3.799 \(\int \cot ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x)) \, dx\)
Optimal. Leaf size=184 \[ \frac{(a-b) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{(a-b) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{(a+b) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{(a+b) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d}-\frac{2 a \cot ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{2 b \sqrt{\cot (c+d x)}}{d} \]
[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*Sqrt[Cot[c + d*x]])/d - (2*a*Cot[c + d*x]^(3/2))/(3*d) + ((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]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(
2*Sqrt[2]*d)
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Rubi [A] time = 0.160141, antiderivative size = 184, normalized size of antiderivative = 1.,
number of steps used = 13, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used
= {3673, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{(a-b) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{(a-b) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{(a+b) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}+\frac{(a+b) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d}-\frac{2 a \cot ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{2 b \sqrt{\cot (c+d x)}}{d} \]
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]^(5/2)*(a + b*Tan[c + d*x]),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*Sqrt[Cot[c + d*x]])/d - (2*a*Cot[c + d*x]^(3/2))/(3*d) + ((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]*Sqrt[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 3528
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
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 \cot ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x)) \, dx &=\int \cot ^{\frac{3}{2}}(c+d x) (b+a \cot (c+d x)) \, dx\\ &=-\frac{2 a \cot ^{\frac{3}{2}}(c+d x)}{3 d}+\int \sqrt{\cot (c+d x)} (-a+b \cot (c+d x)) \, dx\\ &=-\frac{2 b \sqrt{\cot (c+d x)}}{d}-\frac{2 a \cot ^{\frac{3}{2}}(c+d x)}{3 d}+\int \frac{-b-a \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{2 b \sqrt{\cot (c+d x)}}{d}-\frac{2 a \cot ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 \operatorname{Subst}\left (\int \frac{b+a x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=-\frac{2 b \sqrt{\cot (c+d x)}}{d}-\frac{2 a \cot ^{\frac{3}{2}}(c+d x)}{3 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{(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 \sqrt{\cot (c+d x)}}{d}-\frac{2 a \cot ^{\frac{3}{2}}(c+d x)}{3 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{(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{2 b \sqrt{\cot (c+d x)}}{d}-\frac{2 a \cot ^{\frac{3}{2}}(c+d x)}{3 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) \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 \sqrt{\cot (c+d x)}}{d}-\frac{2 a \cot ^{\frac{3}{2}}(c+d x)}{3 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) \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt{2} d}\\ \end{align*}
Mathematica [C] time = 0.14224, size = 65, normalized size = 0.35 \[ -\frac{2 \sqrt{\cot (c+d x)} \left (a \cot (c+d x) \, _2F_1\left (-\frac{3}{4},1;\frac{1}{4};-\tan ^2(c+d x)\right )+3 b \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};-\tan ^2(c+d x)\right )\right )}{3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[c + d*x]^(5/2)*(a + b*Tan[c + d*x]),x]
[Out]
(-2*Sqrt[Cot[c + d*x]]*(a*Cot[c + d*x]*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[c + d*x]^2] + 3*b*Hypergeometric2F
1[-1/4, 1, 3/4, -Tan[c + d*x]^2]))/(3*d)
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Maple [C] time = 0.258, size = 2217, normalized size = 12.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(d*x+c)^(5/2)*(a+b*tan(d*x+c)),x)
[Out]
-1/6/d*2^(1/2)*(3*I*sin(d*x+c)*((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)-1)/sin(d*x+c))^(1/2)*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+3*I*sin(d*x+c)*cos(d*x+c)*((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)-1)/sin(d*x+c))^(1/2)*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-3*I*sin(d*x+c)*((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)-1)/sin(d*x+c))^(1/2)*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-3*I*sin(d*x+c)*cos(d*x+c)*((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)-1)/sin(d*x+c))^(1/2)*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-3*I*sin(d*x+c)*((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)-1)/sin(d*x+c))^(1/2)*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-3*I*sin(d*x+c)*cos(d*x+c)*((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)-1)/sin(d*x+c))^(1/2)*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+3*I*sin(d*x+c)*((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)-1)/sin(d*x+c))^(1/2)*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+3*I*sin(d*x+c)*cos(d*x+c)*((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)-1)/sin(d*x+c))^(1/2)*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+3*sin(d*x+c)*cos(d*x+c)*((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)-1)/sin(d*x+c))^(1/2)*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-3*sin(d*x+c)*cos(d*x+c)*((1-cos(d*x+c)+si
n(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)-1)/sin(d*x+c))^(1/2)*Ell
ipticPi(((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+3*sin(d*x+c)*cos(d*x+c)*((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)-1)/sin(d*x+c))^
(1/2)*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-3*sin(d*x+c)*cos(d*x+c)
*((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)-1)/sin
(d*x+c))^(1/2)*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-6*sin(d*x+c)*c
os(d*x+c)*((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)-1)/sin(d*x+c))^(1/2)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*a+3*sin(d*x+c)*((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)-1)/sin(d*x
+c))^(1/2)*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-3*sin(d*x+c)*((1-c
os(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)-1)/sin(d*x+c
))^(1/2)*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+3*sin(d*x+c)*((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)-1)/sin(d*x+c))
^(1/2)*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-3*sin(d*x+c)*((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)-1)/sin(d*x+c))^(
1/2)*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-6*sin(d*x+c)*((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)-1)/sin(d*x+c))^(1/
2)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*a+6*2^(1/2)*sin(d*x+c)*cos(d*x+c)*b+2*2
^(1/2)*cos(d*x+c)^2*a)*sin(d*x+c)*(cos(d*x+c)/sin(d*x+c))^(5/2)/cos(d*x+c)^3
________________________________________________________________________________________
Maxima [A] time = 1.74936, size = 204, normalized size = 1.11 \begin{align*} \frac{6 \, \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 ) + 6 \, \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 ) - 3 \, \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 ) + 3 \, \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 ) - \frac{24 \, b}{\sqrt{\tan \left (d x + c\right )}} - \frac{8 \, a}{\tan \left (d x + c\right )^{\frac{3}{2}}}}{12 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(5/2)*(a+b*tan(d*x+c)),x, algorithm="maxima")
[Out]
1/12*(6*sqrt(2)*(a + b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 6*sqrt(2)*(a + b)*arctan(-1/2*s
qrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) - 3*sqrt(2)*(a - b)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) +
1) + 3*sqrt(2)*(a - b)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - 24*b/sqrt(tan(d*x + c)) - 8*a/
tan(d*x + c)^(3/2))/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(cot(d*x+c)^(5/2)*(a+b*tan(d*x+c)),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(cot(d*x+c)**(5/2)*(a+b*tan(d*x+c)),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )} \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(cot(d*x+c)^(5/2)*(a+b*tan(d*x+c)),x, algorithm="giac")
[Out]
integrate((b*tan(d*x + c) + a)*cot(d*x + c)^(5/2), x)