\(\int \frac {1}{\sqrt {1+\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [366]

Optimal result

Integrand size = 23, antiderivative size = 125 \[ \int \frac {1}{\sqrt {1+\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{1+\cos (c+d x)}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}-\frac {\arcsin \left (\frac {\sin (c+d x)}{\sqrt {1+\cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {\sin (c+d x)}{d \sqrt {1+\cos (c+d x)} \sqrt {\sec (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 125, 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 = {4307, 2857, 3061, 2860, 222, 2853} \[ \int \frac {1}{\sqrt {1+\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arcsin \left (\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}-\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arcsin \left (\frac {\sin (c+d x)}{\sqrt {\cos (c+d x)+1}}\right )}{d}+\frac {\sin (c+d x)}{d \sqrt {\cos (c+d x)+1} \sqrt {\sec (c+d x)}} \]

[In]

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

[Out]

(Sqrt[2]*ArcSin[Sin[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/d - (ArcSin[Sin[c + d*
x]/Sqrt[1 + Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/d + Sin[c + d*x]/(d*Sqrt[1 + Cos[c + d*x]]*S
qrt[Sec[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 4307

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sin[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.66 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.06 \[ \int \frac {1}{\sqrt {1+\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {i e^{-2 i (c+d x)} \left (1+e^{i (c+d x)}\right ) \left (1-e^{i (c+d x)}+e^{2 i (c+d x)}-e^{3 i (c+d x)}+e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arcsinh}\left (e^{i (c+d x)}\right )+2 \sqrt {2} e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )-e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {\sec (c+d x)}}{4 d \sqrt {1+\cos (c+d x)}} \]

[In]

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

[Out]

((I/4)*(1 + E^(I*(c + d*x)))*(1 - E^(I*(c + d*x)) + E^((2*I)*(c + d*x)) - E^((3*I)*(c + d*x)) + E^(I*(c + d*x)
)*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))] + 2*Sqrt[2]*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*
x))]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] - E^(I*(c + d*x))*Sqrt[1 + E^((2*I
)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])*Sqrt[Sec[c + d*x]])/(d*E^((2*I)*(c + d*x))*Sqrt[1 + Cos[
c + d*x]])

Maple [A] (verified)

Time = 13.54 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06

method	result	size
default	\(-\frac {\left (\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {2}-\sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {2+2 \cos \left (d x +c \right )}\, \sqrt {2}}{2 d \sqrt {\sec \left (d x +c \right )}\, \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(133\)

[In]

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

[Out]

-1/2/d/sec(d*x+c)^(1/2)*(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)+ar
ctan(tan(d*x+c)*(cos(d*x+c)/(1+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 = 125, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {1+\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {{\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {\sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right ) + d} \]

[In]

integrate(1/sec(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)

Sympy [F]

\[ \int \frac {1}{\sqrt {1+\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {1}{\sqrt {\cos {\left (c + d x \right )} + 1} \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\sqrt {1+\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {1}{\sqrt {\cos \left (d x + c\right ) + 1} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {1+\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {1+\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {1}{\sqrt {\cos \left (c+d\,x\right )+1}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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