∫ cot(e+fx)____ (a+b sec2(e+fx))5∕2 dx

3.431 \(\int \frac {\cot (e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=137 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{a^{5/2} f}-\frac {b (2 a+b)}{a^2 f (a+b)^2 \sqrt {a+b \sec ^2(e+f x)}}-\frac {b}{3 a f (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{f (a+b)^{5/2}} \]

[Out]

arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2))/a^(5/2)/f-arctanh((a+b*sec(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/

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

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Rubi [A] time = 0.21, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4139, 446, 85, 152, 156, 63, 208} \[ -\frac {b (2 a+b)}{a^2 f (a+b)^2 \sqrt {a+b \sec ^2(e+f x)}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{a^{5/2} f}-\frac {b}{3 a f (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a+b}}\right )}{f (a+b)^{5/2}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]]/(a^(5/2)*f) - ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a + b]]/((a

+ b)^(5/2)*f) - b/(3*a*(a + b)*f*(a + b*Sec[e + f*x]^2)^(3/2)) - (b*(2*a + b))/(a^2*(a + b)^2*f*Sqrt[a + b*Sec

[e + f*x]^2])

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 85

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[(f*(e + f*x)^(p +

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

*d*f*x)*(e + f*x)^(p + 1))/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[p, -1]

Rule 152

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] && IntegersQ[2*m, 2*n, 2*p]

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 4139

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With

[{ff = FreeFactors[Sec[e + f*x], x]}, Dist[1/f, Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p)/x

, x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[

n, 2] || EqQ[n, 4] || IGtQ[p, 0] || IntegersQ[2*n, p])

Rubi steps

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

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Mathematica [F] time = 8.76, size = 0, normalized size = 0.00 \[ \int \frac {\cot (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx \]

Veriﬁcation is Not applicable to the result.

[In]

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

[Out]

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

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fricas [B] time = 2.61, size = 2279, normalized size = 16.64 \[ \text {result too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(3*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5 + (a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cos(f*x + e)^4 + 2*(a^4*

b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(f*x + e)^2)*sqrt(a)*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b*cos(f*x + e)

^6 + 160*a^2*b^2*cos(f*x + e)^4 + 32*a*b^3*cos(f*x + e)^2 + b^4 + 8*(16*a^3*cos(f*x + e)^8 + 24*a^2*b*cos(f*x

+ e)^6 + 10*a*b^2*cos(f*x + e)^4 + b^3*cos(f*x + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) +

6*(a^5*cos(f*x + e)^4 + 2*a^4*b*cos(f*x + e)^2 + a^3*b^2)*sqrt(a + b)*log(2*((8*a^2 + 8*a*b + b^2)*cos(f*x + e

)^4 + 2*(4*a*b + 3*b^2)*cos(f*x + e)^2 + b^2 - 4*((2*a + b)*cos(f*x + e)^4 + b*cos(f*x + e)^2)*sqrt(a + b)*sqr

t((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)) - 8*((7*a^4*b + 11*a^3*b^2

+ 4*a^2*b^3)*cos(f*x + e)^4 + 3*(2*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/co

s(f*x + e)^2))/((a^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3)*f*cos(f*x + e)^4 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^

4*b^4)*f*cos(f*x + e)^2 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*f), 1/24*(12*(a^5*cos(f*x + e)^4 + 2*a^4

*b*cos(f*x + e)^2 + a^3*b^2)*sqrt(-a - b)*arctan(1/2*((2*a + b)*cos(f*x + e)^2 + b)*sqrt(-a - b)*sqrt((a*cos(f

*x + e)^2 + b)/cos(f*x + e)^2)/((a^2 + a*b)*cos(f*x + e)^2 + a*b + b^2)) + 3*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 +

b^5 + (a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cos(f*x + e)^4 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(f*x

+ e)^2)*sqrt(a)*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b*cos(f*x + e)^6 + 160*a^2*b^2*cos(f*x + e)^4 + 32*a*b^3

*cos(f*x + e)^2 + b^4 + 8*(16*a^3*cos(f*x + e)^8 + 24*a^2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f*x + e)^4 + b^3*cos

(f*x + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) - 8*((7*a^4*b + 11*a^3*b^2 + 4*a^2*b^3)*cos(

f*x + e)^4 + 3*(2*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((

a^8 + 3*a^7*b + 3*a^6*b^2 + a^5*b^3)*f*cos(f*x + e)^4 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)*f*cos(f*x

+ e)^2 + (a^6*b^2 + 3*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*f), -1/12*(3*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5 + (a^5

+ 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*cos(f*x + e)^4 + 2*(a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(f*x + e)^2)*sq

rt(-a)*arctan(1/4*(8*a^2*cos(f*x + e)^4 + 8*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos

(f*x + e)^2)/(2*a^3*cos(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2)) - 3*(a^5*cos(f*x + e)^4 + 2*a^4*b*cos(f*

x + e)^2 + a^3*b^2)*sqrt(a + b)*log(2*((8*a^2 + 8*a*b + b^2)*cos(f*x + e)^4 + 2*(4*a*b + 3*b^2)*cos(f*x + e)^2

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

2))/(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)) + 4*((7*a^4*b + 11*a^3*b^2 + 4*a^2*b^3)*cos(f*x + e)^4 + 3*(2*a^3

*b^2 + 3*a^2*b^3 + a*b^4)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^8 + 3*a^7*b + 3*a^6

*b^2 + a^5*b^3)*f*cos(f*x + e)^4 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)*f*cos(f*x + e)^2 + (a^6*b^2 + 3

*a^5*b^3 + 3*a^4*b^4 + a^3*b^5)*f), -1/12*(3*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5 + (a^5 + 3*a^4*b + 3*a^3*b^2

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

a^2*cos(f*x + e)^4 + 8*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(2*a^3*c

os(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2)) - 6*(a^5*cos(f*x + e)^4 + 2*a^4*b*cos(f*x + e)^2 + a^3*b^2)*s

qrt(-a - b)*arctan(1/2*((2*a + b)*cos(f*x + e)^2 + b)*sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)

/((a^2 + a*b)*cos(f*x + e)^2 + a*b + b^2)) + 4*((7*a^4*b + 11*a^3*b^2 + 4*a^2*b^3)*cos(f*x + e)^4 + 3*(2*a^3*b

^2 + 3*a^2*b^3 + a*b^4)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^8 + 3*a^7*b + 3*a^6*b

^2 + a^5*b^3)*f*cos(f*x + e)^4 + 2*(a^7*b + 3*a^6*b^2 + 3*a^5*b^3 + a^4*b^4)*f*cos(f*x + e)^2 + (a^6*b^2 + 3*a

^5*b^3 + 3*a^4*b^4 + a^3*b^5)*f)]

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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)/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab