3.39 \(\int \frac {\cot (d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx\)
Optimal. Leaf size=142 \[ \frac {\tanh ^{-1}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt {a-b+c}}-\frac {\tanh ^{-1}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt {a} e} \]
[Out]
-1/2*arctanh(1/2*(2*a+b*tan(e*x+d)^2)/a^(1/2)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2))/e/a^(1/2)+1/2*arctanh(1
/2*(2*a-b+(b-2*c)*tan(e*x+d)^2)/(a-b+c)^(1/2)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2))/e/(a-b+c)^(1/2)
________________________________________________________________________________________
Rubi [A] time = 0.23, antiderivative size = 142, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used =
{3700, 1251, 960, 724, 206} \[ \frac {\tanh ^{-1}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt {a-b+c}}-\frac {\tanh ^{-1}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt {a} e} \]
Antiderivative was successfully verified.
[In]
Int[Cot[d + e*x]/Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]
[Out]
-ArcTanh[(2*a + b*Tan[d + e*x]^2)/(2*Sqrt[a]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])]/(2*Sqrt[a]*e) + A
rcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])]
/(2*Sqrt[a - b + c]*e)
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 724
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]
Rule 960
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] &&
ILtQ[n, 0])) && !(IGtQ[m, 0] || IGtQ[n, 0])
Rule 1251
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]
Rule 3700
Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.
) + (e_.)*(x_)])^(n2_.))^(p_), x_Symbol] :> Dist[f/e, Subst[Int[((x/f)^m*(a + b*x^n + c*x^(2*n))^p)/(f^2 + x^2
), x], x, f*Tan[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]
Rubi steps
\begin {align*} \int \frac {\cot (d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x \left (1+x^2\right ) \sqrt {a+b x^2+c x^4}} \, dx,x,\tan (d+e x)\right )}{e}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{x (1+x) \sqrt {a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{2 e}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{(-1-x) \sqrt {a+b x+c x^2}}+\frac {1}{x \sqrt {a+b x+c x^2}}\right ) \, dx,x,\tan ^2(d+e x)\right )}{2 e}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{(-1-x) \sqrt {a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{2 e}+\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{2 e}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b \tan ^2(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{e}-\frac {\operatorname {Subst}\left (\int \frac {1}{4 a-4 b+4 c-x^2} \, dx,x,\frac {-2 a+b-(b-2 c) \tan ^2(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{e}\\ &=-\frac {\tanh ^{-1}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt {a} e}+\frac {\tanh ^{-1}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt {a-b+c} e}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.75, size = 136, normalized size = 0.96 \[ -\frac {\frac {\tanh ^{-1}\left (\frac {2 a+b \tan ^2(d+e x)}{2 \sqrt {a} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {a}}+\frac {\tanh ^{-1}\left (\frac {-2 a-\left ((b-2 c) \tan ^2(d+e x)\right )+b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt {a-b+c}}}{2 e} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[d + e*x]/Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]
[Out]
-1/2*(ArcTanh[(2*a + b*Tan[d + e*x]^2)/(2*Sqrt[a]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])]/Sqrt[a] + Ar
cTanh[(-2*a + b - (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])]
/Sqrt[a - b + c])/e
________________________________________________________________________________________
fricas [A] time = 1.92, size = 1015, normalized size = 7.15 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="fricas")
[Out]
[1/4*(sqrt(a - b + c)*a*log(((b^2 + 4*(a - 2*b)*c + 8*c^2)*tan(e*x + d)^4 + 2*(4*a*b - 3*b^2 - 4*(a - b)*c)*ta
n(e*x + d)^2 + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(a - b
+ c) + 8*a^2 - 8*a*b + b^2 + 4*a*c)/(tan(e*x + d)^4 + 2*tan(e*x + d)^2 + 1)) + (a - b + c)*sqrt(a)*log(((b^2
+ 4*a*c)*tan(e*x + d)^4 + 8*a*b*tan(e*x + d)^2 - 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(b*tan(e*x +
d)^2 + 2*a)*sqrt(a) + 8*a^2)/tan(e*x + d)^4))/((a^2 - a*b + a*c)*e), 1/4*(2*sqrt(-a)*(a - b + c)*arctan(1/2*sq
rt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(b*tan(e*x + d)^2 + 2*a)*sqrt(-a)/(a*c*tan(e*x + d)^4 + a*b*tan(e*
x + d)^2 + a^2)) + sqrt(a - b + c)*a*log(((b^2 + 4*(a - 2*b)*c + 8*c^2)*tan(e*x + d)^4 + 2*(4*a*b - 3*b^2 - 4*
(a - b)*c)*tan(e*x + d)^2 + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a -
b)*sqrt(a - b + c) + 8*a^2 - 8*a*b + b^2 + 4*a*c)/(tan(e*x + d)^4 + 2*tan(e*x + d)^2 + 1)))/((a^2 - a*b + a*c)
*e), 1/4*(2*a*sqrt(-a + b - c)*arctan(-1/2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x +
d)^2 + 2*a - b)*sqrt(-a + b - c)/(((a - b)*c + c^2)*tan(e*x + d)^4 + (a*b - b^2 + b*c)*tan(e*x + d)^2 + a^2 -
a*b + a*c)) + (a - b + c)*sqrt(a)*log(((b^2 + 4*a*c)*tan(e*x + d)^4 + 8*a*b*tan(e*x + d)^2 - 4*sqrt(c*tan(e*x
+ d)^4 + b*tan(e*x + d)^2 + a)*(b*tan(e*x + d)^2 + 2*a)*sqrt(a) + 8*a^2)/tan(e*x + d)^4))/((a^2 - a*b + a*c)*e
), 1/2*(sqrt(-a)*(a - b + c)*arctan(1/2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(b*tan(e*x + d)^2 + 2*a)
*sqrt(-a)/(a*c*tan(e*x + d)^4 + a*b*tan(e*x + d)^2 + a^2)) + a*sqrt(-a + b - c)*arctan(-1/2*sqrt(c*tan(e*x + d
)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(-a + b - c)/(((a - b)*c + c^2)*tan(e*x +
d)^4 + (a*b - b^2 + b*c)*tan(e*x + d)^2 + a^2 - a*b + a*c)))/((a^2 - a*b + a*c)*e)]
________________________________________________________________________________________
giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep-1)]Evaluation time: 12.64Error: Bad Argument Type
________________________________________________________________________________________
maple [F] time = 1.73, size = 0, normalized size = 0.00 \[ \int \frac {\cot \left (e x +d \right )}{\sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x)
[Out]
int(cot(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x)
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot \left (e x + d\right )}{\sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(e*x+d)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="maxima")
[Out]
integrate(cot(e*x + d)/sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a), x)
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {cot}\left (d+e\,x\right )}{\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(d + e*x)/(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(1/2),x)
[Out]
int(cot(d + e*x)/(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(1/2), x)
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot {\left (d + e x \right )}}{\sqrt {a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(e*x+d)/(a+b*tan(e*x+d)**2+c*tan(e*x+d)**4)**(1/2),x)
[Out]
Integral(cot(d + e*x)/sqrt(a + b*tan(d + e*x)**2 + c*tan(d + e*x)**4), x)
________________________________________________________________________________________