∫ cot5(e + fx)(a + b tan2(e + fx))3∕2 dx

3.311 \(\int \cot ^5(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=161 \[ -\frac {\left (8 a^2-12 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{8 \sqrt {a} f}+\frac {(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}-\frac {a \cot ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}+\frac {(4 a-5 b) \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f} \]

[Out]

(a-b)^(3/2)*arctanh((a+b*tan(f*x+e)^2)^(1/2)/(a-b)^(1/2))/f-1/8*(8*a^2-12*a*b+3*b^2)*arctanh((a+b*tan(f*x+e)^2

)^(1/2)/a^(1/2))/f/a^(1/2)+1/8*(4*a-5*b)*cot(f*x+e)^2*(a+b*tan(f*x+e)^2)^(1/2)/f-1/4*a*cot(f*x+e)^4*(a+b*tan(f

*x+e)^2)^(1/2)/f

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Rubi [A] time = 0.22, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3670, 446, 98, 151, 156, 63, 208} \[ -\frac {\left (8 a^2-12 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{8 \sqrt {a} f}+\frac {(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}-\frac {a \cot ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}+\frac {(4 a-5 b) \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f} \]

Antiderivative was successfully veriﬁed.

[In]

Int[Cot[e + f*x]^5*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

-((8*a^2 - 12*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]])/(8*Sqrt[a]*f) + ((a - b)^(3/2)*ArcTanh

[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]])/f + ((4*a - 5*b)*Cot[e + f*x]^2*Sqrt[a + b*Tan[e + f*x]^2])/(8*f) -

(a*Cot[e + f*x]^4*Sqrt[a + b*Tan[e + f*x]^2])/(4*f)

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 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -

a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*

f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d

*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;

FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p

] || IntegersQ[p, m + n])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb

ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*

f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*

g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p

+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

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 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]

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 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]))

Rubi steps

\begin {align*} \int \cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^{3/2}}{x^5 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^3 (1+x)} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac {a \cot ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} a (4 a-5 b)+\frac {1}{2} (3 a-4 b) b x}{x^2 (1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{4 f}\\ &=\frac {(4 a-5 b) \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}-\frac {a \cot ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{4} a \left (8 a^2-12 a b+3 b^2\right )+\frac {1}{4} a (4 a-5 b) b x}{x (1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{4 a f}\\ &=\frac {(4 a-5 b) \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}-\frac {a \cot ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}-\frac {(a-b)^2 \operatorname {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 f}+\frac {\left (8 a^2-12 a b+3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{16 f}\\ &=\frac {(4 a-5 b) \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}-\frac {a \cot ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}-\frac {(a-b)^2 \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 {\left (8 a^2-12 a b+3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \tan ^2(e+f x)}\right )}{8 b f}\\ &=-\frac {\left (8 a^2-12 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{8 \sqrt {a} f}+\frac {(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}+\frac {(4 a-5 b) \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}-\frac {a \cot ^4(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}\\ \end {align*}

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Mathematica [A] time = 1.40, size = 140, normalized size = 0.87 \[ \frac {\left (-8 a^2+12 a b-3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )+\sqrt {a} \left (8 (a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )+\cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)} \left (-2 a \cot ^2(e+f x)+4 a-5 b\right )\right )}{8 \sqrt {a} f} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cot[e + f*x]^5*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

((-8*a^2 + 12*a*b - 3*b^2)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]] + Sqrt[a]*(8*(a - b)^(3/2)*ArcTanh[Sqrt

[a + b*Tan[e + f*x]^2]/Sqrt[a - b]] + Cot[e + f*x]^2*(4*a - 5*b - 2*a*Cot[e + f*x]^2)*Sqrt[a + b*Tan[e + f*x]^

2]))/(8*Sqrt[a]*f)

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fricas [A] time = 0.49, size = 748, normalized size = 4.65 \[ \left [-\frac {8 \, {\left (a^{2} - a b\right )} \sqrt {a - b} \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 )^{4} - {\left (8 \, a^{2} - 12 \, a b + 3 \, b^{2}\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 )^{4} - 2 \, {\left ({\left (4 \, a^{2} - 5 \, a b\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{16 \, a f \tan \left (f x + e\right )^{4}}, \frac {16 \, {\left (a^{2} - a b\right )} \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 )^{4} + {\left (8 \, a^{2} - 12 \, a b + 3 \, b^{2}\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 )^{4} + 2 \, {\left ({\left (4 \, a^{2} - 5 \, a b\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{16 \, a f \tan \left (f x + e\right )^{4}}, \frac {{\left (8 \, a^{2} - 12 \, a b + 3 \, b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a}}{a}\right ) \tan \left (f x + e\right )^{4} - 4 \, {\left (a^{2} - a b\right )} \sqrt {a - b} \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 )^{4} + {\left ({\left (4 \, a^{2} - 5 \, a b\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{8 \, a f \tan \left (f x + e\right )^{4}}, \frac {{\left (8 \, a^{2} - 12 \, a b + 3 \, b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a}}{a}\right ) \tan \left (f x + e\right )^{4} + 8 \, {\left (a^{2} - a b\right )} \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 )^{4} + {\left ({\left (4 \, a^{2} - 5 \, a b\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2}\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{8 \, a f \tan \left (f x + e\right )^{4}}\right ] \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(8*(a^2 - a*b)*sqrt(a - b)*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)^4 - (8*a^2 - 12*a*b + 3*b^2)*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)^4 - 2*((4*a^2 - 5*a*b)*tan(f*x + e)^2 - 2*a^2)*sqrt

(b*tan(f*x + e)^2 + a))/(a*f*tan(f*x + e)^4), 1/16*(16*(a^2 - a*b)*sqrt(-a + b)*arctan(-sqrt(b*tan(f*x + e)^2

+ a)*sqrt(-a + b)/(a - b))*tan(f*x + e)^4 + (8*a^2 - 12*a*b + 3*b^2)*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)^4 + 2*((4*a^2 - 5*a*b)*tan(f*x + e)^2 - 2*a^2)

*sqrt(b*tan(f*x + e)^2 + a))/(a*f*tan(f*x + e)^4), 1/8*((8*a^2 - 12*a*b + 3*b^2)*sqrt(-a)*arctan(sqrt(b*tan(f*

x + e)^2 + a)*sqrt(-a)/a)*tan(f*x + e)^4 - 4*(a^2 - a*b)*sqrt(a - b)*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)^4 + ((4*a^2 - 5*a*b)*tan(f*x + e)^2 - 2*

a^2)*sqrt(b*tan(f*x + e)^2 + a))/(a*f*tan(f*x + e)^4), 1/8*((8*a^2 - 12*a*b + 3*b^2)*sqrt(-a)*arctan(sqrt(b*ta

n(f*x + e)^2 + a)*sqrt(-a)/a)*tan(f*x + e)^4 + 8*(a^2 - a*b)*sqrt(-a + b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)*s

qrt(-a + b)/(a - b))*tan(f*x + e)^4 + ((4*a^2 - 5*a*b)*tan(f*x + e)^2 - 2*a^2)*sqrt(b*tan(f*x + e)^2 + a))/(a*

f*tan(f*x + e)^4)]

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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(f*x+e)^5*(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,x):;OUTPUT:Unab

le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p