan (e+f x) \, dx\) [194]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 192, normalized size of antiderivative = 1.00, 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, 3557, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {\sqrt {d} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}-\frac {\sqrt {d} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}+1\right )}{\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} \log \left (\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*Cot[e + f*x]]*Tan[e + f*x],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) + (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[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 217

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 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 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*} \int \sqrt {d \cot (e+f x)} \tan (e+f x) \, dx &=d \int \frac {1}{\sqrt {d \cot (e+f x)}} \, dx\\ &=-\frac {d^2 \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (d^2+x^2\right )} \, dx,x,d \cot (e+f x)\right )}{f}\\ &=-\frac {\left (2 d^2\right ) \text {Subst}\left (\int \frac {1}{d^2+x^4} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{f}\\ &=-\frac {d \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{f}-\frac {d \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{f}\\ &=\frac {\sqrt {d} \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 {\sqrt {d} \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 \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 \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 {\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} \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 {\sqrt {d} \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 {\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 {\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 [A]
time = 0.20, size = 132, normalized size = 0.69 \begin {gather*} \frac {d \sqrt {\cot (e+f x)} \left (2 \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (e+f x)}\right )-2 \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (e+f x)}\right )+\log \left (1-\sqrt {2} \sqrt {\cot (e+f x)}+\cot (e+f x)\right )-\log \left (1+\sqrt {2} \sqrt {\cot (e+f x)}+\cot (e+f x)\right )\right )}{2 \sqrt {2} f \sqrt {d \cot (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*Sqrt[Cot[e + f*x]]*(2*ArcTan[1 - Sqrt[2]*Sqrt[Cot[e + f*x]]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[Cot[e + f*x]]] + L
og[1 - Sqrt[2]*Sqrt[Cot[e + f*x]] + Cot[e + f*x]] - Log[1 + Sqrt[2]*Sqrt[Cot[e + f*x]] + Cot[e + f*x]]))/(2*Sq
rt[2]*f*Sqrt[d*Cot[e + f*x]])

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.36, size = 292, normalized size = 1.52

method	result	size
default	\(-\frac {\sqrt {\frac {d \cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \left (\cos \left (f x +e \right )-1\right ) \sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}}\, \left (i \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )\right ) \left (\cos \left (f x +e \right )+1\right )^{2} \sqrt {2}}{2 f \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )}\)	\(292\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*(d*cos(f*x+e)/sin(f*x+e))^(1/2)*(cos(f*x+e)-1)*(-(-1+cos(f*x+e)-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)*((cos(f*x+e)-1)/sin(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))-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)))/sin(f*x+e)^2/cos(f*x+e)*(cos(f*x+e)+1)^
2*2^(1/2)

________________________________________________________________________________________

Maxima [A]
time = 0.53, size = 173, normalized size = 0.90 \begin {gather*} -\frac {d^{2} {\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 )}{d^{\frac {3}{2}}} + \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 )}{d^{\frac {3}{2}}} + \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 )}{d^{\frac {3}{2}}} - \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 )}{d^{\frac {3}{2}}}\right )}}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*d^2*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d/tan(f*x + e)))/sqrt(d))/d^(3/2) + 2*sqrt(2)
*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d/tan(f*x + e)))/sqrt(d))/d^(3/2) + sqrt(2)*log(sqrt(2)*sqrt(d)
*sqrt(d/tan(f*x + e)) + d + d/tan(f*x + e))/d^(3/2) - sqrt(2)*log(-sqrt(2)*sqrt(d)*sqrt(d/tan(f*x + e)) + d +
d/tan(f*x + e))/d^(3/2))/f

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 515 vs. \(2 (151) = 302\).
time = 0.38, size = 515, normalized size = 2.68 \begin {gather*} \sqrt {2} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} f^{3} \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {3}{4}} - \sqrt {2} f^{3} \sqrt {\frac {f^{2} \sqrt {\frac {d^{2}}{f^{4}}} \sin \left (f x + e\right ) + \sqrt {2} f \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} \sin \left (f x + e\right ) + d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {3}{4}} + d^{2}}{d^{2}}\right ) + \sqrt {2} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} f^{3} \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {3}{4}} - \sqrt {2} f^{3} \sqrt {\frac {f^{2} \sqrt {\frac {d^{2}}{f^{4}}} \sin \left (f x + e\right ) - \sqrt {2} f \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} \sin \left (f x + e\right ) + d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {3}{4}} - d^{2}}{d^{2}}\right ) - \frac {1}{4} \, \sqrt {2} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} \log \left (\frac {f^{2} \sqrt {\frac {d^{2}}{f^{4}}} \sin \left (f x + e\right ) + \sqrt {2} f \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} \sin \left (f x + e\right ) + d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}\right ) + \frac {1}{4} \, \sqrt {2} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} \log \left (\frac {f^{2} \sqrt {\frac {d^{2}}{f^{4}}} \sin \left (f x + e\right ) - \sqrt {2} f \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} \sin \left (f x + e\right ) + d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {d \cot {\left (e + f x \right )}} \tan {\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cot(f*x+e))**(1/2)*tan(f*x+e),x)

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 0.21, size = 61, normalized size = 0.32 \begin {gather*} \frac {{\left (-1\right )}^{1/4}\,\sqrt {d}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {\frac {d}{\mathrm {tan}\left (e+f\,x\right )}}}{\sqrt {d}}\right )\,1{}\mathrm {i}}{f}+\frac {{\left (-1\right )}^{1/4}\,\sqrt {d}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {\frac {d}{\mathrm {tan}\left (e+f\,x\right )}}}{\sqrt {d}}\right )\,1{}\mathrm {i}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________