3.446 \(\int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=148 \[ -\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}+\frac {4 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{15 d}+\frac {61 a \cos (c+d x)}{15 d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d} \]

[Out]

-2/5*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/a/d-arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))*a^(1/2)/d+61/15*
a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+4/15*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d-cot(d*x+c)*(a+a*sin(d*x+c))^(1/
2)/d

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Rubi [A]  time = 0.48, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {2881, 2759, 2751, 2646, 3044, 2981, 2773, 206} \[ -\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}+\frac {4 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{15 d}+\frac {61 a \cos (c+d x)}{15 d \sqrt {a \sin (c+d x)+a}}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*Cot[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

-((Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d) + (61*a*Cos[c + d*x])/(15*d*Sqrt[a + a
*Sin[c + d*x]]) + (4*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(15*d) - (Cot[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/d
- (2*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(5*a*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 2646

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

Rule 2751

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

Rule 2759

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(Cos[e + f*x]*(a
+ b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*(b*(m + 1) - a*
Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

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 2881

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/d^4, Int[(d*Sin[e + f*x])^(n + 4)*(a + b*Sin[e + f*x])^m, x], x] + Int[(d*Sin[e + f*x])^
n*(a + b*Sin[e + f*x])^m*(1 - 2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&
!IGtQ[m, 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]

Rule 3044

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

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx &=\int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx+\int \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)} \left (1-2 \sin ^2(c+d x)\right ) \, dx\\ &=-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 a d}+\frac {2 \int \left (\frac {3 a}{2}-a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{5 a}+\frac {\int \csc (c+d x) \left (\frac {a}{2}-\frac {5}{2} a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{a}\\ &=\frac {5 a \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 d}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 a d}+\frac {7}{15} \int \sqrt {a+a \sin (c+d x)} \, dx+\frac {1}{2} \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=\frac {61 a \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 d}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 a d}-\frac {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 {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}+\frac {61 a \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 d}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 a d}\\ \end {align*}

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Mathematica [A]  time = 0.74, size = 258, normalized size = 1.74 \[ \frac {\csc ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sin (c+d x)+1)} \left (155 \sin \left (\frac {1}{2} (c+d x)\right )+87 \sin \left (\frac {3}{2} (c+d x)\right )-5 \sin \left (\frac {5}{2} (c+d x)\right )+3 \sin \left (\frac {7}{2} (c+d x)\right )-155 \cos \left (\frac {1}{2} (c+d x)\right )+87 \cos \left (\frac {3}{2} (c+d x)\right )+5 \cos \left (\frac {5}{2} (c+d x)\right )+3 \cos \left (\frac {7}{2} (c+d x)\right )-30 \sin (c+d x) \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )+30 \sin (c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )\right )}{30 d \left (\cot \left (\frac {1}{2} (c+d x)\right )+1\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )-\sec \left (\frac {1}{4} (c+d x)\right )\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )+\sec \left (\frac {1}{4} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*Cot[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Csc[(c + d*x)/2]^4*Sqrt[a*(1 + Sin[c + d*x])]*(-155*Cos[(c + d*x)/2] + 87*Cos[(3*(c + d*x))/2] + 5*Cos[(5*(c
+ d*x))/2] + 3*Cos[(7*(c + d*x))/2] + 155*Sin[(c + d*x)/2] - 30*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*S
in[c + d*x] + 30*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] + 87*Sin[(3*(c + d*x))/2] - 5*Sin[(
5*(c + d*x))/2] + 3*Sin[(7*(c + d*x))/2]))/(30*d*(1 + Cot[(c + d*x)/2])*(Csc[(c + d*x)/4] - Sec[(c + d*x)/4])*
(Csc[(c + d*x)/4] + Sec[(c + d*x)/4]))

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fricas [B]  time = 0.49, size = 320, normalized size = 2.16 \[ \frac {15 \, {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (6 \, \cos \left (d x + c\right )^{4} + 8 \, \cos \left (d x + c\right )^{3} + 40 \, \cos \left (d x + c\right )^{2} + {\left (6 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} + 38 \, \cos \left (d x + c\right ) + 61\right )} \sin \left (d x + c\right ) - 23 \, \cos \left (d x + c\right ) - 61\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{60 \, {\left (d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right ) + d\right )} \sin \left (d x + c\right ) - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*(cos(d*x + c)^2 - (cos(d*x + c) + 1)*sin(d*x + c) - 1)*sqrt(a)*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*(6*cos(d*x + c)^4 + 8*cos(d*x + c)^3 + 40
*cos(d*x + c)^2 + (6*cos(d*x + c)^3 - 2*cos(d*x + c)^2 + 38*cos(d*x + c) + 61)*sin(d*x + c) - 23*cos(d*x + c)
- 61)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^2 - (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)^4*csc(d*x+c)^2*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.28, size = 162, normalized size = 1.09 \[ -\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (\sin \left (d x +c \right ) \left (6 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} a^{\frac {3}{2}}-20 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} a^{\frac {5}{2}}-30 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {7}{2}}+15 \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right ) a^{4}\right )+15 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {7}{2}}\right )}{15 a^{\frac {7}{2}} \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)/a^(7/2)*(sin(d*x+c)*(6*(a-a*sin(d*x+c))^(5/2)*a^(3/2)-20*(a-a*s
in(d*x+c))^(3/2)*a^(5/2)-30*(a-a*sin(d*x+c))^(1/2)*a^(7/2)+15*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))*a^4)+15*
(a-a*sin(d*x+c))^(1/2)*a^(7/2))/sin(d*x+c)/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 \[ \int \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4} \csc \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a*sin(d*x + c) + a)*cos(d*x + c)^4*csc(d*x + c)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^4\,\sqrt {a+a\,\sin \left (c+d\,x\right )}}{{\sin \left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [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)**4*csc(d*x+c)**2*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Timed out

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