3.40 \(\int \frac{\cot ^3(d+e x)}{\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}} \, dx\)

Optimal. Leaf size=249 \[ \frac{b \tanh ^{-1}\left (\frac{2 a+b \tan ^2(d+e x)}{2 \sqrt{a} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{4 a^{3/2} e}+\frac{\tanh ^{-1}\left (\frac{2 a+b \tan ^2(d+e x)}{2 \sqrt{a} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt{a} e}-\frac{\tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt{a-b+c}}-\frac{\cot ^2(d+e x) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{2 a e} \]

[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) + (b
*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])])/(4*a^(3/2)*e) -
ArcTanh[(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) - (Cot[d + e*x]^2*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])/(2*a*e)

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Rubi [A]  time = 0.317719, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {3700, 1251, 960, 730, 724, 206} \[ \frac{b \tanh ^{-1}\left (\frac{2 a+b \tan ^2(d+e x)}{2 \sqrt{a} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{4 a^{3/2} e}+\frac{\tanh ^{-1}\left (\frac{2 a+b \tan ^2(d+e x)}{2 \sqrt{a} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 \sqrt{a} e}-\frac{\tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e \sqrt{a-b+c}}-\frac{\cot ^2(d+e x) \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{2 a e} \]

Antiderivative was successfully verified.

[In]

Int[Cot[d + e*x]^3/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) + (b
*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])])/(4*a^(3/2)*e) -
ArcTanh[(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) - (Cot[d + e*x]^2*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])/(2*a*e)

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]

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 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 730

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)
*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e
^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c,
0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 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 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])

Rubi steps

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

Mathematica [A]  time = 4.13599, size = 188, normalized size = 0.76 \[ \frac{(2 a+b) \tanh ^{-1}\left (\frac{2 a+b \tan ^2(d+e x)}{2 \sqrt{a} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )+2 \sqrt{a} \left (\cot ^2(d+e x) \left (-\sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}\right )-\frac{a \tanh ^{-1}\left (\frac{2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt{a-b+c} \sqrt{a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{\sqrt{a-b+c}}\right )}{4 a^{3/2} e} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[d + e*x]^3/Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]

[Out]

((2*a + b)*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*Sqr
t[a]*(-((a*ArcTanh[(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]) - Cot[d + e*x]^2*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4]))/(4*a^(3/2)*e)

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Maple [F]  time = 0.545, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cot \left ( ex+d \right ) \right ) ^{3}{\frac{1}{\sqrt{a+b \left ( \tan \left ( ex+d \right ) \right ) ^{2}+c \left ( \tan \left ( ex+d \right ) \right ) ^{4}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e*x+d)^3/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x)

[Out]

int(cot(e*x+d)^3/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x)

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Maxima [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(e*x+d)^3/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 16.0172, size = 3267, normalized size = 13.12 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(e*x+d)^3/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(2*sqrt(a - b + c)*a^2*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))*tan(e*x + d)^2 + (2*a^2 - a*
b - b^2 + (2*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)*tan(e*x + d)^2 - 4*sqrt(
c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(a^2 - a*b + a*c))/((a^3 - a^2*b + a^2*c)*e*tan(e*x + d)^2), 1/4*(sqr
t(a - b + c)*a^2*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))*tan(e*x + d)^2 - (2*a^2 - a*b - b^2 + (
2*a + b)*c)*sqrt(-a)*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))*tan(e*x + d)^2 - 2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2
+ a)*(a^2 - a*b + a*c))/((a^3 - a^2*b + a^2*c)*e*tan(e*x + d)^2), -1/8*(4*a^2*sqrt(-a + b - c)*arctan(-1/2*sq
rt(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))*tan(e*x + d)^2 - (2*a^2 - a*b - b
^2 + (2*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)*tan(e*x + d)^2 + 4*sqrt(c*tan
(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(a^2 - a*b + a*c))/((a^3 - a^2*b + a^2*c)*e*tan(e*x + d)^2), -1/4*(2*a^2*s
qrt(-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))*tan
(e*x + d)^2 + (2*a^2 - a*b - b^2 + (2*a + b)*c)*sqrt(-a)*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))*tan(e*x + d)^2 + 2*sqrt
(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(a^2 - a*b + a*c))/((a^3 - a^2*b + a^2*c)*e*tan(e*x + d)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{3}{\left (d + e x \right )}}{\sqrt{a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(e*x+d)**3/(a+b*tan(e*x+d)**2+c*tan(e*x+d)**4)**(1/2),x)

[Out]

Integral(cot(d + e*x)**3/sqrt(a + b*tan(d + e*x)**2 + c*tan(d + e*x)**4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (e x + d\right )^{3}}{\sqrt{c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(e*x+d)^3/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2),x, algorithm="giac")

[Out]

integrate(cot(e*x + d)^3/sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a), x)