3.6.26 \(\int \frac {\cot ^5(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [526]

Optimal. Leaf size=167 \[ -\frac {\left (8 a^2+24 a b+15 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{7/2} f}+\frac {8 a^2+24 a b+15 b^2}{8 a^3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(8 a+5 b) \csc ^2(e+f x)}{8 a^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\csc ^4(e+f x)}{4 a f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

-1/8*(8*a^2+24*a*b+15*b^2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/a^(1/2))/a^(7/2)/f+1/8*(8*a^2+24*a*b+15*b^2)/a^3/f
/(a+b*sin(f*x+e)^2)^(1/2)+1/8*(8*a+5*b)*csc(f*x+e)^2/a^2/f/(a+b*sin(f*x+e)^2)^(1/2)-1/4*csc(f*x+e)^4/a/f/(a+b*
sin(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.11, antiderivative size = 167, 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 = {3273, 91, 79, 53, 65, 214} \begin {gather*} \frac {(8 a+5 b) \csc ^2(e+f x)}{8 a^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (8 a^2+24 a b+15 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{7/2} f}+\frac {8 a^2+24 a b+15 b^2}{8 a^3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\csc ^4(e+f x)}{4 a f \sqrt {a+b \sin ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^5/(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

-1/8*((8*a^2 + 24*a*b + 15*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/(a^(7/2)*f) + (8*a^2 + 24*a*b + 1
5*b^2)/(8*a^3*f*Sqrt[a + b*Sin[e + f*x]^2]) + ((8*a + 5*b)*Csc[e + f*x]^2)/(8*a^2*f*Sqrt[a + b*Sin[e + f*x]^2]
) - Csc[e + f*x]^4/(4*a*f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 53

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 65

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 79

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] || L
tQ[p, n]))))

Rule 91

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 3273

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)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {(1-x)^2}{x^3 (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=-\frac {\csc ^4(e+f x)}{4 a f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {\frac {1}{2} (-8 a-5 b)+2 a x}{x^2 (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 a f}\\ &=\frac {(8 a+5 b) \csc ^2(e+f x)}{8 a^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\csc ^4(e+f x)}{4 a f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\left (8 a^2+24 a b+15 b^2\right ) \text {Subst}\left (\int \frac {1}{x (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{16 a^2 f}\\ &=\frac {8 a^2+24 a b+15 b^2}{8 a^3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(8 a+5 b) \csc ^2(e+f x)}{8 a^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\csc ^4(e+f x)}{4 a f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\left (8 a^2+24 a b+15 b^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{16 a^3 f}\\ &=\frac {8 a^2+24 a b+15 b^2}{8 a^3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(8 a+5 b) \csc ^2(e+f x)}{8 a^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\csc ^4(e+f x)}{4 a f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\left (8 a^2+24 a b+15 b^2\right ) \text {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^3 b f}\\ &=-\frac {\left (8 a^2+24 a b+15 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{8 a^{7/2} f}+\frac {8 a^2+24 a b+15 b^2}{8 a^3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(8 a+5 b) \csc ^2(e+f x)}{8 a^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\csc ^4(e+f x)}{4 a f \sqrt {a+b \sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.22, size = 94, normalized size = 0.56 \begin {gather*} \frac {a \csc ^2(e+f x) \left (8 a+5 b-2 a \csc ^2(e+f x)\right )+\left (8 a^2+24 a b+15 b^2\right ) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};1+\frac {b \sin ^2(e+f x)}{a}\right )}{8 a^3 f \sqrt {a+b \sin ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^5/(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

(a*Csc[e + f*x]^2*(8*a + 5*b - 2*a*Csc[e + f*x]^2) + (8*a^2 + 24*a*b + 15*b^2)*Hypergeometric2F1[-1/2, 1, 1/2,
1 + (b*Sin[e + f*x]^2)/a])/(8*a^3*f*Sqrt[a + b*Sin[e + f*x]^2])

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Maple [A]
time = 10.39, size = 265, normalized size = 1.59

method	result	size
default	\(\frac {\frac {1}{a \sin \left (f x +e \right )^{2} \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}+\frac {3 b}{a^{2} \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}-\frac {3 b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{a^{\frac {5}{2}}}+\frac {1}{a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{a^{\frac {3}{2}}}-\frac {1}{4 a \sin \left (f x +e \right )^{4} \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}+\frac {5 b}{8 a^{2} \sin \left (f x +e \right )^{2} \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}+\frac {15 b^{2}}{8 a^{3} \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}-\frac {15 b^{2} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{8 a^{\frac {7}{2}}}}{f}\)	\(265\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/a/sin(f*x+e)^2/(a+b*sin(f*x+e)^2)^(1/2)+3/a^2*b/(a+b*sin(f*x+e)^2)^(1/2)-3/a^(5/2)*b*ln((2*a+2*a^(1/2)*(a+b
*sin(f*x+e)^2)^(1/2))/sin(f*x+e))+1/a/(a+b*sin(f*x+e)^2)^(1/2)-1/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/a/sin(f*x+e)^4/(a+b*sin(f*x+e)^2)^(1/2)+5/8/a^2*b/sin(f*x+e)^2/(a+b*sin(f*x+e)^2)^(1/2)
+15/8/a^3*b^2/(a+b*sin(f*x+e)^2)^(1/2)-15/8/a^(7/2)*b^2*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e)
))/f

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Maxima [A]
time = 0.29, size = 231, normalized size = 1.38 \begin {gather*} -\frac {\frac {8 \, \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {3}{2}}} + \frac {24 \, b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {5}{2}}} + \frac {15 \, b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right )}{a^{\frac {7}{2}}} - \frac {8}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a} - \frac {24 \, b}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} - \frac {15 \, b^{2}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{3}} - \frac {8}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a \sin \left (f x + e\right )^{2}} - \frac {5 \, b}{\sqrt {b \sin \left (f x + e\right )^{2} + a} 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

-1/8*(8*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/a^(3/2) + 24*b*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/a^(5/
2) + 15*b^2*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))/a^(7/2) - 8/(sqrt(b*sin(f*x + e)^2 + a)*a) - 24*b/(sqrt(b
*sin(f*x + e)^2 + a)*a^2) - 15*b^2/(sqrt(b*sin(f*x + e)^2 + a)*a^3) - 8/(sqrt(b*sin(f*x + e)^2 + a)*a*sin(f*x
+ e)^2) - 5*b/(sqrt(b*sin(f*x + e)^2 + a)*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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (147) = 294\).
time = 0.53, size = 652, normalized size = 3.90 \begin {gather*} \left [\frac {{\left ({\left (8 \, a^{2} b + 24 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (8 \, a^{3} + 48 \, a^{2} b + 87 \, a b^{2} + 45 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 8 \, a^{3} - 32 \, a^{2} b - 39 \, a b^{2} - 15 \, b^{3} + {\left (16 \, a^{3} + 72 \, a^{2} b + 102 \, a b^{2} + 45 \, 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 ({\left (8 \, a^{3} + 24 \, a^{2} b + 15 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} + 14 \, a^{3} + 29 \, a^{2} b + 15 \, a b^{2} - {\left (24 \, a^{3} + 53 \, a^{2} b + 30 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a^{4} b f \cos \left (f x + e\right )^{6} - {\left (a^{5} + 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{4} + {\left (2 \, a^{5} + 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{5} + a^{4} b\right )} f\right )}}, \frac {{\left ({\left (8 \, a^{2} b + 24 \, a b^{2} + 15 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (8 \, a^{3} + 48 \, a^{2} b + 87 \, a b^{2} + 45 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 8 \, a^{3} - 32 \, a^{2} b - 39 \, a b^{2} - 15 \, b^{3} + {\left (16 \, a^{3} + 72 \, a^{2} b + 102 \, a b^{2} + 45 \, 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 ({\left (8 \, a^{3} + 24 \, a^{2} b + 15 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} + 14 \, a^{3} + 29 \, a^{2} b + 15 \, a b^{2} - {\left (24 \, a^{3} + 53 \, a^{2} b + 30 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{8 \, {\left (a^{4} b f \cos \left (f x + e\right )^{6} - {\left (a^{5} + 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{4} + {\left (2 \, a^{5} + 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{5} + a^{4} b\right )} f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(((8*a^2*b + 24*a*b^2 + 15*b^3)*cos(f*x + e)^6 - (8*a^3 + 48*a^2*b + 87*a*b^2 + 45*b^3)*cos(f*x + e)^4 -
8*a^3 - 32*a^2*b - 39*a*b^2 - 15*b^3 + (16*a^3 + 72*a^2*b + 102*a*b^2 + 45*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*((8*a^3 +
24*a^2*b + 15*a*b^2)*cos(f*x + e)^4 + 14*a^3 + 29*a^2*b + 15*a*b^2 - (24*a^3 + 53*a^2*b + 30*a*b^2)*cos(f*x +
e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^4*b*f*cos(f*x + e)^6 - (a^5 + 3*a^4*b)*f*cos(f*x + e)^4 + (2*a^5 + 3
*a^4*b)*f*cos(f*x + e)^2 - (a^5 + a^4*b)*f), 1/8*(((8*a^2*b + 24*a*b^2 + 15*b^3)*cos(f*x + e)^6 - (8*a^3 + 48*
a^2*b + 87*a*b^2 + 45*b^3)*cos(f*x + e)^4 - 8*a^3 - 32*a^2*b - 39*a*b^2 - 15*b^3 + (16*a^3 + 72*a^2*b + 102*a*
b^2 + 45*b^3)*cos(f*x + e)^2)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/a) - ((8*a^3 + 24*a^2*b
+ 15*a*b^2)*cos(f*x + e)^4 + 14*a^3 + 29*a^2*b + 15*a*b^2 - (24*a^3 + 53*a^2*b + 30*a*b^2)*cos(f*x + e)^2)*sq
rt(-b*cos(f*x + e)^2 + a + b))/(a^4*b*f*cos(f*x + e)^6 - (a^5 + 3*a^4*b)*f*cos(f*x + e)^4 + (2*a^5 + 3*a^4*b)*
f*cos(f*x + e)^2 - (a^5 + a^4*b)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1150 vs. \(2 (153) = 306\).
time = 1.05, size = 1150, normalized size = 6.89 \begin {gather*} -\frac {\frac {{\left ({\left (\frac {{\left (a^{8} b + a^{7} b^{2}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{9} b + a^{8} b^{2}} - \frac {11 \, a^{8} b + 21 \, a^{7} b^{2} + 10 \, a^{6} b^{3}}{a^{9} b + a^{8} b^{2}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \frac {89 \, a^{8} b + 297 \, a^{7} b^{2} + 328 \, a^{6} b^{3} + 120 \, a^{5} b^{4}}{a^{9} b + a^{8} b^{2}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \frac {77 \, a^{8} b + 219 \, a^{7} b^{2} + 206 \, a^{6} b^{3} + 64 \, a^{5} b^{4}}{a^{9} b + a^{8} b^{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}} - \frac {8 \, {\left (8 \, a^{2} + 24 \, a b + 15 \, b^{2}\right )} \arctan \left (-\frac {\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}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} - \frac {4 \, {\left (8 \, a^{\frac {5}{2}} + 24 \, a^{\frac {3}{2}} b + 15 \, \sqrt {a} b^{2}\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 {4 \, {\left (6 \, {\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 )}^{3} a^{2} + 20 \, {\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 )}^{3} a b + 14 \, {\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 )}^{3} b^{2} + 5 \, {\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} a^{\frac {5}{2}} + 4 \, {\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} a^{\frac {3}{2}} b - 8 \, {\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^{3} - 24 \, {\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^{2} b - 18 \, {\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 b^{2} - 7 \, a^{\frac {7}{2}} - 8 \, a^{\frac {5}{2}} b\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} - a\right )}^{2} a^{3}}}{64 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

-1/64*(((((a^8*b + a^7*b^2)*tan(1/2*f*x + 1/2*e)^2/(a^9*b + a^8*b^2) - (11*a^8*b + 21*a^7*b^2 + 10*a^6*b^3)/(a
^9*b + a^8*b^2))*tan(1/2*f*x + 1/2*e)^2 - (89*a^8*b + 297*a^7*b^2 + 328*a^6*b^3 + 120*a^5*b^4)/(a^9*b + a^8*b^
2))*tan(1/2*f*x + 1/2*e)^2 - (77*a^8*b + 219*a^7*b^2 + 206*a^6*b^3 + 64*a^5*b^4)/(a^9*b + a^8*b^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) - 8*(8*a^2 + 24*a*b + 15*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^3) - 4*(8*a^(5/2) + 24*a^(3/2)*b + 15*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^4 + 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 + 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))^3*a*b
+ 14*(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) + 4*(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^(3/2)*b - 8*(sqrt(a)*t
an(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 - 24*(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 - 18*(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) - 8*a^(5/2)*b)/(((sqrt(a)*ta
n(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^3))/f

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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