3.478 \(\int \frac{\cot ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=144 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{8 a^{3/2} d}-\frac{\cot (c+d x) \csc ^2(c+d x) \sqrt{a \sin (c+d x)+a}}{3 a^2 d}-\frac{\cot (c+d x)}{8 a d \sqrt{a \sin (c+d x)+a}}+\frac{11 \cot (c+d x) \csc (c+d x)}{12 a d \sqrt{a \sin (c+d x)+a}} \]

[Out]

-ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]]/(8*a^(3/2)*d) - Cot[c + d*x]/(8*a*d*Sqrt[a + a*Sin[c
+ d*x]]) + (11*Cot[c + d*x]*Csc[c + d*x])/(12*a*d*Sqrt[a + a*Sin[c + d*x]]) - (Cot[c + d*x]*Csc[c + d*x]^2*Sq
rt[a + a*Sin[c + d*x]])/(3*a^2*d)

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Rubi [A]  time = 0.539332, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2717, 2772, 2773, 206, 3044, 2980} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{8 a^{3/2} d}-\frac{\cot (c+d x) \csc ^2(c+d x) \sqrt{a \sin (c+d x)+a}}{3 a^2 d}-\frac{\cot (c+d x)}{8 a d \sqrt{a \sin (c+d x)+a}}+\frac{11 \cot (c+d x) \csc (c+d x)}{12 a d \sqrt{a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

-ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]]/(8*a^(3/2)*d) - Cot[c + d*x]/(8*a*d*Sqrt[a + a*Sin[c
+ d*x]]) + (11*Cot[c + d*x]*Csc[c + d*x])/(12*a*d*Sqrt[a + a*Sin[c + d*x]]) - (Cot[c + d*x]*Csc[c + d*x]^2*Sq
rt[a + a*Sin[c + d*x]])/(3*a^2*d)

Rule 2717

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> Dist[-2/(a*b), Int[(a
+ b*Sin[e + f*x])^(m + 2)/Sin[e + f*x]^3, x], x] + Dist[1/a^2, Int[((a + b*Sin[e + f*x])^(m + 2)*(1 + Sin[e +
f*x]^2))/Sin[e + f*x]^4, x], x] /; FreeQ[{a, b, e, f}, 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 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 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
])

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]

Rubi steps

\begin{align*} \int \frac{\cot ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac{\int \csc ^4(c+d x) \sqrt{a+a \sin (c+d x)} \left (1+\sin ^2(c+d x)\right ) \, dx}{a^2}-\frac{2 \int \csc ^3(c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{a^2}\\ &=\frac{\cot (c+d x) \csc (c+d x)}{a 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 a^2 d}+\frac{\int \csc ^3(c+d x) \sqrt{a+a \sin (c+d x)} \left (\frac{a}{2}+\frac{9}{2} a \sin (c+d x)\right ) \, dx}{3 a^3}-\frac{3 \int \csc ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{2 a^2}\\ &=\frac{3 \cot (c+d x)}{2 a d \sqrt{a+a \sin (c+d x)}}+\frac{11 \cot (c+d x) \csc (c+d x)}{12 a 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 a^2 d}-\frac{3 \int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{4 a^2}+\frac{13 \int \csc ^2(c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{8 a^2}\\ &=-\frac{\cot (c+d x)}{8 a d \sqrt{a+a \sin (c+d x)}}+\frac{11 \cot (c+d x) \csc (c+d x)}{12 a 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 a^2 d}+\frac{13 \int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{16 a^2}+\frac{3 \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 )}{2 a d}\\ &=\frac{3 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{2 a^{3/2} d}-\frac{\cot (c+d x)}{8 a d \sqrt{a+a \sin (c+d x)}}+\frac{11 \cot (c+d x) \csc (c+d x)}{12 a 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 a^2 d}-\frac{13 \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 a d}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{8 a^{3/2} d}-\frac{\cot (c+d x)}{8 a d \sqrt{a+a \sin (c+d x)}}+\frac{11 \cot (c+d x) \csc (c+d x)}{12 a 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 a^2 d}\\ \end{align*}

Mathematica [B]  time = 0.767357, size = 294, normalized size = 2.04 \[ \frac{\csc ^9\left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3 \left (132 \sin \left (\frac{1}{2} (c+d x)\right )+62 \sin \left (\frac{3}{2} (c+d x)\right )-6 \sin \left (\frac{5}{2} (c+d x)\right )-132 \cos \left (\frac{1}{2} (c+d x)\right )+62 \cos \left (\frac{3}{2} (c+d x)\right )+6 \cos \left (\frac{5}{2} (c+d x)\right )-9 \sin (c+d x) \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )+9 \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )+3 \sin (3 (c+d x)) \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-3 \sin (3 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )\right )}{24 d (a (\sin (c+d x)+1))^{3/2} \left (\csc ^2\left (\frac{1}{4} (c+d x)\right )-\sec ^2\left (\frac{1}{4} (c+d x)\right )\right )^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[c + d*x]^4/(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(Csc[(c + d*x)/2]^9*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(-132*Cos[(c + d*x)/2] + 62*Cos[(3*(c + d*x))/2] +
6*Cos[(5*(c + d*x))/2] + 132*Sin[(c + d*x)/2] - 9*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] +
9*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] + 62*Sin[(3*(c + d*x))/2] - 6*Sin[(5*(c + d*x))/2
] + 3*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[3*(c + d*x)] - 3*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*
x)/2]]*Sin[3*(c + d*x)]))/(24*d*(Csc[(c + d*x)/4]^2 - Sec[(c + d*x)/4]^2)^3*(a*(1 + Sin[c + d*x]))^(3/2))

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Maple [A]  time = 1.036, size = 144, normalized size = 1. \begin{align*}{\frac{1+\sin \left ( dx+c \right ) }{24\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) d}\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) } \left ( 3\,\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{7/2}-8\, \left ( -a \left ( \sin \left ( dx+c \right ) -1 \right ) \right ) ^{3/2}{a}^{5/2}-3\, \left ( -a \left ( \sin \left ( dx+c \right ) -1 \right ) \right ) ^{5/2}{a}^{3/2}-3\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }}{\sqrt{a}}} \right ){a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{3} \right ){a}^{-{\frac{11}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

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))^(3/2),x)

[Out]

1/24*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(3*(-a*(sin(d*x+c)-1))^(1/2)*a^(7/2)-8*(-a*(sin(d*x+c)-1))^(3/2)
*a^(5/2)-3*(-a*(sin(d*x+c)-1))^(5/2)*a^(3/2)-3*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*a^4*sin(d*x+c)^3)/a^
(11/2)/sin(d*x+c)^3/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.15885, size = 1003, normalized size = 6.97 \begin{align*} \frac{3 \,{\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 (3 \, \cos \left (d x + c\right )^{3} + 17 \, \cos \left (d x + c\right )^{2} -{\left (3 \, \cos \left (d x + c\right )^{2} - 14 \, \cos \left (d x + c\right ) - 25\right )} \sin \left (d x + c\right ) - 11 \, \cos \left (d x + c\right ) - 25\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{96 \,{\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d -{\left (a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

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))^(3/2),x, algorithm="fricas")

[Out]

1/96*(3*(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*(3*cos(d*x + c)^3 + 17*cos(d*x + c)^2 - (3*cos(d*x + c)^2 - 14*cos(d*x + c) - 25)*sin(d*x + c) - 11
*cos(d*x + c) - 25)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d - (a^2*d*
cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2 - a^2*d*cos(d*x + c) - a^2*d)*sin(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 2.56644, size = 795, normalized size = 5.52 \begin{align*} \text{result too large to display} \end{align*}

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))^(3/2),x, algorithm="giac")

[Out]

1/48*(sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*((2*tan(1/2*d*x + 1/2*c)/(a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 9/(a^2
*sgn(tan(1/2*d*x + 1/2*c) + 1)))*tan(1/2*d*x + 1/2*c) + 14/(a^2*sgn(tan(1/2*d*x + 1/2*c) + 1))) - (30*sqrt(2)*
sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 15*sqrt(2)*sqrt(-a)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 42
*sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 21*sqrt(-a)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 280*sqrt(
2)*sqrt(-a) + 402*sqrt(-a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(5*sqrt(2)*sqrt(-a)*a^(3/2) + 7*sqrt(-a)*a^(3/2)) +
6*arctan(-(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))/sqrt(-a))/(sqrt(-a)*a*sgn(tan(1/
2*d*x + 1/2*c) + 1)) - 3*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/(a^(3/2)
*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 2*(9*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^5 -
18*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*sqrt(a) + 24*(sqrt(a)*tan(1/2*d*x +
1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a^(3/2) - 9*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x
+ 1/2*c)^2 + a))*a^2 - 14*a^(5/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a
)^3*a*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d