3.257 \(\int \cot (c+d x) \sqrt{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=51 \[ \frac{2 \sqrt{a \sin (c+d x)+a}}{d}-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{a}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.0616903, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2707, 50, 63, 207} \[ \frac{2 \sqrt{a \sin (c+d x)+a}}{d}-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2707

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^(m - (p + 1)/2))/(a - x)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \cot (c+d x) \sqrt{a+a \sin (c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+x}}{x} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{2 \sqrt{a+a \sin (c+d x)}}{d}+\frac{a \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+x}} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{2 \sqrt{a+a \sin (c+d x)}}{d}+\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{-a+x^2} \, dx,x,\sqrt{a+a \sin (c+d x)}\right )}{d}\\ &=-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+a \sin (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{2 \sqrt{a+a \sin (c+d x)}}{d}\\ \end{align*}

Mathematica [B]  time = 0.145009, size = 118, normalized size = 2.31 \[ \frac{\sqrt{a (\sin (c+d x)+1)} \left (2 \sin \left (\frac{1}{2} (c+d x)\right )+2 \cos \left (\frac{1}{2} (c+d x)\right )+\log \left (-\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )\right )}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Cos[(c + d*x)/2] + Log[1 - Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[1 + Cos[(c + d*x)/2] + Sin[(c + d*x)
/2]] + 2*Sin[(c + d*x)/2])*Sqrt[a*(1 + Sin[c + d*x])])/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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Maple [A]  time = 0.103, size = 42, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( 2\,\sqrt{a+a\sin \left ( dx+c \right ) }-2\,\sqrt{a}{\it Artanh} \left ({\frac{\sqrt{a+a\sin \left ( dx+c \right ) }}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*csc(d*x+c)*(a+a*sin(d*x+c))^(1/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*csc(d*x+c)*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.44299, size = 232, normalized size = 4.55 \begin{align*} \frac{\sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{2} + 4 \, \sqrt{a \sin \left (d x + c\right ) + a} \sqrt{a}{\left (\sin \left (d x + c\right ) + 2\right )} - 8 \, a \sin \left (d x + c\right ) - 9 \, a}{\cos \left (d x + c\right )^{2} - 1}\right ) + 4 \, \sqrt{a \sin \left (d x + c\right ) + a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*csc(d*x+c)*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/2*(sqrt(a)*log((a*cos(d*x + c)^2 + 4*sqrt(a*sin(d*x + c) + a)*sqrt(a)*(sin(d*x + c) + 2) - 8*a*sin(d*x + c)
- 9*a)/(cos(d*x + c)^2 - 1)) + 4*sqrt(a*sin(d*x + c) + a))/d

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (\sin{\left (c + d x \right )} + 1\right )} \cos{\left (c + d x \right )} \csc{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*csc(d*x+c)*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(c + d*x) + 1))*cos(c + d*x)*csc(c + d*x), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*csc(d*x+c)*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError