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

Optimal. Leaf size=119 \[ -\frac{2 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{3 a d}+\frac{4 \cos (c+d x)}{3 d \sqrt{a \sin (c+d x)+a}}-\frac{\cot (c+d x)}{d \sqrt{a \sin (c+d x)+a}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{\sqrt{a} d} \]

[Out]

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

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Rubi [A]  time = 0.503369, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {2881, 2759, 2751, 2649, 206, 3044, 2985, 2773} \[ -\frac{2 \cos (c+d x) \sqrt{a \sin (c+d x)+a}}{3 a d}+\frac{4 \cos (c+d x)}{3 d \sqrt{a \sin (c+d x)+a}}-\frac{\cot (c+d x)}{d \sqrt{a \sin (c+d x)+a}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{\sqrt{a} d} \]

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]

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

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 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 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 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^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 2985

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), 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]

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]

Rubi steps

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

Mathematica [A]  time = 0.361274, size = 190, normalized size = 1.6 \[ \frac{\left (\tan \left (\frac{1}{2} (c+d x)\right )+1\right ) \csc \left (\frac{1}{4} (c+d x)\right ) \sec \left (\frac{1}{4} (c+d x)\right ) \left (10 \sin \left (\frac{1}{2} (c+d x)\right )+3 \sin \left (\frac{3}{2} (c+d x)\right )-\sin \left (\frac{5}{2} (c+d x)\right )-10 \cos \left (\frac{1}{2} (c+d x)\right )+3 \cos \left (\frac{3}{2} (c+d x)\right )+\cos \left (\frac{5}{2} (c+d x)\right )+3 \sin (c+d x) \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-3 \sin (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )\right )}{24 d \sqrt{a (\sin (c+d x)+1)}} \]

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)/4]*Sec[(c + d*x)/4]*(-10*Cos[(c + d*x)/2] + 3*Cos[(3*(c + d*x))/2] + Cos[(5*(c + d*x))/2] + 10*
Sin[(c + d*x)/2] + 3*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[c + d*x] - 3*Log[1 - Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]]*Sin[c + d*x] + 3*Sin[(3*(c + d*x))/2] - Sin[(5*(c + d*x))/2])*(1 + Tan[(c + d*x)/2]))/(24*d*
Sqrt[a*(1 + Sin[c + d*x])])

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Maple [A]  time = 0.967, size = 126, normalized size = 1.1 \begin{align*} -{\frac{1+\sin \left ( dx+c \right ) }{3\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) d}\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) } \left ( \sin \left ( dx+c \right ) \left ( -2\,\sqrt{a} \left ( a-a\sin \left ( dx+c \right ) \right ) ^{3/2}-3\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( dx+c \right ) }}{\sqrt{a}}} \right ){a}^{2} \right ) +3\,\sqrt{a-a\sin \left ( dx+c \right ) }{a}^{3/2} \right ){a}^{-{\frac{5}{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)^2/(a+a*sin(d*x+c))^(1/2),x)

[Out]

-1/3*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(sin(d*x+c)*(-2*a^(1/2)*(a-a*sin(d*x+c))^(3/2)-3*arctanh((a-a*si
n(d*x+c))^(1/2)/a^(1/2))*a^2)+3*(a-a*sin(d*x+c))^(1/2)*a^(3/2))/a^(5/2)/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., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{4} \csc \left (d x + c\right )^{2}}{\sqrt{a \sin \left (d x + c\right ) + a}}\,{d x} \end{align*}

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

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Fricas [B]  time = 1.21375, size = 818, normalized size = 6.87 \begin{align*} \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 )^{3} + 4 \, \cos \left (d x + c\right )^{2} -{\left (2 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 7\right )} \sin \left (d x + c\right ) - 5 \, \cos \left (d x + c\right ) - 7\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{12 \,{\left (a d \cos \left (d x + c\right )^{2} - a d -{\left (a d \cos \left (d x + c\right ) + a 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)^2/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/12*(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)^3 + 4*cos(d*x + c)^2 - (2*
cos(d*x + c)^2 - 2*cos(d*x + c) - 7)*sin(d*x + c) - 5*cos(d*x + c) - 7)*sqrt(a*sin(d*x + c) + a))/(a*d*cos(d*x
+ c)^2 - a*d - (a*d*cos(d*x + c) + a*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)**2/(a+a*sin(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 2.24795, size = 649, normalized size = 5.45 \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)^2/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/6*((6*sqrt(2)*a^(15/2)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) + 6*a^(15/2)*arctan((sqrt(2)*sqrt(a) + s
qrt(a))/sqrt(-a)) - 3*sqrt(-a)*a^7 - 3072*sqrt(2)*sqrt(-a)*log(sqrt(2)*sqrt(a) + sqrt(a)) - 3072*sqrt(-a)*log(
sqrt(2)*sqrt(a) + sqrt(a)) - 11264*sqrt(2)*sqrt(-a) - 22528*sqrt(-a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(sqrt(2)*s
qrt(-a)*a^(15/2) + sqrt(-a)*a^(15/2)) + 1024*((((3*sgn(tan(1/2*d*x + 1/2*c) + 1)*tan(1/2*d*x + 1/2*c)/a^6 - 4*
sgn(tan(1/2*d*x + 1/2*c) + 1)/a^6)*tan(1/2*d*x + 1/2*c) + 18*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^6)*tan(1/2*d*x +
1/2*c) - 12*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^6)*tan(1/2*d*x + 1/2*c) + 7*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^6)/(a*
tan(1/2*d*x + 1/2*c)^2 + 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)*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 6*sqrt(a)/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*ta
n(1/2*d*x + 1/2*c)^2 + a))^2 - a)*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 3072*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c)
+ sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))*sgn(tan(1/2*d*x + 1/2*c) + 1)/a^(15/2))/d