\(\int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx\) [231]

Optimal result

Integrand size = 23, antiderivative size = 126 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=-\frac {\sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{1+\cos (c+d x)}\right )}{d}+\frac {7 \arcsin \left (\frac {\sin (c+d x)}{\sqrt {1+\cos (c+d x)}}\right )}{4 d}-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{4 d \sqrt {1+\cos (c+d x)}}+\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {1+\cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2857, 3062, 3061, 2860, 222, 2853} \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=-\frac {\sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}+\frac {7 \arcsin \left (\frac {\sin (c+d x)}{\sqrt {\cos (c+d x)+1}}\right )}{4 d}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {\cos (c+d x)+1}}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{4 d \sqrt {\cos (c+d x)+1}} \]

[In]

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

[Out]

-((Sqrt[2]*ArcSin[Sin[c + d*x]/(1 + Cos[c + d*x])])/d) + (7*ArcSin[Sin[c + d*x]/Sqrt[1 + Cos[c + d*x]]])/(4*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*Sqr
t[1 + Cos[c + d*x]])

Rule 222

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

Rule 2853

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 2857

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)/Sqrt[a + b*Sin[e + f*x]])*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], 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 2860

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[-Sqr
t[2]/(Sqrt[a]*f), Subst[Int[1/Sqrt[1 - x^2], x], x, b*(Cos[e + f*x]/(a + b*Sin[e + f*x]))], x] /; FreeQ[{a, b,
d, e, f}, x] && EqQ[a^2 - b^2, 0] && EqQ[d, a/b] && GtQ[a, 0]

Rule 3061

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 3062

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] &&
(IntegerQ[n] || EqQ[m + 1/2, 0])

Rubi steps \begin{align*} \text {integral}& = \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 {(-3+\cos (c+d x)) \sqrt {\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 {\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 {7}{8} \int \frac {\sqrt {1+\cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx-\int \frac {1}{\sqrt {\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 {7 \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,-\frac {\sin (c+d x)}{\sqrt {1+\cos (c+d x)}}\right )}{4 d}+\frac {\sqrt {2} \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,-\frac {\sin (c+d x)}{1+\cos (c+d x)}\right )}{d} \\ & = -\frac {\sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{1+\cos (c+d x)}\right )}{d}+\frac {7 \arcsin \left (\frac {\sin (c+d x)}{\sqrt {1+\cos (c+d x)}}\right )}{4 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 [A] (verified)

Time = 0.17 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=-\frac {\left (\arcsin \left (\sqrt {1-\cos (c+d x)}\right )+8 \arcsin \left (\sqrt {\cos (c+d x)}\right )-4 \sqrt {2} \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right )-2 \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)+\sqrt {-((-1+\cos (c+d x)) \cos (c+d x))}\right ) \sin (c+d x)}{4 d \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

-1/4*((ArcSin[Sqrt[1 - Cos[c + d*x]]] + 8*ArcSin[Sqrt[Cos[c + d*x]]] - 4*Sqrt[2]*ArcTan[Sqrt[Cos[c + d*x]]/Sqr
t[Sin[(c + d*x)/2]^2]] - 2*Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(3/2) + Sqrt[-((-1 + Cos[c + d*x])*Cos[c + d*x]
)])*Sin[c + d*x])/(d*Sqrt[Sin[c + d*x]^2])

Maple [A] (verified)

Time = 12.56 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.34

method	result	size
default	\(\frac {\left (2 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+4 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {2}-\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+7 \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {2+2 \cos \left (d x +c \right )}\, \sqrt {2}}{8 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(169\)

[In]

int(cos(d*x+c)^(5/2)/(1+cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/8/d*(2*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+4*arcsin(cot(d*x+c)-csc(d*x+c))*2^(1/2)-sin(d
*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+7*arctan(tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)))*cos(d*x+c)^(1/
2)*(2+2*cos(d*x+c))^(1/2)/(1+cos(d*x+c))/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=\frac {{\left (2 \, \cos \left (d x + c\right ) - 1\right )} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 4 \, {\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - 7 \, {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {\sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right )}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

1/4*((2*cos(d*x + c) - 1)*sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c))*sin(d*x + c) + 4*(sqrt(2)*cos(d*x + c) + s
qrt(2))*arctan(sqrt(2)*sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c))/sin(d*x + c)) - 7*(cos(d*x + c) + 1)*arctan(s
qrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c))/sin(d*x + c)))/(d*cos(d*x + c) + d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {\cos \left (d x + c\right ) + 1}} \,d x } \]

[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)

Giac [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}}{\sqrt {\cos \left (c+d\,x\right )+1}} \,d x \]

[In]

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

[Out]

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