3.36 \(\int \frac{1}{(a+b \cot ^2(c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=135 \[ \frac{b (5 a-2 b) \cot (c+d x)}{3 a^2 d (a-b)^2 \sqrt{a+b \cot ^2(c+d x)}}+\frac{b \cot (c+d x)}{3 a d (a-b) \left (a+b \cot ^2(c+d x)\right )^{3/2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)}}\right )}{d (a-b)^{5/2}} \]

[Out]

-(ArcTan[(Sqrt[a - b]*Cot[c + d*x])/Sqrt[a + b*Cot[c + d*x]^2]]/((a - b)^(5/2)*d)) + (b*Cot[c + d*x])/(3*a*(a
- b)*d*(a + b*Cot[c + d*x]^2)^(3/2)) + ((5*a - 2*b)*b*Cot[c + d*x])/(3*a^2*(a - b)^2*d*Sqrt[a + b*Cot[c + d*x]
^2])

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Rubi [A]  time = 0.109509, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3661, 414, 527, 12, 377, 203} \[ \frac{b (5 a-2 b) \cot (c+d x)}{3 a^2 d (a-b)^2 \sqrt{a+b \cot ^2(c+d x)}}+\frac{b \cot (c+d x)}{3 a d (a-b) \left (a+b \cot ^2(c+d x)\right )^{3/2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \cot (c+d x)}{\sqrt{a+b \cot ^2(c+d x)}}\right )}{d (a-b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cot[c + d*x]^2)^(-5/2),x]

[Out]

-(ArcTan[(Sqrt[a - b]*Cot[c + d*x])/Sqrt[a + b*Cot[c + d*x]^2]]/((a - b)^(5/2)*d)) + (b*Cot[c + d*x])/(3*a*(a
- b)*d*(a + b*Cot[c + d*x]^2)^(3/2)) + ((5*a - 2*b)*b*Cot[c + d*x])/(3*a^2*(a - b)^2*d*Sqrt[a + b*Cot[c + d*x]
^2])

Rule 3661

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])

Rule 414

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[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

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 203

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

Rubi steps

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

Mathematica [C]  time = 7.90356, size = 367, normalized size = 2.72 \[ -\frac{\cot ^5(c+d x) \left (24 (a-b)^3 \cos ^2(c+d x) \left (a \tan ^2(c+d x)+b\right )^2 \text{HypergeometricPFQ}\left (\{2,2,2\},\left \{1,\frac{9}{2}\right \},\frac{(a-b) \cos ^2(c+d x)}{a}\right )+24 (a-b)^3 \cos ^2(c+d x) \left (4 a^2 \tan ^4(c+d x)+7 a b \tan ^2(c+d x)+3 b^2\right ) \text{Hypergeometric2F1}\left (2,2,\frac{9}{2},\frac{(a-b) \cos ^2(c+d x)}{a}\right )-\frac{35 a \left (15 a^2 \tan ^4(c+d x)+20 a b \tan ^2(c+d x)+8 b^2\right ) \left (a \sec ^2(c+d x) \left (a \left (3 \tan ^2(c+d x)-1\right )+4 b\right ) \sqrt{\frac{(a-b) \cos ^4(c+d x) \left (a \tan ^2(c+d x)+b\right )}{a^2}}-3 \left (a \tan ^2(c+d x)+b\right )^2 \sin ^{-1}\left (\sqrt{\frac{(a-b) \cos ^2(c+d x)}{a}}\right )\right )}{\sqrt{\frac{(a-b) \cos ^4(c+d x) \left (a \tan ^2(c+d x)+b\right )}{a^2}}}\right )}{315 a^5 d (a-b)^2 \left (\cot ^2(c+d x)+1\right ) \sqrt{a+b \cot ^2(c+d x)} \left (\frac{b \cot ^2(c+d x)}{a}+1\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Cot[c + d*x]^2)^(-5/2),x]

[Out]

-(Cot[c + d*x]^5*(24*(a - b)^3*Cos[c + d*x]^2*HypergeometricPFQ[{2, 2, 2}, {1, 9/2}, ((a - b)*Cos[c + d*x]^2)/
a]*(b + a*Tan[c + d*x]^2)^2 + 24*(a - b)^3*Cos[c + d*x]^2*Hypergeometric2F1[2, 2, 9/2, ((a - b)*Cos[c + d*x]^2
)/a]*(3*b^2 + 7*a*b*Tan[c + d*x]^2 + 4*a^2*Tan[c + d*x]^4) - (35*a*(8*b^2 + 20*a*b*Tan[c + d*x]^2 + 15*a^2*Tan
[c + d*x]^4)*(-3*ArcSin[Sqrt[((a - b)*Cos[c + d*x]^2)/a]]*(b + a*Tan[c + d*x]^2)^2 + a*Sec[c + d*x]^2*Sqrt[((a
- b)*Cos[c + d*x]^4*(b + a*Tan[c + d*x]^2))/a^2]*(4*b + a*(-1 + 3*Tan[c + d*x]^2))))/Sqrt[((a - b)*Cos[c + d*
x]^4*(b + a*Tan[c + d*x]^2))/a^2]))/(315*a^5*(a - b)^2*d*(1 + Cot[c + d*x]^2)*Sqrt[a + b*Cot[c + d*x]^2]*(1 +
(b*Cot[c + d*x]^2)/a))

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Maple [A]  time = 0.027, size = 176, normalized size = 1.3 \begin{align*} -{\frac{1}{d \left ( a-b \right ) ^{3}{b}^{2}}\sqrt{{b}^{4} \left ( a-b \right ) }\arctan \left ({ \left ( a-b \right ){b}^{2}\cot \left ( dx+c \right ){\frac{1}{\sqrt{{b}^{4} \left ( a-b \right ) }}}{\frac{1}{\sqrt{a+b \left ( \cot \left ( dx+c \right ) \right ) ^{2}}}}} \right ) }+{\frac{b\cot \left ( dx+c \right ) }{d \left ( a-b \right ) ^{2}a}{\frac{1}{\sqrt{a+b \left ( \cot \left ( dx+c \right ) \right ) ^{2}}}}}+{\frac{b\cot \left ( dx+c \right ) }{3\,a \left ( a-b \right ) d} \left ( a+b \left ( \cot \left ( dx+c \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,b\cot \left ( dx+c \right ) }{3\,d \left ( a-b \right ){a}^{2}}{\frac{1}{\sqrt{a+b \left ( \cot \left ( dx+c \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cot(d*x+c)^2)^(5/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cot(d*x+c)^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.18914, size = 1963, normalized size = 14.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cot(d*x+c)^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(a^4 + 2*a^3*b + a^2*b^2 + (a^4 - 2*a^3*b + a^2*b^2)*cos(2*d*x + 2*c)^2 - 2*(a^4 - a^2*b^2)*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)) - 8*(3*a^3*b - 2*a^2*b^2 - 2*a*b^3 + b^4 - (3*a^3*b - 7*a^2*b^2 + 5*a*b^3 - b^4)*cos
(2*d*x + 2*c))*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^7 - 5*a^6
*b + 10*a^5*b^2 - 10*a^4*b^3 + 5*a^3*b^4 - a^2*b^5)*d*cos(2*d*x + 2*c)^2 - 2*(a^7 - 3*a^6*b + 2*a^5*b^2 + 2*a^
4*b^3 - 3*a^3*b^4 + a^2*b^5)*d*cos(2*d*x + 2*c) + (a^7 - a^6*b - 2*a^5*b^2 + 2*a^4*b^3 + a^3*b^4 - a^2*b^5)*d)
, -1/6*(3*(a^4 + 2*a^3*b + a^2*b^2 + (a^4 - 2*a^3*b + a^2*b^2)*cos(2*d*x + 2*c)^2 - 2*(a^4 - a^2*b^2)*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)) - 4*(3*a^3*b - 2*a^2*b^2 - 2*a*b^3 + b^4 - (3*a^3*b - 7*a^2*b^2 +
5*a*b^3 - b^4)*cos(2*d*x + 2*c))*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^7 - 5*a^6*b + 10*a^5*b^2 - 10*a^4*b^3 + 5*a^3*b^4 - a^2*b^5)*d*cos(2*d*x + 2*c)^2 - 2*(a^7 - 3*a^6*b
+ 2*a^5*b^2 + 2*a^4*b^3 - 3*a^3*b^4 + a^2*b^5)*d*cos(2*d*x + 2*c) + (a^7 - a^6*b - 2*a^5*b^2 + 2*a^4*b^3 + a^
3*b^4 - a^2*b^5)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \cot ^{2}{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cot(d*x+c)**2)**(5/2),x)

[Out]

Integral((a + b*cot(c + d*x)**2)**(-5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cot \left (d x + c\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cot(d*x+c)^2)^(5/2),x, algorithm="giac")

[Out]

integrate((b*cot(d*x + c)^2 + a)^(-5/2), x)