3.362 \(\int \frac{\sec ^{\frac{5}{2}}(c+d x)}{\sqrt{1+\cos (c+d x)}} \, dx\)

Optimal. Leaf size=118 \[ \frac{2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 d \sqrt{\cos (c+d x)+1}}+\frac{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}-\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d \sqrt{\cos (c+d x)+1}} \]

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

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Rubi [A]  time = 0.201423, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4222, 2779, 2984, 12, 2781, 216} \[ \frac{2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 d \sqrt{\cos (c+d x)+1}}+\frac{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}-\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)}}{3 d \sqrt{\cos (c+d x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(5/2)/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])/(3*d*Sqrt[1 + Cos[c + d*x]]) + (2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*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 2984

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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{5}{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{5}{2}}(c+d x) \sqrt{1+\cos (c+d x)}} \, dx\\ &=\frac{2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt{1+\cos (c+d x)}}-\frac{1}{3} \left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1-2 \cos (c+d x)}{\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)}{3 d \sqrt{1+\cos (c+d x)}}+\frac{2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt{1+\cos (c+d x)}}-\frac{1}{3} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int -\frac{3}{2 \sqrt{\cos (c+d x)} \sqrt{1+\cos (c+d x)}} \, dx\\ &=-\frac{2 \sqrt{\sec (c+d x)} \sin (c+d x)}{3 d \sqrt{1+\cos (c+d x)}}+\frac{2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 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)}{3 d \sqrt{1+\cos (c+d x)}}+\frac{2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 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)}{3 d \sqrt{1+\cos (c+d x)}}+\frac{2 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d \sqrt{1+\cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.63411, size = 473, normalized size = 4.01 \[ -\frac{2 \left (\frac{1}{1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}\right )^{7/2} \cot \left (\frac{c}{2}+\frac{d x}{2}\right ) \csc ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (12 \sin ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \, _3F_2\left (2,2,\frac{7}{2};1,\frac{9}{2};\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}\right )+12 \left (3 \sin ^4\left (\frac{c}{2}+\frac{d x}{2}\right )-7 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )+4\right ) \sin ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}\right )+7 \left (1-2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right )^3 \sqrt{\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}} \left (8 \sin ^4\left (\frac{c}{2}+\frac{d x}{2}\right )-20 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )+15\right ) \left (\sqrt{\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}} \left (7 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-3\right )+\left (3-6 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )\right ) \tanh ^{-1}\left (\sqrt{\frac{\sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 \sin ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-1}}\right )\right )\right )}{63 d \sqrt{\cos (c+d x)+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^(5/2)/Sqrt[1 + Cos[c + d*x]],x]

[Out]

(-2*Cot[c/2 + (d*x)/2]*Csc[c/2 + (d*x)/2]^4*((1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1))^(7/2)*(12*Cos[(c + d*x)/2]^4*H
ypergeometricPFQ[{2, 2, 7/2}, {1, 9/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]
^8 + 12*Hypergeometric2F1[2, 7/2, 9/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^
8*(4 - 7*Sin[c/2 + (d*x)/2]^2 + 3*Sin[c/2 + (d*x)/2]^4) + 7*(1 - 2*Sin[c/2 + (d*x)/2]^2)^3*Sqrt[Sin[c/2 + (d*x
)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(15 - 20*Sin[c/2 + (d*x)/2]^2 + 8*Sin[c/2 + (d*x)/2]^4)*(ArcTanh[Sqrt[Si
n[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*(3 - 6*Sin[c/2 + (d*x)/2]^2) + Sqrt[Sin[c/2 + (d*x)/2]^2/(-
1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-3 + 7*Sin[c/2 + (d*x)/2]^2))))/(63*d*Sqrt[1 + Cos[c + d*x]])

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Maple [B]  time = 0.332, size = 228, normalized size = 1.9 \begin{align*}{\frac{\sqrt{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{6\,d \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( 1+\cos \left ( dx+c \right ) \right ) ^{2}} \left ( 3\, \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +6\, \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}\cos \left ( dx+c \right ) \sqrt{2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +3\, \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{3/2}\sqrt{2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -2\,\sin \left ( dx+c \right ) \right ) \sqrt{2+2\,\cos \left ( dx+c \right ) } \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(5/2)/(1+cos(d*x+c))^(1/2),x)

[Out]

1/6/d*2^(1/2)*(3*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^2*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+6*(
cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+3*(cos(d*x+c)/(1+cos(d*
x+c)))^(3/2)*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+2*cos(d*x+c)*sin(d*x+c)-2*sin(d*x+c))*cos(d*x+c)*(2+2*
cos(d*x+c))^(1/2)*sin(d*x+c)^2*(1/cos(d*x+c))^(5/2)/(-1+cos(d*x+c))/(1+cos(d*x+c))^2

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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)^(5/2)/(1+cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.84904, size = 328, normalized size = 2.78 \begin{align*} -\frac{3 \,{\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 )}\right ) + \frac{2 \, \sqrt{\cos \left (d x + c\right ) + 1}{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{3 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/2)/(1+cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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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)**(5/2)/(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{5}{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)^(5/2)/(1+cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^(5/2)/sqrt(cos(d*x + c) + 1), x)