3.445 \(\int \cos ^3(c+d x) \cot (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=159 \[ -\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {12 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{35 a d}+\frac {164 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{105 d}+\frac {8 a \cos (c+d x)}{15 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d} \]

[Out]

-12/35*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/a/d-2*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))*a^(1/2)/d+8/
15*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-2/7*a*cos(d*x+c)*sin(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)+164/105*cos(d*
x+c)*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.49, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2881, 2770, 2759, 2751, 2646, 3046, 2981, 2773, 206} \[ -\frac {2 a \sin ^3(c+d x) \cos (c+d x)}{7 d \sqrt {a \sin (c+d x)+a}}-\frac {12 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{35 a d}+\frac {164 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{105 d}+\frac {8 a \cos (c+d x)}{15 d \sqrt {a \sin (c+d x)+a}}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*Cot[c + d*x]*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-2*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d + (8*a*Cos[c + d*x])/(15*d*Sqrt[a + a*
Sin[c + d*x]]) - (2*a*Cos[c + d*x]*Sin[c + d*x]^3)/(7*d*Sqrt[a + a*Sin[c + d*x]]) + (164*Cos[c + d*x]*Sqrt[a +
a*Sin[c + d*x]])/(105*d) - (12*Cos[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(35*a*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 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 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 2770

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(2*n*(b*c + a*d)
)/(b*(2*n + 1)), 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] && GtQ[n, 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 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 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 \cos ^3(c+d x) \cot (c+d x) \sqrt {a+a \sin (c+d x)} \, dx &=\int \sin ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx+\int \csc (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) \sin ^3(c+d x)}{7 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}+\frac {6}{7} \int \sin ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx+\frac {2 \int \csc (c+d x) \left (\frac {3 a}{2}-a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{3 a}\\ &=\frac {4 a \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos (c+d x) \sin ^3(c+d x)}{7 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}-\frac {12 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{35 a d}+\frac {12 \int \left (\frac {3 a}{2}-a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx}{35 a}+\int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=\frac {4 a \cos (c+d x)}{3 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos (c+d x) \sin ^3(c+d x)}{7 d \sqrt {a+a \sin (c+d x)}}+\frac {164 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{105 d}-\frac {12 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{35 a d}+\frac {2}{5} \int \sqrt {a+a \sin (c+d x)} \, dx-\frac {(2 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 )}{d}\\ &=-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}+\frac {8 a \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}-\frac {2 a \cos (c+d x) \sin ^3(c+d x)}{7 d \sqrt {a+a \sin (c+d x)}}+\frac {164 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{105 d}-\frac {12 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{35 a d}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 195, normalized size = 1.23 \[ \frac {\sqrt {a (\sin (c+d x)+1)} \left (-525 \sin \left (\frac {1}{2} (c+d x)\right )+175 \sin \left (\frac {3}{2} (c+d x)\right )-21 \sin \left (\frac {5}{2} (c+d x)\right )+15 \sin \left (\frac {7}{2} (c+d x)\right )+525 \cos \left (\frac {1}{2} (c+d x)\right )+175 \cos \left (\frac {3}{2} (c+d x)\right )+21 \cos \left (\frac {5}{2} (c+d x)\right )+15 \cos \left (\frac {7}{2} (c+d x)\right )-420 \log \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )+1\right )+420 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )+1\right )\right )}{420 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*Cot[c + d*x]*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Sqrt[a*(1 + Sin[c + d*x])]*(525*Cos[(c + d*x)/2] + 175*Cos[(3*(c + d*x))/2] + 21*Cos[(5*(c + d*x))/2] + 15*Co
s[(7*(c + d*x))/2] - 420*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 420*Log[1 - Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2]] - 525*Sin[(c + d*x)/2] + 175*Sin[(3*(c + d*x))/2] - 21*Sin[(5*(c + d*x))/2] + 15*Sin[(7*(c + d*x))/
2]))/(420*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

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fricas [B]  time = 0.46, size = 294, normalized size = 1.85 \[ \frac {105 \, \sqrt {a} {\left (\cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (15 \, \cos \left (d x + c\right )^{4} + 18 \, \cos \left (d x + c\right )^{3} + 34 \, \cos \left (d x + c\right )^{2} + {\left (15 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )^{2} + 31 \, \cos \left (d x + c\right ) - 43\right )} \sin \left (d x + c\right ) + 74 \, \cos \left (d x + c\right ) + 43\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{210 \, {\left (d \cos \left (d x + c\right ) + d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(105*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*(15*cos(d*x + c)^4 + 18*cos(d*x + c)^3 + 34*cos(d*x + c)^2 +
(15*cos(d*x + c)^3 - 3*cos(d*x + c)^2 + 31*cos(d*x + c) - 43)*sin(d*x + c) + 74*cos(d*x + c) + 43)*sqrt(a*sin
(d*x + c) + a))/(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)*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.33, size = 141, normalized size = 0.89 \[ -\frac {2 \left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (105 a^{\frac {7}{2}} \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right )-15 \left (a -a \sin \left (d x +c \right )\right )^{\frac {7}{2}}+63 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} a -35 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} a^{2}-105 a^{3} \sqrt {a -a \sin \left (d x +c \right )}\right )}{105 a^{3} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

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))^(1/2),x)

[Out]

-2/105*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(105*a^(7/2)*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))-15*(a-a*s
in(d*x+c))^(7/2)+63*(a-a*sin(d*x+c))^(5/2)*a-35*(a-a*sin(d*x+c))^(3/2)*a^2-105*a^3*(a-a*sin(d*x+c))^(1/2))/a^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 \[ \int \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4} \csc \left (d x + c\right )\,{d x} \]

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))^(1/2),x, algorithm="maxima")

[Out]

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

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

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),x)

[Out]

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

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

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))**(1/2),x)

[Out]

Integral(sqrt(a*(sin(c + d*x) + 1))*cos(c + d*x)**4*csc(c + d*x), x)

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