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

Optimal. Leaf size=317 \[ -\frac {(283 A-163 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(157 A-85 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{80 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(787 A-475 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{240 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(2671 A-1495 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{240 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(21 A-13 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[Out]

-1/4*(A-B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)-1/16*(21*A-13*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/a
/d/(a+a*cos(d*x+c))^(3/2)-1/240*(787*A-475*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)+1/80*(1
57*A-85*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c))^(1/2)-1/32*(283*A-163*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(5/2)/d*2^(1/2)+1/
240*(2671*A-1495*B)*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 = 1.12, antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2961, 2978, 2984, 12, 2782, 205} \[ \frac {(157 A-85 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{80 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(787 A-475 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{240 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(2671 A-1495 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{240 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(283 A-163 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tan ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(21 A-13 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((283*A - 163*B)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Co
s[c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) + ((2671*A - 1495*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(
240*a^2*d*Sqrt[a + a*Cos[c + d*x]]) - ((787*A - 475*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(240*a^2*d*Sqrt[a + a*
Cos[c + d*x]]) - ((A - B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - ((21*A - 13*B)*S
ec[c + d*x]^(5/2)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) + ((157*A - 85*B)*Sec[c + d*x]^(5/2)*Sin[c
+ d*x])/(80*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 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 2961

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Sin[e + f*x])^m*(
c + d*Sin[e + f*x])^n)/(g*Sin[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

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)) \sec ^{\frac {7}{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 {A+B \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))^{5/2}} \, dx\\ &=-\frac {(A-B) \sec ^{\frac {5}{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 {1}{2} a (13 A-5 B)-4 a (A-B) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(21 A-13 B) \sec ^{\frac {5}{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 {1}{4} a^2 (157 A-85 B)-\frac {3}{2} a^2 (21 A-13 B) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(21 A-13 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(157 A-85 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{80 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 {1}{8} a^3 (787 A-475 B)+\frac {1}{2} a^3 (157 A-85 B) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{20 a^5}\\ &=-\frac {(787 A-475 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(21 A-13 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(157 A-85 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{80 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 {1}{16} a^4 (2671 A-1495 B)-\frac {1}{8} a^4 (787 A-475 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx}{30 a^6}\\ &=\frac {(2671 A-1495 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(787 A-475 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(21 A-13 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(157 A-85 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{80 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 {15 a^5 (283 A-163 B)}{32 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{15 a^7}\\ &=\frac {(2671 A-1495 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(787 A-475 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(21 A-13 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(157 A-85 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {\left ((283 A-163 B) \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 {(2671 A-1495 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(787 A-475 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(21 A-13 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(157 A-85 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\left ((283 A-163 B) \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 {(283 A-163 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 ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}+\frac {(2671 A-1495 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(787 A-475 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{240 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(21 A-13 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(157 A-85 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 8.46, size = 261, normalized size = 0.82 \[ \frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) (10 (2605 A-1381 B) \cos (c+d x)+108 (157 A-85 B) \cos (2 (c+d x))+9110 A \cos (3 (c+d x))+2671 A \cos (4 (c+d x))+15053 A-5030 B \cos (3 (c+d x))-1495 B \cos (4 (c+d x))-7685 B)-240 i (283 A-163 B) e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )\right )}{960 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^5*(((-240*I)*(283*A - 163*B)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*
I)*(c + d*x))]*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/E^((I/2)*(c + d*x)) + (
15053*A - 7685*B + 10*(2605*A - 1381*B)*Cos[c + d*x] + 108*(157*A - 85*B)*Cos[2*(c + d*x)] + 9110*A*Cos[3*(c +
d*x)] - 5030*B*Cos[3*(c + d*x)] + 2671*A*Cos[4*(c + d*x)] - 1495*B*Cos[4*(c + d*x)])*Sec[(c + d*x)/2]^3*Sec[c
+ d*x]^(5/2)*Tan[(c + d*x)/2]))/(960*d*(a*(1 + Cos[c + d*x]))^(5/2))

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fricas [A]  time = 0.61, size = 266, normalized size = 0.84 \[ \frac {15 \, \sqrt {2} {\left ({\left (283 \, A - 163 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (283 \, A - 163 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (283 \, A - 163 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (283 \, A - 163 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (2671 \, A - 1495 \, B\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (911 \, A - 503 \, B\right )} \cos \left (d x + c\right )^{3} + 32 \, {\left (49 \, A - 25 \, B\right )} \cos \left (d x + c\right )^{2} - 160 \, {\left (A - B\right )} \cos \left (d x + c\right ) + 96 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{480 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(15*sqrt(2)*((283*A - 163*B)*cos(d*x + c)^5 + 3*(283*A - 163*B)*cos(d*x + c)^4 + 3*(283*A - 163*B)*cos(d
*x + c)^3 + (283*A - 163*B)*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))
/(sqrt(a)*sin(d*x + c))) + 2*((2671*A - 1495*B)*cos(d*x + c)^4 + 5*(911*A - 503*B)*cos(d*x + c)^3 + 32*(49*A -
25*B)*cos(d*x + c)^2 - 160*(A - B)*cos(d*x + c) + 96*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x +
c)))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.52, size = 729, normalized size = 2.30 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/480/d*(4245*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-
2445*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+16980*A*ar
csin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-9780*B*arcsin((-1+c
os(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+25470*A*arcsin((-1+cos(d*x+c)
)/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-14670*B*arcsin((-1+cos(d*x+c))/sin(d*x
+c))*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+16980*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(
d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-9780*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*cos(d
*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+4245*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c)/(1+co
s(d*x+c)))^(5/2)-2445*B*arcsin((-1+cos(d*x+c))/sin(d*x+c))*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-2671*A
*2^(1/2)*cos(d*x+c)^5+1495*B*2^(1/2)*cos(d*x+c)^5-1884*A*2^(1/2)*cos(d*x+c)^4+1020*B*2^(1/2)*cos(d*x+c)^4+2987
*A*2^(1/2)*cos(d*x+c)^3-1715*B*2^(1/2)*cos(d*x+c)^3+1728*A*2^(1/2)*cos(d*x+c)^2-960*B*2^(1/2)*cos(d*x+c)^2-256
*A*2^(1/2)*cos(d*x+c)+160*B*2^(1/2)*cos(d*x+c)+96*A*2^(1/2))*cos(d*x+c)*sin(d*x+c)*(1/cos(d*x+c))^(7/2)*(a*(1+
cos(d*x+c)))^(1/2)/(-1+cos(d*x+c))/(1+cos(d*x+c))^3*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((A+B*cos(d*x+c))*sec(d*x+c)^(7/2)/(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 \[ \int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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