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]

-((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]
])

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Rubi [A]  time = 0.182391, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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 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])

Rule 12

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

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 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])

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*}

Mathematica [C]  time = 0.286337, 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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Maple [A]  time = 0.321, size = 170, normalized size = 1.4 \begin{align*} -{\frac{\sqrt{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{3\,d \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2} \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( 3\,\cos \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}{\it Artanh} \left ( 1/2\,{\sqrt{2}{\frac{1}{\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}}}} \right ) \sqrt{2}+3\, \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}{\it Artanh} \left ( 1/2\,{\sqrt{2}{\frac{1}{\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}}}} \right ) \sqrt{2}-2\,\cos \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{2-2\,\cos \left ( dx+c \right ) }}}} \end{align*}

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

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Maxima [C]  time = 1.90369, size = 760, normalized size = 6.23 \begin{align*} \text{result too large to display} \end{align*}

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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Fricas [A]  time = 2.1414, size = 429, normalized size = 3.52 \begin{align*} \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 )} \end{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.81366, size = 120, normalized size = 0.98 \begin{align*} -\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} \end{align*}

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