3.448 \(\int \cot ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=163 \[ -\frac {2 a \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}+\frac {11 a \cot (c+d x)}{8 d \sqrt {a \sin (c+d x)+a}}+\frac {11 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{8 d}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {a \cot (c+d x) \csc (c+d x)}{12 d \sqrt {a \sin (c+d x)+a}} \]

[Out]

11/8*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))*a^(1/2)/d-2*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+11/8
*a*cot(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-1/12*a*cot(d*x+c)*csc(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-1/3*cot(d*x+c)*cs
c(d*x+c)^2*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.39, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2718, 2646, 3044, 2980, 2772, 2773, 206} \[ -\frac {2 a \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}+\frac {11 a \cot (c+d x)}{8 d \sqrt {a \sin (c+d x)+a}}+\frac {11 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{8 d}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {a \cot (c+d x) \csc (c+d x)}{12 d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(8*d) - (2*a*Cos[c + d*x])/(d*Sqrt[a + a
*Sin[c + d*x]]) + (11*a*Cot[c + d*x])/(8*d*Sqrt[a + a*Sin[c + d*x]]) - (a*Cot[c + d*x]*Csc[c + d*x])/(12*d*Sqr
t[a + a*Sin[c + d*x]]) - (Cot[c + d*x]*Csc[c + d*x]^2*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 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 2718

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

Rule 2772

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[((b*c - a*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x]
+ Dist[((2*n + 3)*(b*c - a*d))/(2*b*(n + 1)*(c^2 - d^2)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), 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] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

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 2980

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n + 1)
*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, 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 \cot ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx &=\int \sqrt {a+a \sin (c+d x)} \, dx+\int \csc ^4(c+d x) \sqrt {a+a \sin (c+d x)} \left (1-2 \sin ^2(c+d x)\right ) \, dx\\ &=-\frac {2 a \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\frac {\int \csc ^3(c+d x) \left (\frac {a}{2}-\frac {9}{2} a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{3 a}\\ &=-\frac {2 a \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}-\frac {a \cot (c+d x) \csc (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}-\frac {11}{8} \int \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {2 a \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\frac {11 a \cot (c+d x)}{8 d \sqrt {a+a \sin (c+d x)}}-\frac {a \cot (c+d x) \csc (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}-\frac {11}{16} \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=-\frac {2 a \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\frac {11 a \cot (c+d x)}{8 d \sqrt {a+a \sin (c+d x)}}-\frac {a \cot (c+d x) \csc (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\frac {(11 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 )}{8 d}\\ &=\frac {11 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{8 d}-\frac {2 a \cos (c+d x)}{d \sqrt {a+a \sin (c+d x)}}+\frac {11 a \cot (c+d x)}{8 d \sqrt {a+a \sin (c+d x)}}-\frac {a \cot (c+d x) \csc (c+d x)}{12 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}\\ \end {align*}

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Mathematica [A]  time = 1.42, size = 309, normalized size = 1.90 \[ \frac {\csc ^{10}\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sin (c+d x)+1)} \left (-252 \sin \left (\frac {1}{2} (c+d x)\right )-250 \sin \left (\frac {3}{2} (c+d x)\right )+114 \sin \left (\frac {5}{2} (c+d x)\right )+48 \sin \left (\frac {7}{2} (c+d x)\right )+252 \cos \left (\frac {1}{2} (c+d x)\right )-250 \cos \left (\frac {3}{2} (c+d x)\right )-114 \cos \left (\frac {5}{2} (c+d x)\right )+48 \cos \left (\frac {7}{2} (c+d x)\right )+99 \sin (c+d x) \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-99 \sin (c+d x) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-33 \sin (3 (c+d x)) \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )+33 \sin (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )\right )}{24 d \left (\cot \left (\frac {1}{2} (c+d x)\right )+1\right ) \left (\csc ^2\left (\frac {1}{4} (c+d x)\right )-\sec ^2\left (\frac {1}{4} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[(c + d*x)/2]^10*Sqrt[a*(1 + Sin[c + d*x])]*(252*Cos[(c + d*x)/2] - 250*Cos[(3*(c + d*x))/2] - 114*Cos[(5*
(c + d*x))/2] + 48*Cos[(7*(c + d*x))/2] - 252*Sin[(c + d*x)/2] + 99*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2
]]*Sin[c + d*x] - 99*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] - 250*Sin[(3*(c + d*x))/2] + 11
4*Sin[(5*(c + d*x))/2] - 33*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 33*Log[1 - Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] + 48*Sin[(7*(c + d*x))/2]))/(24*d*(1 + Cot[(c + d*x)/2])*(Csc[(
c + d*x)/4]^2 - Sec[(c + d*x)/4]^2)^3)

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fricas [B]  time = 0.49, size = 380, normalized size = 2.33 \[ \frac {33 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (48 \, \cos \left (d x + c\right )^{4} - 33 \, \cos \left (d x + c\right )^{3} - 139 \, \cos \left (d x + c\right )^{2} + {\left (48 \, \cos \left (d x + c\right )^{3} + 81 \, \cos \left (d x + c\right )^{2} - 58 \, \cos \left (d x + c\right ) - 83\right )} \sin \left (d x + c\right ) + 25 \, \cos \left (d x + c\right ) + 83\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{96 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(33*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^3 + 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*(48*cos(d*x + c)^4 - 33*cos(d*x + c)^3 - 139*cos(d*x + c)^2 + (48*cos(d*x + c)^3 + 81*cos(d*x + c)
^2 - 58*cos(d*x + c) - 83)*sin(d*x + c) + 25*cos(d*x + c) + 83)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^4 -
2*d*cos(d*x + c)^2 - (d*cos(d*x + c)^3 + 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)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.38, size = 170, normalized size = 1.04 \[ -\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (48 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {7}{2}} \left (\sin ^{3}\left (d x +c \right )\right )-15 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {7}{2}}+56 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} a^{\frac {5}{2}}-33 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {5}{2}} a^{\frac {3}{2}}-33 \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) a^{4} \left (\sin ^{3}\left (d x +c \right )\right )\right )}{24 a^{\frac {7}{2}} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(48*(-a*(sin(d*x+c)-1))^(1/2)*a^(7/2)*sin(d*x+c)^3-15*(-a*(sin(
d*x+c)-1))^(1/2)*a^(7/2)+56*(-a*(sin(d*x+c)-1))^(3/2)*a^(5/2)-33*(-a*(sin(d*x+c)-1))^(5/2)*a^(3/2)-33*arctanh(
(-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*a^4*sin(d*x+c)^3)/a^(7/2)/sin(d*x+c)^3/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 \[ \int \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4} \csc \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^4*(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)^4, x)

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

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)^4,x)

[Out]

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

[Out]

Timed out

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