3.3.32 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx\) [232]

Optimal. Leaf size=85 \[ \frac {\sqrt {2} \text {ArcSin}\left (\frac {\sin (c+d x)}{1+\cos (c+d x)}\right )}{d}-\frac {\text {ArcSin}\left (\frac {\sin (c+d x)}{\sqrt {1+\cos (c+d x)}}\right )}{d}+\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {1+\cos (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2857, 3061, 2860, 222, 2853} \begin {gather*} \frac {\sqrt {2} \text {ArcSin}\left (\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}-\frac {\text {ArcSin}\left (\frac {\sin (c+d x)}{\sqrt {\cos (c+d x)+1}}\right )}{d}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {\cos (c+d x)+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(3/2)/Sqrt[1 + Cos[c + d*x]],x]

[Out]

(Sqrt[2]*ArcSin[Sin[c + d*x]/(1 + Cos[c + d*x])])/d - ArcSin[Sin[c + d*x]/Sqrt[1 + Cos[c + d*x]]]/d + (Sqrt[Co
s[c + d*x]]*Sin[c + d*x])/(d*Sqrt[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]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.86, size = 231, normalized size = 2.72 \begin {gather*} \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (-\frac {i \sqrt {2} e^{\frac {1}{2} i (c+d x)} \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (-i d x-\sinh ^{-1}\left (e^{i (c+d x)}\right )+2 \sqrt {2} \log \left (1+e^{i (c+d x)}\right )+\log \left (1+\sqrt {1+e^{2 i (c+d x)}}\right )-2 \sqrt {2} \log \left (1-e^{i (c+d x)}+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )\right )}{d \sqrt {1+e^{2 i (c+d x)}}}+\frac {4 \sqrt {\cos (c+d x)} \sin \left (\frac {1}{2} (c+d x)\right )}{d}\right )}{2 \sqrt {1+\cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(3/2)/Sqrt[1 + Cos[c + d*x]],x]

[Out]

(Cos[(c + d*x)/2]*(((-I)*Sqrt[2]*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*((-I)*d*x
- ArcSinh[E^(I*(c + d*x))] + 2*Sqrt[2]*Log[1 + E^(I*(c + d*x))] + Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]] - 2*
Sqrt[2]*Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]]))/(d*Sqrt[1 + E^((2*I)*(c + d*x))]) +
(4*Sqrt[Cos[c + d*x]]*Sin[(c + d*x)/2])/d))/(2*Sqrt[1 + Cos[c + d*x]])

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Maple [A]
time = 0.16, size = 151, normalized size = 1.78

method	result	size
default	\(-\frac {\left (\cos ^{\frac {3}{2}}\left (d x +c \right )\right ) \sqrt {2+2 \cos \left (d x +c \right )}\, \left (-1+\cos \left (d x +c \right )\right )^{2} \left (\arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sqrt {2}-\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )\right ) \sqrt {2}}{2 d \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )^{4}}\)	\(151\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(3/2)/(1+cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)/(1+cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^(3/2)/sqrt(cos(d*x + c) + 1), x)

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Fricas [A]
time = 0.42, size = 125, normalized size = 1.47 \begin {gather*} -\frac {{\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 ) - {\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 ) - \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right ) + d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)/(1+cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-((sqrt(2)*cos(d*x + c) + sqrt(2))*arctan(sqrt(2)*sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c))/sin(d*x + c)) - (c
os(d*x + c) + 1)*arctan(sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c))/sin(d*x + c)) - sqrt(cos(d*x + c) + 1)*sqrt(
cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{\frac {3}{2}}{\left (c + d x \right )}}{\sqrt {\cos {\left (c + d x \right )} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**(3/2)/(1+cos(d*x+c))**(1/2),x)

[Out]

Integral(cos(c + d*x)**(3/2)/sqrt(cos(c + d*x) + 1), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)/(1+cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^{3/2}}{\sqrt {\cos \left (c+d\,x\right )+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(3/2)/(cos(c + d*x) + 1)^(1/2),x)

[Out]

int(cos(c + d*x)^(3/2)/(cos(c + d*x) + 1)^(1/2), x)

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