3.379 \(\int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=237 \[ \frac {163 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {95 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{48 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {299 \sin (c+d x) \sqrt {\sec (c+d x)}}{48 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {17 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[Out]

-1/4*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)-17/16*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*cos(d*x+c
))^(3/2)+95/48*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)+163/32*arctan(1/2*sin(d*x+c)*a^(1/2)*2
^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(5/2)/d*2^(1/2)-299/48*sin
(d*x+c)*sec(d*x+c)^(1/2)/a^2/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.64, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4222, 2766, 2978, 2984, 12, 2782, 205} \[ \frac {95 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{48 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {299 \sin (c+d x) \sqrt {\sec (c+d x)}}{48 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {163 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {17 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(5/2)/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(163*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[c + d*x]]*S
qrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - (299*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(48*a^2*d*Sqrt[a + a*Cos[c +
d*x]]) - (Sec[c + d*x]^(3/2)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - (17*Sec[c + d*x]^(3/2)*Sin[c +
d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) + (95*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(48*a^2*d*Sqrt[a + a*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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2766

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

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*Sin[e + f*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]

Rule 2978

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
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 {5}{2}}(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx\\ &=-\frac {\sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {11 a}{2}-3 a \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {\sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {17 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {95 a^2}{4}-17 a^2 \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {\sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {17 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {95 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {299 a^3}{8}+\frac {95}{4} a^3 \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{12 a^5}\\ &=-\frac {299 \sqrt {\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {\sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {17 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {95 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {489 a^4}{16 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{6 a^6}\\ &=-\frac {299 \sqrt {\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {\sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {17 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {95 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\left (163 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {299 \sqrt {\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {\sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {17 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {95 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {\left (163 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{2 a^2+a x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 a d}\\ &=\frac {163 \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {299 \sqrt {\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {\sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {17 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {95 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{48 a^2 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 8.11, size = 641, normalized size = 2.70 \[ -\frac {\left (\frac {1}{1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right )^{7/2} \cot ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (640 \sin ^{12}\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos ^8\left (\frac {1}{2} (c+d x)\right ) \, _5F_4\left (2,2,2,2,\frac {7}{2};1,1,1,\frac {13}{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 )-1280 \left (5 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-6\right ) \sin ^{12}\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos ^6\left (\frac {1}{2} (c+d x)\right ) \, _4F_3\left (2,2,2,\frac {7}{2};1,1,\frac {13}{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 )+33 \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 (\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 (4344400 \sin ^{14}\left (\frac {c}{2}+\frac {d x}{2}\right )-26448512 \sin ^{12}\left (\frac {c}{2}+\frac {d x}{2}\right )[phone] \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right )-96281836 \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right )[phone] \sin ^6\left (\frac {c}{2}+\frac {d x}{2}\right )-38990350 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )[phone] \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-1148175\right )-105 \left (33208 \sin ^{10}\left (\frac {c}{2}+\frac {d x}{2}\right )-140732 \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right )+234156 \sin ^6\left (\frac {c}{2}+\frac {d x}{2}\right )-188110 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )+72902 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-10935\right ) \cos ^4\left (\frac {1}{2} (c+d x)\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 )}{41580 d (a (\cos (c+d x)+1))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/41580*(Cot[c/2 + (d*x)/2]^5*Csc[c/2 + (d*x)/2]^4*Sec[(c + d*x)/2]^4*((1 - 2*Sin[c/2 + (d*x)/2]^2)^(-1))^(7/
2)*(640*Cos[(c + d*x)/2]^8*HypergeometricPFQ[{2, 2, 2, 2, 7/2}, {1, 1, 1, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*
Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12 - 1280*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 7/2}, {1, 1
, 13/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^12*(-6 + 5*Sin[c/2 + (d*x)/2]^
2) + 33*(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)]*(-105*ArcTanh[
Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*Cos[(c + d*x)/2]^4*(-10935 + 72902*Sin[c/2 + (d*x)/2
]^2 - 188110*Sin[c/2 + (d*x)/2]^4 + 234156*Sin[c/2 + (d*x)/2]^6 - 140732*Sin[c/2 + (d*x)/2]^8 + 33208*Sin[c/2
+ (d*x)/2]^10) + Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-1148175 + 10333785*Sin[c/2 + (d*x)
/2]^2 - 38990350*Sin[c/2 + (d*x)/2]^4 + 79946462*Sin[c/2 + (d*x)/2]^6 - 96281836*Sin[c/2 + (d*x)/2]^8 + 682435
96*Sin[c/2 + (d*x)/2]^10 - 26448512*Sin[c/2 + (d*x)/2]^12 + 4344400*Sin[c/2 + (d*x)/2]^14))))/(d*(a*(1 + Cos[c
+ d*x]))^(5/2))

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fricas [A]  time = 0.98, size = 195, normalized size = 0.82 \[ -\frac {489 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (299 \, \cos \left (d x + c\right )^{3} + 503 \, \cos \left (d x + c\right )^{2} + 160 \, \cos \left (d x + c\right ) - 32\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{96 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(489*sqrt(2)*(cos(d*x + c)^4 + 3*cos(d*x + c)^3 + 3*cos(d*x + c)^2 + cos(d*x + c))*sqrt(a)*arctan(sqrt(2
)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*(299*cos(d*x + c)^3 + 503*cos(d*x +
c)^2 + 160*cos(d*x + c) - 32)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^4
+ 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.24, size = 316, normalized size = 1.33 \[ -\frac {\left (489 \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+1467 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+1467 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}+489 \sin \left (d x +c \right ) \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}}-299 \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}-204 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}+343 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+192 \cos \left (d x +c \right ) \sqrt {2}-32 \sqrt {2}\right ) \cos \left (d x +c \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{96 d \sin \left (d x +c \right ) \left (1+\cos \left (d x +c \right )\right )^{2} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96/d*(489*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3*sin(d*x+c)+1467
*cos(d*x+c)^2*sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+1467*cos(d*x+c)*
sin(d*x+c)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+489*sin(d*x+c)*arcsin((-1+cos(
d*x+c))/sin(d*x+c))*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-299*cos(d*x+c)^4*2^(1/2)-204*cos(d*x+c)^3*2^(1/2)+343*co
s(d*x+c)^2*2^(1/2)+192*cos(d*x+c)*2^(1/2)-32*2^(1/2))*cos(d*x+c)*(1/cos(d*x+c))^(5/2)*(a*(1+cos(d*x+c)))^(1/2)
/sin(d*x+c)/(1+cos(d*x+c))^2*2^(1/2)/a^3

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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