\(\int \cot ^3(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\) [503]

Optimal result

Integrand size = 25, antiderivative size = 140 \[ \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {a} (2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 f}-\frac {(2 a-3 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}-\frac {(2 a-3 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 a f}-\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3273, 79, 52, 65, 214} \[ \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {\sqrt {a} (2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 f}-\frac {(2 a-3 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 a f}-\frac {(2 a-3 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}-\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a f} \]

[In]

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

[Out]

(Sqrt[a]*(2*a - 3*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/(2*f) - ((2*a - 3*b)*Sqrt[a + b*Sin[e + f*x]
^2])/(2*f) - ((2*a - 3*b)*(a + b*Sin[e + f*x]^2)^(3/2))/(6*a*f) - (Csc[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(5/2)
)/(2*a*f)

Rule 52

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 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 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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(1-x) (a+b x)^{3/2}}{x^2} \, dx,x,\sin ^2(e+f x)\right )}{2 f} \\ & = -\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a f}-\frac {(2 a-3 b) \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x} \, dx,x,\sin ^2(e+f x)\right )}{4 a f} \\ & = -\frac {(2 a-3 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 a f}-\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a f}-\frac {(2 a-3 b) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\sin ^2(e+f x)\right )}{4 f} \\ & = -\frac {(2 a-3 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}-\frac {(2 a-3 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 a f}-\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a f}-\frac {(a (2 a-3 b)) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 f} \\ & = -\frac {(2 a-3 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}-\frac {(2 a-3 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 a f}-\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a f}-\frac {(a (2 a-3 b)) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sin ^2(e+f x)}\right )}{2 b f} \\ & = \frac {\sqrt {a} (2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )}{2 f}-\frac {(2 a-3 b) \sqrt {a+b \sin ^2(e+f x)}}{2 f}-\frac {(2 a-3 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 a f}-\frac {\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.64 \[ \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {3 \sqrt {a} (2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )+\left (-8 a+5 b+b \cos (2 (e+f x))-3 a \csc ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}{6 f} \]

[In]

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

[Out]

(3*Sqrt[a]*(2*a - 3*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]] + (-8*a + 5*b + b*Cos[2*(e + f*x)] - 3*a*Cs
c[e + f*x]^2)*Sqrt[a + b*Sin[e + f*x]^2])/(6*f)

Maple [A] (verified)

Time = 1.14 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.18

method	result	size
default	\(\frac {-\frac {b \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{3}-\frac {4 a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{3}+b \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}-\frac {a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{2 \sin \left (f x +e \right )^{2}}-\frac {3 \sqrt {a}\, b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{2}+a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{f}\)	\(165\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.97 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.01 \[ \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\left [-\frac {3 \, {\left ({\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a + 3 \, b\right )} \sqrt {a} \log \left (\frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left (2 \, b \cos \left (f x + e\right )^{4} - 2 \, {\left (4 \, a - b\right )} \cos \left (f x + e\right )^{2} + 11 \, a - 4 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{12 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )}}, -\frac {3 \, {\left ({\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a + 3 \, b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{a}\right ) - {\left (2 \, b \cos \left (f x + e\right )^{4} - 2 \, {\left (4 \, a - b\right )} \cos \left (f x + e\right )^{2} + 11 \, a - 4 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )}}\right ] \]

[In]

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

[Out]

[-1/12*(3*((2*a - 3*b)*cos(f*x + e)^2 - 2*a + 3*b)*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*(2*b*cos(f*x + e)^4 - 2*(4*a - b)*cos(f*x + e)^2 + 11*a
- 4*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2 - f), -1/6*(3*((2*a - 3*b)*cos(f*x + e)^2 - 2*a + 3*
b)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/a) - (2*b*cos(f*x + e)^4 - 2*(4*a - b)*cos(f*x + e
)^2 + 11*a - 4*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2 - f)]

Sympy [F]

\[ \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int \left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \cot ^{3}{\left (e + f x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.06 \[ \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\frac {6 \, a^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right ) - 9 \, \sqrt {a} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right ) - 2 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} - 6 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} a + 9 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b + \frac {3 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b}{a} - \frac {3 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}{a \sin \left (f x + e\right )^{2}}}{6 \, f} \]

[In]

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

[Out]

1/6*(6*a^(3/2)*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e)))) - 9*sqrt(a)*b*arcsinh(a/(sqrt(a*b)*abs(sin(f*x + e))))
- 2*(b*sin(f*x + e)^2 + a)^(3/2) - 6*sqrt(b*sin(f*x + e)^2 + a)*a + 9*sqrt(b*sin(f*x + e)^2 + a)*b + 3*(b*sin
(f*x + e)^2 + a)^(3/2)*b/a - 3*(b*sin(f*x + e)^2 + a)^(5/2)/(a*sin(f*x + e)^2))/f

Giac [F(-1)]

Timed out. \[ \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \cot ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^3\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]

[In]

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

[Out]

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