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

Optimal. Leaf size=98 \[ \frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}+\frac {\sqrt {2} \sin ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}-\frac {2 \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}} \]

[Out]

arcsin(sin(d*x+c)/(1+cos(d*x+c)))*2^(1/2)/d+2/3*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(1+cos(d*x+c))^(1/2)-2/3*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 = 0.17, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2779, 2984, 12, 2781, 216} \[ \frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x)+1}}+\frac {\sqrt {2} \sin ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}-\frac {2 \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x)+1}} \]

Antiderivative was successfully verified.

[In]

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

Rule 12

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

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])

Rubi steps

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

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Mathematica [C]  time = 6.63, size = 471, normalized size = 4.81 \[ -\frac {2 \cot \left (\frac {c}{2}+\frac {d x}{2}\right ) \csc ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (12 \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \, _3F_2\left (2,2,\frac {7}{2};1,\frac {9}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-1}\right )+12 \left (3 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )-7 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+4\right ) \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \, _2F_1\left (2,\frac {7}{2};\frac {9}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-1}\right )+7 \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3 \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-1}} \left (8 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )-20 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+15\right ) \left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-1}} \left (7 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-3\right )+\left (3-6 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \tanh ^{-1}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-1}}\right )\right )\right )}{63 d \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{7/2} \sqrt {\cos (c+d x)+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Cos[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*(12*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2, 7/2}, {1, 9/2}, Si
n[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[Sin[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]]*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2))

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fricas [A]  time = 3.61, size = 134, normalized size = 1.37 \[ -\frac {2 \, \sqrt {\cos \left (d x + c\right ) + 1} {\left (\cos \left (d x + c\right ) - 1\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 3 \, {\left (\sqrt {2} \cos \left (d x + c\right )^{3} + \sqrt {2} \cos \left (d x + c\right )^{2}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )}}\right )}{3 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\cos \left (d x + c\right ) + 1} \cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.14, size = 278, normalized size = 2.84 \[ -\frac {\left (3 \sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+9 \sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+9 \sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \cos \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+3 \sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+2 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \left (\sin ^{4}\left (d x +c \right )\right ) \sqrt {2+2 \cos \left (d x +c \right )}\, \sqrt {2}}{6 d \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{3} \cos \left (d x +c \right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d*(3*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+9*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)))^(5/2)+9*2^(1/2)*arcsin((-1+cos(d*x
+c))/sin(d*x+c))*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+3*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(co
s(d*x+c)/(1+cos(d*x+c)))^(5/2)+2*cos(d*x+c)^2*sin(d*x+c)-2*cos(d*x+c)*sin(d*x+c))*sin(d*x+c)^4*(2+2*cos(d*x+c)
)^(1/2)/(-1+cos(d*x+c))^2/(1+cos(d*x+c))^3/cos(d*x+c)^(5/2)*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)^(5/2)/(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 = -0.01 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {\cos \left (c+d\,x\right )+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\cos {\left (c + d x \right )} + 1} \cos ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(cos(c + d*x) + 1)*cos(c + d*x)**(5/2)), x)

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