3.804 \(\int \frac{a+b \tan (c+d x)}{\cot ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=202 \[ -\frac{(a+b) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{(a+b) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{(a-b) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{(a-b) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 a}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 b}{5 d \cot ^{\frac{5}{2}}(c+d x)}-\frac{2 b}{d \sqrt{\cot (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.178445, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3673, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{(a+b) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{(a+b) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{(a-b) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{(a-b) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 a}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 b}{5 d \cot ^{\frac{5}{2}}(c+d x)}-\frac{2 b}{d \sqrt{\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])/Cot[c + d*x]^(5/2),x]

[Out]

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

Rule 3673

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

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

Mathematica [C]  time = 0.849819, size = 207, normalized size = 1.02 \[ \frac{\sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \left (3 b \left (8 \tan ^{\frac{5}{2}}(c+d x)-10 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )+10 \sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )-40 \sqrt{\tan (c+d x)}-5 \sqrt{2} \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )+5 \sqrt{2} \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )\right )-40 a \tan ^{\frac{3}{2}}(c+d x) \left (\, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\tan ^2(c+d x)\right )-1\right )\right )}{60 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])/Cot[c + d*x]^(5/2),x]

[Out]

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-40*a*(-1 + Hypergeometric2F1[3/4, 1, 7/4, -Tan[c + d*x]^2])*Tan[c + d
*x]^(3/2) + 3*b*(-10*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] + 10*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c
+ d*x]]] - 5*Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + 5*Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[
c + d*x]] + Tan[c + d*x]] - 40*Sqrt[Tan[c + d*x]] + 8*Tan[c + d*x]^(5/2))))/(60*d)

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Maple [C]  time = 0.257, size = 1254, normalized size = 6.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30/d*2^(1/2)*(cos(d*x+c)-1)*(-15*I*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/s
in(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi(((1-cos(d*x+c)+sin(d*x+c
))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*a+15*I*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((1-cos(d*x+c)
+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi(((1-cos(d*
x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*b+15*I*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*
((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*Ellipti
cPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a-15*I*sin(d*x+c)*((cos(d*x+c)-1)/sin(
d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*
x+c)^2*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*b-15*sin(d*x+c)*((cos(d*
x+c)-1)/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^
(1/2)*cos(d*x+c)^2*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*a-15*sin(d*x
+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/
sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*
b-15*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+
sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1
/2*2^(1/2))*a-15*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((c
os(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)
,1/2-1/2*I,1/2*2^(1/2))*b+30*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c
))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*
x+c))^(1/2),1/2*2^(1/2))*b+10*sin(d*x+c)*2^(1/2)*cos(d*x+c)^2*a-36*2^(1/2)*cos(d*x+c)^3*b-10*sin(d*x+c)*cos(d*
x+c)*2^(1/2)*a+36*2^(1/2)*cos(d*x+c)^2*b+6*cos(d*x+c)*2^(1/2)*b-6*2^(1/2)*b)*(cos(d*x+c)+1)^2/(cos(d*x+c)/sin(
d*x+c))^(5/2)/sin(d*x+c)^6

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Maxima [A]  time = 1.71183, size = 223, normalized size = 1.1 \begin{align*} \frac{8 \,{\left (3 \, b + \frac{5 \, a}{\tan \left (d x + c\right )} - \frac{15 \, b}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac{5}{2}} + 30 \, \sqrt{2}{\left (a - b\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + 30 \, \sqrt{2}{\left (a - b\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + 15 \, \sqrt{2}{\left (a + b\right )} \log \left (\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right ) - 15 \, \sqrt{2}{\left (a + b\right )} \log \left (-\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(8*(3*b + 5*a/tan(d*x + c) - 15*b/tan(d*x + c)^2)*tan(d*x + c)^(5/2) + 30*sqrt(2)*(a - b)*arctan(1/2*sqrt
(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 30*sqrt(2)*(a - b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c))
)) + 15*sqrt(2)*(a + b)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - 15*sqrt(2)*(a + b)*log(-sqrt(2)
/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))/d

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

[Out]

Timed out

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Sympy [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((a+b*tan(d*x+c))/cot(d*x+c)**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(d*x + c) + a)/cot(d*x + c)^(5/2), x)