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

Optimal. Leaf size=82 \[ \frac{2 \sin (c+d x) \sqrt{\sec (c+d x)}}{d \sqrt{\cos (c+d x)+1}}-\frac{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)+1}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.118816, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4222, 2779, 2781, 216} \[ \frac{2 \sin (c+d x) \sqrt{\sec (c+d x)}}{d \sqrt{\cos (c+d x)+1}}-\frac{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)+1}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

Rule 4222

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]

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]

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]

Rubi steps

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

Mathematica [C]  time = 1.85072, size = 178, normalized size = 2.17 \[ \frac{2 \sin \left (\frac{1}{2} (c+d x)\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{3}{2}}(c+d x) \left (\frac{1}{2} \cos (c+d x) (\cos (c+d x)+2) \csc ^4\left (\frac{1}{2} (c+d x)\right ) \left (-\cos (c+d x)+\cos (c+d x) \sqrt{2-2 \sec (c+d x)} \tanh ^{-1}\left (\sqrt{\sin ^2\left (\frac{1}{2} (c+d x)\right ) (-\sec (c+d x))}\right )+1\right )-\frac{1}{10} \sin (c+d x) \tan (c+d x) \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};-\sec (c+d x) \sin ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )}{d \sqrt{\cos (c+d x)+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*Sec[c + d*x]^(3/2)*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
- (Hypergeometric2F1[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*Sqrt
[1 + Cos[c + d*x]])

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Maple [A]  time = 0.318, size = 144, normalized size = 1.8 \begin{align*}{\frac{\sqrt{2}\cos \left ( dx+c \right ) }{2\,d \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ( \sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \cos \left ( dx+c \right ) +\sqrt{2}\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +2\,\sin \left ( dx+c \right ) \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{3}{2}}}\sqrt{2+2\,\cos \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.79921, size = 252, normalized size = 3.07 \begin{align*} \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 ) + \frac{2 \, \sqrt{\cos \left (d x + c\right ) + 1} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{d \cos \left (d x + c\right ) + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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)) + 2*s
qrt(cos(d*x + c) + 1)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{\frac{3}{2}}}{\sqrt{\cos \left (d x + c\right ) + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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