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

Optimal. Leaf size=298 \[ -\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right ) \tan ^{\frac{3}{2}}(c+d x)}{\sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}+\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right ) \tan ^{\frac{3}{2}}(c+d x)}{\sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}+\frac{2}{3 b d \sqrt{b \tan ^3(c+d x)}}-\frac{\tan ^{\frac{3}{2}}(c+d x) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}+\frac{\tan ^{\frac{3}{2}}(c+d x) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}-\frac{2 \cot ^2(c+d x)}{7 b d \sqrt{b \tan ^3(c+d x)}} \]

[Out]

2/(3*b*d*Sqrt[b*Tan[c + d*x]^3]) - (2*Cot[c + d*x]^2)/(7*b*d*Sqrt[b*Tan[c + d*x]^3]) - (ArcTan[1 - Sqrt[2]*Sqr
t[Tan[c + d*x]]]*Tan[c + d*x]^(3/2))/(Sqrt[2]*b*d*Sqrt[b*Tan[c + d*x]^3]) + (ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d
*x]]]*Tan[c + d*x]^(3/2))/(Sqrt[2]*b*d*Sqrt[b*Tan[c + d*x]^3]) - (Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c +
 d*x]]*Tan[c + d*x]^(3/2))/(2*Sqrt[2]*b*d*Sqrt[b*Tan[c + d*x]^3]) + (Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[
c + d*x]]*Tan[c + d*x]^(3/2))/(2*Sqrt[2]*b*d*Sqrt[b*Tan[c + d*x]^3])

________________________________________________________________________________________

Rubi [A]  time = 0.130004, antiderivative size = 298, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.714, Rules used = {3658, 3474, 3476, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right ) \tan ^{\frac{3}{2}}(c+d x)}{\sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}+\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right ) \tan ^{\frac{3}{2}}(c+d x)}{\sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}+\frac{2}{3 b d \sqrt{b \tan ^3(c+d x)}}-\frac{\tan ^{\frac{3}{2}}(c+d x) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}+\frac{\tan ^{\frac{3}{2}}(c+d x) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}-\frac{2 \cot ^2(c+d x)}{7 b d \sqrt{b \tan ^3(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x]^3)^(-3/2),x]

[Out]

2/(3*b*d*Sqrt[b*Tan[c + d*x]^3]) - (2*Cot[c + d*x]^2)/(7*b*d*Sqrt[b*Tan[c + d*x]^3]) - (ArcTan[1 - Sqrt[2]*Sqr
t[Tan[c + d*x]]]*Tan[c + d*x]^(3/2))/(Sqrt[2]*b*d*Sqrt[b*Tan[c + d*x]^3]) + (ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d
*x]]]*Tan[c + d*x]^(3/2))/(Sqrt[2]*b*d*Sqrt[b*Tan[c + d*x]^3]) - (Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c +
 d*x]]*Tan[c + d*x]^(3/2))/(2*Sqrt[2]*b*d*Sqrt[b*Tan[c + d*x]^3]) + (Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[
c + d*x]]*Tan[c + d*x]^(3/2))/(2*Sqrt[2]*b*d*Sqrt[b*Tan[c + d*x]^3])

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

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 211

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 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]

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])

Rubi steps

\begin{align*} \int \frac{1}{\left (b \tan ^3(c+d x)\right )^{3/2}} \, dx &=\frac{\tan ^{\frac{3}{2}}(c+d x) \int \frac{1}{\tan ^{\frac{9}{2}}(c+d x)} \, dx}{b \sqrt{b \tan ^3(c+d x)}}\\ &=-\frac{2 \cot ^2(c+d x)}{7 b d \sqrt{b \tan ^3(c+d x)}}-\frac{\tan ^{\frac{3}{2}}(c+d x) \int \frac{1}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{b \sqrt{b \tan ^3(c+d x)}}\\ &=\frac{2}{3 b d \sqrt{b \tan ^3(c+d x)}}-\frac{2 \cot ^2(c+d x)}{7 b d \sqrt{b \tan ^3(c+d x)}}+\frac{\tan ^{\frac{3}{2}}(c+d x) \int \frac{1}{\sqrt{\tan (c+d x)}} \, dx}{b \sqrt{b \tan ^3(c+d x)}}\\ &=\frac{2}{3 b d \sqrt{b \tan ^3(c+d x)}}-\frac{2 \cot ^2(c+d x)}{7 b d \sqrt{b \tan ^3(c+d x)}}+\frac{\tan ^{\frac{3}{2}}(c+d x) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b d \sqrt{b \tan ^3(c+d x)}}\\ &=\frac{2}{3 b d \sqrt{b \tan ^3(c+d x)}}-\frac{2 \cot ^2(c+d x)}{7 b d \sqrt{b \tan ^3(c+d x)}}+\frac{\left (2 \tan ^{\frac{3}{2}}(c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{b d \sqrt{b \tan ^3(c+d x)}}\\ &=\frac{2}{3 b d \sqrt{b \tan ^3(c+d x)}}-\frac{2 \cot ^2(c+d x)}{7 b d \sqrt{b \tan ^3(c+d x)}}+\frac{\tan ^{\frac{3}{2}}(c+d x) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{b d \sqrt{b \tan ^3(c+d x)}}+\frac{\tan ^{\frac{3}{2}}(c+d x) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{b d \sqrt{b \tan ^3(c+d x)}}\\ &=\frac{2}{3 b d \sqrt{b \tan ^3(c+d x)}}-\frac{2 \cot ^2(c+d x)}{7 b d \sqrt{b \tan ^3(c+d x)}}+\frac{\tan ^{\frac{3}{2}}(c+d x) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 b d \sqrt{b \tan ^3(c+d x)}}+\frac{\tan ^{\frac{3}{2}}(c+d x) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 b d \sqrt{b \tan ^3(c+d x)}}-\frac{\tan ^{\frac{3}{2}}(c+d x) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}-\frac{\tan ^{\frac{3}{2}}(c+d x) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}\\ &=\frac{2}{3 b d \sqrt{b \tan ^3(c+d x)}}-\frac{2 \cot ^2(c+d x)}{7 b d \sqrt{b \tan ^3(c+d x)}}-\frac{\log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right ) \tan ^{\frac{3}{2}}(c+d x)}{2 \sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}+\frac{\log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right ) \tan ^{\frac{3}{2}}(c+d x)}{2 \sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}+\frac{\tan ^{\frac{3}{2}}(c+d x) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}-\frac{\tan ^{\frac{3}{2}}(c+d x) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}\\ &=\frac{2}{3 b d \sqrt{b \tan ^3(c+d x)}}-\frac{2 \cot ^2(c+d x)}{7 b d \sqrt{b \tan ^3(c+d x)}}-\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right ) \tan ^{\frac{3}{2}}(c+d x)}{\sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}+\frac{\tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right ) \tan ^{\frac{3}{2}}(c+d x)}{\sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}-\frac{\log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right ) \tan ^{\frac{3}{2}}(c+d x)}{2 \sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}+\frac{\log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right ) \tan ^{\frac{3}{2}}(c+d x)}{2 \sqrt{2} b d \sqrt{b \tan ^3(c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.0667278, size = 45, normalized size = 0.15 \[ -\frac{2 \tan (c+d x) \, _2F_1\left (-\frac{7}{4},1;-\frac{3}{4};-\tan ^2(c+d x)\right )}{7 d \left (b \tan ^3(c+d x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x]^3)^(-3/2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.022, size = 235, normalized size = 0.8 \begin{align*}{\frac{\tan \left ( dx+c \right ) }{84\,d{b}^{4}} \left ( 21\,\sqrt [4]{{b}^{2}}\sqrt{2} \left ( b\tan \left ( dx+c \right ) \right ) ^{7/2}\ln \left ( -{\frac{b\tan \left ( dx+c \right ) +\sqrt [4]{{b}^{2}}\sqrt{b\tan \left ( dx+c \right ) }\sqrt{2}+\sqrt{{b}^{2}}}{\sqrt [4]{{b}^{2}}\sqrt{b\tan \left ( dx+c \right ) }\sqrt{2}-b\tan \left ( dx+c \right ) -\sqrt{{b}^{2}}}} \right ) +42\,\sqrt [4]{{b}^{2}}\sqrt{2} \left ( b\tan \left ( dx+c \right ) \right ) ^{7/2}\arctan \left ({\frac{\sqrt{2}\sqrt{b\tan \left ( dx+c \right ) }+\sqrt [4]{{b}^{2}}}{\sqrt [4]{{b}^{2}}}} \right ) -42\,\sqrt [4]{{b}^{2}}\sqrt{2} \left ( b\tan \left ( dx+c \right ) \right ) ^{7/2}\arctan \left ({\frac{-\sqrt{2}\sqrt{b\tan \left ( dx+c \right ) }+\sqrt [4]{{b}^{2}}}{\sqrt [4]{{b}^{2}}}} \right ) +56\,{b}^{4} \left ( \tan \left ( dx+c \right ) \right ) ^{2}-24\,{b}^{4} \right ) \left ( b \left ( \tan \left ( dx+c \right ) \right ) ^{3} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/84/d*tan(d*x+c)/b^4*(21*(b^2)^(1/4)*2^(1/2)*(b*tan(d*x+c))^(7/2)*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^2)^(1/4)*(b*tan(d*x+c))^(1/2)*2^(1/2)-b*tan(d*x+c)-(b^2)^(1/2)))+42*(b^2)^(1/
4)*2^(1/2)*(b*tan(d*x+c))^(7/2)*arctan((2^(1/2)*(b*tan(d*x+c))^(1/2)+(b^2)^(1/4))/(b^2)^(1/4))-42*(b^2)^(1/4)*
2^(1/2)*(b*tan(d*x+c))^(7/2)*arctan((-2^(1/2)*(b*tan(d*x+c))^(1/2)+(b^2)^(1/4))/(b^2)^(1/4))+56*b^4*tan(d*x+c)
^2-24*b^4)/(b*tan(d*x+c)^3)^(3/2)

________________________________________________________________________________________

Maxima [A]  time = 1.4358, size = 220, normalized size = 0.74 \begin{align*} \frac{\frac{21 \,{\left (2 \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + \sqrt{2} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt{2} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )}}{b^{\frac{3}{2}}} + \frac{8 \,{\left (21 \, \sqrt{\tan \left (d x + c\right )} + \frac{7}{\tan \left (d x + c\right )^{\frac{3}{2}}} - \frac{3}{\tan \left (d x + c\right )^{\frac{7}{2}}}\right )}}{b^{\frac{3}{2}}} - \frac{168 \, \sqrt{\tan \left (d x + c\right )}}{b^{\frac{3}{2}}}}{84 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/84*(21*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt
(2) - 2*sqrt(tan(d*x + c)))) + sqrt(2)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2)*log(-sqrt(
2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/b^(3/2) + 8*(21*sqrt(tan(d*x + c)) + 7/tan(d*x + c)^(3/2) - 3/tan(d
*x + c)^(7/2))/b^(3/2) - 168*sqrt(tan(d*x + c))/b^(3/2))/d

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*tan(c + d*x)**3)**(-3/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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