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]

-2/15*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(1+cos(d*x+c))^(1/2)+2/5*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(1+cos(d*x+c))^(1/2
)-arcsin(sin(d*x+c)/(1+cos(d*x+c)))*2^(1/2)*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+26/15*sin(d*x+c)*sec(d*x+c)^(1
/2)/d/(1+cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.28, antiderivative size = 154, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [C]  time = 7.81, size = 1540, normalized size = 10.00 \[ \text {result too large to display} \]

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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fricas [A]  time = 1.25, size = 130, normalized size = 0.84 \[ \frac {15 \, {\left (\sqrt {2} \cos \left (d x + c\right )^{3} + \sqrt {2} \cos \left (d x + c\right )^{2}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (13 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 3\right )} \sqrt {\cos \left (d x + c\right ) + 1} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{\frac {7}{2}}}{\sqrt {\cos \left (d x + c\right ) + 1}}\,{d x} \]

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)

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maple [B]  time = 0.22, size = 294, normalized size = 1.91 \[ \frac {\left (15 \sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+45 \sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+45 \sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \cos \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+15 \sqrt {2}\, \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}+26 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )+6 \sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {2+2 \cos \left (d x +c \right )}\, \left (\sin ^{4}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {7}{2}} \sqrt {2}}{30 d \left (-1+\cos \left (d x +c \right )\right )^{2} \left (1+\cos \left (d x +c \right )\right )^{3}} \]

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*(15*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)+45*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)+45*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)+15*2^(1/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))
*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+26*cos(d*x+c)^2*sin(d*x+c)-2*cos(d*x+c)*sin(d*x+c)+6*sin(d*x+c))*cos(d*x+c)
*(2+2*cos(d*x+c))^(1/2)*sin(d*x+c)^4*(1/cos(d*x+c))^(7/2)/(-1+cos(d*x+c))^2/(1+cos(d*x+c))^3*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(sec(d*x+c)^(7/2)/(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 \[ \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{\sqrt {\cos \left (c+d\,x\right )+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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