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

Optimal. Leaf size=161 \[ \frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{1-\cos (c+d x)}}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{4 d \sqrt{1-\cos (c+d x)}}+\frac{7 \tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{4 d}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{2} \sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{d} \]

[Out]

(7*ArcTanh[Sin[c + d*x]/(Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])])/(4*d) - (Sqrt[2]*ArcTanh[Sin[c + d*x]/(S
qrt[2]*Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])])/d + (Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*d*Sqrt[1 - Cos[c
+ d*x]]) + (Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[1 - Cos[c + d*x]])

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Rubi [A]  time = 0.301612, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2778, 2983, 2982, 2782, 206, 2775, 207} \[ \frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{1-\cos (c+d x)}}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{4 d \sqrt{1-\cos (c+d x)}}+\frac{7 \tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{4 d}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{2} \sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*ArcTanh[Sin[c + d*x]/(Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])])/(4*d) - (Sqrt[2]*ArcTanh[Sin[c + d*x]/(S
qrt[2]*Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])])/d + (Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*d*Sqrt[1 - Cos[c
+ d*x]]) + (Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[1 - Cos[c + d*x]])

Rule 2778

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp
[(-2*d*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n - 1))/(f*(2*n - 1)*Sqrt[a + b*Sin[e + f*x]]), x] - Dist[1/(b*(2*n
- 1)), Int[((c + d*Sin[e + f*x])^(n - 2)*Simp[a*c*d - b*(2*d^2*(n - 1) + c^2*(2*n - 1)) + d*(a*d - b*c*(4*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] && GtQ[n, 1] && IntegerQ[2*n]

Rule 2983

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*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c*(
m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*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] && GtQ[n, 0] && (I
ntegerQ[n] || EqQ[m + 1/2, 0])

Rule 2982

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

Rule 2775

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*b)/f, Subst[Int[1/(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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\cos ^{\frac{5}{2}}(c+d x)}{\sqrt{1-\cos (c+d x)}} \, dx &=\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt{1-\cos (c+d x)}}+\frac{1}{4} \int \frac{\sqrt{\cos (c+d x)} (3+\cos (c+d x))}{\sqrt{1-\cos (c+d x)}} \, dx\\ &=\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{4 d \sqrt{1-\cos (c+d x)}}+\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt{1-\cos (c+d x)}}-\frac{1}{4} \int \frac{-\frac{1}{2}-\frac{7}{2} \cos (c+d x)}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}} \, dx\\ &=\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{4 d \sqrt{1-\cos (c+d x)}}+\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt{1-\cos (c+d x)}}-\frac{7}{8} \int \frac{\sqrt{1-\cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx+\int \frac{1}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}} \, dx\\ &=\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{4 d \sqrt{1-\cos (c+d x)}}+\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt{1-\cos (c+d x)}}-\frac{7 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\frac{\sin (c+d x)}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{4 d}-\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{7 \tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{4 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{\sqrt{\cos (c+d x)} \sin (c+d x)}{4 d \sqrt{1-\cos (c+d x)}}+\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt{1-\cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.41225, size = 255, normalized size = 1.58 \[ -\frac{i e^{-2 i (c+d x)} \left (-1+e^{i (c+d x)}\right ) \sqrt{\cos (c+d x)} \left (7 \sqrt{2} e^{2 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )-16 e^{2 i (c+d x)} \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+\sqrt{2} \left (\sqrt{1+e^{2 i (c+d x)}} \left (2 e^{i (c+d x)}+2 e^{2 i (c+d x)}+e^{3 i (c+d x)}+1\right )+7 e^{2 i (c+d x)} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )\right )}{8 \sqrt{2} d \sqrt{1+e^{2 i (c+d x)}} \sqrt{1-\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/8)*(-1 + E^(I*(c + d*x)))*(7*Sqrt[2]*E^((2*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))] - 16*E^((2*I)*(c + d*x)
)*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] + Sqrt[2]*(Sqrt[1 + E^((2*I)*(c + d*x
))]*(1 + 2*E^(I*(c + d*x)) + 2*E^((2*I)*(c + d*x)) + E^((3*I)*(c + d*x))) + 7*E^((2*I)*(c + d*x))*ArcTanh[Sqrt
[1 + E^((2*I)*(c + d*x))]]))*Sqrt[Cos[c + d*x]])/(Sqrt[2]*d*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*
Sqrt[1 - Cos[c + d*x]])

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Maple [A]  time = 0.329, size = 194, normalized size = 1.2 \begin{align*} -{\frac{\sqrt{2} \left ( -1+\cos \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}} \left ( \cos \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}} \left ( 2\,\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+3\,\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\cos \left ( dx+c \right ) -4\,{\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}+\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+7\,{\it Artanh} \left ( \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}} \right ) \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\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(cos(d*x+c)^(5/2)/(1-cos(d*x+c))^(1/2),x)

[Out]

-1/4/d*2^(1/2)*cos(d*x+c)^(5/2)*(-1+cos(d*x+c))^3*(2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+3*(cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)-4*arctanh(1/2*2^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*2^(1/2)+(cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)+7*arctanh((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)))/(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)/sin(
d*x+c)^5/(2-2*cos(d*x+c))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{\frac{5}{2}}}{\sqrt{-\cos \left (d x + c\right ) + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.00554, size = 653, normalized size = 4.06 \begin{align*} \frac{4 \, \sqrt{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 ) + 2 \,{\left (2 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt{-\cos \left (d x + c\right ) + 1} \sqrt{\cos \left (d x + c\right )} + 7 \, \log \left (\frac{2 \,{\left (\sqrt{-\cos \left (d x + c\right ) + 1} \sqrt{\cos \left (d x + c\right )} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 7 \, \log \left (\frac{2 \,{\left (\sqrt{-\cos \left (d x + c\right ) + 1} \sqrt{\cos \left (d x + c\right )} - \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right )}{8 \, d \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*sqrt(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) + 2*(2*cos(d*x + c)^2 + 3*cos(d*x +
c) + 1)*sqrt(-cos(d*x + c) + 1)*sqrt(cos(d*x + c)) + 7*log(2*(sqrt(-cos(d*x + c) + 1)*sqrt(cos(d*x + c)) + sin
(d*x + c))/sin(d*x + c))*sin(d*x + c) - 7*log(2*(sqrt(-cos(d*x + c) + 1)*sqrt(cos(d*x + c)) - sin(d*x + c))/si
n(d*x + c))*sin(d*x + c))/(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)**(5/2)/(1-cos(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 2.0876, size = 219, normalized size = 1.36 \begin{align*} -\frac{\sqrt{2}{\left (7 \, \sqrt{2} \log \left (\frac{\sqrt{2} - \sqrt{-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1}}{\sqrt{2} + \sqrt{-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1}}\right ) - \frac{4 \,{\left ({\left (-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{\frac{3}{2}} + 2 \, \sqrt{-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}} + 8 \, \log \left (\sqrt{-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + 1\right ) - 8 \, \log \left (-\sqrt{-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + 1\right )\right )}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*sqrt(2)*(7*sqrt(2)*log((sqrt(2) - sqrt(-tan(1/2*d*x + 1/2*c)^2 + 1))/(sqrt(2) + sqrt(-tan(1/2*d*x + 1/2*
c)^2 + 1))) - 4*((-tan(1/2*d*x + 1/2*c)^2 + 1)^(3/2) + 2*sqrt(-tan(1/2*d*x + 1/2*c)^2 + 1))/(tan(1/2*d*x + 1/2
*c)^2 + 1)^2 + 8*log(sqrt(-tan(1/2*d*x + 1/2*c)^2 + 1) + 1) - 8*log(-sqrt(-tan(1/2*d*x + 1/2*c)^2 + 1) + 1))/d