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

Optimal. Leaf size=94 \[ \frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{a^2 d}-\frac {\cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {2880, 2773, 206, 3044, 21, 2763} \[ -\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{a^2 d}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}-\frac {\cos (c+d x)}{a d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 2763

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1] && (IntegersQ[2*m, 2*
n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 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 2880

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[-2/(a*b*d), Int[(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 2), x], x] + Dist[1/a^2
, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 2)*(1 + Sin[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}
, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -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 \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac {\int \csc ^2(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 (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{a^2}\\ &=-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{a^2 d}+\frac {\int \csc (c+d x) \left (\frac {a}{2}+\frac {1}{2} a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{a^3}+\frac {4 \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 )}{a d}\\ &=\frac {4 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{a^2 d}+\frac {\int \csc (c+d x) (a+a \sin (c+d x))^{3/2} \, dx}{2 a^3}\\ &=\frac {4 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}-\frac {\cos (c+d x)}{a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{a^2 d}+\frac {\int \frac {\csc (c+d x) \left (\frac {a^2}{2}+\frac {1}{2} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{a^3}\\ &=\frac {4 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}-\frac {\cos (c+d x)}{a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{a^2 d}+\frac {\int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{2 a^2}\\ &=\frac {4 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}-\frac {\cos (c+d x)}{a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{a^2 d}-\frac {\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 )}{a d}\\ &=\frac {3 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}-\frac {\cos (c+d x)}{a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{a^2 d}\\ \end {align*}

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Mathematica [B]  time = 0.70, size = 220, normalized size = 2.34 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (8 \sin \left (\frac {1}{2} (c+d x)\right )-8 \cos \left (\frac {1}{2} (c+d x)\right )-\tan \left (\frac {1}{4} (c+d x)\right )-\cot \left (\frac {1}{4} (c+d x)\right )+\frac {2 \sin \left (\frac {1}{4} (c+d x)\right )}{\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )}-\frac {2 \sin \left (\frac {1}{4} (c+d x)\right )}{\sin \left (\frac {1}{4} (c+d x)\right )+\cos \left (\frac {1}{4} (c+d x)\right )}+6 \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )-6 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )+2\right )}{4 d (a (\sin (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(2 - 8*Cos[(c + d*x)/2] - Cot[(c + d*x)/4] + 6*Log[1 + Cos[(c + d*x)/
2] - Sin[(c + d*x)/2]] - 6*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (2*Sin[(c + d*x)/4])/(Cos[(c + d*x)/
4] - Sin[(c + d*x)/4]) - (2*Sin[(c + d*x)/4])/(Cos[(c + d*x)/4] + Sin[(c + d*x)/4]) + 8*Sin[(c + d*x)/2] - Tan
[(c + d*x)/4]))/(4*d*(a*(1 + Sin[c + d*x]))^(3/2))

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fricas [B]  time = 0.48, size = 291, normalized size = 3.10 \[ \frac {3 \, {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + {\left (2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right ) - 1\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d - {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]

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))^(3/2),x, algorithm="fricas")

[Out]

1/4*(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*(2*cos(d*x + c)^2 + (2*cos(d*x + c) + 1)*si
n(d*x + c) + cos(d*x + c) - 1)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c)^2 - a^2*d - (a^2*d*cos(d*x + c) +
a^2*d)*sin(d*x + c))

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giac [B]  time = 0.85, size = 424, normalized size = 4.51 \[ \frac {\frac {{\left (6 \, \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a} + \sqrt {a}}{\sqrt {-a}}\right ) - 3 \, \sqrt {2} \sqrt {-a} \log \left (\sqrt {2} \sqrt {a} + \sqrt {a}\right ) + 6 \, \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a} + \sqrt {a}}{\sqrt {-a}}\right ) - 3 \, \sqrt {-a} \log \left (\sqrt {2} \sqrt {a} + \sqrt {a}\right ) + 3 \, \sqrt {2} \sqrt {-a} + 5 \, \sqrt {-a}\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {2} \sqrt {-a} a^{\frac {3}{2}} + \sqrt {-a} a^{\frac {3}{2}}} + \frac {{\left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {4}{a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {3}{a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {6 \, \arctan \left (-\frac {\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {3 \, \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {2}{{\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a\right )} \sqrt {a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{2 \, d} \]

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))^(3/2),x, algorithm="giac")

[Out]

1/2*((6*sqrt(2)*sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 3*sqrt(2)*sqrt(-a)*log(sqrt(2)*sqrt(a)
+ sqrt(a)) + 6*sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 3*sqrt(-a)*log(sqrt(2)*sqrt(a) + sqrt(a)
) + 3*sqrt(2)*sqrt(-a) + 5*sqrt(-a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(sqrt(2)*sqrt(-a)*a^(3/2) + sqrt(-a)*a^(3/2
)) + ((tan(1/2*d*x + 1/2*c)/(a*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 4/(a*sgn(tan(1/2*d*x + 1/2*c) + 1)))*tan(1/2*d
*x + 1/2*c) - 3/(a*sgn(tan(1/2*d*x + 1/2*c) + 1)))/sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) - 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/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a)*sqrt(a)*sgn(tan(1/2
*d*x + 1/2*c) + 1)))/d

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maple [A]  time = 1.10, size = 123, normalized size = 1.31 \[ -\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (\sin \left (d x +c \right ) \left (2 \sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {a}-3 \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right ) a \right )+\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {a}\right )}{a^{\frac {5}{2}} \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

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))^(3/2),x)

[Out]

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

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maxima [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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^4}{{\sin \left (c+d\,x\right )}^2\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(sin(c + d*x)^2*(a + a*sin(c + d*x))^(3/2)),x)

[Out]

int(cos(c + d*x)^4/(sin(c + d*x)^2*(a + a*sin(c + d*x))^(3/2)), 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))**(3/2),x)

[Out]

Timed out

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