3.339 \(\int \frac{\cot ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=236 \[ \frac{2 b^4}{a d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{a^{3/2} d}+\frac{\sqrt{a+b \sec (c+d x)}}{4 d (a+b)^2 (1-\sec (c+d x))}+\frac{\sqrt{a+b \sec (c+d x)}}{4 d (a-b)^2 (\sec (c+d x)+1)}+\frac{(4 a-7 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a-b}}\right )}{4 d (a-b)^{5/2}}+\frac{(4 a+7 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )}{4 d (a+b)^{5/2}} \]

[Out]

(-2*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/(a^(3/2)*d) + ((4*a - 7*b)*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqr
t[a - b]])/(4*(a - b)^(5/2)*d) + ((4*a + 7*b)*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]])/(4*(a + b)^(5/2)*
d) + (2*b^4)/(a*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]]) + Sqrt[a + b*Sec[c + d*x]]/(4*(a + b)^2*d*(1 - Sec[c
+ d*x])) + Sqrt[a + b*Sec[c + d*x]]/(4*(a - b)^2*d*(1 + Sec[c + d*x]))

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Rubi [A]  time = 0.384817, antiderivative size = 316, normalized size of antiderivative = 1.34, number of steps used = 11, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3885, 898, 1335, 206, 199} \[ \frac{2 b^4}{a d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{a^{3/2} d}+\frac{\sqrt{a+b \sec (c+d x)}}{4 d (a+b)^2 (1-\sec (c+d x))}+\frac{\sqrt{a+b \sec (c+d x)}}{4 d (a-b)^2 (\sec (c+d x)+1)}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a-b}}\right )}{4 d (a-b)^{5/2}}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )}{4 d (a+b)^{5/2}}+\frac{(2 a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a-b}}\right )}{2 d (a-b)^{5/2}}+\frac{(2 a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )}{2 d (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(-2*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/(a^(3/2)*d) + ((2*a - 3*b)*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqr
t[a - b]])/(2*(a - b)^(5/2)*d) - (b*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a - b]])/(4*(a - b)^(5/2)*d) + (b*Ar
cTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]])/(4*(a + b)^(5/2)*d) + ((2*a + 3*b)*ArcTanh[Sqrt[a + b*Sec[c + d*x
]]/Sqrt[a + b]])/(2*(a + b)^(5/2)*d) + (2*b^4)/(a*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]]) + Sqrt[a + b*Sec[c
+ d*x]]/(4*(a + b)^2*d*(1 - Sec[c + d*x])) + Sqrt[a + b*Sec[c + d*x]]/(4*(a - b)^2*d*(1 + Sec[c + d*x]))

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 898

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 + a*e^2)/e^2 - (2*c
*d*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]

Rule 1335

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x]
&& NeQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0] || IntegersQ[m, q])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rubi steps

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

Mathematica [B]  time = 7.16121, size = 1114, normalized size = 4.72 \[ \frac{(b+a \cos (c+d x))^2 \left (-\frac{2 b^5}{a^2 \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}+\frac{\left (-a^2+2 b \cos (c+d x) a-b^2\right ) \csc ^2(c+d x)}{2 \left (b^2-a^2\right )^2}+\frac{a^4+b^2 a^2+4 b^4}{2 a^2 \left (b^2-a^2\right )^2}\right ) \sec ^2(c+d x)}{d (a+b \sec (c+d x))^{3/2}}-\frac{(b+a \cos (c+d x))^{3/2} \left (-\frac{\left (2 a^4-6 b^2 a^2-2 b^4\right ) \sqrt{-a \cos (c+d x)} \sqrt{\sec (c+d x)} \left (\sqrt{a-b} (a+b) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{b+a \cos (c+d x)}}{\sqrt{a-b} \sqrt{-a \cos (c+d x)}}\right )+(a-b) \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{b+a \cos (c+d x)}}{\sqrt{a+b} \sqrt{-a \cos (c+d x)}}\right )\right )}{\sqrt{a} (a-b) (a+b)}-\frac{a \left (2 a^4-4 b^2 a^2+2 b^4\right ) \left (4 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{b+a \cos (c+d x)}}{\sqrt{-a \cos (c+d x)}}\right )-\sqrt{a} \left (\sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{b+a \cos (c+d x)}}{\sqrt{a-b} \sqrt{-a \cos (c+d x)}}\right )+\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{b+a \cos (c+d x)}}{\sqrt{a+b} \sqrt{-a \cos (c+d x)}}\right )\right )\right ) \sqrt{-a \cos (c+d x)} \cos (2 (c+d x)) \sqrt{\sec (c+d x)}}{\sqrt{a-b} \sqrt{a+b} \left (a^2-2 b^2-2 (b+a \cos (c+d x))^2+4 b (b+a \cos (c+d x))\right )}-\frac{a \left (7 a b^3-a^3 b\right ) \left (-\sqrt{-a^2} \sqrt{a+b} \log \left (\sqrt{b+a \cos (c+d x)}-\sqrt{b-a}\right )+\sqrt{-a^2} \sqrt{a+b} \log \left (\sqrt{b-a}+\sqrt{b+a \cos (c+d x)}\right )-a \sqrt{b-a} \log \left (\sqrt{b+a \cos (c+d x)}-\sqrt{a+b}\right )+a \sqrt{b-a} \log \left (\sqrt{a+b}+\sqrt{b+a \cos (c+d x)}\right )+\sqrt{-a^2} \sqrt{a+b} \log \left (b+\sqrt{a} \sqrt{-a \cos (c+d x)}-\sqrt{b-a} \sqrt{b+a \cos (c+d x)}\right )-\sqrt{-a^2} \sqrt{a+b} \log \left (b+\sqrt{a} \sqrt{-a \cos (c+d x)}+\sqrt{b-a} \sqrt{b+a \cos (c+d x)}\right )+a \sqrt{b-a} \log \left (b+\sqrt{-a} \sqrt{-a \cos (c+d x)}-\sqrt{a+b} \sqrt{b+a \cos (c+d x)}\right )-a \sqrt{b-a} \log \left (b+\sqrt{-a} \sqrt{-a \cos (c+d x)}+\sqrt{a+b} \sqrt{b+a \cos (c+d x)}\right )\right )}{2 (-a)^{3/2} \sqrt{b-a} \sqrt{a+b} \sqrt{-a \cos (c+d x)} \sqrt{\sec (c+d x)}}\right ) \sec ^{\frac{3}{2}}(c+d x)}{4 a (a-b)^2 (a+b)^2 d (a+b \sec (c+d x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[c + d*x]^3/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

-((b + a*Cos[c + d*x])^(3/2)*(-(a*(-(a^3*b) + 7*a*b^3)*(-(Sqrt[-a^2]*Sqrt[a + b]*Log[-Sqrt[-a + b] + Sqrt[b +
a*Cos[c + d*x]]]) + Sqrt[-a^2]*Sqrt[a + b]*Log[Sqrt[-a + b] + Sqrt[b + a*Cos[c + d*x]]] - a*Sqrt[-a + b]*Log[-
Sqrt[a + b] + Sqrt[b + a*Cos[c + d*x]]] + a*Sqrt[-a + b]*Log[Sqrt[a + b] + Sqrt[b + a*Cos[c + d*x]]] + Sqrt[-a
^2]*Sqrt[a + b]*Log[b + Sqrt[a]*Sqrt[-(a*Cos[c + d*x])] - Sqrt[-a + b]*Sqrt[b + a*Cos[c + d*x]]] - Sqrt[-a^2]*
Sqrt[a + b]*Log[b + Sqrt[a]*Sqrt[-(a*Cos[c + d*x])] + Sqrt[-a + b]*Sqrt[b + a*Cos[c + d*x]]] + a*Sqrt[-a + b]*
Log[b + Sqrt[-a]*Sqrt[-(a*Cos[c + d*x])] - Sqrt[a + b]*Sqrt[b + a*Cos[c + d*x]]] - a*Sqrt[-a + b]*Log[b + Sqrt
[-a]*Sqrt[-(a*Cos[c + d*x])] + Sqrt[a + b]*Sqrt[b + a*Cos[c + d*x]]]))/(2*(-a)^(3/2)*Sqrt[-a + b]*Sqrt[a + b]*
Sqrt[-(a*Cos[c + d*x])]*Sqrt[Sec[c + d*x]]) - ((2*a^4 - 6*a^2*b^2 - 2*b^4)*(Sqrt[a - b]*(a + b)*ArcTan[(Sqrt[a
]*Sqrt[b + a*Cos[c + d*x]])/(Sqrt[a - b]*Sqrt[-(a*Cos[c + d*x])])] + (a - b)*Sqrt[a + b]*ArcTan[(Sqrt[a]*Sqrt[
b + a*Cos[c + d*x]])/(Sqrt[a + b]*Sqrt[-(a*Cos[c + d*x])])])*Sqrt[-(a*Cos[c + d*x])]*Sqrt[Sec[c + d*x]])/(Sqrt
[a]*(a - b)*(a + b)) - (a*(2*a^4 - 4*a^2*b^2 + 2*b^4)*(4*Sqrt[a - b]*Sqrt[a + b]*ArcTan[Sqrt[b + a*Cos[c + d*x
]]/Sqrt[-(a*Cos[c + d*x])]] - Sqrt[a]*(Sqrt[a + b]*ArcTan[(Sqrt[a]*Sqrt[b + a*Cos[c + d*x]])/(Sqrt[a - b]*Sqrt
[-(a*Cos[c + d*x])])] + Sqrt[a - b]*ArcTan[(Sqrt[a]*Sqrt[b + a*Cos[c + d*x]])/(Sqrt[a + b]*Sqrt[-(a*Cos[c + d*
x])])]))*Sqrt[-(a*Cos[c + d*x])]*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(Sqrt[a - b]*Sqrt[a + b]*(a^2 - 2*b^2 +
4*b*(b + a*Cos[c + d*x]) - 2*(b + a*Cos[c + d*x])^2)))*Sec[c + d*x]^(3/2))/(4*a*(a - b)^2*(a + b)^2*d*(a + b*S
ec[c + d*x])^(3/2)) + ((b + a*Cos[c + d*x])^2*((a^4 + a^2*b^2 + 4*b^4)/(2*a^2*(-a^2 + b^2)^2) - (2*b^5)/(a^2*(
a^2 - b^2)^2*(b + a*Cos[c + d*x])) + ((-a^2 - b^2 + 2*a*b*Cos[c + d*x])*Csc[c + d*x]^2)/(2*(-a^2 + b^2)^2))*Se
c[c + d*x]^2)/(d*(a + b*Sec[c + d*x])^(3/2))

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Maple [B]  time = 0.76, size = 10977, normalized size = 46.5 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x)

[Out]

result too large to display

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Maxima [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(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [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(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3/(a+b*sec(d*x+c))**(3/2),x)

[Out]

Integral(cot(c + d*x)**3/(a + b*sec(c + d*x))**(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(cot(d*x + c)^3/(b*sec(d*x + c) + a)^(3/2), x)