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

Optimal. Leaf size=98 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}-\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 a^2 d}+\frac {10 \cos (c+d x)}{3 a d \sqrt {a \sin (c+d x)+a}} \]

[Out]

-2*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(3/2)/d+10/3*cos(d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)-2/3
*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/a^2/d

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Rubi [A]  time = 0.36, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2880, 2646, 3046, 2981, 2773, 206} \[ -\frac {2 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 a^2 d}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}+\frac {10 \cos (c+d x)}{3 a d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(a^(3/2)*d) + (10*Cos[c + d*x])/(3*a*d*Sqrt[a +
a*Sin[c + d*x]]) - (2*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*a^2*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 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 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 3046

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

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx &=\frac {\int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \left (1+\sin ^2(c+d x)\right ) \, dx}{a^2}-\frac {2 \int \sqrt {a+a \sin (c+d x)} \, dx}{a^2}\\ &=\frac {4 \cos (c+d x)}{a d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 a^2 d}+\frac {2 \int \csc (c+d x) \left (\frac {3 a}{2}+\frac {1}{2} a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{3 a^3}\\ &=\frac {10 \cos (c+d x)}{3 a d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 a^2 d}+\frac {\int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx}{a^2}\\ &=\frac {10 \cos (c+d x)}{3 a d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 a^2 d}-\frac {2 \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 {2 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}+\frac {10 \cos (c+d x)}{3 a d \sqrt {a+a \sin (c+d x)}}-\frac {2 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 147, normalized size = 1.50 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (-9 \sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )+9 \cos \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {3}{2} (c+d x)\right )-3 \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )+3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )\right )}{3 d (a (\sin (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(9*Cos[(c + d*x)/2] - Cos[(3*(c + d*x))/2] - 3*Log[1 + Cos[(c + d*x)/
2] - Sin[(c + d*x)/2]] + 3*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 9*Sin[(c + d*x)/2] - Sin[(3*(c + d*x
))/2]))/(3*d*(a*(1 + Sin[c + d*x]))^(3/2))

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fricas [B]  time = 0.49, size = 260, normalized size = 2.65 \[ \frac {3 \, \sqrt {a} {\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right ) - 4 \, \cos \left (d x + c\right ) - 5\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{6 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/6*(3*sqrt(a)*(cos(d*x + c) + sin(d*x + c) + 1)*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*(cos(d*x + c)^2 + (cos(d*x + c) + 5)*sin(d*x + c) - 4*cos(d*x +
c) - 5)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c) + a^2*d*sin(d*x + c) + a^2*d)

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giac [B]  time = 0.84, size = 313, normalized size = 3.19 \[ -\frac {\frac {{\left (6 \, \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a} + \sqrt {a}}{\sqrt {-a}}\right ) - 3 \, \sqrt {-a} \log \left (\sqrt {2} \sqrt {a} + \sqrt {a}\right ) + 10 \, \sqrt {2} \sqrt {-a}\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {-a} a^{\frac {3}{2}}} + \frac {4 \, {\left ({\left ({\left (\frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {3}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {3}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {2}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}} - \frac {6 \, \arctan \left (-\frac {\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {3 \, \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-1/3*((6*sqrt(a)*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 3*sqrt(-a)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 10
*sqrt(2)*sqrt(-a))*sgn(tan(1/2*d*x + 1/2*c) + 1)/(sqrt(-a)*a^(3/2)) + 4*(((2*tan(1/2*d*x + 1/2*c)/sgn(tan(1/2*
d*x + 1/2*c) + 1) - 3/sgn(tan(1/2*d*x + 1/2*c) + 1))*tan(1/2*d*x + 1/2*c) + 3/sgn(tan(1/2*d*x + 1/2*c) + 1))*t
an(1/2*d*x + 1/2*c) - 2/sgn(tan(1/2*d*x + 1/2*c) + 1))/(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)*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)))/d

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maple [A]  time = 1.10, size = 103, normalized size = 1.05 \[ \frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (-3 a^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )+\left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}}+3 a \sqrt {a -a \sin \left (d x +c \right )}\right )}{3 a^{3} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/a^3*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(-3*a^(3/2)*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))+(a-a*sin(
d*x+c))^(3/2)+3*a*(a-a*sin(d*x+c))^(1/2))/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 \[ \int \frac {\cos \left (d x + c\right )^{4} \csc \left (d x + c\right )}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{4}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*csc(d*x+c)/(a+a*sin(d*x+c))**(3/2),x)

[Out]

Integral(cos(c + d*x)**4*csc(c + d*x)/(a*(sin(c + d*x) + 1))**(3/2), x)

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