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

Optimal. Leaf size=93 \[ \frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}+\frac {2 a \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d} \]

[Out]

-2*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))*a^(1/2)/d+2/3*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+2/3*
cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.34, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2874, 2976, 2981, 2773, 206} \[ \frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}+\frac {2 a \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d + (2*a*Cos[c + d*x])/(3*d*Sqrt[a + a*S
in[c + d*x]]) + (2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d)

Rule 206

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

Rule 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2874

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[n, 0])

Rule 2976

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&
NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rubi steps

\begin {align*} \int \cos (c+d x) \cot (c+d x) \sqrt {a+a \sin (c+d x)} \, dx &=\frac {\int \csc (c+d x) (a-a \sin (c+d x)) (a+a \sin (c+d x))^{3/2} \, dx}{a^2}\\ &=\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\frac {2 \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \left (\frac {3 a^2}{2}-\frac {1}{2} a^2 \sin (c+d x)\right ) \, dx}{3 a^2}\\ &=\frac {2 a \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=\frac {2 a \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}\\ &=-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}+\frac {2 a \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 143, normalized size = 1.54 \[ \frac {\sqrt {a (\sin (c+d x)+1)} \left (-3 \sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )+3 \cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )-3 \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )+3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )\right )}{3 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*(1 + Sin[c + d*x])]*(3*Cos[(c + d*x)/2] + Cos[(3*(c + d*x))/2] - 3*Log[1 + Cos[(c + d*x)/2] - Sin[(c +
d*x)/2]] + 3*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 3*Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2]))/(3*d*
(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [B]  time = 0.48, size = 250, normalized size = 2.69 \[ \frac {3 \, \sqrt {a} {\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{6 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*sqrt(a)*(cos(d*x + c) + sin(d*x + c) + 1)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^
2 + (cos(d*x + c) + 3)*sin(d*x + c) - 2*cos(d*x + c) - 3)*sqrt(a*sin(d*x + c) + a)*sqrt(a) - 9*a*cos(d*x + c)
+ (a*cos(d*x + c)^2 + 8*a*cos(d*x + c) - a)*sin(d*x + c) - a)/(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)
^2 - 1)*sin(d*x + c) - cos(d*x + c) - 1)) + 4*(cos(d*x + c)^2 + (cos(d*x + c) - 1)*sin(d*x + c) + 2*cos(d*x +
c) + 1)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c) + d*sin(d*x + c) + d)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 1.20, size = 103, normalized size = 1.11 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (3 a^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )+\left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}-3 a \sqrt {a -a \sin \left (d x +c \right )}\right )}{3 a \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(3*a^(3/2)*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))+(a-a*sin(d*x+
c))^(3/2)-3*a*(a-a*sin(d*x+c))^(1/2))/a/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{2} \csc \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*sin(d*x + c) + a)*cos(d*x + c)^2*csc(d*x + c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^2\,\sqrt {a+a\,\sin \left (c+d\,x\right )}}{\sin \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \cos ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*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)**2*csc(c + d*x), x)

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