\(\int \frac {1}{(b \tan (c+d x))^{3/2}} \, dx\) [14]

Optimal result

Integrand size = 12, antiderivative size = 212 \[ \int \frac {1}{(b \tan (c+d x))^{3/2}} \, dx=\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} b^{3/2} d}-\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} b^{3/2} d}-\frac {\log \left (\sqrt {b}+\sqrt {b} \tan (c+d x)-\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{2 \sqrt {2} b^{3/2} d}+\frac {\log \left (\sqrt {b}+\sqrt {b} \tan (c+d x)+\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{2 \sqrt {2} b^{3/2} d}-\frac {2}{b d \sqrt {b \tan (c+d x)}} \]

[Out]

1/2*arctan(1-2^(1/2)*(b*tan(d*x+c))^(1/2)/b^(1/2))/b^(3/2)/d*2^(1/2)-1/2*arctan(1+2^(1/2)*(b*tan(d*x+c))^(1/2)
/b^(1/2))/b^(3/2)/d*2^(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))/b^(3/2)/d*2^(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))/b^(3/2)/d*2^(1/2)-2/b/d/(b*tan(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3555, 3557, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {1}{(b \tan (c+d x))^{3/2}} \, dx=\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} b^{3/2} d}-\frac {\arctan \left (\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}+1\right )}{\sqrt {2} b^{3/2} d}-\frac {\log \left (\sqrt {b} \tan (c+d x)-\sqrt {2} \sqrt {b \tan (c+d x)}+\sqrt {b}\right )}{2 \sqrt {2} b^{3/2} d}+\frac {\log \left (\sqrt {b} \tan (c+d x)+\sqrt {2} \sqrt {b \tan (c+d x)}+\sqrt {b}\right )}{2 \sqrt {2} b^{3/2} d}-\frac {2}{b d \sqrt {b \tan (c+d x)}} \]

[In]

Int[(b*Tan[c + d*x])^(-3/2),x]

[Out]

ArcTan[1 - (Sqrt[2]*Sqrt[b*Tan[c + d*x]])/Sqrt[b]]/(Sqrt[2]*b^(3/2)*d) - ArcTan[1 + (Sqrt[2]*Sqrt[b*Tan[c + d*
x]])/Sqrt[b]]/(Sqrt[2]*b^(3/2)*d) - Log[Sqrt[b] + Sqrt[b]*Tan[c + d*x] - Sqrt[2]*Sqrt[b*Tan[c + d*x]]]/(2*Sqrt
[2]*b^(3/2)*d) + Log[Sqrt[b] + Sqrt[b]*Tan[c + d*x] + Sqrt[2]*Sqrt[b*Tan[c + d*x]]]/(2*Sqrt[2]*b^(3/2)*d) - 2/
(b*d*Sqrt[b*Tan[c + d*x]])

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 3555

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 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 {2}{b d \sqrt {b \tan (c+d x)}}-\frac {\int \sqrt {b \tan (c+d x)} \, dx}{b^2} \\ & = -\frac {2}{b d \sqrt {b \tan (c+d x)}}-\frac {\text {Subst}\left (\int \frac {\sqrt {x}}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{b d} \\ & = -\frac {2}{b d \sqrt {b \tan (c+d x)}}-\frac {2 \text {Subst}\left (\int \frac {x^2}{b^2+x^4} \, dx,x,\sqrt {b \tan (c+d x)}\right )}{b d} \\ & = -\frac {2}{b d \sqrt {b \tan (c+d x)}}+\frac {\text {Subst}\left (\int \frac {b-x^2}{b^2+x^4} \, dx,x,\sqrt {b \tan (c+d x)}\right )}{b d}-\frac {\text {Subst}\left (\int \frac {b+x^2}{b^2+x^4} \, dx,x,\sqrt {b \tan (c+d x)}\right )}{b d} \\ & = -\frac {2}{b d \sqrt {b \tan (c+d x)}}-\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} b^{3/2} 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} b^{3/2} 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 b 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 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} b^{3/2} d}+\frac {\log \left (\sqrt {b}+\sqrt {b} \tan (c+d x)+\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{2 \sqrt {2} b^{3/2} d}-\frac {2}{b d \sqrt {b \tan (c+d x)}}-\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} b^{3/2} 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} b^{3/2} d} \\ & = \frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} b^{3/2} d}-\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} b^{3/2} d}-\frac {\log \left (\sqrt {b}+\sqrt {b} \tan (c+d x)-\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{2 \sqrt {2} b^{3/2} d}+\frac {\log \left (\sqrt {b}+\sqrt {b} \tan (c+d x)+\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{2 \sqrt {2} b^{3/2} d}-\frac {2}{b d \sqrt {b \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.39 \[ \int \frac {1}{(b \tan (c+d x))^{3/2}} \, dx=\frac {-2-\arctan \left (\sqrt [4]{-\tan ^2(c+d x)}\right ) \sqrt [4]{-\tan ^2(c+d x)}+\text {arctanh}\left (\sqrt [4]{-\tan ^2(c+d x)}\right ) \sqrt [4]{-\tan ^2(c+d x)}}{b d \sqrt {b \tan (c+d x)}} \]

[In]

Integrate[(b*Tan[c + d*x])^(-3/2),x]

[Out]

(-2 - ArcTan[(-Tan[c + d*x]^2)^(1/4)]*(-Tan[c + d*x]^2)^(1/4) + ArcTanh[(-Tan[c + d*x]^2)^(1/4)]*(-Tan[c + d*x
]^2)^(1/4))/(b*d*Sqrt[b*Tan[c + d*x]])

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {2 b \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {b \tan \left (d x +c \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 b^{2} \left (b^{2}\right )^{\frac {1}{4}}}-\frac {1}{b^{2} \sqrt {b \tan \left (d x +c \right )}}\right )}{d}\)	\(157\)
default	\(\frac {2 b \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {b \tan \left (d x +c \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 b^{2} \left (b^{2}\right )^{\frac {1}{4}}}-\frac {1}{b^{2} \sqrt {b \tan \left (d x +c \right )}}\right )}{d}\)	\(157\)

[In]

int(1/(b*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/d*b*(-1/8/b^2/(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))-1/b^2/(b*tan(d*x+c))^(1/2))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.11 \[ \int \frac {1}{(b \tan (c+d x))^{3/2}} \, dx=-\frac {b^{2} d \left (-\frac {1}{b^{6} d^{4}}\right )^{\frac {1}{4}} \log \left (b^{5} d^{3} \left (-\frac {1}{b^{6} d^{4}}\right )^{\frac {3}{4}} + \sqrt {b \tan \left (d x + c\right )}\right ) \tan \left (d x + c\right ) - i \, b^{2} d \left (-\frac {1}{b^{6} d^{4}}\right )^{\frac {1}{4}} \log \left (i \, b^{5} d^{3} \left (-\frac {1}{b^{6} d^{4}}\right )^{\frac {3}{4}} + \sqrt {b \tan \left (d x + c\right )}\right ) \tan \left (d x + c\right ) + i \, b^{2} d \left (-\frac {1}{b^{6} d^{4}}\right )^{\frac {1}{4}} \log \left (-i \, b^{5} d^{3} \left (-\frac {1}{b^{6} d^{4}}\right )^{\frac {3}{4}} + \sqrt {b \tan \left (d x + c\right )}\right ) \tan \left (d x + c\right ) - b^{2} d \left (-\frac {1}{b^{6} d^{4}}\right )^{\frac {1}{4}} \log \left (-b^{5} d^{3} \left (-\frac {1}{b^{6} d^{4}}\right )^{\frac {3}{4}} + \sqrt {b \tan \left (d x + c\right )}\right ) \tan \left (d x + c\right ) + 4 \, \sqrt {b \tan \left (d x + c\right )}}{2 \, b^{2} d \tan \left (d x + c\right )} \]

[In]

integrate(1/(b*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/2*(b^2*d*(-1/(b^6*d^4))^(1/4)*log(b^5*d^3*(-1/(b^6*d^4))^(3/4) + sqrt(b*tan(d*x + c)))*tan(d*x + c) - I*b^2
*d*(-1/(b^6*d^4))^(1/4)*log(I*b^5*d^3*(-1/(b^6*d^4))^(3/4) + sqrt(b*tan(d*x + c)))*tan(d*x + c) + I*b^2*d*(-1/
(b^6*d^4))^(1/4)*log(-I*b^5*d^3*(-1/(b^6*d^4))^(3/4) + sqrt(b*tan(d*x + c)))*tan(d*x + c) - b^2*d*(-1/(b^6*d^4
))^(1/4)*log(-b^5*d^3*(-1/(b^6*d^4))^(3/4) + sqrt(b*tan(d*x + c)))*tan(d*x + c) + 4*sqrt(b*tan(d*x + c)))/(b^2
*d*tan(d*x + c))

Sympy [F]

\[ \int \frac {1}{(b \tan (c+d x))^{3/2}} \, dx=\int \frac {1}{\left (b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(b*tan(d*x+c))**(3/2),x)

[Out]

Integral((b*tan(c + d*x))**(-3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(b \tan (c+d x))^{3/2}} \, dx=-\frac {\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} + 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right )}{\sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} - 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right )}{\sqrt {b}} - \frac {\sqrt {2} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right )}{\sqrt {b}} + \frac {\sqrt {2} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right )}{\sqrt {b}} + \frac {8}{\sqrt {b \tan \left (d x + c\right )}}}{4 \, b d} \]

[In]

integrate(1/(b*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-1/4*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(b) + 2*sqrt(b*tan(d*x + c)))/sqrt(b))/sqrt(b) + 2*sqrt(2)*arc
tan(-1/2*sqrt(2)*(sqrt(2)*sqrt(b) - 2*sqrt(b*tan(d*x + c)))/sqrt(b))/sqrt(b) - sqrt(2)*log(b*tan(d*x + c) + sq
rt(2)*sqrt(b*tan(d*x + c))*sqrt(b) + b)/sqrt(b) + sqrt(2)*log(b*tan(d*x + c) - sqrt(2)*sqrt(b*tan(d*x + c))*sq
rt(b) + b)/sqrt(b) + 8/sqrt(b*tan(d*x + c)))/(b*d)

Giac [F(-1)]

Timed out. \[ \int \frac {1}{(b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(b*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 2.97 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.36 \[ \int \frac {1}{(b \tan (c+d x))^{3/2}} \, dx=\frac {{\left (-1\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{3/2}\,d}-\frac {{\left (-1\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{3/2}\,d}-\frac {2}{b\,d\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}} \]

[In]

int(1/(b*tan(c + d*x))^(3/2),x)

[Out]

((-1)^(1/4)*atanh(((-1)^(1/4)*(b*tan(c + d*x))^(1/2))/b^(1/2)))/(b^(3/2)*d) - ((-1)^(1/4)*atan(((-1)^(1/4)*(b*
tan(c + d*x))^(1/2))/b^(1/2)))/(b^(3/2)*d) - 2/(b*d*(b*tan(c + d*x))^(1/2))