\(\int \frac {1}{\sqrt {b \tan (c+d x)}} \, dx\) [13]

Optimal result

Integrand size = 12, antiderivative size = 192 \[ \int \frac {1}{\sqrt {b \tan (c+d x)}} \, dx=-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} \sqrt {b} d}+\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} \sqrt {b} d}-\frac {\log \left (\sqrt {b}+\sqrt {b} \tan (c+d x)-\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{2 \sqrt {2} \sqrt {b} d}+\frac {\log \left (\sqrt {b}+\sqrt {b} \tan (c+d x)+\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{2 \sqrt {2} \sqrt {b} d} \]

[Out]

-1/2*arctan(1-2^(1/2)*(b*tan(d*x+c))^(1/2)/b^(1/2))/d*2^(1/2)/b^(1/2)+1/2*arctan(1+2^(1/2)*(b*tan(d*x+c))^(1/2
)/b^(1/2))/d*2^(1/2)/b^(1/2)-1/4*ln(b^(1/2)-2^(1/2)*(b*tan(d*x+c))^(1/2)+b^(1/2)*tan(d*x+c))/d*2^(1/2)/b^(1/2)
+1/4*ln(b^(1/2)+2^(1/2)*(b*tan(d*x+c))^(1/2)+b^(1/2)*tan(d*x+c))/d*2^(1/2)/b^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3557, 335, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {1}{\sqrt {b \tan (c+d x)}} \, dx=-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} \sqrt {b} d}+\frac {\arctan \left (\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}+1\right )}{\sqrt {2} \sqrt {b} d}-\frac {\log \left (\sqrt {b} \tan (c+d x)-\sqrt {2} \sqrt {b \tan (c+d x)}+\sqrt {b}\right )}{2 \sqrt {2} \sqrt {b} d}+\frac {\log \left (\sqrt {b} \tan (c+d x)+\sqrt {2} \sqrt {b \tan (c+d x)}+\sqrt {b}\right )}{2 \sqrt {2} \sqrt {b} d} \]

[In]

Int[1/Sqrt[b*Tan[c + d*x]],x]

[Out]

-(ArcTan[1 - (Sqrt[2]*Sqrt[b*Tan[c + d*x]])/Sqrt[b]]/(Sqrt[2]*Sqrt[b]*d)) + ArcTan[1 + (Sqrt[2]*Sqrt[b*Tan[c +
d*x]])/Sqrt[b]]/(Sqrt[2]*Sqrt[b]*d) - Log[Sqrt[b] + Sqrt[b]*Tan[c + d*x] - Sqrt[2]*Sqrt[b*Tan[c + d*x]]]/(2*S
qrt[2]*Sqrt[b]*d) + Log[Sqrt[b] + Sqrt[b]*Tan[c + d*x] + Sqrt[2]*Sqrt[b*Tan[c + d*x]]]/(2*Sqrt[2]*Sqrt[b]*d)

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.68 \[ \int \frac {1}{\sqrt {b \tan (c+d x)}} \, dx=\frac {\left (-2 \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )+2 \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )-\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )+\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right ) \sqrt {\tan (c+d x)}}{2 \sqrt {2} d \sqrt {b \tan (c+d x)}} \]

[In]

Integrate[1/Sqrt[b*Tan[c + d*x]],x]

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(\frac {\left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (d x +c \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d b}\)	\(138\)
default	\(\frac {\left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (d x +c \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d b}\)	\(138\)

[In]

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

[Out]

1/4/d/b*(b^2)^(1/4)*2^(1/2)*(ln((b*tan(d*x+c)+(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)*2^(1/2)+(b^2)^(1/2))/(b*tan(d*x
+c)-(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)*2^(1/2)+(b^2)^(1/2)))+2*arctan(2^(1/2)/(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)+1
)-2*arctan(-2^(1/2)/(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)+1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\sqrt {b \tan (c+d x)}} \, dx=\frac {1}{2} \, \left (-\frac {1}{b^{2} d^{4}}\right )^{\frac {1}{4}} \log \left (b d \left (-\frac {1}{b^{2} d^{4}}\right )^{\frac {1}{4}} + \sqrt {b \tan \left (d x + c\right )}\right ) + \frac {1}{2} i \, \left (-\frac {1}{b^{2} d^{4}}\right )^{\frac {1}{4}} \log \left (i \, b d \left (-\frac {1}{b^{2} d^{4}}\right )^{\frac {1}{4}} + \sqrt {b \tan \left (d x + c\right )}\right ) - \frac {1}{2} i \, \left (-\frac {1}{b^{2} d^{4}}\right )^{\frac {1}{4}} \log \left (-i \, b d \left (-\frac {1}{b^{2} d^{4}}\right )^{\frac {1}{4}} + \sqrt {b \tan \left (d x + c\right )}\right ) - \frac {1}{2} \, \left (-\frac {1}{b^{2} d^{4}}\right )^{\frac {1}{4}} \log \left (-b d \left (-\frac {1}{b^{2} d^{4}}\right )^{\frac {1}{4}} + \sqrt {b \tan \left (d x + c\right )}\right ) \]

[In]

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

[Out]

1/2*(-1/(b^2*d^4))^(1/4)*log(b*d*(-1/(b^2*d^4))^(1/4) + sqrt(b*tan(d*x + c))) + 1/2*I*(-1/(b^2*d^4))^(1/4)*log
(I*b*d*(-1/(b^2*d^4))^(1/4) + sqrt(b*tan(d*x + c))) - 1/2*I*(-1/(b^2*d^4))^(1/4)*log(-I*b*d*(-1/(b^2*d^4))^(1/
4) + sqrt(b*tan(d*x + c))) - 1/2*(-1/(b^2*d^4))^(1/4)*log(-b*d*(-1/(b^2*d^4))^(1/4) + sqrt(b*tan(d*x + c)))

Sympy [F]

\[ \int \frac {1}{\sqrt {b \tan (c+d x)}} \, dx=\int \frac {1}{\sqrt {b \tan {\left (c + d x \right )}}}\, dx \]

[In]

integrate(1/(b*tan(d*x+c))**(1/2),x)

[Out]

Integral(1/sqrt(b*tan(c + d*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\sqrt {b \tan (c+d x)}} \, dx=\frac {2 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} + 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right ) + 2 \, \sqrt {2} \sqrt {b} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} - 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right ) + \sqrt {2} \sqrt {b} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right ) - \sqrt {2} \sqrt {b} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right )}{4 \, b d} \]

[In]

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

[Out]

1/4*(2*sqrt(2)*sqrt(b)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(b) + 2*sqrt(b*tan(d*x + c)))/sqrt(b)) + 2*sqrt(2)*sqrt
(b)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(b) - 2*sqrt(b*tan(d*x + c)))/sqrt(b)) + sqrt(2)*sqrt(b)*log(b*tan(d*x +
c) + sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(b) + b) - sqrt(2)*sqrt(b)*log(b*tan(d*x + c) - sqrt(2)*sqrt(b*tan(d*x +
c))*sqrt(b) + b))/(b*d)

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\sqrt {b \tan (c+d x)}} \, dx=\frac {\sqrt {2} \sqrt {{\left | b \right |}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} + 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{2 \, b d} + \frac {\sqrt {2} \sqrt {{\left | b \right |}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | b \right |}} - 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {{\left | b \right |}}}\right )}{2 \, b d} + \frac {\sqrt {2} \sqrt {{\left | b \right |}} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{4 \, b d} - \frac {\sqrt {2} \sqrt {{\left | b \right |}} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {{\left | b \right |}} + {\left | b \right |}\right )}{4 \, b d} \]

[In]

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

[Out]

1/2*sqrt(2)*sqrt(abs(b))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) + 2*sqrt(b*tan(d*x + c)))/sqrt(abs(b)))/(b*d
) + 1/2*sqrt(2)*sqrt(abs(b))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) - 2*sqrt(b*tan(d*x + c)))/sqrt(abs(b)))
/(b*d) + 1/4*sqrt(2)*sqrt(abs(b))*log(b*tan(d*x + c) + sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(abs(b)) + abs(b))/(b*
d) - 1/4*sqrt(2)*sqrt(abs(b))*log(b*tan(d*x + c) - sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(abs(b)) + abs(b))/(b*d)

Mupad [B] (verification not implemented)

Time = 3.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.31 \[ \int \frac {1}{\sqrt {b \tan (c+d x)}} \, dx=-\frac {{\left (-1\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {b}}\right )\,1{}\mathrm {i}}{\sqrt {b}\,d}-\frac {{\left (-1\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {b}}\right )\,1{}\mathrm {i}}{\sqrt {b}\,d} \]

[In]

int(1/(b*tan(c + d*x))^(1/2),x)

[Out]

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