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

Optimal. Leaf size=62 \[ \frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}-\frac {\sqrt {2} \sin ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2779, 2781, 216} \[ \frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}}-\frac {\sqrt {2} \sin ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[1/(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) + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*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 2779

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Sim
p[(d*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x] - Dist[1/
(2*b*(n + 1)*(c^2 - d^2)), Int[((c + d*Sin[e + f*x])^(n + 1)*Simp[a*d - 2*b*c*(n + 1) + b*d*(2*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] && LtQ[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]

Rubi steps

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

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Mathematica [C]  time = 1.79, size = 178, normalized size = 2.87 \[ \frac {2 \sin \left (\frac {1}{2} (c+d x)\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (\frac {1}{2} \cos (c+d x) (\cos (c+d x)+2) \csc ^4\left (\frac {1}{2} (c+d x)\right ) \left (-\cos (c+d x)+\cos (c+d x) \sqrt {2-2 \sec (c+d x)} \tanh ^{-1}\left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right ) (-\sec (c+d x))}\right )+1\right )-\frac {1}{10} \sin (c+d x) \tan (c+d x) \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )}{d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*Sin[(c + d*x)/2]*((Cos[c + d*x]*(2 + Cos[c + d*x])*Csc[(c + d*x)/2]^4*(1 - Cos[c + d*x] +
ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]]*Cos[c + d*x]*Sqrt[2 - 2*Sec[c + d*x]]))/2 - (Hypergeometric2
F1[2, 5/2, 7/2, -(Sec[c + d*x]*Sin[(c + d*x)/2]^2)]*Sin[c + d*x]*Tan[c + d*x])/10))/(d*Cos[c + d*x]^(3/2)*Sqrt
[1 + Cos[c + d*x]])

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fricas [B]  time = 2.13, size = 121, normalized size = 1.95 \[ -\frac {{\left (\sqrt {2} \cos \left (d x + c\right )^{2} + \sqrt {2} \cos \left (d x + c\right )\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )}}\right ) - 2 \, \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.15, size = 210, normalized size = 3.39 \[ -\frac {\left (\sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+2 \sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \cos \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+\sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right ) \sqrt {2+2 \cos \left (d x +c \right )}\, \sqrt {2}}{2 d \left (-1+\cos \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \cos \left (d x +c \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*(2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+2*2^(1/2)*ar
csin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+2^(1/2)*arcsin((-1+cos(d*x+c))/s
in(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+2*cos(d*x+c)*sin(d*x+c))*sin(d*x+c)^2*(2+2*cos(d*x+c))^(1/2)/(-1+
cos(d*x+c))/(1+cos(d*x+c))^2/cos(d*x+c)^(3/2)*2^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {\cos \left (c+d\,x\right )+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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