3.4.9 \(\int \cot (e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\) [309]

Optimal. Leaf size=95 \[ -\frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}+\frac {b \sqrt {a+b \tan ^2(e+f x)}}{f} \]

[Out]

-a^(3/2)*arctanh((a+b*tan(f*x+e)^2)^(1/2)/a^(1/2))/f+(a-b)^(3/2)*arctanh((a+b*tan(f*x+e)^2)^(1/2)/(a-b)^(1/2))
/f+b*(a+b*tan(f*x+e)^2)^(1/2)/f

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Rubi [A]
time = 0.09, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3751, 457, 86, 162, 65, 214} \begin {gather*} -\frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {b \sqrt {a+b \tan ^2(e+f x)}}{f}+\frac {(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^(3/2)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]])/f) + ((a - b)^(3/2)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]
/Sqrt[a - b]])/f + (b*Sqrt[a + b*Tan[e + f*x]^2])/f

Rule 65

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 86

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

Rule 162

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 214

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

Rule 457

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 3751

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]))

Rubi steps

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

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Mathematica [A]
time = 0.17, size = 90, normalized size = 0.95 \begin {gather*} \frac {-a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )+(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )+b \sqrt {a+b \tan ^2(e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1764\) vs. \(2(81)=162\).
time = 0.30, size = 1765, normalized size = 18.58

method	result	size
default	\(\text {Expression too large to display}\)	\(1765\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/f*(cos(f*x+e)-1)^3*(2*cos(f*x+e)*ln(4*(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*(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))*a^(9/2)-4*cos(f*x+e)*ln(4*(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*(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))
*a^(7/2)*b+2*cos(f*x+e)*ln(4*(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*(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))*a^(5/
2)*b^2+2*cos(f*x+e)*a^(5/2)*(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)*b+2*b*((a*c
os(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(5/2)*(a-b)^(1/2)+3*cos(f*x+e)*ln(-4*(cos(f*x+e)-1)*(c
os(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(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/a^(1/2))*(a-b)^(1/2)*a^3*b-6*cos(f
*x+e)*ln(-4*(cos(f*x+e)-1)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(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/a^(1/
2))*(a-b)^(1/2)*a^2*b^2+3*cos(f*x+e)*ln(-4*(cos(f*x+e)-1)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+
b)/(cos(f*x+e)+1)^2)^(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/a^(1/2))*(a-b)^(1/2)*a*b^3+cos(f*x+e)*ln(-2*(cos(f*x+e)-1)*(cos(f*x+e)*a^(1/2)*((a*c
os(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(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/a^(1/2))*(a-b)^(1/2)*a^4-3*cos(f*x+e)*ln(-2*(cos(f*x+e)-1
)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(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/a^(1/2))*(a-b)^(1/2)*a^3*b+6*c
os(f*x+e)*ln(-2*(cos(f*x+e)-1)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(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/a
^(1/2))*(a-b)^(1/2)*a^2*b^2-3*cos(f*x+e)*ln(-2*(cos(f*x+e)-1)*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^
2*b+b)/(cos(f*x+e)+1)^2)^(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/a^(1/2))*(a-b)^(1/2)*a*b^3-cos(f*x+e)*ln(-4*(cos(f*x+e)*a^(1/2)*((a*cos(f*x+e)^2
-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)+((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)+b)/(cos(f*x+e)-1))*(a-b)^(1/2)*a^4)*cos(f*x+e)^2*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/
cos(f*x+e)^2)^(3/2)*4^(1/2)/sin(f*x+e)^6/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)/a^(5/2)/(a
-b)^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 5.92, size = 619, normalized size = 6.52 \begin {gather*} \left [-\frac {{\left (a - b\right )}^{\frac {3}{2}} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, a^{\frac {3}{2}} \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 ) - 4 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b}{4 \, f}, \frac {4 \, \sqrt {-a} a \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a}}{a}\right ) - {\left (a - b\right )}^{\frac {3}{2}} \log \left (-\frac {b^{2} \tan \left (f x + e\right )^{4} + 2 \, {\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \, {\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b}{4 \, f}, \frac {{\left (-a + b\right )}^{\frac {3}{2}} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + a^{\frac {3}{2}} \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 ) + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b}{2 \, f}, \frac {2 \, \sqrt {-a} a \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a}}{a}\right ) + {\left (-a + b\right )}^{\frac {3}{2}} \arctan \left (\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b}{2 \, f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((a - b)^(3/2)*log(-(b^2*tan(f*x + e)^4 + 2*(4*a*b - 3*b^2)*tan(f*x + e)^2 - 4*(b*tan(f*x + e)^2 + 2*a -
b)*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b) + 8*a^2 - 8*a*b + b^2)/(tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1)) - 2
*a^(3/2)*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) - 4*sqrt(b*tan(f*
x + e)^2 + a)*b)/f, 1/4*(4*sqrt(-a)*a*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a) - (a - b)^(3/2)*log(-(b^2*
tan(f*x + e)^4 + 2*(4*a*b - 3*b^2)*tan(f*x + e)^2 - 4*(b*tan(f*x + e)^2 + 2*a - b)*sqrt(b*tan(f*x + e)^2 + a)*
sqrt(a - b) + 8*a^2 - 8*a*b + b^2)/(tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1)) + 4*sqrt(b*tan(f*x + e)^2 + a)*b)/
f, 1/2*((-a + b)^(3/2)*arctan(2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(b*tan(f*x + e)^2 + 2*a - b)) + a^(3/2
)*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) + 2*sqrt(b*tan(f*x + e)^
2 + a)*b)/f, 1/2*(2*sqrt(-a)*a*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a) + (-a + b)^(3/2)*arctan(2*sqrt(b*
tan(f*x + e)^2 + a)*sqrt(-a + b)/(b*tan(f*x + e)^2 + 2*a - b)) + 2*sqrt(b*tan(f*x + e)^2 + a)*b)/f]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \cot {\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(t_

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Mupad [B]
time = 12.03, size = 546, normalized size = 5.75 \begin {gather*} \frac {b\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{f}+\frac {\mathrm {atanh}\left (\frac {6\,a^3\,b^3\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3-3\,a^2\,b+3\,a\,b^2-b^3}}{6\,a^5\,b^3-18\,a^4\,b^4+20\,a^3\,b^5-10\,a^2\,b^6+2\,a\,b^7}-\frac {6\,a^2\,b^4\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3-3\,a^2\,b+3\,a\,b^2-b^3}}{6\,a^5\,b^3-18\,a^4\,b^4+20\,a^3\,b^5-10\,a^2\,b^6+2\,a\,b^7}+\frac {2\,a\,b^5\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3-3\,a^2\,b+3\,a\,b^2-b^3}}{6\,a^5\,b^3-18\,a^4\,b^4+20\,a^3\,b^5-10\,a^2\,b^6+2\,a\,b^7}\right )\,\sqrt {{\left (a-b\right )}^3}}{f}-\frac {\mathrm {atanh}\left (\frac {2\,b^6\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3}}{-6\,a^5\,b^3+12\,a^4\,b^4-8\,a^3\,b^5+2\,a^2\,b^6}-\frac {8\,a\,b^5\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3}}{-6\,a^5\,b^3+12\,a^4\,b^4-8\,a^3\,b^5+2\,a^2\,b^6}+\frac {12\,a^2\,b^4\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3}}{-6\,a^5\,b^3+12\,a^4\,b^4-8\,a^3\,b^5+2\,a^2\,b^6}-\frac {6\,a^3\,b^3\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}\,\sqrt {a^3}}{-6\,a^5\,b^3+12\,a^4\,b^4-8\,a^3\,b^5+2\,a^2\,b^6}\right )\,\sqrt {a^3}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*(a + b*tan(e + f*x)^2)^(1/2))/f + (atanh((6*a^3*b^3*(a + b*tan(e + f*x)^2)^(1/2)*(3*a*b^2 - 3*a^2*b + a^3 -
b^3)^(1/2))/(2*a*b^7 - 10*a^2*b^6 + 20*a^3*b^5 - 18*a^4*b^4 + 6*a^5*b^3) - (6*a^2*b^4*(a + b*tan(e + f*x)^2)^
(1/2)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)^(1/2))/(2*a*b^7 - 10*a^2*b^6 + 20*a^3*b^5 - 18*a^4*b^4 + 6*a^5*b^3) + (2
*a*b^5*(a + b*tan(e + f*x)^2)^(1/2)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)^(1/2))/(2*a*b^7 - 10*a^2*b^6 + 20*a^3*b^5
- 18*a^4*b^4 + 6*a^5*b^3))*((a - b)^3)^(1/2))/f - (atanh((2*b^6*(a + b*tan(e + f*x)^2)^(1/2)*(a^3)^(1/2))/(2*a
^2*b^6 - 8*a^3*b^5 + 12*a^4*b^4 - 6*a^5*b^3) - (8*a*b^5*(a + b*tan(e + f*x)^2)^(1/2)*(a^3)^(1/2))/(2*a^2*b^6 -
8*a^3*b^5 + 12*a^4*b^4 - 6*a^5*b^3) + (12*a^2*b^4*(a + b*tan(e + f*x)^2)^(1/2)*(a^3)^(1/2))/(2*a^2*b^6 - 8*a^
3*b^5 + 12*a^4*b^4 - 6*a^5*b^3) - (6*a^3*b^3*(a + b*tan(e + f*x)^2)^(1/2)*(a^3)^(1/2))/(2*a^2*b^6 - 8*a^3*b^5
+ 12*a^4*b^4 - 6*a^5*b^3))*(a^3)^(1/2))/f

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