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]

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))

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Rubi [A]  time = 0.150489, antiderivative size = 214, 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, 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 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}{(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*}

Mathematica [C]  time = 0.0805243, 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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Maple [A]  time = 0.017, size = 184, normalized size = 0.9 \begin{align*} -{\frac{\sqrt{2}}{4\,d{b}^{3}}\sqrt [4]{{b}^{2}}\ln \left ({ \left ( b\tan \left ( dx+c \right ) +\sqrt [4]{{b}^{2}}\sqrt{b\tan \left ( dx+c \right ) }\sqrt{2}+\sqrt{{b}^{2}} \right ) \left ( b\tan \left ( dx+c \right ) -\sqrt [4]{{b}^{2}}\sqrt{b\tan \left ( dx+c \right ) }\sqrt{2}+\sqrt{{b}^{2}} \right ) ^{-1}} \right ) }-{\frac{\sqrt{2}}{2\,d{b}^{3}}\sqrt [4]{{b}^{2}}\arctan \left ({\sqrt{2}\sqrt{b\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{b}^{2}}}}}+1 \right ) }+{\frac{\sqrt{2}}{2\,d{b}^{3}}\sqrt [4]{{b}^{2}}\arctan \left ( -{\sqrt{2}\sqrt{b\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [4]{{b}^{2}}}}}+1 \right ) }-{\frac{2}{3\,bd} \left ( b\tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.82421, size = 1717, normalized size = 8.02 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \tan{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

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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Giac [A]  time = 1.48354, size = 288, normalized size = 1.35 \begin{align*} -\frac{1}{12} \, b{\left (\frac{6 \, \sqrt{2} \sqrt{{\left | b \right |}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | b \right |}} + 2 \, \sqrt{b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt{{\left | b \right |}}}\right )}{b^{4} d} + \frac{6 \, \sqrt{2} \sqrt{{\left | b \right |}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \sqrt{{\left | b \right |}} - 2 \, \sqrt{b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt{{\left | b \right |}}}\right )}{b^{4} d} + \frac{3 \, \sqrt{2} \sqrt{{\left | b \right |}} \log \left (b \tan \left (d x + c\right ) + \sqrt{2} \sqrt{b \tan \left (d x + c\right )} \sqrt{{\left | b \right |}} +{\left | b \right |}\right )}{b^{4} d} - \frac{3 \, \sqrt{2} \sqrt{{\left | b \right |}} \log \left (b \tan \left (d x + c\right ) - \sqrt{2} \sqrt{b \tan \left (d x + c\right )} \sqrt{{\left | b \right |}} +{\left | b \right |}\right )}{b^{4} d} + \frac{8}{\sqrt{b \tan \left (d x + c\right )} b^{3} d \tan \left (d x + c\right )}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*b*(6*sqrt(2)*sqrt(abs(b))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) + 2*sqrt(b*tan(d*x + c)))/sqrt(abs(b)
))/(b^4*d) + 6*sqrt(2)*sqrt(abs(b))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(b)) - 2*sqrt(b*tan(d*x + c)))/sqrt(a
bs(b)))/(b^4*d) + 3*sqrt(2)*sqrt(abs(b))*log(b*tan(d*x + c) + sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(abs(b)) + abs(
b))/(b^4*d) - 3*sqrt(2)*sqrt(abs(b))*log(b*tan(d*x + c) - sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(abs(b)) + abs(b))/
(b^4*d) + 8/(sqrt(b*tan(d*x + c))*b^3*d*tan(d*x + c)))