3.232 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1+\cos (c+d x)}} \, dx\)

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

[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 = 0.19, 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 = {2778, 2982, 2781, 216, 2774} \[ \frac {\sqrt {2} \sin ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}-\frac {\sin ^{-1}\left (\frac {\sin (c+d x)}{\sqrt {\cos (c+d x)+1}}\right )}{d}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {\cos (c+d x)+1}} \]

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 216

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 2774

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 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 2781

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 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]

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 {\operatorname {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} \operatorname {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]  time = 0.86, size = 224, normalized size = 2.64 \[ -\frac {i e^{-\frac {1}{2} i (c+d x)} \cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}} \left (-\sqrt {2} e^{i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )-4 e^{i (c+d x)} \tanh ^{-1}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\sqrt {2} \left (\sqrt {1+e^{2 i (c+d x)}} \left (-1+e^{i (c+d x)}\right )+e^{i (c+d x)} \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )\right )}{\sqrt {2} d \sqrt {1+e^{2 i (c+d x)}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 1.47, size = 125, normalized size = 1.47 \[ -\frac {{\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {\sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right ) + d} \]

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

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]

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

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maple [A]  time = 0.14, size = 151, normalized size = 1.78 \[ -\frac {\left (\cos ^{\frac {3}{2}}\left (d x +c \right )\right ) \sqrt {2+2 \cos \left (d x +c \right )}\, \left (-1+\cos \left (d x +c \right )\right )^{2} \left (\sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{\cos \left (d x +c \right )}\right )\right ) \sqrt {2}}{2 d \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )^{4}} \]

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)

[Out]

-1/2/d*cos(d*x+c)^(3/2)*(2+2*cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-(
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 \[ \int \frac {\cos \left (d x + c\right )^{\frac {3}{2}}}{\sqrt {\cos \left (d x + c\right ) + 1}}\,{d x} \]

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

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

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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