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

Optimal. Leaf size=126 \[ \frac {(8 a+3 b) \csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 a^2 f}-\frac {\left (8 a^2+8 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{5/2} f}-\frac {\csc ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 a f} \]

[Out]

-1/8*(8*a^2+8*a*b+3*b^2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/2))/a^(5/2)/f+1/8*(8*a+3*b)*csc(f*x+e)^2*(a+b*s
in(f*x+e)^2)^(1/2)/a^2/f-1/4*csc(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2)/a/f

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Rubi [A]  time = 0.13, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3194, 89, 78, 63, 208} \[ -\frac {\left (8 a^2+8 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{5/2} f}+\frac {(8 a+3 b) \csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 a^2 f}-\frac {\csc ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*a^2 + 8*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/(8*a^(5/2)*f) + ((8*a + 3*b)*Csc[e + f*
x]^2*Sqrt[a + b*Sin[e + f*x]^2])/(8*a^2*f) - (Csc[e + f*x]^4*Sqrt[a + b*Sin[e + f*x]^2])/(4*a*f)

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 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 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 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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.36, size = 101, normalized size = 0.80 \[ \frac {\sqrt {a} \csc ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)} \left (-2 a \csc ^2(e+f x)+8 a+3 b\right )-\left (8 a^2+8 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{5/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((8*a^2 + 8*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]]) + Sqrt[a]*Csc[e + f*x]^2*(8*a + 3*b -
2*a*Csc[e + f*x]^2)*Sqrt[a + b*Sin[e + f*x]^2])/(8*a^(5/2)*f)

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fricas [A]  time = 0.61, size = 388, normalized size = 3.08 \[ \left [\frac {{\left ({\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left ({\left (8 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - 6 \, a^{2} - 3 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )}}, \frac {{\left ({\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) - {\left ({\left (8 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - 6 \, a^{2} - 3 \, a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{8 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(((8*a^2 + 8*a*b + 3*b^2)*cos(f*x + e)^4 - 2*(8*a^2 + 8*a*b + 3*b^2)*cos(f*x + e)^2 + 8*a^2 + 8*a*b + 3*
b^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*((8*a^2 + 3*a*b)*cos(f*x + e)^2 - 6*a^2 - 3*a*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^3*f*cos(f*x + e)^
4 - 2*a^3*f*cos(f*x + e)^2 + a^3*f), 1/8*(((8*a^2 + 8*a*b + 3*b^2)*cos(f*x + e)^4 - 2*(8*a^2 + 8*a*b + 3*b^2)*
cos(f*x + e)^2 + 8*a^2 + 8*a*b + 3*b^2)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/a) - ((8*a^2
+ 3*a*b)*cos(f*x + e)^2 - 6*a^2 - 3*a*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^3*f*cos(f*x + e)^4 - 2*a^3*f*cos(
f*x + e)^2 + a^3*f)]

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giac [B]  time = 1.03, size = 898, normalized size = 7.13 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

-1/64*(sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a)*(tan(1/2*f
*x + 1/2*e)^2/a - (13*a + 6*b)/a^2) - 8*(8*a^2 + 8*a*b + 3*b^2)*arctan(-(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt
(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))/sqrt(-a))/(sqrt(-a)*
a^2) - 4*(8*a^(5/2) + 8*a^(3/2)*b + 3*sqrt(a)*b^2)*log(abs(-(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f
*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a - a^(3/2) - 2*sqrt(a)*b))/a^3
+ 4*(6*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(
1/2*f*x + 1/2*e)^2 + a))^3*a^2 + 16*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(
1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*a*b + 6*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1
/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*b^2 + 5*(sqrt(a)*tan(1/2*f
*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^
2*a^(5/2) - 8*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4
*b*tan(1/2*f*x + 1/2*e)^2 + a))*a^3 - 20*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a
*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a^2*b - 10*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a
*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a*b^2 - 7*a^(7/2) - 4*
a^(5/2)*b)/(((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*
b*tan(1/2*f*x + 1/2*e)^2 + a))^2 - a)^2*a^2))/f

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maple [A]  time = 1.97, size = 219, normalized size = 1.74 \[ -\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{\sqrt {a}\, f}+\frac {\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{f a \sin \left (f x +e \right )^{2}}-\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{f \,a^{\frac {3}{2}}}-\frac {\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{4 f a \sin \left (f x +e \right )^{4}}+\frac {3 b \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{8 f \,a^{2} \sin \left (f x +e \right )^{2}}-\frac {3 b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{8 f \,a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^5/(a+b*sin(f*x+e)^2)^(1/2),x)

[Out]

-1/a^(1/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e))/f+1/f/a/sin(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/
2)-1/f*b/a^(3/2)*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e))-1/4/f/a/sin(f*x+e)^4*(a+b*sin(f*x+e)^
2)^(1/2)+3/8/f/a^2*b/sin(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2)-3/8/f/a^(5/2)*b^2*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)
^2)^(1/2))/sin(f*x+e))

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maxima [A]  time = 0.32, size = 158, normalized size = 1.25 \[ -\frac {\frac {8 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{\sqrt {a}} + \frac {8 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {3}{2}}} + \frac {3 \, b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {5}{2}}} - \frac {8 \, \sqrt {b \sin \left (f x + e\right )^{2} + a}}{a \sin \left (f x + e\right )^{2}} - \frac {3 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b}{a^{2} \sin \left (f x + e\right )^{2}} + \frac {2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a}}{a \sin \left (f x + e\right )^{4}}}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/8*(8*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/sqrt(a) + 8*b*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/a^(3/2
) + 3*b^2*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/a^(5/2) - 8*sqrt(b*sin(f*x + e)^2 + a)/(a*sin(f*x + e)^2) -
3*sqrt(b*sin(f*x + e)^2 + a)*b/(a^2*sin(f*x + e)^2) + 2*sqrt(b*sin(f*x + e)^2 + a)/(a*sin(f*x + e)^4))/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cot}\left (e+f\,x\right )}^5}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^5/(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

int(cot(e + f*x)^5/(a + b*sin(e + f*x)^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{5}{\left (e + f x \right )}}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**5/(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

Integral(cot(e + f*x)**5/sqrt(a + b*sin(e + f*x)**2), x)

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