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

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

[Out]

1/2*(2*a+5*b)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/2))/a^(7/2)/f+1/6*(-2*a-5*b)/a^2/f/(a+b*sin(f*x+e)^2)^(3/2
)-1/2*csc(f*x+e)^2/a/f/(a+b*sin(f*x+e)^2)^(3/2)+1/2*(-2*a-5*b)/a^3/f/(a+b*sin(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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 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 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 ^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*}

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Mathematica [C]  time = 0.25, 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]

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

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fricas [B]  time = 0.61, size = 666, normalized size = 4.66 \[ \left [\frac {3 \, {\left ({\left (2 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (4 \, a^{2} b + 16 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 2 \, a^{3} - 9 \, a^{2} b - 12 \, a b^{2} - 5 \, b^{3} + {\left (2 \, a^{3} + 13 \, a^{2} b + 26 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{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 (3 \, {\left (2 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} + 11 \, a^{3} + 26 \, a^{2} b + 15 \, a b^{2} - 2 \, {\left (4 \, a^{3} + 16 \, a^{2} b + 15 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{12 \, {\left (a^{4} b^{2} f \cos \left (f x + e\right )^{6} - {\left (2 \, a^{5} b + 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{6} + 4 \, a^{5} b + 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} f\right )}}, -\frac {3 \, {\left ({\left (2 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (4 \, a^{2} b + 16 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 2 \, a^{3} - 9 \, a^{2} b - 12 \, a b^{2} - 5 \, b^{3} + {\left (2 \, a^{3} + 13 \, a^{2} b + 26 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) - {\left (3 \, {\left (2 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} + 11 \, a^{3} + 26 \, a^{2} b + 15 \, a b^{2} - 2 \, {\left (4 \, a^{3} + 16 \, a^{2} b + 15 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, {\left (a^{4} b^{2} f \cos \left (f x + e\right )^{6} - {\left (2 \, a^{5} b + 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{6} + 4 \, a^{5} b + 3 \, a^{4} b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{6} + 2 \, a^{5} b + a^{4} b^{2}\right )} f\right )}}\right ] \]

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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giac [B]  time = 1.36, size = 751, normalized size = 5.25 \[ -\frac {\frac {{\left ({\left ({\left (\frac {3 \, {\left (a^{12} b^{2} + 2 \, a^{11} b^{3} + a^{10} b^{4}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{13} b^{2} + 2 \, a^{12} b^{3} + a^{11} b^{4}} + \frac {4 \, {\left (11 \, a^{12} b^{2} + 42 \, a^{11} b^{3} + 51 \, a^{10} b^{4} + 20 \, a^{9} b^{5}\right )}}{a^{13} b^{2} + 2 \, a^{12} b^{3} + a^{11} b^{4}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \frac {6 \, {\left (19 \, a^{12} b^{2} + 90 \, a^{11} b^{3} + 163 \, a^{10} b^{4} + 132 \, a^{9} b^{5} + 40 \, a^{8} b^{6}\right )}}{a^{13} b^{2} + 2 \, a^{12} b^{3} + a^{11} b^{4}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \frac {12 \, {\left (9 \, a^{12} b^{2} + 42 \, a^{11} b^{3} + 73 \, a^{10} b^{4} + 56 \, a^{9} b^{5} + 16 \, a^{8} b^{6}\right )}}{a^{13} b^{2} + 2 \, a^{12} b^{3} + a^{11} b^{4}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \frac {7 \, {\left (5 \, a^{12} b^{2} + 18 \, a^{11} b^{3} + 21 \, a^{10} b^{4} + 8 \, a^{9} b^{5}\right )}}{a^{13} b^{2} + 2 \, a^{12} b^{3} + a^{11} b^{4}}}{{\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a\right )}^{\frac {3}{2}}} + \frac {6 \, {\left (2 \, a^{\frac {3}{2}} + 5 \, \sqrt {a} b\right )} \log \left ({\left | -{\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )} a - a^{\frac {3}{2}} - 2 \, \sqrt {a} b \right |}\right )}{a^{4}} - \frac {6 \, {\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )} a^{\frac {3}{2}} + 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )} \sqrt {a} b + a^{2}\right )}}{{\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} \sqrt {a} - a^{\frac {3}{2}}\right )} a^{3}}}{24 \, f} \]

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]

-1/24*(((((3*(a^12*b^2 + 2*a^11*b^3 + a^10*b^4)*tan(1/2*f*x + 1/2*e)^2/(a^13*b^2 + 2*a^12*b^3 + a^11*b^4) + 4*
(11*a^12*b^2 + 42*a^11*b^3 + 51*a^10*b^4 + 20*a^9*b^5)/(a^13*b^2 + 2*a^12*b^3 + a^11*b^4))*tan(1/2*f*x + 1/2*e
)^2 + 6*(19*a^12*b^2 + 90*a^11*b^3 + 163*a^10*b^4 + 132*a^9*b^5 + 40*a^8*b^6)/(a^13*b^2 + 2*a^12*b^3 + a^11*b^
4))*tan(1/2*f*x + 1/2*e)^2 + 12*(9*a^12*b^2 + 42*a^11*b^3 + 73*a^10*b^4 + 56*a^9*b^5 + 16*a^8*b^6)/(a^13*b^2 +
2*a^12*b^3 + a^11*b^4))*tan(1/2*f*x + 1/2*e)^2 + 7*(5*a^12*b^2 + 18*a^11*b^3 + 21*a^10*b^4 + 8*a^9*b^5)/(a^13
*b^2 + 2*a^12*b^3 + a^11*b^4))/(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)^(3/2) + 6*(2*a^(3/2) + 5*sqrt(a)*b)*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^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))*a^(3/2) + 2*(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)*b + a^2)/(((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*ta
n(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*sqrt(a) - a^(3/2))*a^3)
)/f

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maple [B]  time = 4.28, size = 1038, normalized size = 7.26 \[ \text {result too large to display} \]

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*cos(f*x+e)^2*a*b+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*a^6*b^2*ln(2/sin(f*x+e)*(a^(1/2)
*(a+b-b*cos(f*x+e)^2)^(1/2)+a))+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*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^(9/2)*b^3+6*(a+b-b*cos(f*x+e)^2)^(1/2)*a^(9/2)*
b^3+12*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^(7/2)*b^4-27*a^5*b^3*ln(2/sin(f*x+e)*(a^(1/2)*(a+b-b*cos(f*x+
e)^2)^(1/2)+a))-36*a^4*b^4*ln(2/sin(f*x+e)*(a^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+a))-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*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^(9/2)+(a+b-b*cos(f*x+e)^2)^
(1/2)*a^(7/2)*b+4*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^(7/2)*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*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^(11/2
)+6*(a+b-b*cos(f*x+e)^2)^(1/2)*a^(9/2)*b+26*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^(9/2)*b+6*(a+b-b*cos(f*x
+e)^2)^(1/2)*a^(7/2)*b^2+24*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^(7/2)*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 [A]  time = 0.44, size = 156, normalized size = 1.09 \[ \frac {\frac {6 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {5}{2}}} + \frac {15 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {7}{2}}} - \frac {6}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} - \frac {2}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a} - \frac {15 \, b}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{3}} - \frac {5 \, b}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {3}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \sin \left (f x + e\right )^{2}}}{6 \, f} \]

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]

1/6*(6*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/a^(5/2) + 15*b*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/a^(7/2
) - 6/(sqrt(b*sin(f*x + e)^2 + a)*a^2) - 2/((b*sin(f*x + e)^2 + a)^(3/2)*a) - 15*b/(sqrt(b*sin(f*x + e)^2 + a)
*a^3) - 5*b/((b*sin(f*x + e)^2 + a)^(3/2)*a^2) - 3/((b*sin(f*x + e)^2 + a)^(3/2)*a*sin(f*x + e)^2))/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cot}\left (e+f\,x\right )}^3}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{3}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

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]

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

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