3.381 \(\int \cot ^5(e+f x) \sqrt{a+b \sec ^2(e+f x)} \, dx\)

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

[Out]

(Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]])/f - ((8*a^2 + 12*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sec[e +
f*x]^2]/Sqrt[a + b]])/(8*(a + b)^(3/2)*f) + ((4*a + 3*b)*Cot[e + f*x]^2*Sqrt[a + b*Sec[e + f*x]^2])/(8*(a + b
)*f) - (Cot[e + f*x]^4*Sqrt[a + b*Sec[e + f*x]^2])/(4*f)

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Rubi [A]  time = 0.229269, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {4139, 446, 99, 151, 156, 63, 208} \[ -\frac{\left (8 a^2+12 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 f (a+b)^{3/2}}-\frac{\cot ^4(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{4 f}+\frac{(4 a+3 b) \cot ^2(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{8 f (a+b)}+\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]])/f - ((8*a^2 + 12*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sec[e +
f*x]^2]/Sqrt[a + b]])/(8*(a + b)^(3/2)*f) + ((4*a + 3*b)*Cot[e + f*x]^2*Sqrt[a + b*Sec[e + f*x]^2])/(8*(a + b
)*f) - (Cot[e + f*x]^4*Sqrt[a + b*Sec[e + f*x]^2])/(4*f)

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])

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

Mathematica [F]  time = 5.31879, size = 0, normalized size = 0. \[ \int \cot ^5(e+f x) \sqrt{a+b \sec ^2(e+f x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [B]  time = 0.447, size = 7346, normalized size = 45.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (f x + e\right )^{2} + a} \cot \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sec(f*x + e)^2 + a)*cot(f*x + e)^5, x)

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Fricas [B]  time = 4.82482, size = 4782, normalized size = 29.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/32*(4*((a^2 + 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^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)*s
qrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) + ((8*a^2 + 12*a*b + 3*b^2)*cos(f*x + e)^4 - 2*(8*a^2 + 12
*a*b + 3*b^2)*cos(f*x + e)^2 + 8*a^2 + 12*a*b + 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*((6*a^2 + 11*a*b + 5*b^2)*
cos(f*x + e)^4 - (4*a^2 + 7*a*b + 3*b^2)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^2 +
2*a*b + b^2)*f*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 + (a^2 + 2*a*b + b^2)*f), 1/16*(((8*a^2
+ 12*a*b + 3*b^2)*cos(f*x + e)^4 - 2*(8*a^2 + 12*a*b + 3*b^2)*cos(f*x + e)^2 + 8*a^2 + 12*a*b + 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)) + 2*((a^2 + 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x
+ e)^2 + a^2 + 2*a*b + b^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)) - 2*((6*a^2 + 11*a*b + 5*
b^2)*cos(f*x + e)^4 - (4*a^2 + 7*a*b + 3*b^2)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a
^2 + 2*a*b + b^2)*f*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 + (a^2 + 2*a*b + b^2)*f), -1/32*(8
*((a^2 + 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(-a)*arct
an(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)) - ((8*a^2 + 12*a*b + 3*b^2)*cos(f*x + e)^4 - 2*(8*a
^2 + 12*a*b + 3*b^2)*cos(f*x + e)^2 + 8*a^2 + 12*a*b + 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*((6*a^2 + 11*a*b +
5*b^2)*cos(f*x + e)^4 - (4*a^2 + 7*a*b + 3*b^2)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(
(a^2 + 2*a*b + b^2)*f*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 + (a^2 + 2*a*b + b^2)*f), -1/16*
(4*((a^2 + 2*a*b + b^2)*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(-a)*ar
ctan(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)) - ((8*a^2 + 12*a*b + 3*b^2)*cos(f*x + e)^4 - 2*(8
*a^2 + 12*a*b + 3*b^2)*cos(f*x + e)^2 + 8*a^2 + 12*a*b + 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)) +
2*((6*a^2 + 11*a*b + 5*b^2)*cos(f*x + e)^4 - (4*a^2 + 7*a*b + 3*b^2)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 +
b)/cos(f*x + e)^2))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^4 - 2*(a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 + (a^2 + 2*
a*b + b^2)*f)]

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Sympy [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(f*x+e)**5*(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (f x + e\right )^{2} + a} \cot \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sec(f*x + e)^2 + a)*cot(f*x + e)^5, x)