3.1.35 \(\int \frac {1}{(a+b \cot ^2(c+d x))^{3/2}} \, dx\) [35]

Optimal. Leaf size=85 \[ -\frac {\text {ArcTan}\left (\frac {\sqrt {a-b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)}}\right )}{(a-b)^{3/2} d}+\frac {b \cot (c+d x)}{a (a-b) d \sqrt {a+b \cot ^2(c+d x)}} \]

[Out]

-arctan(cot(d*x+c)*(a-b)^(1/2)/(a+b*cot(d*x+c)^2)^(1/2))/(a-b)^(3/2)/d+b*cot(d*x+c)/a/(a-b)/d/(a+b*cot(d*x+c)^
2)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3742, 390, 385, 209} \begin {gather*} \frac {b \cot (c+d x)}{a d (a-b) \sqrt {a+b \cot ^2(c+d x)}}-\frac {\text {ArcTan}\left (\frac {\sqrt {a-b} \cot (c+d x)}{\sqrt {a+b \cot ^2(c+d x)}}\right )}{d (a-b)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cot[c + d*x]^2)^(-3/2),x]

[Out]

-(ArcTan[(Sqrt[a - b]*Cot[c + d*x])/Sqrt[a + b*Cot[c + d*x]^2]]/((a - b)^(3/2)*d)) + (b*Cot[c + d*x])/(a*(a -
b)*d*Sqrt[a + b*Cot[c + d*x]^2])

Rule 209

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

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a
*d)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && Eq
Q[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 3742

Int[((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[(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, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 3.76, size = 231, normalized size = 2.72 \begin {gather*} -\frac {\cos ^2(c+d x) \cot (c+d x) \left (4 (a-b)^2 \cos ^2(c+d x) \, _2F_1\left (2,2;\frac {7}{2};\frac {(a-b) \cos ^2(c+d x)}{a}\right ) \left (b+a \tan ^2(c+d x)\right )-\frac {15 a \left (2 b+3 a \tan ^2(c+d x)\right ) \left (\text {ArcSin}\left (\sqrt {\frac {(a-b) \cos ^2(c+d x)}{a}}\right ) \left (b+a \tan ^2(c+d x)\right )-a \sec ^2(c+d x) \sqrt {\frac {(a-b) \cos ^4(c+d x) \left (b+a \tan ^2(c+d x)\right )}{a^2}}\right )}{\sqrt {\frac {(a-b) \cos ^4(c+d x) \left (b+a \tan ^2(c+d x)\right )}{a^2}}}\right )}{15 a^3 (a-b) d \sqrt {a+b \cot ^2(c+d x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/15*(Cos[c + d*x]^2*Cot[c + d*x]*(4*(a - b)^2*Cos[c + d*x]^2*Hypergeometric2F1[2, 2, 7/2, ((a - b)*Cos[c + d
*x]^2)/a]*(b + a*Tan[c + d*x]^2) - (15*a*(2*b + 3*a*Tan[c + d*x]^2)*(ArcSin[Sqrt[((a - b)*Cos[c + d*x]^2)/a]]*
(b + a*Tan[c + d*x]^2) - a*Sec[c + d*x]^2*Sqrt[((a - b)*Cos[c + d*x]^4*(b + a*Tan[c + d*x]^2))/a^2]))/Sqrt[((a
- b)*Cos[c + d*x]^4*(b + a*Tan[c + d*x]^2))/a^2]))/(a^3*(a - b)*d*Sqrt[a + b*Cot[c + d*x]^2])

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Maple [A]
time = 0.20, size = 102, normalized size = 1.20

method	result	size
derivativedivides	\(\frac {\frac {b \cot \left (d x +c \right )}{\left (a -b \right ) a \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \cot \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}\right )}{\left (a -b \right )^{2} b^{2}}}{d}\)	\(102\)
default	\(\frac {\frac {b \cot \left (d x +c \right )}{\left (a -b \right ) a \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \cot \left (d x +c \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\cot ^{2}\left (d x +c \right )\right )}}\right )}{\left (a -b \right )^{2} b^{2}}}{d}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cot(d*x+c)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/(a-b)*b*cot(d*x+c)/a/(a+b*cot(d*x+c)^2)^(1/2)-1/(a-b)^2*(b^4*(a-b))^(1/2)/b^2*arctan(b^2*(a-b)/(b^4*(a-
b))^(1/2)/(a+b*cot(d*x+c)^2)^(1/2)*cot(d*x+c)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cot(d*x+c)^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (77) = 154\).
time = 3.32, size = 526, normalized size = 6.19 \begin {gather*} \left [-\frac {{\left (a^{2} + a b - {\left (a^{2} - a b\right )} \cos \left (2 \, d x + 2 \, c\right )\right )} \sqrt {-a + b} \log \left (-2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )^{2} + 2 \, {\left ({\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - b\right )} \sqrt {-a + b} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - a - b}{\cos \left (2 \, d x + 2 \, c\right ) - 1}} \sin \left (2 \, d x + 2 \, c\right ) + a^{2} - 2 \, b^{2} + 4 \, {\left (a b - b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )\right ) + 4 \, {\left (a b - b^{2}\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - a - b}{\cos \left (2 \, d x + 2 \, c\right ) - 1}} \sin \left (2 \, d x + 2 \, c\right )}{4 \, {\left ({\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} d \cos \left (2 \, d x + 2 \, c\right ) - {\left (a^{4} - a^{3} b - a^{2} b^{2} + a b^{3}\right )} d\right )}}, \frac {{\left (a^{2} + a b - {\left (a^{2} - a b\right )} \cos \left (2 \, d x + 2 \, c\right )\right )} \sqrt {a - b} \arctan \left (-\frac {\sqrt {a - b} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - a - b}{\cos \left (2 \, d x + 2 \, c\right ) - 1}} \sin \left (2 \, d x + 2 \, c\right )}{{\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - b}\right ) - 2 \, {\left (a b - b^{2}\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (2 \, d x + 2 \, c\right ) - a - b}{\cos \left (2 \, d x + 2 \, c\right ) - 1}} \sin \left (2 \, d x + 2 \, c\right )}{2 \, {\left ({\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} d \cos \left (2 \, d x + 2 \, c\right ) - {\left (a^{4} - a^{3} b - a^{2} b^{2} + a b^{3}\right )} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cot(d*x+c)^2)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*((a^2 + a*b - (a^2 - a*b)*cos(2*d*x + 2*c))*sqrt(-a + b)*log(-2*(a^2 - 2*a*b + b^2)*cos(2*d*x + 2*c)^2 +
2*((a - b)*cos(2*d*x + 2*c) - b)*sqrt(-a + b)*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))
*sin(2*d*x + 2*c) + a^2 - 2*b^2 + 4*(a*b - b^2)*cos(2*d*x + 2*c)) + 4*(a*b - b^2)*sqrt(((a - b)*cos(2*d*x + 2*
c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x + 2*c))/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d*cos(2*d*x + 2*c)
- (a^4 - a^3*b - a^2*b^2 + a*b^3)*d), 1/2*((a^2 + a*b - (a^2 - a*b)*cos(2*d*x + 2*c))*sqrt(a - b)*arctan(-sqrt
(a - b)*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x + 2*c)/((a - b)*cos(2*d*x +
2*c) - b)) - 2*(a*b - b^2)*sqrt(((a - b)*cos(2*d*x + 2*c) - a - b)/(cos(2*d*x + 2*c) - 1))*sin(2*d*x + 2*c))/(
(a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d*cos(2*d*x + 2*c) - (a^4 - a^3*b - a^2*b^2 + a*b^3)*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \cot ^{2}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cot(d*x+c)**2)**(3/2),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (77) = 154\).
time = 0.96, size = 300, normalized size = 3.53 \begin {gather*} -\frac {\frac {\frac {{\left (a^{2} b \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 2 \, a b^{2} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + b^{3} \mathrm {sgn}\left (\sin \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}} - \frac {a^{2} b \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - 2 \, a b^{2} \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) + b^{3} \mathrm {sgn}\left (\sin \left (d x + c\right )\right )}{a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}}}{\sqrt {b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b}} - \frac {2 \, \arctan \left (-\frac {\sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + b} + \sqrt {b}}{2 \, \sqrt {a - b}}\right )}{{\left (a \mathrm {sgn}\left (\sin \left (d x + c\right )\right ) - b \mathrm {sgn}\left (\sin \left (d x + c\right )\right )\right )} \sqrt {a - b}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cot(d*x+c)^2)^(3/2),x, algorithm="giac")

[Out]

-(((a^2*b*sgn(sin(d*x + c)) - 2*a*b^2*sgn(sin(d*x + c)) + b^3*sgn(sin(d*x + c)))*tan(1/2*d*x + 1/2*c)^2/(a^4 -
3*a^3*b + 3*a^2*b^2 - a*b^3) - (a^2*b*sgn(sin(d*x + c)) - 2*a*b^2*sgn(sin(d*x + c)) + b^3*sgn(sin(d*x + c)))/
(a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3))/sqrt(b*tan(1/2*d*x + 1/2*c)^4 + 4*a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*
d*x + 1/2*c)^2 + b) - 2*arctan(-1/2*(sqrt(b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(b*tan(1/2*d*x + 1/2*c)^4 + 4*a*tan(
1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c)^2 + b) + sqrt(b))/sqrt(a - b))/((a*sgn(sin(d*x + c)) - b*sgn(sin
(d*x + c)))*sqrt(a - b)))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (b\,{\mathrm {cot}\left (c+d\,x\right )}^2+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cot(c + d*x)^2)^(3/2),x)

[Out]

int(1/(a + b*cot(c + d*x)^2)^(3/2), x)

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