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

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

[Out]

2/3/b/d/(b*tan(d*x+c)^3)^(1/2)-2/7*cot(d*x+c)^2/b/d/(b*tan(d*x+c)^3)^(1/2)+1/2*arctan(-1+2^(1/2)*tan(d*x+c)^(1
/2))*tan(d*x+c)^(3/2)/b/d*2^(1/2)/(b*tan(d*x+c)^3)^(1/2)+1/2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(3/
2)/b/d*2^(1/2)/(b*tan(d*x+c)^3)^(1/2)-1/4*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))*tan(d*x+c)^(3/2)/b/d*2^(1/
2)/(b*tan(d*x+c)^3)^(1/2)+1/4*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))*tan(d*x+c)^(3/2)/b/d*2^(1/2)/(b*tan(d*
x+c)^3)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 298, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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 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 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
]])

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*}

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Mathematica [C]  time = 0.07, 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))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [A]  time = 0.10, size = 236, normalized size = 0.79 \[ \frac {\tan \left (d x +c \right ) \left (21 \left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (b \tan \left (d x +c \right )\right )^{\frac {7}{2}} \ln \left (-\frac {b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}-b \tan \left (d x +c \right )-\sqrt {b^{2}}}\right )+42 \left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (b \tan \left (d x +c \right )\right )^{\frac {7}{2}} \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}+\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+42 \left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (b \tan \left (d x +c \right )\right )^{\frac {7}{2}} \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}-\left (b^{2}\right )^{\frac {1}{4}}}{\left (b^{2}\right )^{\frac {1}{4}}}\right )+56 b^{4} \left (\tan ^{2}\left (d x +c \right )\right )-24 b^{4}\right )}{84 d \,b^{4} \left (b \left (\tan ^{3}\left (d x +c \right )\right )\right )^{\frac {3}{2}}} \]

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)

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maxima [A]  time = 0.63, size = 163, normalized size = 0.55 \[ \frac {\frac {21 \, {\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )}}{b^{\frac {3}{2}}} + \frac {8 \, {\left (21 \, \sqrt {\tan \left (d x + c\right )} + \frac {7}{\tan \left (d x + c\right )^{\frac {3}{2}}} - \frac {3}{\tan \left (d x + c\right )^{\frac {7}{2}}}\right )}}{b^{\frac {3}{2}}} - \frac {168 \, \sqrt {\tan \left (d x + c\right )}}{b^{\frac {3}{2}}}}{84 \, d} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(c + d*x)^3)^(3/2),x)

[Out]

int(1/(b*tan(c + d*x)^3)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tan ^{3}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

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)

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