3.287 \(\int \frac {1}{\sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=122 \[ \frac {2 \sin (c+d x)}{3 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x)}{3 d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sin (c+d x)}{\sqrt {2} \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2779, 2984, 12, 2782, 206} \[ \frac {2 \sin (c+d x)}{3 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x)}{3 d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sin (c+d x)}{\sqrt {2} \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(5/2)),x]

[Out]

-((Sqrt[2]*ArcTanh[Sin[c + d*x]/(Sqrt[2]*Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])])/d) + (2*Sin[c + d*x])/(3
*d*Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(3/2)) + (2*Sin[c + d*x])/(3*d*Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]
])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[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 2779

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Sim
p[(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[1/
(2*b*(n + 1)*(c^2 - d^2)), Int[((c + d*Sin[e + f*x])^(n + 1)*Simp[a*d - 2*b*c*(n + 1) + b*d*(2*n + 3)*Sin[e +
f*x], x])/Sqrt[a + b*Sin[e + f*x]], 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] && IntegerQ[2*n]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*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 2984

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

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)} \, dx &=\frac {2 \sin (c+d x)}{3 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} \int \frac {1+2 \cos (c+d x)}{\sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 \sin (c+d x)}{3 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x)}{3 d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}-\frac {2}{3} \int -\frac {3}{2 \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \sin (c+d x)}{3 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x)}{3 d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}+\int \frac {1}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \sin (c+d x)}{3 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x)}{3 d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}\\ &=-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sin (c+d x)}{\sqrt {2} \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}+\frac {2 \sin (c+d x)}{3 d \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x)}{3 d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.28, size = 170, normalized size = 1.39 \[ \frac {2 \sin \left (\frac {1}{2} (c+d x)\right ) \left (2 \sqrt {1+e^{2 i (c+d x)}} \cos \left (\frac {1}{2} (c+d x)\right ) (\cos (c+d x)+1)-\frac {3 e^{-\frac {3}{2} i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^2 \tanh ^{-1}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )}{2 \sqrt {2}}\right )}{3 d \sqrt {1+e^{2 i (c+d x)}} \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(5/2)),x]

[Out]

(2*((-3*(1 + E^((2*I)*(c + d*x)))^2*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/(2
*Sqrt[2]*E^(((3*I)/2)*(c + d*x))) + 2*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[(c + d*x)/2]*(1 + Cos[c + d*x]))*Sin[(
c + d*x)/2])/(3*d*Sqrt[1 + E^((2*I)*(c + d*x))]*Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(3/2))

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fricas [A]  time = 0.89, size = 157, normalized size = 1.29 \[ \frac {3 \, \sqrt {2} \cos \left (d x + c\right )^{2} \log \left (-\frac {2 \, {\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 4 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )}}{6 \, d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(d*x+c))^(1/2)/cos(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.79, size = 89, normalized size = 0.73 \[ -\frac {\sqrt {2} {\left (\frac {8}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} \sqrt {-\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}} + 3 \, \log \left (\sqrt {-\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 1\right ) - 3 \, \log \left (-\sqrt {-\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 1\right )\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(d*x+c))^(1/2)/cos(d*x+c)^(5/2),x, algorithm="giac")

[Out]

-1/6*sqrt(2)*(8/((tan(1/2*d*x + 1/2*c)^2 - 1)*sqrt(-tan(1/2*d*x + 1/2*c)^2 + 1)) + 3*log(sqrt(-tan(1/2*d*x + 1
/2*c)^2 + 1) + 1) - 3*log(-sqrt(-tan(1/2*d*x + 1/2*c)^2 + 1) + 1))/d

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maple [A]  time = 0.15, size = 170, normalized size = 1.39 \[ -\frac {\left (\sin ^{5}\left (d x +c \right )\right ) \left (3 \sqrt {2}\, \cos \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+3 \sqrt {2}\, \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-2 \cos \left (d x +c \right )\right ) \sqrt {2}}{3 d \left (-1+\cos \left (d x +c \right )\right )^{2} \sqrt {2-2 \cos \left (d x +c \right )}\, \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-cos(d*x+c))^(1/2)/cos(d*x+c)^(5/2),x)

[Out]

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

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maxima [C]  time = 1.01, size = 563, normalized size = 4.61 \[ \frac {3 \, {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \log \left (\frac {4 \, {\left ({\left | i \, e^{\left (i \, d x + i \, c\right )} - i \right |}^{2} + 2 \, \sqrt {\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1} {\left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )^{2}\right )} - 2 \, {\left (\sqrt {2} {\left | i \, e^{\left (i \, d x + i \, c\right )} - i \right |} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )\right )} {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} + 4\right )}}{{\left | i \, e^{\left (i \, d x + i \, c\right )} - i \right |}^{2}}\right ) - 2 \, {\left (\sqrt {2} \sin \left (d x + c\right ) \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + {\left (\sqrt {2} \cos \left (d x + c\right ) + 3 \, \sqrt {2}\right )} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )\right )} {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {3}{4}} - 4 \, {\left (\sqrt {2} \sin \left (d x + c\right ) \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + {\left (\sqrt {2} \cos \left (d x + c\right ) - \sqrt {2}\right )} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )\right )} {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}}}{3 \, {\left (\sqrt {2} \cos \left (2 \, d x + 2 \, c\right )^{2} + \sqrt {2} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \sqrt {2} \cos \left (2 \, d x + 2 \, c\right ) + \sqrt {2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(d*x+c))^(1/2)/cos(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

1/3*(3*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(4*(abs(I*e^(I*d*x + I*c) - I)^2
+ 2*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c) + 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2) - 2*(sqrt(2)*abs(I*e^(I
*d*x + I*c) - I)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 2*sqrt(2)*cos(1/2*arctan2(sin(2*d*
x + 2*c), cos(2*d*x + 2*c) + 1)))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4) + 4
)/abs(I*e^(I*d*x + I*c) - I)^2) - 2*(sqrt(2)*sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) +
1)) + (sqrt(2)*cos(d*x + c) + 3*sqrt(2))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*(cos(2*d*x
+ 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4) - 4*(sqrt(2)*sin(d*x + c)*sin(3/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (sqrt(2)*cos(d*x + c) - sqrt(2))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c) + 1)))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4))/((sqrt(2)*cos(2*d
*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^(5/2)*(1 - cos(c + d*x))^(1/2)),x)

[Out]

int(1/(cos(c + d*x)^(5/2)*(1 - cos(c + d*x))^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {1 - \cos {\left (c + d x \right )}} \cos ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(d*x+c))**(1/2)/cos(d*x+c)**(5/2),x)

[Out]

Integral(1/(sqrt(1 - cos(c + d*x))*cos(c + d*x)**(5/2)), x)

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