3.15 \(\int \frac {1}{(b \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=214 \[ \frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} b^{5/2} d}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}+1\right )}{\sqrt {2} b^{5/2} d}+\frac {\log \left (\sqrt {b} \tan (c+d x)-\sqrt {2} \sqrt {b \tan (c+d x)}+\sqrt {b}\right )}{2 \sqrt {2} b^{5/2} d}-\frac {\log \left (\sqrt {b} \tan (c+d x)+\sqrt {2} \sqrt {b \tan (c+d x)}+\sqrt {b}\right )}{2 \sqrt {2} b^{5/2} d}-\frac {2}{3 b d (b \tan (c+d x))^{3/2}} \]

[Out]

1/2*arctan(1-2^(1/2)*(b*tan(d*x+c))^(1/2)/b^(1/2))/b^(5/2)/d*2^(1/2)-1/2*arctan(1+2^(1/2)*(b*tan(d*x+c))^(1/2)
/b^(1/2))/b^(5/2)/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^(5/2)/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^(5/2)/d*2^(1/2)-2/3/b/d/(b*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.15, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3474, 3476, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} b^{5/2} d}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}+1\right )}{\sqrt {2} b^{5/2} d}+\frac {\log \left (\sqrt {b} \tan (c+d x)-\sqrt {2} \sqrt {b \tan (c+d x)}+\sqrt {b}\right )}{2 \sqrt {2} b^{5/2} d}-\frac {\log \left (\sqrt {b} \tan (c+d x)+\sqrt {2} \sqrt {b \tan (c+d x)}+\sqrt {b}\right )}{2 \sqrt {2} b^{5/2} d}-\frac {2}{3 b d (b \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x])^(-5/2),x]

[Out]

ArcTan[1 - (Sqrt[2]*Sqrt[b*Tan[c + d*x]])/Sqrt[b]]/(Sqrt[2]*b^(5/2)*d) - ArcTan[1 + (Sqrt[2]*Sqrt[b*Tan[c + d*
x]])/Sqrt[b]]/(Sqrt[2]*b^(5/2)*d) + Log[Sqrt[b] + Sqrt[b]*Tan[c + d*x] - Sqrt[2]*Sqrt[b*Tan[c + d*x]]]/(2*Sqrt
[2]*b^(5/2)*d) - Log[Sqrt[b] + Sqrt[b]*Tan[c + d*x] + Sqrt[2]*Sqrt[b*Tan[c + d*x]]]/(2*Sqrt[2]*b^(5/2)*d) - 2/
(3*b*d*(b*Tan[c + d*x])^(3/2))

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]

Rubi steps

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

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Mathematica [C]  time = 0.09, size = 40, normalized size = 0.19 \[ -\frac {2 \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\tan ^2(c+d x)\right )}{3 b d (b \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x])^(-5/2),x]

[Out]

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

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fricas [B]  time = 0.81, size = 653, normalized size = 3.05 \[ \frac {8 \, \sqrt {\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{2} + 12 \, {\left (\sqrt {2} b^{3} d \cos \left (d x + c\right )^{2} - \sqrt {2} b^{3} d\right )} \left (\frac {1}{b^{10} d^{4}}\right )^{\frac {1}{4}} \arctan \left (-\sqrt {2} b^{7} d^{3} \sqrt {\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \left (\frac {1}{b^{10} d^{4}}\right )^{\frac {3}{4}} + \sqrt {2} b^{7} d^{3} \sqrt {\frac {b^{6} d^{2} \sqrt {\frac {1}{b^{10} d^{4}}} \cos \left (d x + c\right ) + \sqrt {2} b^{3} d \sqrt {\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \left (\frac {1}{b^{10} d^{4}}\right )^{\frac {1}{4}} \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \left (\frac {1}{b^{10} d^{4}}\right )^{\frac {3}{4}} - 1\right ) + 12 \, {\left (\sqrt {2} b^{3} d \cos \left (d x + c\right )^{2} - \sqrt {2} b^{3} d\right )} \left (\frac {1}{b^{10} d^{4}}\right )^{\frac {1}{4}} \arctan \left (-\sqrt {2} b^{7} d^{3} \sqrt {\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \left (\frac {1}{b^{10} d^{4}}\right )^{\frac {3}{4}} + \sqrt {2} b^{7} d^{3} \sqrt {\frac {b^{6} d^{2} \sqrt {\frac {1}{b^{10} d^{4}}} \cos \left (d x + c\right ) - \sqrt {2} b^{3} d \sqrt {\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \left (\frac {1}{b^{10} d^{4}}\right )^{\frac {1}{4}} \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \left (\frac {1}{b^{10} d^{4}}\right )^{\frac {3}{4}} + 1\right ) - 3 \, {\left (\sqrt {2} b^{3} d \cos \left (d x + c\right )^{2} - \sqrt {2} b^{3} d\right )} \left (\frac {1}{b^{10} d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {b^{6} d^{2} \sqrt {\frac {1}{b^{10} d^{4}}} \cos \left (d x + c\right ) + \sqrt {2} b^{3} d \sqrt {\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \left (\frac {1}{b^{10} d^{4}}\right )^{\frac {1}{4}} \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right ) + 3 \, {\left (\sqrt {2} b^{3} d \cos \left (d x + c\right )^{2} - \sqrt {2} b^{3} d\right )} \left (\frac {1}{b^{10} d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {b^{6} d^{2} \sqrt {\frac {1}{b^{10} d^{4}}} \cos \left (d x + c\right ) - \sqrt {2} b^{3} d \sqrt {\frac {b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}} \left (\frac {1}{b^{10} d^{4}}\right )^{\frac {1}{4}} \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right )}{12 \, {\left (b^{3} d \cos \left (d x + c\right )^{2} - b^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/12*(8*sqrt(b*sin(d*x + c)/cos(d*x + c))*cos(d*x + c)^2 + 12*(sqrt(2)*b^3*d*cos(d*x + c)^2 - sqrt(2)*b^3*d)*(
1/(b^10*d^4))^(1/4)*arctan(-sqrt(2)*b^7*d^3*sqrt(b*sin(d*x + c)/cos(d*x + c))*(1/(b^10*d^4))^(3/4) + sqrt(2)*b
^7*d^3*sqrt((b^6*d^2*sqrt(1/(b^10*d^4))*cos(d*x + c) + sqrt(2)*b^3*d*sqrt(b*sin(d*x + c)/cos(d*x + c))*(1/(b^1
0*d^4))^(1/4)*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*(1/(b^10*d^4))^(3/4) - 1) + 12*(sqrt(2)*b^3*d*cos(d
*x + c)^2 - sqrt(2)*b^3*d)*(1/(b^10*d^4))^(1/4)*arctan(-sqrt(2)*b^7*d^3*sqrt(b*sin(d*x + c)/cos(d*x + c))*(1/(
b^10*d^4))^(3/4) + sqrt(2)*b^7*d^3*sqrt((b^6*d^2*sqrt(1/(b^10*d^4))*cos(d*x + c) - sqrt(2)*b^3*d*sqrt(b*sin(d*
x + c)/cos(d*x + c))*(1/(b^10*d^4))^(1/4)*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*(1/(b^10*d^4))^(3/4) +
1) - 3*(sqrt(2)*b^3*d*cos(d*x + c)^2 - sqrt(2)*b^3*d)*(1/(b^10*d^4))^(1/4)*log((b^6*d^2*sqrt(1/(b^10*d^4))*cos
(d*x + c) + sqrt(2)*b^3*d*sqrt(b*sin(d*x + c)/cos(d*x + c))*(1/(b^10*d^4))^(1/4)*cos(d*x + c) + b*sin(d*x + c)
)/cos(d*x + c)) + 3*(sqrt(2)*b^3*d*cos(d*x + c)^2 - sqrt(2)*b^3*d)*(1/(b^10*d^4))^(1/4)*log((b^6*d^2*sqrt(1/(b
^10*d^4))*cos(d*x + c) - sqrt(2)*b^3*d*sqrt(b*sin(d*x + c)/cos(d*x + c))*(1/(b^10*d^4))^(1/4)*cos(d*x + c) + b
*sin(d*x + c))/cos(d*x + c)))/(b^3*d*cos(d*x + c)^2 - b^3*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tan \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c))^(-5/2), x)

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maple [A]  time = 0.06, size = 184, normalized size = 0.86 \[ -\frac {2}{3 b d \left (b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {\left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (d x +c \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )}{4 d \,b^{3}}-\frac {\left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )}{2 d \,b^{3}}+\frac {\left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (-\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )}{2 d \,b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(d*x+c))^(5/2),x)

[Out]

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

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maxima [A]  time = 0.73, size = 168, normalized size = 0.79 \[ -\frac {\frac {6 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} + 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right )}{b^{\frac {3}{2}}} + \frac {6 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} - 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right )}{b^{\frac {3}{2}}} + \frac {3 \, \sqrt {2} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right )}{b^{\frac {3}{2}}} - \frac {3 \, \sqrt {2} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right )}{b^{\frac {3}{2}}} + \frac {8}{\left (b \tan \left (d x + c\right )\right )^{\frac {3}{2}}}}{12 \, b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-1/12*(6*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(b) + 2*sqrt(b*tan(d*x + c)))/sqrt(b))/b^(3/2) + 6*sqrt(2)*ar
ctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(b) - 2*sqrt(b*tan(d*x + c)))/sqrt(b))/b^(3/2) + 3*sqrt(2)*log(b*tan(d*x + c) +
sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(b) + b)/b^(3/2) - 3*sqrt(2)*log(b*tan(d*x + c) - sqrt(2)*sqrt(b*tan(d*x + c
))*sqrt(b) + b)/b^(3/2) + 8/(b*tan(d*x + c))^(3/2))/(b*d)

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mupad [B]  time = 3.07, size = 75, normalized size = 0.35 \[ -\frac {2}{3\,b\,d\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}+\frac {{\left (-1\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {b}}\right )\,1{}\mathrm {i}}{b^{5/2}\,d}+\frac {{\left (-1\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {b}}\right )\,1{}\mathrm {i}}{b^{5/2}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*tan(c + d*x))^(5/2),x)

[Out]

((-1)^(1/4)*atan(((-1)^(1/4)*(b*tan(c + d*x))^(1/2))/b^(1/2))*1i)/(b^(5/2)*d) - 2/(3*b*d*(b*tan(c + d*x))^(3/2
)) + ((-1)^(1/4)*atanh(((-1)^(1/4)*(b*tan(c + d*x))^(1/2))/b^(1/2))*1i)/(b^(5/2)*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*tan(d*x+c))**(5/2),x)

[Out]

Integral((b*tan(c + d*x))**(-5/2), x)

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