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

Optimal. Leaf size=154 \[ \frac{2 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{\cos (c+d x)+1}}-\frac{2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 d \sqrt{\cos (c+d x)+1}}-\frac{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}+\frac{26 \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{\cos (c+d x)+1}} \]

[Out]

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

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Rubi [A]  time = 0.284152, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{2 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d \sqrt{\cos (c+d x)+1}}-\frac{2 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{15 d \sqrt{\cos (c+d x)+1}}-\frac{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sin ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)+1}\right )}{d}+\frac{26 \sin (c+d x) \sqrt{\sec (c+d x)}}{15 d \sqrt{\cos (c+d x)+1}} \]

Antiderivative was successfully verified.

[In]

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

Mathematica [C]  time = 7.76938, size = 1540, normalized size = 10. \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[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*((1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1))^(7/2)*(4725*Sin[c/2 + (d*x)/2]
^2 - 48825*Sin[c/2 + (d*x)/2]^4 + 210105*Sin[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*HypergeometricPFQ[{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*Sin[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 + 440*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)] + 29106
0*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)] - 833760*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)] - 41472
0*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)] + 69120*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*Hyperg
eometricPFQ[{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*Sqrt[1 + Cos[c + d*x]]*(-1 + 2*Sin[c/2 + (d*x)/2]^2))

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Maple [B]  time = 0.356, size = 294, normalized size = 1.9 \begin{align*}{\frac{\sqrt{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{30\,d \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2} \left ( 1+\cos \left ( dx+c \right ) \right ) ^{3}} \left ( 15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}\sqrt{2}+45\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}\sqrt{2}+45\,\cos \left ( dx+c \right ) \arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}\sqrt{2}+15\,\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{5/2}\sqrt{2}+26\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +6\,\sin \left ( dx+c \right ) \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{7}{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)^(7/2)/(1+cos(d*x+c))^(1/2),x)

[Out]

1/30/d*2^(1/2)*(15*cos(d*x+c)^3*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*2^(1/2)+4
5*cos(d*x+c)^2*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*2^(1/2)+45*cos(d*x+c)*arcs
in((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*2^(1/2)+15*arcsin((-1+cos(d*x+c))/sin(d*x+c))
*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*2^(1/2)+26*cos(d*x+c)^2*sin(d*x+c)-2*cos(d*x+c)*sin(d*x+c)+6*sin(d*x+c))*co
s(d*x+c)*(1/cos(d*x+c))^(7/2)*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

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.81763, size = 362, normalized size = 2.35 \begin{align*} \frac{15 \,{\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 )}\right ) + \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} \sin \left (d x + c\right )}{\sqrt{\cos \left (d x + c\right )}}}{15 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*(sqrt(2)*cos(d*x + c)^3 + sqrt(2)*cos(d*x + c)^2)*arctan(sqrt(2)*sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x
+ c))/sin(d*x + c)) + 2*(13*cos(d*x + c)^2 - cos(d*x + c) + 3)*sqrt(cos(d*x + c) + 1)*sin(d*x + c)/sqrt(cos(d*
x + c)))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)

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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)**(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{\sec \left (d x + c\right )^{\frac{7}{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)^(7/2)/(1+cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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