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

Optimal. Leaf size=297 \[ \frac {(1015 A-363 B) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{64 \sqrt {2} a^{7/2} d}+\frac {(579 A-199 B) \sin (c+d x)}{192 a^3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {(1887 A-691 B) \sin (c+d x)}{192 a^3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {(109 A-41 B) \sin (c+d x)}{64 a^2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac {(23 A-11 B) \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}-\frac {(A-B) \sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}} \]

[Out]

-1/6*(A-B)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(7/2)-1/48*(23*A-11*B)*sin(d*x+c)/a/d/cos(d*x+c)^(3/
2)/(a+a*cos(d*x+c))^(5/2)-1/64*(109*A-41*B)*sin(d*x+c)/a^2/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(3/2)+1/128*(10
15*A-363*B)*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)+1
/192*(579*A-199*B)*sin(d*x+c)/a^3/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)-1/192*(1887*A-691*B)*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 = 1.03, antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2978, 2984, 12, 2782, 205} \[ \frac {(579 A-199 B) \sin (c+d x)}{192 a^3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {(109 A-41 B) \sin (c+d x)}{64 a^2 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac {(1887 A-691 B) \sin (c+d x)}{192 a^3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}+\frac {(1015 A-363 B) \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{64 \sqrt {2} a^{7/2} d}-\frac {(23 A-11 B) \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{5/2}}-\frac {(A-B) \sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1015*A - 363*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(64*Sq
rt[2]*a^(7/2)*d) - ((A - B)*Sin[c + d*x])/(6*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(7/2)) - ((23*A - 11*B)
*Sin[c + d*x])/(48*a*d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(5/2)) - ((109*A - 41*B)*Sin[c + d*x])/(64*a^2*
d*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)) + ((579*A - 199*B)*Sin[c + d*x])/(192*a^3*d*Cos[c + d*x]^(3/2
)*Sqrt[a + a*Cos[c + d*x]]) - ((1887*A - 691*B)*Sin[c + d*x])/(192*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 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 {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}} \, dx &=-\frac {(A-B) \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 {3}{2} a (5 A-B)-4 a (A-B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx}{6 a^2}\\ &=-\frac {(A-B) \sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac {(23 A-11 B) \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 {3}{4} a^2 (63 A-19 B)-\frac {3}{2} a^2 (23 A-11 B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{24 a^4}\\ &=-\frac {(A-B) \sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac {(23 A-11 B) \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {(109 A-41 B) \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 {3}{8} a^3 (579 A-199 B)-\frac {3}{2} a^3 (109 A-41 B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{48 a^6}\\ &=-\frac {(A-B) \sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac {(23 A-11 B) \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {(109 A-41 B) \sin (c+d x)}{64 a^2 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {(579 A-199 B) \sin (c+d x)}{192 a^3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {-\frac {3}{16} a^4 (1887 A-691 B)+\frac {3}{8} a^4 (579 A-199 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{72 a^7}\\ &=-\frac {(A-B) \sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac {(23 A-11 B) \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {(109 A-41 B) \sin (c+d x)}{64 a^2 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {(579 A-199 B) \sin (c+d x)}{192 a^3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {(1887 A-691 B) \sin (c+d x)}{192 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {9 a^5 (1015 A-363 B)}{32 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{36 a^8}\\ &=-\frac {(A-B) \sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac {(23 A-11 B) \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {(109 A-41 B) \sin (c+d x)}{64 a^2 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {(579 A-199 B) \sin (c+d x)}{192 a^3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {(1887 A-691 B) \sin (c+d x)}{192 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {(1015 A-363 B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{128 a^3}\\ &=-\frac {(A-B) \sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac {(23 A-11 B) \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {(109 A-41 B) \sin (c+d x)}{64 a^2 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {(579 A-199 B) \sin (c+d x)}{192 a^3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {(1887 A-691 B) \sin (c+d x)}{192 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}-\frac {(1015 A-363 B) \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 A-363 B) \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 {(A-B) \sin (c+d x)}{6 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{7/2}}-\frac {(23 A-11 B) \sin (c+d x)}{48 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}}-\frac {(109 A-41 B) \sin (c+d x)}{64 a^2 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}}+\frac {(579 A-199 B) \sin (c+d x)}{192 a^3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {(1887 A-691 B) \sin (c+d x)}{192 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 5.40, size = 262, normalized size = 0.88 \[ \frac {\cos ^7\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) (4 (9415 A-3579 B) \cos (c+d x)+8 (3069 A-1145 B) \cos (2 (c+d x))+10164 A \cos (3 (c+d x))+1887 A \cos (4 (c+d x))+21641 A-3748 B \cos (3 (c+d x))-691 B \cos (4 (c+d x))-8469 B)}{32 \cos ^{\frac {3}{2}}(c+d x)}+\frac {3 i (1015 A-363 B) e^{\frac {1}{2} i (c+d x)} \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \tanh ^{-1}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )}{\sqrt {1+e^{2 i (c+d x)}}}\right )}{24 d (a (\cos (c+d x)+1))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^7*(((3*I)*(1015*A - 363*B)*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x)
)]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/Sqrt[1 + E^((2*I)*(c + d*x))] - ((2
1641*A - 8469*B + 4*(9415*A - 3579*B)*Cos[c + d*x] + 8*(3069*A - 1145*B)*Cos[2*(c + d*x)] + 10164*A*Cos[3*(c +
d*x)] - 3748*B*Cos[3*(c + d*x)] + 1887*A*Cos[4*(c + d*x)] - 691*B*Cos[4*(c + d*x)])*Sec[(c + d*x)/2]^5*Tan[(c
+ d*x)/2])/(32*Cos[c + d*x]^(3/2))))/(24*d*(a*(1 + Cos[c + d*x]))^(7/2))

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fricas [A]  time = 0.83, size = 319, normalized size = 1.07 \[ \frac {3 \, \sqrt {2} {\left ({\left (1015 \, A - 363 \, B\right )} \cos \left (d x + c\right )^{6} + 4 \, {\left (1015 \, A - 363 \, B\right )} \cos \left (d x + c\right )^{5} + 6 \, {\left (1015 \, A - 363 \, B\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (1015 \, A - 363 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (1015 \, A - 363 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) - 2 \, {\left ({\left (1887 \, A - 691 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (2541 \, A - 937 \, B\right )} \cos \left (d x + c\right )^{3} + 39 \, {\left (109 \, A - 41 \, B\right )} \cos \left (d x + c\right )^{2} + 128 \, {\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right ) - 128 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{384 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} + 4 \, a^{4} d \cos \left (d x + c\right )^{5} + 6 \, a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + a^{4} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(3*sqrt(2)*((1015*A - 363*B)*cos(d*x + c)^6 + 4*(1015*A - 363*B)*cos(d*x + c)^5 + 6*(1015*A - 363*B)*cos
(d*x + c)^4 + 4*(1015*A - 363*B)*cos(d*x + c)^3 + (1015*A - 363*B)*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*((18
87*A - 691*B)*cos(d*x + c)^4 + 2*(2541*A - 937*B)*cos(d*x + c)^3 + 39*(109*A - 41*B)*cos(d*x + c)^2 + 128*(7*A
- 3*B)*cos(d*x + c) - 128*A)*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 \[ \int \frac {B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.24, size = 715, normalized size = 2.41 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384/d*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))*(-3045*A*cos(d*x+c)^4*2^(1/2)*sin(d*x+c)*(cos(d*x+c)/(1+cos(
d*x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))+1089*B*cos(d*x+c)^4*2^(1/2)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*
x+c)))^(3/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))-12180*A*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3
/2)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)+4356*B*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2
)*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)-18270*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^
2*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+6534*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^2*
2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-12180*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)*2^(
1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+4356*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)*2^(1/2)
*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-3045*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*2^(1/2)*(cos(d*x+c)/(1
+cos(d*x+c)))^(3/2)+1089*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(
3/2)+3774*A*cos(d*x+c)^5-1382*B*cos(d*x+c)^5+6390*A*cos(d*x+c)^4-2366*B*cos(d*x+c)^4-1662*A*cos(d*x+c)^3+550*B
*cos(d*x+c)^3-6710*A*cos(d*x+c)^2+2430*B*cos(d*x+c)^2-2048*A*cos(d*x+c)+768*B*cos(d*x+c)+256*A)/a^4/sin(d*x+c)
^3/(1+cos(d*x+c))^2/cos(d*x+c)^(3/2)

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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((A+B*cos(d*x+c))/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 \[ \int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((A + B*cos(c + d*x))/(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((A+B*cos(d*x+c))/cos(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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