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

Optimal. Leaf size=165 \[ \frac {\left (8 a^2-8 a b-b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 a^2 f}+\frac {(8 a+b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 a^2 f}-\frac {\left (8 a^2-8 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{3/2} f}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 a f} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3194, 89, 78, 50, 63, 208} \[ \frac {\left (8 a^2-8 a b-b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 a^2 f}-\frac {\left (8 a^2-8 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{3/2} f}+\frac {(8 a+b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 a^2 f}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 a f} \]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 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 \cot ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(1-x)^2 \sqrt {a+b x}}{x^3} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 a f}+\frac {\operatorname {Subst}\left (\int \frac {\left (\frac {1}{2} (-8 a-b)+2 a x\right ) \sqrt {a+b x}}{x^2} \, dx,x,\sin ^2(e+f x)\right )}{4 a f}\\ &=\frac {(8 a+b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 a^2 f}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 a f}+\frac {\left (8 a^2-8 a b-b^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\sin ^2(e+f x)\right )}{16 a^2 f}\\ &=\frac {\left (8 a^2-8 a b-b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 a^2 f}+\frac {(8 a+b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 a^2 f}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 a f}+\frac {\left (8 a^2-8 a b-b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{16 a f}\\ &=\frac {\left (8 a^2-8 a b-b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 a^2 f}+\frac {(8 a+b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 a^2 f}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 a f}+\frac {\left (8 a^2-8 a b-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 b f}\\ &=-\frac {\left (8 a^2-8 a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{3/2} f}+\frac {\left (8 a^2-8 a b-b^2\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 a^2 f}+\frac {(8 a+b) \csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 a^2 f}-\frac {\csc ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 a f}\\ \end {align*}

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Mathematica [A]  time = 0.57, size = 103, normalized size = 0.62 \[ \frac {\left (-8 a^2+8 a b+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )+\sqrt {a} \sqrt {a+b \sin ^2(e+f x)} \left ((8 a-b) \csc ^2(e+f x)-2 a \csc ^4(e+f x)+8 a\right )}{8 a^{3/2} f} \]

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

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fricas [A]  time = 2.06, size = 415, normalized size = 2.52 \[ \left [-\frac {{\left ({\left (8 \, a^{2} - 8 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (8 \, a^{2} - 8 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} - 8 \, a b - b^{2}\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left (8 \, a^{2} \cos \left (f x + e\right )^{4} - {\left (24 \, a^{2} - a b\right )} \cos \left (f x + e\right )^{2} + 14 \, a^{2} - a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )}}, \frac {{\left ({\left (8 \, a^{2} - 8 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (8 \, a^{2} - 8 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} - 8 \, a b - b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) + {\left (8 \, a^{2} \cos \left (f x + e\right )^{4} - {\left (24 \, a^{2} - a b\right )} \cos \left (f x + e\right )^{2} + 14 \, a^{2} - a b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{8 \, {\left (a^{2} f \cos \left (f x + e\right )^{4} - 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f\right )}}\right ] \]

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 - b^2)*cos(f*x + e)^4 - 2*(8*a^2 - 8*a*b - b^2)*cos(f*x + e)^2 + 8*a^2 - 8*a*b - 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*cos(f*x + e)^4 - (24*a^2 - a*b)*cos(f*x + e)^2 + 14*a^2 - a*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^2
*f*cos(f*x + e)^4 - 2*a^2*f*cos(f*x + e)^2 + a^2*f), 1/8*(((8*a^2 - 8*a*b - b^2)*cos(f*x + e)^4 - 2*(8*a^2 - 8
*a*b - b^2)*cos(f*x + e)^2 + 8*a^2 - 8*a*b - b^2)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/a)
+ (8*a^2*cos(f*x + e)^4 - (24*a^2 - a*b)*cos(f*x + e)^2 + 14*a^2 - a*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^2*
f*cos(f*x + e)^4 - 2*a^2*f*cos(f*x + e)^2 + a^2*f)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/
2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_n
ostep/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to chec
k sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2
*pi/x/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*
pi/t_nostep/2)Evaluation time: 0.85Unable to divide, perhaps due to rounding error%%%{1,[4,0]%%%}+%%%{%%{[-4,0
]:[1,0,%%%{-1,[1]%%%}]%%},[3,0]%%%}+%%%{8,[2,1]%%%}+%%%{%%%{6,[1]%%%},[2,0]%%%}+%%%{%%{[-16,0]:[1,0,%%%{-1,[1]
%%%}]%%},[1,1]%%%}+%%%{%%{[%%%{-4,[1]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,0]%%%}+%%%{16,[0,2]%%%}+%%%{%%%{8,[1]%
%%},[0,1]%%%}+%%%{%%%{1,[2]%%%},[0,0]%%%} / %%%{%%%{1,[1]%%%},[4,0]%%%}+%%%{%%{poly1[%%%{-4,[1]%%%},0]:[1,0,%%
%{-1,[1]%%%}]%%},[3,0]%%%}+%%%{%%%{8,[1]%%%},[2,1]%%%}+%%%{%%%{6,[2]%%%},[2,0]%%%}+%%%{%%{poly1[%%%{-16,[1]%%%
},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,1]%%%}+%%%{%%{poly1[%%%{-4,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,0]%%%}+%%%{%%
%{16,[1]%%%},[0,2]%%%}+%%%{%%%{8,[2]%%%},[0,1]%%%}+%%%{%%%{1,[3]%%%},[0,0]%%%} Error: Bad Argument Value

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maple [A]  time = 1.87, size = 230, normalized size = 1.39 \[ \frac {\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{f}-\frac {\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{f}+\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right ) b}{f \sqrt {a}}-\frac {b \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{8 f a \sin \left (f x +e \right )^{2}}+\frac {b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{8 f \,a^{\frac {3}{2}}}+\frac {\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{f \sin \left (f x +e \right )^{2}}-\frac {\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{4 f \sin \left (f x +e \right )^{4}} \]

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]

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

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maxima [A]  time = 0.38, size = 215, normalized size = 1.30 \[ -\frac {8 \, \sqrt {a} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right ) - \frac {8 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{\sqrt {a}} - \frac {b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {3}{2}}} - 8 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} + \frac {8 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b}{a} + \frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} b^{2}}{a^{2}} - \frac {8 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{a \sin \left (f x + e\right )^{2}} - \frac {{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b}{a^{2} \sin \left (f x + e\right )^{2}} + \frac {2 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{a \sin \left (f x + e\right )^{4}}}{8 \, f} \]

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

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

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 \[ \int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \cot ^{5}{\left (e + f x \right )}\, dx \]

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

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