3.3.29 \(\int \sqrt {d \tan (e+f x)} \, dx\) [229]

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

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3557, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {\sqrt {d} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}+\frac {\sqrt {d} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} f}+\frac {\sqrt {d} \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f}-\frac {\sqrt {d} \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*Tan[e + f*x]],x]

[Out]

-((Sqrt[d]*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f)) + (Sqrt[d]*ArcTan[1 + (Sqrt[2]*Sqr
t[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) + (Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] - Sqrt[2]*Sqrt[d*Tan[e
+ f*x]]])/(2*Sqrt[2]*f) - (Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(2*Sqrt
[2]*f)

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 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 \tan (e+f x)} \, dx &=\frac {d \text {Subst}\left (\int \frac {\sqrt {x}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}\\ &=\frac {(2 d) \text {Subst}\left (\int \frac {x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}\\ &=-\frac {d \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{f}+\frac {d \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (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 \tan (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 \tan (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 \tan (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 \tan (e+f x)}\right )}{2 f}\\ &=\frac {\sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{2 \sqrt {2} f}-\frac {\sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (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 \tan (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 \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}\\ &=-\frac {\sqrt {d} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}+\frac {\sqrt {d} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}+\frac {\sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{2 \sqrt {2} f}-\frac {\sqrt {d} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{2 \sqrt {2} f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.05, size = 40, normalized size = 0.21 \begin {gather*} \frac {2 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{3/2}}{3 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*Tan[e + f*x]],x]

[Out]

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

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Maple [A]
time = 0.18, size = 136, normalized size = 0.71

method	result	size
derivativedivides	\(\frac {d \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 f \left (d^{2}\right )^{\frac {1}{4}}}\)	\(136\)
default	\(\frac {d \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 f \left (d^{2}\right )^{\frac {1}{4}}}\)	\(136\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 159, normalized size = 0.83 \begin {gather*} \frac {d {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {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 {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*d*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + 2*sqrt(2)*ar
ctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - sqrt(2)*log(d*tan(f*x + e) + s
qrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt(d) + sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*s
qrt(d) + d)/sqrt(d))/f

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 547 vs. \(2 (151) = 302\).
time = 0.37, size = 547, normalized size = 2.85 \begin {gather*} -\sqrt {2} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} d f \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} - \sqrt {2} f \sqrt {\frac {\sqrt {2} d f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {3}{4}} \cos \left (f x + e\right ) + d^{2} f^{2} \sqrt {\frac {d^{2}}{f^{4}}} \cos \left (f x + e\right ) + d^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} + d^{2}}{d^{2}}\right ) - \sqrt {2} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} d f \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} - \sqrt {2} f \sqrt {-\frac {\sqrt {2} d f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {3}{4}} \cos \left (f x + e\right ) - d^{2} f^{2} \sqrt {\frac {d^{2}}{f^{4}}} \cos \left (f x + e\right ) - d^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} - d^{2}}{d^{2}}\right ) - \frac {1}{4} \, \sqrt {2} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} \log \left (\frac {\sqrt {2} d f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {3}{4}} \cos \left (f x + e\right ) + d^{2} f^{2} \sqrt {\frac {d^{2}}{f^{4}}} \cos \left (f x + e\right ) + d^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) + \frac {1}{4} \, \sqrt {2} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {1}{4}} \log \left (-\frac {\sqrt {2} d f^{3} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \left (\frac {d^{2}}{f^{4}}\right )^{\frac {3}{4}} \cos \left (f x + e\right ) - d^{2} f^{2} \sqrt {\frac {d^{2}}{f^{4}}} \cos \left (f x + e\right ) - d^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(d*tan(e + f*x)), x)

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Giac [A]
time = 0.47, size = 182, normalized size = 0.95 \begin {gather*} \frac {\frac {2 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{f} + \frac {2 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{f} - \frac {\sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{f} + \frac {\sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{f}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.50, size = 49, normalized size = 0.26 \begin {gather*} \frac {{\left (-1\right )}^{1/4}\,\sqrt {d}\,\left (\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )-\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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