3.237 \(\int \frac{1}{\cos ^{\frac{7}{2}}(c+d x) \sqrt{1+\cos (c+d x)}} \, dx\)

Optimal. Leaf size=134 \[ -\frac{2 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{\cos (c+d x)+1}}+\frac{2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{\cos (c+d x)+1}}-\frac{\sqrt{2} \sin ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}+\frac{26 \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{\cos (c+d x)+1}} \]

[Out]

-((Sqrt[2]*ArcSin[Sin[c + d*x]/(1 + Cos[c + d*x])])/d) + (2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[1 + Cos
[c + d*x]]) - (2*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)*Sqrt[1 + Cos[c + d*x]]) + (26*Sin[c + d*x])/(15*d*Sqrt
[Cos[c + d*x]]*Sqrt[1 + Cos[c + d*x]])

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Rubi [A]  time = 0.242472, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{2 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{\cos (c+d x)+1}}+\frac{2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{\cos (c+d x)+1}}-\frac{\sqrt{2} \sin ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}+\frac{26 \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{\cos (c+d x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(7/2)*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])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[1 + Cos
[c + d*x]]) - (2*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)*Sqrt[1 + Cos[c + d*x]]) + (26*Sin[c + d*x])/(15*d*Sqrt
[Cos[c + d*x]]*Sqrt[1 + Cos[c + d*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{1}{\cos ^{\frac{7}{2}}(c+d x) \sqrt{1+\cos (c+d x)}} \, dx &=\frac{2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{1+\cos (c+d x)}}-\frac{1}{5} \int \frac{1-4 \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{1+\cos (c+d x)}} \, dx\\ &=\frac{2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{1+\cos (c+d x)}}-\frac{2 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{1+\cos (c+d x)}}-\frac{2}{15} \int \frac{-\frac{13}{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)}{5 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{1+\cos (c+d x)}}-\frac{2 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{1+\cos (c+d x)}}+\frac{26 \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{1+\cos (c+d x)}}-\frac{4}{15} \int \frac{15}{4 \sqrt{\cos (c+d x)} \sqrt{1+\cos (c+d x)}} \, dx\\ &=\frac{2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{1+\cos (c+d x)}}-\frac{2 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{1+\cos (c+d x)}}+\frac{26 \sin (c+d x)}{15 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)}{5 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{1+\cos (c+d x)}}-\frac{2 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{1+\cos (c+d x)}}+\frac{26 \sin (c+d x)}{15 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)}{5 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{1+\cos (c+d x)}}-\frac{2 \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{1+\cos (c+d x)}}+\frac{26 \sin (c+d x)}{15 d \sqrt{\cos (c+d x)} \sqrt{1+\cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 7.95131, size = 1538, normalized size = 11.48 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Cos[c + d*x]^(7/2)*Sqrt[1 + Cos[c + d*x]]),x]

[Out]

(-2*Cot[c/2 + (d*x)/2]*Csc[c/2 + (d*x)/2]^6*(4725*Sin[c/2 + (d*x)/2]^2 - 48825*Sin[c/2 + (d*x)/2]^4 + 210105*S
in[c/2 + (d*x)/2]^6 - 486630*Sin[c/2 + (d*x)/2]^8 + 655812*Sin[c/2 + (d*x)/2]^10 - 710*Hypergeometric2F1[2, 9/
2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^10 - 40*Cos[(c + d*x)/2]^6*Hyp
ergeometricPFQ[{2, 2, 2, 9/2}, {1, 1, 11/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*
x)/2]^10 - 518760*Sin[c/2 + (d*x)/2]^12 + 1770*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Si
n[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 + 226656*Sin[c/2 + (d*x)/2]^14 - 1500*Hypergeometric2F1[2, 9/2, 11/
2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^14 - 42048*Sin[c/2 + (d*x)/2]^16 + 4
40*Hypergeometric2F1[2, 9/2, 11/2, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^16 +
4725*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[
c/2 + (d*x)/2]^2)] - 56700*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2
]^2*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 291060*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 +
2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^4*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 8337
60*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^6*Sqrt[Sin[c/2 + (d*x)
/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 1458000*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)
]]*Sin[c/2 + (d*x)/2]^8*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 1598400*ArcTanh[Sqrt[Sin[c/
2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^10*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2
+ (d*x)/2]^2)] + 1080000*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]
^12*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] - 414720*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 +
2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^14*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 691
20*ArcTanh[Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Sin[c/2 + (d*x)/2]^16*Sqrt[Sin[c/2 + (d*x
)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)] + 60*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2, 9/2}, {1, 11/2}, Sin[c/
2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^10*(-5 + 4*Sin[c/2 + (d*x)/2]^2)))/(675*d*Sqr
t[1 + Cos[c + d*x]]*(1 - 2*Sin[c/2 + (d*x)/2]^2)^(7/2)*(-1 + 2*Sin[c/2 + (d*x)/2]^2))

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Maple [B]  time = 0.316, size = 344, normalized size = 2.6 \begin{align*} -{\frac{\sqrt{2} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{30\,d \left ( -1+\cos \left ( dx+c \right ) \right ) ^{3} \left ( 1+\cos \left ( dx+c \right ) \right ) ^{4}} \left ( 15\,\sqrt{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{7/2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +60\,\sqrt{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{7/2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +90\,\sqrt{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{7/2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +60\,\sqrt{2}\cos \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{7/2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +15\,\sqrt{2} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{7/2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +26\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +6\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) \sqrt{2+2\,\cos \left ( dx+c \right ) } \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(d*x+c)^(7/2)/(1+cos(d*x+c))^(1/2),x)

[Out]

-1/30/d*2^(1/2)*(15*2^(1/2)*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+
60*2^(1/2)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+90*2^(1/2)*cos(d*
x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+60*2^(1/2)*cos(d*x+c)*(cos(d*x+c)/
(1+cos(d*x+c)))^(7/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+15*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arcsin((
-1+cos(d*x+c))/sin(d*x+c))+26*cos(d*x+c)^3*sin(d*x+c)-2*cos(d*x+c)^2*sin(d*x+c)+6*cos(d*x+c)*sin(d*x+c))*sin(d
*x+c)^6*(2+2*cos(d*x+c))^(1/2)/(-1+cos(d*x+c))^3/(1+cos(d*x+c))^4/cos(d*x+c)^(7/2)

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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(1/cos(d*x+c)^(7/2)/(1+cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.13795, size = 410, normalized size = 3.06 \begin{align*} \frac{2 \,{\left (13 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 3\right )} \sqrt{\cos \left (d x + c\right ) + 1} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 15 \,{\left (\sqrt{2} \cos \left (d x + c\right )^{4} + \sqrt{2} \cos \left (d x + c\right )^{3}\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 )}{15 \,{\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(7/2)/(1+cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/15*(2*(13*cos(d*x + c)^2 - cos(d*x + c) + 3)*sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c))*sin(d*x + c) - 15*(sq
rt(2)*cos(d*x + c)^4 + sqrt(2)*cos(d*x + c)^3)*arctan(1/2*sqrt(2)*sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c))*si
n(d*x + c)/(cos(d*x + c)^2 + cos(d*x + c))))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)

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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(1/cos(d*x+c)**(7/2)/(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{1}{\sqrt{\cos \left (d x + c\right ) + 1} \cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(7/2)/(1+cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(cos(d*x + c) + 1)*cos(d*x + c)^(7/2)), x)