3.536 \(\int \frac{\cot ^3(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=143 \[ -\frac{2 a+5 b}{2 a^3 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 a+5 b}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{(2 a+5 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{2 a^{7/2} f}-\frac{\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[Out]

((2*a + 5*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/(2*a^(7/2)*f) - (2*a + 5*b)/(6*a^2*f*(a + b*Sin[e +
f*x]^2)^(3/2)) - Csc[e + f*x]^2/(2*a*f*(a + b*Sin[e + f*x]^2)^(3/2)) - (2*a + 5*b)/(2*a^3*f*Sqrt[a + b*Sin[e +
f*x]^2])

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Rubi [A]  time = 0.13723, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3194, 78, 51, 63, 208} \[ -\frac{2 a+5 b}{2 a^3 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 a+5 b}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{(2 a+5 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{2 a^{7/2} f}-\frac{\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^3/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

((2*a + 5*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/(2*a^(7/2)*f) - (2*a + 5*b)/(6*a^2*f*(a + b*Sin[e +
f*x]^2)^(3/2)) - Csc[e + f*x]^2/(2*a*f*(a + b*Sin[e + f*x]^2)^(3/2)) - (2*a + 5*b)/(2*a^3*f*Sqrt[a + b*Sin[e +
f*x]^2])

Rule 3194

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, 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 \frac{\cot ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1-x}{x^2 (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=-\frac{\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(2 a+5 b) \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 a f}\\ &=-\frac{2 a+5 b}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(2 a+5 b) \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 a^2 f}\\ &=-\frac{2 a+5 b}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{2 a+5 b}{2 a^3 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(2 a+5 b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 a^3 f}\\ &=-\frac{2 a+5 b}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{2 a+5 b}{2 a^3 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(2 a+5 b) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sin ^2(e+f x)}\right )}{2 a^3 b f}\\ &=\frac{(2 a+5 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{2 a^{7/2} f}-\frac{2 a+5 b}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\csc ^2(e+f x)}{2 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{2 a+5 b}{2 a^3 f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.261647, size = 69, normalized size = 0.48 \[ -\frac{(2 a+5 b) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b \sin ^2(e+f x)}{a}+1\right )+3 a \csc ^2(e+f x)}{6 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^3/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

-(3*a*Csc[e + f*x]^2 + (2*a + 5*b)*Hypergeometric2F1[-3/2, 1, -1/2, 1 + (b*Sin[e + f*x]^2)/a])/(6*a^2*f*(a + b
*Sin[e + f*x]^2)^(3/2))

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Maple [B]  time = 4.118, size = 1038, normalized size = 7.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^3/(a+b*sin(f*x+e)^2)^(5/2),x)

[Out]

1/6/a^(13/2)/b^2/(b^2*cos(f*x+e)^6-2*a*b*cos(f*x+e)^4-3*b^2*cos(f*x+e)^4+a^2*cos(f*x+e)^2+4*a*b*cos(f*x+e)^2+3
*b^2*cos(f*x+e)^2-a^2-2*a*b-b^2)*(3*(a+b-b*cos(f*x+e)^2)^(1/2)*a^(11/2)*b^2-6*ln(2/sin(f*x+e)*(a^(1/2)*(a+b-b*
cos(f*x+e)^2)^(1/2)+a))*a^6*b^2+3*(a+b-b*cos(f*x+e)^2)^(1/2)*a^(7/2)*b^4+8*a^(11/2)*b^2*(-b*cos(f*x+e)^2+(a*b^
2+b^3)/b^2)^(1/2)+20*a^(9/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^3+6*(a+b-b*cos(f*x+e)^2)^(1/2)*a^(9/2)*
b^3+12*a^(7/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^4-27*ln(2/sin(f*x+e)*(a^(1/2)*(a+b-b*cos(f*x+e)^2)^(1
/2)+a))*a^5*b^3-36*ln(2/sin(f*x+e)*(a^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^4*b^4-15*ln(2/sin(f*x+e)*(a^(1/2)
*(a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^3*b^5+3*ln(2/sin(f*x+e)*(a^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^3*b^4*(2*a
+5*b)*cos(f*x+e)^6-3*cos(f*x+e)^4*b^3*(-2*a^(9/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)-(a+b-b*cos(f*x+e)^2)
^(1/2)*a^(7/2)*b-4*a^(7/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b+4*ln(2/sin(f*x+e)*(a^(1/2)*(a+b-b*cos(f*x
+e)^2)^(1/2)+a))*a^5+16*ln(2/sin(f*x+e)*(a^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^4*b+15*ln(2/sin(f*x+e)*(a^(1
/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^3*b^2)+cos(f*x+e)^2*b^2*(-8*a^(11/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1
/2)-6*(a+b-b*cos(f*x+e)^2)^(1/2)*a^(9/2)*b-26*a^(9/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b-6*(a+b-b*cos(f
*x+e)^2)^(1/2)*a^(7/2)*b^2-24*a^(7/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^2+6*ln(2/sin(f*x+e)*(a^(1/2)*(
a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^6+39*ln(2/sin(f*x+e)*(a^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^5*b+78*ln(2/sin
(f*x+e)*(a^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))*a^4*b^2+45*ln(2/sin(f*x+e)*(a^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)
+a))*a^3*b^3))/f

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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(cot(f*x+e)^3/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.77743, size = 1540, normalized size = 10.77 \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)^3/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(3*((2*a*b^2 + 5*b^3)*cos(f*x + e)^6 - (4*a^2*b + 16*a*b^2 + 15*b^3)*cos(f*x + e)^4 - 2*a^3 - 9*a^2*b -
12*a*b^2 - 5*b^3 + (2*a^3 + 13*a^2*b + 26*a*b^2 + 15*b^3)*cos(f*x + e)^2)*sqrt(a)*log(2*(b*cos(f*x + e)^2 - 2*
sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a) - 2*a - b)/(cos(f*x + e)^2 - 1)) + 2*(3*(2*a^2*b + 5*a*b^2)*cos(f*x +
e)^4 + 11*a^3 + 26*a^2*b + 15*a*b^2 - 2*(4*a^3 + 16*a^2*b + 15*a*b^2)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 +
a + b))/(a^4*b^2*f*cos(f*x + e)^6 - (2*a^5*b + 3*a^4*b^2)*f*cos(f*x + e)^4 + (a^6 + 4*a^5*b + 3*a^4*b^2)*f*co
s(f*x + e)^2 - (a^6 + 2*a^5*b + a^4*b^2)*f), -1/6*(3*((2*a*b^2 + 5*b^3)*cos(f*x + e)^6 - (4*a^2*b + 16*a*b^2 +
15*b^3)*cos(f*x + e)^4 - 2*a^3 - 9*a^2*b - 12*a*b^2 - 5*b^3 + (2*a^3 + 13*a^2*b + 26*a*b^2 + 15*b^3)*cos(f*x
+ e)^2)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/a) - (3*(2*a^2*b + 5*a*b^2)*cos(f*x + e)^4 +
11*a^3 + 26*a^2*b + 15*a*b^2 - 2*(4*a^3 + 16*a^2*b + 15*a*b^2)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)
)/(a^4*b^2*f*cos(f*x + e)^6 - (2*a^5*b + 3*a^4*b^2)*f*cos(f*x + e)^4 + (a^6 + 4*a^5*b + 3*a^4*b^2)*f*cos(f*x +
e)^2 - (a^6 + 2*a^5*b + a^4*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)**3/(a+b*sin(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )^{3}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(cot(f*x + e)^3/(b*sin(f*x + e)^2 + a)^(5/2), x)