3.258 \(\int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=257 \[ \frac {1015 \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{64 \sqrt {2} a^{7/2} d}+\frac {193 \sin (c+d x)}{64 a^3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {629 \sin (c+d x)}{64 a^3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {109 \sin (c+d x)}{64 a^2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac {23 \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}-\frac {\sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}} \]

[Out]

-1/6*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(7/2)-23/48*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c
))^(5/2)-109/64*sin(d*x+c)/a^2/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2)+1015/128*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))/a^(7/2)/d*2^(1/2)+193/64*sin(d*x+c)/a^3/d/cos(d*x+c)^(3/2)/
(a+a*cos(d*x+c))^(1/2)-629/64*sin(d*x+c)/a^3/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.70, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2766, 2978, 2984, 12, 2782, 205} \[ \frac {193 \sin (c+d x)}{64 a^3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {109 \sin (c+d x)}{64 a^2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac {629 \sin (c+d x)}{64 a^3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {1015 \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{64 \sqrt {2} a^{7/2} d}-\frac {23 \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}-\frac {\sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(7/2)),x]

[Out]

(1015*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(64*Sqrt[2]*a^(7/2
)*d) - Sin[c + d*x]/(6*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(7/2)) - (23*Sin[c + d*x])/(48*a*d*Cos[c + d*
x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)) - (109*Sin[c + d*x])/(64*a^2*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3
/2)) + (193*Sin[c + d*x])/(64*a^3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - (629*Sin[c + d*x])/(64*a^3*
d*Sqrt[Cos[c + d*x]]*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])

Rubi steps

\begin {align*} \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}} \, dx &=-\frac {\sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}+\frac {\int \frac {\frac {15 a}{2}-4 a \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=-\frac {\sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac {23 \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {\frac {189 a^2}{4}-\frac {69}{2} a^2 \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=-\frac {\sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac {23 \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {109 \sin (c+d x)}{64 a^2 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\frac {1737 a^3}{8}-\frac {327}{2} a^3 \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=-\frac {\sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac {23 \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {109 \sin (c+d x)}{64 a^2 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {193 \sin (c+d x)}{64 a^3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {-\frac {5661 a^4}{16}+\frac {1737}{8} a^4 \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{72 a^7}\\ &=-\frac {\sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac {23 \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {109 \sin (c+d x)}{64 a^2 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {193 \sin (c+d x)}{64 a^3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {629 \sin (c+d x)}{64 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {9135 a^5}{32 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{36 a^8}\\ &=-\frac {\sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac {23 \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {109 \sin (c+d x)}{64 a^2 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {193 \sin (c+d x)}{64 a^3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {629 \sin (c+d x)}{64 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {1015 \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{128 a^3}\\ &=-\frac {\sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac {23 \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {109 \sin (c+d x)}{64 a^2 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {193 \sin (c+d x)}{64 a^3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {629 \sin (c+d x)}{64 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}-\frac {1015 \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 )}{64 a^2 d}\\ &=\frac {1015 \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 )}{64 \sqrt {2} a^{7/2} d}-\frac {\sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac {23 \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {109 \sin (c+d x)}{64 a^2 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {193 \sin (c+d x)}{64 a^3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {629 \sin (c+d x)}{64 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 8.43, size = 273, normalized size = 1.06 \[ \frac {i e^{-\frac {3}{2} i (c+d x)} \cos ^7\left (\frac {1}{2} (c+d x)\right ) \left (3045 \sqrt {2} \left (1+e^{i (c+d x)}\right )^6 \left (1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )-2 \left (8277 e^{i (c+d x)}+14388 e^{2 i (c+d x)}+13108 e^{3 i (c+d x)}+5622 e^{4 i (c+d x)}-5622 e^{5 i (c+d x)}-13108 e^{6 i (c+d x)}-14388 e^{7 i (c+d x)}-8277 e^{8 i (c+d x)}-1887 e^{9 i (c+d x)}+1887\right )\right )}{96 d \left (1+e^{i (c+d x)}\right )^6 \cos ^{\frac {3}{2}}(c+d x) (a (\cos (c+d x)+1))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x])^(7/2)),x]

[Out]

((I/96)*(-2*(1887 + 8277*E^(I*(c + d*x)) + 14388*E^((2*I)*(c + d*x)) + 13108*E^((3*I)*(c + d*x)) + 5622*E^((4*
I)*(c + d*x)) - 5622*E^((5*I)*(c + d*x)) - 13108*E^((6*I)*(c + d*x)) - 14388*E^((7*I)*(c + d*x)) - 8277*E^((8*
I)*(c + d*x)) - 1887*E^((9*I)*(c + d*x))) + 3045*Sqrt[2]*(1 + E^(I*(c + d*x)))^6*(1 + E^((2*I)*(c + d*x)))^(3/
2)*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])*Cos[(c + d*x)/2]^7)/(d*E^(((3*I)/2)
*(c + d*x))*(1 + E^(I*(c + d*x)))^6*Cos[c + d*x]^(3/2)*(a*(1 + Cos[c + d*x]))^(7/2))

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fricas [A]  time = 0.98, size = 253, normalized size = 0.98 \[ \frac {3045 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{6} + 4 \, \cos \left (d x + c\right )^{5} + 6 \, \cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, {\left (1887 \, \cos \left (d x + c\right )^{4} + 5082 \, \cos \left (d x + c\right )^{3} + 4251 \, \cos \left (d x + c\right )^{2} + 896 \, \cos \left (d x + c\right ) - 128\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{384 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} + 4 \, a^{4} d \cos \left (d x + c\right )^{5} + 6 \, a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + a^{4} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/384*(3045*sqrt(2)*(cos(d*x + c)^6 + 4*cos(d*x + c)^5 + 6*cos(d*x + c)^4 + 4*cos(d*x + c)^3 + cos(d*x + c)^2)
*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2
+ a*cos(d*x + c))) - 2*(1887*cos(d*x + c)^4 + 5082*cos(d*x + c)^3 + 4251*cos(d*x + c)^2 + 896*cos(d*x + c) -
128)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c))/(a^4*d*cos(d*x + c)^6 + 4*a^4*d*cos(d*x + c)^5
+ 6*a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + a^4*d*cos(d*x + c)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(1/((a*cos(d*x + c) + a)^(7/2)*cos(d*x + c)^(5/2)), x)

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maple [B]  time = 0.21, size = 435, normalized size = 1.69 \[ \frac {\left (-3045 \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}-15225 \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}-30450 \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{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 )-30450 \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{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 ) \sin \left (d x +c \right )-15225 \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )+1887 \sqrt {2}\, \left (\cos ^{6}\left (d x +c \right )\right )-3045 \left (\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \arcsin \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )+3195 \sqrt {2}\, \left (\cos ^{5}\left (d x +c \right )\right )-831 \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}-3355 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}-1024 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+128 \cos \left (d x +c \right ) \sqrt {2}\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{384 d \sin \left (d x +c \right ) \left (1+\cos \left (d x +c \right )\right )^{3} \cos \left (d x +c \right )^{\frac {5}{2}} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(7/2),x)

[Out]

1/384/d*(-3045*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^5*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-15
225*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^4*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-30450*(cos(d*
x+c)/(1+cos(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^3*sin(d*x+c)-30450*(cos(d*x+c)/(1+cos
(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)-15225*(cos(d*x+c)/(1+cos(d*x+c)))^(
5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*cos(d*x+c)*sin(d*x+c)+1887*2^(1/2)*cos(d*x+c)^6-3045*(cos(d*x+c)/(1+co
s(d*x+c)))^(5/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)+3195*2^(1/2)*cos(d*x+c)^5-831*cos(d*x+c)^4*2^(1
/2)-3355*cos(d*x+c)^3*2^(1/2)-1024*cos(d*x+c)^2*2^(1/2)+128*cos(d*x+c)*2^(1/2))*(a*(1+cos(d*x+c)))^(1/2)/sin(d
*x+c)/(1+cos(d*x+c))^3/cos(d*x+c)^(5/2)*2^(1/2)/a^4

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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(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \]

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))^(7/2)),x)

[Out]

int(1/(cos(c + d*x)^(5/2)*(a + a*cos(c + d*x))^(7/2)), 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(1/cos(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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