\(\int \cot (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx\) [555]

Optimal result

Integrand size = 23, antiderivative size = 59 \[ \int \cot (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )}{2 d}+\frac {\sqrt {a+b \sin ^4(c+d x)}}{2 d} \]

[Out]

-1/2*arctanh((a+b*sin(d*x+c)^4)^(1/2)/a^(1/2))*a^(1/2)/d+1/2*(a+b*sin(d*x+c)^4)^(1/2)/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3308, 272, 52, 65, 214} \[ \int \cot (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\frac {\sqrt {a+b \sin ^4(c+d x)}}{2 d}-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )}{2 d} \]

[In]

Int[Cot[c + d*x]*Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

-1/2*(Sqrt[a]*ArcTanh[Sqrt[a + b*Sin[c + d*x]^4]/Sqrt[a]])/d + Sqrt[a + b*Sin[c + d*x]^4]/(2*d)

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 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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 3308

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[x^((m - 1)/2)*((a + b*ff^(n/2)*x^(n/2))^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] &
& IntegerQ[n/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\sqrt {a+b x^2}}{x} \, dx,x,\sin ^2(c+d x)\right )}{2 d} \\ & = \frac {\text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\sin ^4(c+d x)\right )}{4 d} \\ & = \frac {\sqrt {a+b \sin ^4(c+d x)}}{2 d}+\frac {a \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sin ^4(c+d x)\right )}{4 d} \\ & = \frac {\sqrt {a+b \sin ^4(c+d x)}}{2 d}+\frac {a \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sin ^4(c+d x)}\right )}{2 b d} \\ & = -\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )}{2 d}+\frac {\sqrt {a+b \sin ^4(c+d x)}}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.92 \[ \int \cot (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\frac {-\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )+\sqrt {a+b \sin ^4(c+d x)}}{2 d} \]

[In]

Integrate[Cot[c + d*x]*Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

(-(Sqrt[a]*ArcTanh[Sqrt[a + b*Sin[c + d*x]^4]/Sqrt[a]]) + Sqrt[a + b*Sin[c + d*x]^4])/(2*d)

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {\frac {\sqrt {a +b \left (\sin ^{4}\left (d x +c \right )\right )}}{2}-\frac {\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{4}\left (d x +c \right )\right )}}{\sin \left (d x +c \right )^{2}}\right )}{2}}{d}\)	\(60\)

[In]

int(cot(d*x+c)*(a+b*sin(d*x+c)^4)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/2*(a+b*sin(d*x+c)^4)^(1/2)-1/2*a^(1/2)*ln((2*a+2*a^(1/2)*(a+b*sin(d*x+c)^4)^(1/2))/sin(d*x+c)^2))

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 195, normalized size of antiderivative = 3.31 \[ \int \cot (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\left [\frac {\sqrt {a} \log \left (\frac {8 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} \sqrt {a} + 2 \, a + b\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b}}{4 \, d}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} \sqrt {-a}}{a}\right ) + \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b}}{2 \, d}\right ] \]

[In]

integrate(cot(d*x+c)*(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(sqrt(a)*log(8*(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 - 2*sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a
+ b)*sqrt(a) + 2*a + b)/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)) + 2*sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^
2 + a + b))/d, 1/2*(sqrt(-a)*arctan(sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b)*sqrt(-a)/a) + sqrt(b*c
os(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b))/d]

Sympy [F]

\[ \int \cot (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\int \sqrt {a + b \sin ^{4}{\left (c + d x \right )}} \cot {\left (c + d x \right )}\, dx \]

[In]

integrate(cot(d*x+c)*(a+b*sin(d*x+c)**4)**(1/2),x)

[Out]

Integral(sqrt(a + b*sin(c + d*x)**4)*cot(c + d*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.15 \[ \int \cot (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\frac {\sqrt {a} \log \left (\frac {\sqrt {b \sin \left (d x + c\right )^{4} + a} - \sqrt {a}}{\sqrt {b \sin \left (d x + c\right )^{4} + a} + \sqrt {a}}\right ) + 2 \, \sqrt {b \sin \left (d x + c\right )^{4} + a}}{4 \, d} \]

[In]

integrate(cot(d*x+c)*(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")

[Out]

1/4*(sqrt(a)*log((sqrt(b*sin(d*x + c)^4 + a) - sqrt(a))/(sqrt(b*sin(d*x + c)^4 + a) + sqrt(a))) + 2*sqrt(b*sin
(d*x + c)^4 + a))/d

Giac [F(-1)]

Timed out. \[ \int \cot (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(cot(d*x+c)*(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \cot (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\int \mathrm {cot}\left (c+d\,x\right )\,\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a} \,d x \]

[In]

int(cot(c + d*x)*(a + b*sin(c + d*x)^4)^(1/2),x)

[Out]

int(cot(c + d*x)*(a + b*sin(c + d*x)^4)^(1/2), x)