3.297 \(\int \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=115 \[ \frac{(2 a-b) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a}}\right )}{2 \sqrt{a} f}-\frac{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f}-\frac{\cot ^2(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{2 f} \]

[Out]

((2*a - b)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]])/(2*Sqrt[a]*f) - (Sqrt[a - b]*ArcTanh[Sqrt[a + b*Tan[e
+ f*x]^2]/Sqrt[a - b]])/f - (Cot[e + f*x]^2*Sqrt[a + b*Tan[e + f*x]^2])/(2*f)

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Rubi [A]  time = 0.14774, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3670, 446, 99, 156, 63, 208} \[ \frac{(2 a-b) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a}}\right )}{2 \sqrt{a} f}-\frac{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f}-\frac{\cot ^2(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

((2*a - b)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]])/(2*Sqrt[a]*f) - (Sqrt[a - b]*ArcTanh[Sqrt[a + b*Tan[e
+ f*x]^2]/Sqrt[a - b]])/f - (Cot[e + f*x]^2*Sqrt[a + b*Tan[e + f*x]^2])/(2*f)

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 99

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

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.360486, size = 115, normalized size = 1. \[ \frac{(2 a-b) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a}}\right )-\sqrt{a} \left (2 \sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )+\cot ^2(e+f x) \sqrt{a+b \tan ^2(e+f x)}\right )}{2 \sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

((2*a - b)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]] - Sqrt[a]*(2*Sqrt[a - b]*ArcTanh[Sqrt[a + b*Tan[e + f*x
]^2]/Sqrt[a - b]] + Cot[e + f*x]^2*Sqrt[a + b*Tan[e + f*x]^2]))/(2*Sqrt[a]*f)

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Maple [B]  time = 0.319, size = 2135, normalized size = 18.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/f/a^(3/2)/(a-b)^(1/2)*(cos(f*x+e)-1)*(4*a^(5/2)*cos(f*x+e)^2*ln(4*cos(f*x+e)*(a-b)^(1/2)*((a*cos(f*x+e)^2
-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)+4*cos(f*x+e)*a-4*b*cos(f*x+e)+4*(a-b)^(1/2)*((a*cos(f*x+e)^2-cos(f*
x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2))*4^(1/2)+2*a^(3/2)*(a-b)^(1/2)*cos(f*x+e)^2*((a*cos(f*x+e)^2-cos(f*x+e)^2*
b+b)/(cos(f*x+e)+1)^2)^(1/2)*4^(1/2)-4*a^(3/2)*cos(f*x+e)^2*ln(4*cos(f*x+e)*(a-b)^(1/2)*((a*cos(f*x+e)^2-cos(f
*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)+4*cos(f*x+e)*a-4*b*cos(f*x+e)+4*(a-b)^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2
*b+b)/(cos(f*x+e)+1)^2)^(1/2))*4^(1/2)*b-4*a^(5/2)*ln(4*cos(f*x+e)*(a-b)^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b
+b)/(cos(f*x+e)+1)^2)^(1/2)+4*cos(f*x+e)*a-4*b*cos(f*x+e)+4*(a-b)^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(co
s(f*x+e)+1)^2)^(1/2))*4^(1/2)-2*a^(3/2)*(a-b)^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+
1)^2)^(1/2)*4^(1/2)-8*a^(1/2)*(a-b)^(1/2)*cos(f*x+e)^2*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3
/2)-4*a^(1/2)*(a-b)^(1/2)*cos(f*x+e)^2*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*4^(1/2)*b+2*
(a-b)^(1/2)*cos(f*x+e)^2*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+
e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^
(1/2)+b)/sin(f*x+e)^2)*4^(1/2)*a^2-(a-b)^(1/2)*cos(f*x+e)^2*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f
*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e
)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^(1/2)*a*b-2*(a-b)^(1/2)*cos(f*x+e)^2*ln(-4*(cos(f*
x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+
e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*4^(1/2)*a^2+(a-b)^(1/2)*cos(f*x+e)^2
*ln(-4*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f*x+e)*a-b*cos(f*x+e
)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*4^(1/2)*a*b+4*a^(3/2)*
ln(4*cos(f*x+e)*(a-b)^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)+4*cos(f*x+e)*a-4*b*cos(
f*x+e)+4*(a-b)^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2))*4^(1/2)*b-16*a^(1/2)*(a-b)^(1
/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)-8*a^(1/2)*(a-b)^(1/2)*((a*cos(f*x+e)
^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)+4*a^(1/2)*(a-b)^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x
+e)+1)^2)^(1/2)*4^(1/2)*b-2*(a-b)^(1/2)*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2
*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)
+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^(1/2)*a^2+(a-b)^(1/2)*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos
(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x
+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*4^(1/2)*a*b+2*(a-b)^(1/2)*ln(-4*(cos(f*x+e)*((a*co
s(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*
x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*4^(1/2)*a^2-(a-b)^(1/2)*ln(-4*(cos(f*x+e)*((a*c
os(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f
*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*4^(1/2)*a*b)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f
*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)/sin(f*x+e)^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*tan(f*x + e)^2 + a)*cot(f*x + e)^3, x)

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Fricas [A]  time = 1.79282, size = 1467, normalized size = 12.76 \begin{align*} \left [\frac{2 \, \sqrt{a - b} a \log \left (\frac{b \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{a - b} + 2 \, a - b}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{2} -{\left (2 \, a - b\right )} \sqrt{a} \log \left (\frac{b \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{a} + 2 \, a}{\tan \left (f x + e\right )^{2}}\right ) \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} a}{4 \, a f \tan \left (f x + e\right )^{2}}, -\frac{4 \, a \sqrt{-a + b} \arctan \left (-\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b}}{a - b}\right ) \tan \left (f x + e\right )^{2} +{\left (2 \, a - b\right )} \sqrt{a} \log \left (\frac{b \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{a} + 2 \, a}{\tan \left (f x + e\right )^{2}}\right ) \tan \left (f x + e\right )^{2} + 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} a}{4 \, a f \tan \left (f x + e\right )^{2}}, -\frac{\sqrt{-a}{\left (2 \, a - b\right )} \arctan \left (\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a}}{a}\right ) \tan \left (f x + e\right )^{2} - \sqrt{a - b} a \log \left (\frac{b \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{a - b} + 2 \, a - b}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{2} + \sqrt{b \tan \left (f x + e\right )^{2} + a} a}{2 \, a f \tan \left (f x + e\right )^{2}}, -\frac{\sqrt{-a}{\left (2 \, a - b\right )} \arctan \left (\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a}}{a}\right ) \tan \left (f x + e\right )^{2} + 2 \, a \sqrt{-a + b} \arctan \left (-\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b}}{a - b}\right ) \tan \left (f x + e\right )^{2} + \sqrt{b \tan \left (f x + e\right )^{2} + a} a}{2 \, a f \tan \left (f x + e\right )^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*sqrt(a - b)*a*log((b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b) + 2*a - b)/(tan(f*x + e
)^2 + 1))*tan(f*x + e)^2 - (2*a - b)*sqrt(a)*log((b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a) + 2*
a)/tan(f*x + e)^2)*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*a)/(a*f*tan(f*x + e)^2), -1/4*(4*a*sqrt(-a +
b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(a - b))*tan(f*x + e)^2 + (2*a - b)*sqrt(a)*log((b*tan(f*x
+ e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a) + 2*a)/tan(f*x + e)^2)*tan(f*x + e)^2 + 2*sqrt(b*tan(f*x + e)^2
+ a)*a)/(a*f*tan(f*x + e)^2), -1/2*(sqrt(-a)*(2*a - b)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a)*tan(f*x +
e)^2 - sqrt(a - b)*a*log((b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b) + 2*a - b)/(tan(f*x + e
)^2 + 1))*tan(f*x + e)^2 + sqrt(b*tan(f*x + e)^2 + a)*a)/(a*f*tan(f*x + e)^2), -1/2*(sqrt(-a)*(2*a - b)*arctan
(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a)*tan(f*x + e)^2 + 2*a*sqrt(-a + b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)*s
qrt(-a + b)/(a - b))*tan(f*x + e)^2 + sqrt(b*tan(f*x + e)^2 + a)*a)/(a*f*tan(f*x + e)^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*tan(e + f*x)**2)*cot(e + f*x)**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*tan(f*x + e)^2 + a)*cot(f*x + e)^3, x)