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

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

[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)

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Rubi [A]  time = 0.128517, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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 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)}{\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*}

Mathematica [A]  time = 0.36684, size = 101, normalized size = 0.8 \[ \frac{\sqrt{a} \csc ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)} \left (-2 a \csc ^2(e+f x)+8 a+3 b\right )-\left (8 a^2+8 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{5/2} f} \]

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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Maple [A]  time = 1.662, size = 219, normalized size = 1.7 \begin{align*} -{\frac{1}{f}\ln \left ({\frac{1}{\sin \left ( fx+e \right ) } \left ( 2\,a+2\,\sqrt{a}\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{1}{af \left ( \sin \left ( fx+e \right ) \right ) ^{2}}\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}-{\frac{b}{f}\ln \left ({\frac{1}{\sin \left ( fx+e \right ) } \left ( 2\,a+2\,\sqrt{a}\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{1}{4\,af \left ( \sin \left ( fx+e \right ) \right ) ^{4}}\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}+{\frac{3\,b}{8\,{a}^{2}f \left ( \sin \left ( fx+e \right ) \right ) ^{2}}\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}-{\frac{3\,{b}^{2}}{8\,f}\ln \left ({\frac{1}{\sin \left ( fx+e \right ) } \left ( 2\,a+2\,\sqrt{a}\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \right ) } \right ){a}^{-{\frac{5}{2}}}} \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)^(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 [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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.08619, size = 923, normalized size = 7.33 \begin{align*} \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 ] \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)^(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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{5}{\left (e + f x \right )}}{\sqrt{a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \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)**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )^{5}}{\sqrt{b \sin \left (f x + e\right )^{2} + a}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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