\(\int \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x) \, dx\) [269]
Optimal result
Integrand size = 25, antiderivative size = 114 \[
\int \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x) \, dx=-\frac {3 \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{4 d}+\frac {3 \sqrt {\cos (c+d x)} \sin (c+d x)}{4 d \sqrt {1-\cos (c+d x)}}-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {1-\cos (c+d x)}}
\]
[Out]
-3/4*arctanh(sin(d*x+c)/(1-cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2))/d-1/2*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(1-cos(d*x+
c))^(1/2)+3/4*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(1-cos(d*x+c))^(1/2)
Rubi [A] (verified)
Time = 0.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of
steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2849, 2854, 213}
\[
\int \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x) \, dx=-\frac {3 \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{4 d}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {1-\cos (c+d x)}}+\frac {3 \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d \sqrt {1-\cos (c+d x)}}
\]
[In]
Int[Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(3/2),x]
[Out]
(-3*ArcTanh[Sin[c + d*x]/(Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])])/(4*d) + (3*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 213
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Rule 2849
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[2*n*((b*c + a*d)
/(b*(2*n + 1))), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), 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, 0] && IntegerQ[2*n]
Rule 2854
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]
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 {3}{4} \int \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)} \, dx \\ & = \frac {3 \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 {3}{8} \int \frac {\sqrt {1-\cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {3 \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 {3 \text {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 {3 \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{4 d}+\frac {3 \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] (verified)
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.59
\[
\int \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x) \, dx=-\frac {e^{-\frac {3}{2} i (c+d x)} \left (\sqrt {1+e^{2 i (c+d x)}} \left (1-2 e^{i (c+d x)}-2 e^{2 i (c+d x)}+e^{3 i (c+d x)}\right )+3 e^{2 i (c+d x)} \text {arcsinh}\left (e^{i (c+d x)}\right )+3 e^{2 i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {-((-1+\cos (c+d x)) \cos (c+d x))} \csc \left (\frac {1}{2} (c+d x)\right )}{8 d \sqrt {1+e^{2 i (c+d x)}}}
\]
[In]
Integrate[Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(3/2),x]
[Out]
-1/8*((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))) + 3
*E^((2*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))] + 3*E^((2*I)*(c + d*x))*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])*
Sqrt[-((-1 + Cos[c + d*x])*Cos[c + d*x])]*Csc[(c + d*x)/2])/(d*E^(((3*I)/2)*(c + d*x))*Sqrt[1 + E^((2*I)*(c +
d*x))])
Maple [A] (verified)
Time = 13.34 (sec) , antiderivative size = 154, normalized size of antiderivative =
1.35
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| method | result | size |
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| default |
\(-\frac {\csc \left (d x +c \right ) \left (2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )-\cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3 \,\operatorname {arctanh}\left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-3 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \sqrt {-2 \cos \left (d x +c \right )+2}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {2}}{8 d \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) |
\(154\) |
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[In]
int((1-cos(d*x+c))^(1/2)*cos(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
[Out]
-1/8/d*csc(d*x+c)*(2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2-cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)+3*arctanh((cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*(-2*cos(d*x+c)+2)^(1/2)*c
os(d*x+c)^(1/2)/(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 = 124, normalized size of antiderivative = 1.09
\[
\int \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x) \, dx=-\frac {2 \, {\left (2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 3\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - 3 \, \log \left (-\frac {2 \, {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - {\left (2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right )}{8 \, d \sin \left (d x + c\right )}
\]
[In]
integrate((1-cos(d*x+c))^(1/2)*cos(d*x+c)^(3/2),x, algorithm="fricas")
[Out]
-1/8*(2*(2*cos(d*x + c)^2 - cos(d*x + c) - 3)*sqrt(-cos(d*x + c) + 1)*sqrt(cos(d*x + c)) - 3*log(-(2*(cos(d*x
+ c) + 1)*sqrt(-cos(d*x + c) + 1)*sqrt(cos(d*x + c)) - (2*cos(d*x + c) + 1)*sin(d*x + c))/sin(d*x + c))*sin(d*
x + c))/(d*sin(d*x + c))
Sympy [F]
\[
\int \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x) \, dx=\int \sqrt {1 - \cos {\left (c + d x \right )}} \cos ^{\frac {3}{2}}{\left (c + d x \right )}\, dx
\]
[In]
integrate((1-cos(d*x+c))**(1/2)*cos(d*x+c)**(3/2),x)
[Out]
Integral(sqrt(1 - cos(c + d*x))*cos(c + d*x)**(3/2), x)
Maxima [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 1305 vs. \(2 (96) = 192\).
Time = 0.40 (sec) , antiderivative size = 1305, normalized size of antiderivative = 11.45
\[
\int \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x) \, dx=\text {Too large to display}
\]
[In]
integrate((1-cos(d*x+c))^(1/2)*cos(d*x+c)^(3/2),x, algorithm="maxima")
[Out]
1/32*(4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(((cos(2*d*x + 2*c) - 2)*cos(
1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(2*d*x + 2*c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*
x + 2*c))) + cos(2*d*x + 2*c) - 2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (cos(1/2*arctan2
(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) - (cos(2*d*x + 2*c) - 2)*sin(1/2*arctan2(sin(2*d*x + 2*
c), cos(2*d*x + 2*c))) - sin(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) + 3*log(s
qrt(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 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(si
n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1
)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) - 3*log(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 + sqrt(cos(2
*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c
) + 1))^2 - 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d
*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) + 3*log(((cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(
1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2
+ (cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c
)))^2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2)*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^
2 + 2*cos(2*d*x + 2*c) + 1) + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(
1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + si
n(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) +
1) - 3*log(((cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c)))^2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + (cos(1/2*arctan2(sin(2*d*x + 2*c
), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2)*sin(1/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c) + 1))^2)*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) - 2*(co
s(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos
(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) + 1))/d
Giac [F]
\[
\int \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x) \, dx=\int { \sqrt {-\cos \left (d x + c\right ) + 1} \cos \left (d x + c\right )^{\frac {3}{2}} \,d x }
\]
[In]
integrate((1-cos(d*x+c))^(1/2)*cos(d*x+c)^(3/2),x, algorithm="giac")
[Out]
integrate(sqrt(-cos(d*x + c) + 1)*cos(d*x + c)^(3/2), x)
Mupad [F(-1)]
Timed out. \[
\int \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x) \, dx=\int {\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {1-\cos \left (c+d\,x\right )} \,d x
\]
[In]
int(cos(c + d*x)^(3/2)*(1 - cos(c + d*x))^(1/2),x)
[Out]
int(cos(c + d*x)^(3/2)*(1 - cos(c + d*x))^(1/2), x)