3.261 \(\int \frac{1}{(d \tan (a+b x))^{3/2}} \, dx\)

Optimal. Leaf size=212 \[ \frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}\right )}{\sqrt{2} b d^{3/2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}+1\right )}{\sqrt{2} b d^{3/2}}-\frac{\log \left (\sqrt{d} \tan (a+b x)-\sqrt{2} \sqrt{d \tan (a+b x)}+\sqrt{d}\right )}{2 \sqrt{2} b d^{3/2}}+\frac{\log \left (\sqrt{d} \tan (a+b x)+\sqrt{2} \sqrt{d \tan (a+b x)}+\sqrt{d}\right )}{2 \sqrt{2} b d^{3/2}}-\frac{2}{b d \sqrt{d \tan (a+b x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.141506, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {3474, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}\right )}{\sqrt{2} b d^{3/2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}+1\right )}{\sqrt{2} b d^{3/2}}-\frac{\log \left (\sqrt{d} \tan (a+b x)-\sqrt{2} \sqrt{d \tan (a+b x)}+\sqrt{d}\right )}{2 \sqrt{2} b d^{3/2}}+\frac{\log \left (\sqrt{d} \tan (a+b x)+\sqrt{2} \sqrt{d \tan (a+b x)}+\sqrt{d}\right )}{2 \sqrt{2} b d^{3/2}}-\frac{2}{b d \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Tan[a + b*x])^(-3/2),x]

[Out]

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

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

Mathematica [C]  time = 0.0336544, size = 38, normalized size = 0.18 \[ -\frac{2 \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};-\tan ^2(a+b x)\right )}{b d \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Tan[a + b*x])^(-3/2),x]

[Out]

(-2*Hypergeometric2F1[-1/4, 1, 3/4, -Tan[a + b*x]^2])/(b*d*Sqrt[d*Tan[a + b*x]])

________________________________________________________________________________________

Maple [A]  time = 0.011, size = 184, normalized size = 0.9 \begin{align*} -2\,{\frac{1}{bd\sqrt{d\tan \left ( bx+a \right ) }}}-{\frac{\sqrt{2}}{4\,bd}\ln \left ({ \left ( d\tan \left ( bx+a \right ) -\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( bx+a \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) \left ( d\tan \left ( bx+a \right ) +\sqrt [4]{{d}^{2}}\sqrt{d\tan \left ( bx+a \right ) }\sqrt{2}+\sqrt{{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}-{\frac{\sqrt{2}}{2\,bd}\arctan \left ({\sqrt{2}\sqrt{d\tan \left ( bx+a \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}}+{\frac{\sqrt{2}}{2\,bd}\arctan \left ( -{\sqrt{2}\sqrt{d\tan \left ( bx+a \right ) }{\frac{1}{\sqrt [4]{{d}^{2}}}}}+1 \right ){\frac{1}{\sqrt [4]{{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*tan(b*x+a))^(3/2),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.76502, size = 1693, normalized size = 7.99 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(8*sqrt(d*sin(b*x + a)/cos(b*x + a))*cos(b*x + a)*sin(b*x + a) + 4*(sqrt(2)*b*d^2*cos(b*x + a)^2 - sqrt(2)
*b*d^2)*(1/(b^4*d^6))^(1/4)*arctan(-sqrt(2)*b*d*sqrt(d*sin(b*x + a)/cos(b*x + a))*(1/(b^4*d^6))^(1/4) + sqrt(2
)*b*d*sqrt((sqrt(2)*b^3*d^5*sqrt(d*sin(b*x + a)/cos(b*x + a))*(1/(b^4*d^6))^(3/4)*cos(b*x + a) + b^2*d^4*sqrt(
1/(b^4*d^6))*cos(b*x + a) + d*sin(b*x + a))/cos(b*x + a))*(1/(b^4*d^6))^(1/4) - 1) + 4*(sqrt(2)*b*d^2*cos(b*x
+ a)^2 - sqrt(2)*b*d^2)*(1/(b^4*d^6))^(1/4)*arctan(-sqrt(2)*b*d*sqrt(d*sin(b*x + a)/cos(b*x + a))*(1/(b^4*d^6)
)^(1/4) + sqrt(2)*b*d*sqrt(-(sqrt(2)*b^3*d^5*sqrt(d*sin(b*x + a)/cos(b*x + a))*(1/(b^4*d^6))^(3/4)*cos(b*x + a
) - b^2*d^4*sqrt(1/(b^4*d^6))*cos(b*x + a) - d*sin(b*x + a))/cos(b*x + a))*(1/(b^4*d^6))^(1/4) + 1) + (sqrt(2)
*b*d^2*cos(b*x + a)^2 - sqrt(2)*b*d^2)*(1/(b^4*d^6))^(1/4)*log((sqrt(2)*b^3*d^5*sqrt(d*sin(b*x + a)/cos(b*x +
a))*(1/(b^4*d^6))^(3/4)*cos(b*x + a) + b^2*d^4*sqrt(1/(b^4*d^6))*cos(b*x + a) + d*sin(b*x + a))/cos(b*x + a))
- (sqrt(2)*b*d^2*cos(b*x + a)^2 - sqrt(2)*b*d^2)*(1/(b^4*d^6))^(1/4)*log(-(sqrt(2)*b^3*d^5*sqrt(d*sin(b*x + a)
/cos(b*x + a))*(1/(b^4*d^6))^(3/4)*cos(b*x + a) - b^2*d^4*sqrt(1/(b^4*d^6))*cos(b*x + a) - d*sin(b*x + a))/cos
(b*x + a)))/(b*d^2*cos(b*x + a)^2 - b*d^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \tan{\left (a + b x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d*tan(a + b*x))**(-3/2), x)

________________________________________________________________________________________

Giac [A]  time = 1.2477, size = 275, normalized size = 1.3 \begin{align*} -\frac{1}{4} \, d{\left (\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 (b x + a\right )}\right )}}{2 \, \sqrt{{\left | d \right |}}}\right )}{b d^{4}} + \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 (b x + a\right )}\right )}}{2 \, \sqrt{{\left | d \right |}}}\right )}{b d^{4}} - \frac{\sqrt{2}{\left | d \right |}^{\frac{3}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt{2} \sqrt{d \tan \left (b x + a\right )} \sqrt{{\left | d \right |}} +{\left | d \right |}\right )}{b d^{4}} + \frac{\sqrt{2}{\left | d \right |}^{\frac{3}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt{2} \sqrt{d \tan \left (b x + a\right )} \sqrt{{\left | d \right |}} +{\left | d \right |}\right )}{b d^{4}} + \frac{8}{\sqrt{d \tan \left (b x + a\right )} b d^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*d*(2*sqrt(2)*abs(d)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(b*x + a)))/sqrt(abs(d))
)/(b*d^4) + 2*sqrt(2)*abs(d)^(3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(b*x + a)))/sqrt(ab
s(d)))/(b*d^4) - sqrt(2)*abs(d)^(3/2)*log(d*tan(b*x + a) + sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(abs(d)) + abs(d))
/(b*d^4) + sqrt(2)*abs(d)^(3/2)*log(d*tan(b*x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(abs(d)) + abs(d))/(b*d^
4) + 8/(sqrt(d*tan(b*x + a))*b*d^2))