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

Optimal. Leaf size=208 \[ \frac{8 a^2+40 a b+35 b^2}{8 a^4 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{8 a^2+40 a b+35 b^2}{24 a^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\left (8 a^2+40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{9/2} f}+\frac{(8 a+7 b) \csc ^2(e+f x)}{8 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[Out]

-((8*a^2 + 40*a*b + 35*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/(8*a^(9/2)*f) + (8*a^2 + 40*a*b + 35*
b^2)/(24*a^3*f*(a + b*Sin[e + f*x]^2)^(3/2)) + ((8*a + 7*b)*Csc[e + f*x]^2)/(8*a^2*f*(a + b*Sin[e + f*x]^2)^(3
/2)) - Csc[e + f*x]^4/(4*a*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (8*a^2 + 40*a*b + 35*b^2)/(8*a^4*f*Sqrt[a + b*Sin
[e + f*x]^2])

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Rubi [A]  time = 0.195977, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3194, 89, 78, 51, 63, 208} \[ \frac{8 a^2+40 a b+35 b^2}{8 a^4 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{8 a^2+40 a b+35 b^2}{24 a^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\left (8 a^2+40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{9/2} f}+\frac{(8 a+7 b) \csc ^2(e+f x)}{8 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*a^2 + 40*a*b + 35*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/(8*a^(9/2)*f) + (8*a^2 + 40*a*b + 35*
b^2)/(24*a^3*f*(a + b*Sin[e + f*x]^2)^(3/2)) + ((8*a + 7*b)*Csc[e + f*x]^2)/(8*a^2*f*(a + b*Sin[e + f*x]^2)^(3
/2)) - Csc[e + f*x]^4/(4*a*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (8*a^2 + 40*a*b + 35*b^2)/(8*a^4*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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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 ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(1-x)^2}{x^3 (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=-\frac{\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (-8 a-7 b)+2 a x}{x^2 (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 a f}\\ &=\frac{(8 a+7 b) \csc ^2(e+f x)}{8 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\left (8 a^2+40 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{16 a^2 f}\\ &=\frac{8 a^2+40 a b+35 b^2}{24 a^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{(8 a+7 b) \csc ^2(e+f x)}{8 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\left (8 a^2+40 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{16 a^3 f}\\ &=\frac{8 a^2+40 a b+35 b^2}{24 a^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{(8 a+7 b) \csc ^2(e+f x)}{8 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{8 a^2+40 a b+35 b^2}{8 a^4 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left (8 a^2+40 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{16 a^4 f}\\ &=\frac{8 a^2+40 a b+35 b^2}{24 a^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{(8 a+7 b) \csc ^2(e+f x)}{8 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{8 a^2+40 a b+35 b^2}{8 a^4 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left (8 a^2+40 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sin ^2(e+f x)}\right )}{8 a^4 b f}\\ &=-\frac{\left (8 a^2+40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{9/2} f}+\frac{8 a^2+40 a b+35 b^2}{24 a^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{(8 a+7 b) \csc ^2(e+f x)}{8 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\csc ^4(e+f x)}{4 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{8 a^2+40 a b+35 b^2}{8 a^4 f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.867869, size = 117, normalized size = 0.56 \[ \frac{\left (8 a^2+40 a b+35 b^2\right ) \csc ^2(e+f x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b \sin ^2(e+f x)}{a}+1\right )+3 a \csc ^4(e+f x) \left (-2 a \csc ^2(e+f x)+8 a+7 b\right )}{24 a^3 f \sqrt{a+b \sin ^2(e+f x)} \left (a \csc ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*Csc[e + f*x]^4*(8*a + 7*b - 2*a*Csc[e + f*x]^2) + (8*a^2 + 40*a*b + 35*b^2)*Csc[e + f*x]^2*Hypergeometric
2F1[-3/2, 1, -1/2, 1 + (b*Sin[e + f*x]^2)/a])/(24*a^3*f*(b + a*Csc[e + f*x]^2)*Sqrt[a + b*Sin[e + f*x]^2])

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Maple [B]  time = 4.849, size = 901, normalized size = 4.3 \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)^5/(a+b*sin(f*x+e)^2)^(5/2),x)

[Out]

-7/12/f/a^2/(-a*b)^(1/2)/(sin(f*x+e)+(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-13/6/f/a^3/(-a*b)^(1/
2)*b/(sin(f*x+e)+(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-19/12/f/a^4/(-a*b)^(1/2)/(sin(f*x+e)+(-a*
b)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)*b^2-1/12/f/a^2/b/(sin(f*x+e)-(-a*b)^(1/2)/b)^2*(-b*cos(f*x+e)^
2+(a*b+b^2)/b)^(1/2)-1/6/f/a^3/(sin(f*x+e)-(-a*b)^(1/2)/b)^2*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-1/12/f/a^4*b/
(sin(f*x+e)-(-a*b)^(1/2)/b)^2*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)+7/12/f/a^2/(-a*b)^(1/2)/(sin(f*x+e)-(-a*b)^(
1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)+13/6/f/a^3/(-a*b)^(1/2)*b/(sin(f*x+e)-(-a*b)^(1/2)/b)*(-b*cos(f*x+
e)^2+(a*b+b^2)/b)^(1/2)+19/12/f/a^4/(-a*b)^(1/2)/(sin(f*x+e)-(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/
2)*b^2-1/12/f/a^2/b/(sin(f*x+e)+(-a*b)^(1/2)/b)^2*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-1/6/f/a^3/(sin(f*x+e)+(-
a*b)^(1/2)/b)^2*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-1/12/f/a^4*b/(sin(f*x+e)+(-a*b)^(1/2)/b)^2*(-b*cos(f*x+e)^
2+(a*b+b^2)/b)^(1/2)-1/4/f/a^3/sin(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2)+11/8/f/a^4*b/sin(f*x+e)^2*(a+b*sin(f*x+e)
^2)^(1/2)-35/8/f/a^(9/2)*b^2*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e))-1/f/a^(5/2)*ln((2*a+2*a^(
1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e))-5/f/a^(7/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e))
*b+1/f/a^3/sin(f*x+e)^2*(a+b*sin(f*x+e)^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(cot(f*x+e)^5/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.16792, size = 2334, normalized size = 11.22 \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*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/48*(3*((8*a^2*b^2 + 40*a*b^3 + 35*b^4)*cos(f*x + e)^8 - 2*(8*a^3*b + 56*a^2*b^2 + 115*a*b^3 + 70*b^4)*cos(f
*x + e)^6 + (8*a^4 + 88*a^3*b + 323*a^2*b^2 + 450*a*b^3 + 210*b^4)*cos(f*x + e)^4 + 8*a^4 + 56*a^3*b + 123*a^2
*b^2 + 110*a*b^3 + 35*b^4 - 2*(8*a^4 + 64*a^3*b + 171*a^2*b^2 + 185*a*b^3 + 70*b^4)*cos(f*x + e)^2)*sqrt(a)*lo
g(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*(8*a
^3*b + 40*a^2*b^2 + 35*a*b^3)*cos(f*x + e)^6 - (32*a^4 + 232*a^3*b + 500*a^2*b^2 + 315*a*b^3)*cos(f*x + e)^4 -
50*a^4 - 205*a^3*b - 260*a^2*b^2 - 105*a*b^3 + (88*a^4 + 413*a^3*b + 640*a^2*b^2 + 315*a*b^3)*cos(f*x + e)^2)
*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^5*b^2*f*cos(f*x + e)^8 - 2*(a^6*b + 2*a^5*b^2)*f*cos(f*x + e)^6 + (a^7 +
6*a^6*b + 6*a^5*b^2)*f*cos(f*x + e)^4 - 2*(a^7 + 3*a^6*b + 2*a^5*b^2)*f*cos(f*x + e)^2 + (a^7 + 2*a^6*b + a^5*
b^2)*f), 1/24*(3*((8*a^2*b^2 + 40*a*b^3 + 35*b^4)*cos(f*x + e)^8 - 2*(8*a^3*b + 56*a^2*b^2 + 115*a*b^3 + 70*b^
4)*cos(f*x + e)^6 + (8*a^4 + 88*a^3*b + 323*a^2*b^2 + 450*a*b^3 + 210*b^4)*cos(f*x + e)^4 + 8*a^4 + 56*a^3*b +
123*a^2*b^2 + 110*a*b^3 + 35*b^4 - 2*(8*a^4 + 64*a^3*b + 171*a^2*b^2 + 185*a*b^3 + 70*b^4)*cos(f*x + e)^2)*sq
rt(-a)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/a) - (3*(8*a^3*b + 40*a^2*b^2 + 35*a*b^3)*cos(f*x + e)^
6 - (32*a^4 + 232*a^3*b + 500*a^2*b^2 + 315*a*b^3)*cos(f*x + e)^4 - 50*a^4 - 205*a^3*b - 260*a^2*b^2 - 105*a*b
^3 + (88*a^4 + 413*a^3*b + 640*a^2*b^2 + 315*a*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^5*b^2*
f*cos(f*x + e)^8 - 2*(a^6*b + 2*a^5*b^2)*f*cos(f*x + e)^6 + (a^7 + 6*a^6*b + 6*a^5*b^2)*f*cos(f*x + e)^4 - 2*(
a^7 + 3*a^6*b + 2*a^5*b^2)*f*cos(f*x + e)^2 + (a^7 + 2*a^6*b + a^5*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*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 )^{5}}{{\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)^5/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

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